Advanced Quantum

Advanced Quantum Theory

Advanced Quantum Theory

Oliver Freyermuth

Wintersemester 2010 / 2011

Abstract

These are notes taken in the lecture on “Advanced Quantum Theory”.

Contents

0.1 Information . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10.2 Advanced Quantum: Topics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

1 Scattering Theory 21.1 Preludium . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Isometric Operators: . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Convergence of elements (vectors) in Hilbert–space . . . . . . . . . . . . . . . 4

1.3.1 Repetitorium 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.3.2 Operator limit . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.4 Scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4.1 Quantum scattering . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 61.4.2 Repetitorium 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4.3 Orthogonality and Completeness . . . . . . . . . . . . . . . . . . . . . . 91.4.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 91.4.5 Scattering Operator . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101.4.6 Repetitorium 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111.4.7 Quantum cross–section . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.4.8 Repetitorium 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 131.4.9 Rotational Invariance (central Potentials only) . . . . . . . . . . . . . . 151.4.10 Repetitorium 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.4.11 Repetitorium 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.4.12 Green’s Operator (Resolvent) and Transition Operator . . . . . . . . . . 221.4.13 Repetitorium 7 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.4.14 Relation to the Møller-Operators . . . . . . . . . . . . . . . . . . . . . 241.4.15 Stationary scattering states . . . . . . . . . . . . . . . . . . . . . . . . . 271.4.16 Repetitorium 8 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 281.4.17 Partial wave stationary scattering states . . . . . . . . . . . . . . . . . . 291.4.18 Repetitorium 9 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 301.4.19 Repetitorium 10 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 341.4.20 Repetitorium 11 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 351.4.21 A small excursion on Jost-function . . . . . . . . . . . . . . . . . . . . 381.4.22 Repetitorium 12 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391.4.23 Repetitorium 13 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41

2 Relativistic Wave equations 442.0.24 Repetitorium 14 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 482.0.25 Repetitorium 15 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 502.0.26 Repetitorium 16 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 552.0.27 Repetitorium 17 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 582.0.28 Repetitorium 18 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 632.0.29 Repetitorium 19 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 672.0.30 Repetitorium 20 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73

2.1 Solutions of the free Dirac-equation . . . . . . . . . . . . . . . . . . . . . . . . 742.1.1 Repetitorium 21 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

5

2.1.2 Wave packets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 822.1.3 Repetitorium 22 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 822.1.4 Repetitorium 23 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 842.1.5 Repetitorium 24 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 892.1.6 Repetitorium 25 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92

3 Fundamentals of many-body problems 953.0.7 Repetitorium 26 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 983.0.8 Repetitorium 27 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1013.0.9 Repetitorium 28 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 105

6

0.1 Information

Exercises: 2h in 6 groupsDrop–In Fr. 13-15hhttp://www.itkp.uni-bonn.de/~metsch/AQT1011/aqt1011.html

Assistent: Christoph Ditsche, Kerstin Helfrich (both for organisation of exercises)2 Examinations: after lecturing period (4.2.2011)1. beginning of February2. end of MarchPrerequisite for the examination: Present exercise more than once

0.2 Advanced Quantum: Topics

1. Scattering Theory: J. R. Taylor, “Scattering Theory”, Dover

2. Relativistic Wave Equations (Dirac, Klein–Gordon): Franz Schwabl, “Advanced QuantumMechanics”

3. Many–Particle Systems

http://www.itkp.uni-bonn.de/~metsch/AQT1011/aqt1011.html

1 Scattering Theory

1.1 Preludium

Some definitions on operators:Inverse: Let A be a linear operator (mapping, “Abbildung”)

A(a |ψ〉+ b |ϕ〉

)a, b ∈ C

= a · A |ψ〉+ b · A |ϕ〉

In a space H

A : D(A)→ R

(A)⊂ H

D is the “domain” (Urbild) of the Operator and R is the “range”.A has an inverse A−1 if:

|ψ〉 6= |ϕ〉 ⇒ A |ψ〉 6= A |ϕ〉

mapping “injective”, then

∃A−1 : A−1 : R(A)→ D

(A)

equivalent:

|χ〉 6= 0⇒ A |χ〉 6= 0

Definition of a unitary operator U :

• U is linear & D(U)

= R(U)

= H&∥∥∥Uψ∥∥∥ =‖ψ‖ ∀ψ ∈ H

‖ψ‖ :=√〈ψ|ψ〉 =

√〈ψ,ψ〉

If equivalency is fulfilled, then U−1 exists.IMAGE1every |ψ〉 has a unique image |ψ′〉every |ϕ′〉 is the image of unique |ϕ〉

Lemma: U †U = 1

Argument:∥∥∥Uψ∥∥∥ =‖ψ‖ ⇔ 〈ψ|U †U |ψ〉 = 〈ψ|ψ〉 ∀ψ ∈ H

put

|ψ〉 = |ϕ〉+ |χ〉〈ϕ+ χ|U †U |ϕ+ χ〉 = 〈ϕ+ χ|ϕ+ χ〉

Seite 2 Oliver Freyermuth Advanced Quantum Theory

1.2. ISOMETRIC OPERATORS:

put

⇒ 〈ψ|U †U |ϕ〉+ 〈ϕ|U †U |χ〉 = 〈ϕ|χ〉+ 〈χ|ϕ〉. . .

⇒ 〈ϕ|U †U |χ〉 = 〈ϕ|χ〉 ∀ϕ, χ ∈ H⇒ U †U = 1

also:

U(U †U

)|ψ〉 = U |ψ〉 ∀ψ ∈ H(

U U †)U |ψ〉 = U |ψ〉 ∀ψ ∈ H

U |ψ〉 covers all H. ⇒ U U † = 1

1.2 Isometric Operators:

Definition: An operator Ω is isometric, if

1. Ω is linear on H

2. D (Ω) = H

3. for all |ψ〉 ∈ H: ‖Ωψ‖ =‖ψ‖, norm preserved.

but: R 6= H⇒ unitary operators are isometric.Example: Let |n〉 be a discrete Basis of H, n ∈ N0.

Ω : Ω |n〉 = |n+ 1〉

IMAGE2inverse Ω−1 exists.

D(Ω−1

)6= H

Although Ω†Ω = 1 we have ΩΩ† 6= 1

indeed:

Ω† Ω |ψ〉︸︷︷︸|ϕ〉

= |ψ〉 ∀ψ ∈ H

|ϕ〉 ⇔ Ω† |ϕ〉 = Ω−1 |ϕ〉 ∀ |ϕ〉 ∈ R(Ω)

now if |ϕ〉 /∈ R (Ω)

〈ϕ|Ω|ψ〉 = 0 ∀ψ ∈ H〈ψ|Ω†|ϕ〉 = 0 ∀ψ ∈ H

Ω† |ϕ〉 = 0⇒ ΩΩ† |ϕ〉 = 0

Ω† =

Ω−1 on R (Ω)0 on R (Ω)⊥

Advanced Quantum Theory Oliver Freyermuth Seite 3

CHAPTER 1. SCATTERING THEORY

1.3 Convergence of elements (vectors) in Hilbert–space

limt→∞|ψt〉 = |ψ〉

if:

limt→∞‖ψt − ψ‖ = 0

A more useful criterion is the one by Cauchy:

limt→∞,t′→∞

‖ψt − ψt′‖ = 0

Special case:

|ψt〉 =t∫

0

dτ |ϕτ 〉

then Cauchy:


∥∥∥∥∥∥∥t′∫t

dτ |ϕτ 〉

∥∥∥∥∥∥∥ = 0

This is implied by:


t′∫t

dτ‖ϕτ‖ = 0

(triangle inequality)

‖ψ + ϕ‖ ≤‖ψ‖+‖ϕ‖

so it is sufficient that:∞∫0

dτ‖ϕτ‖ <∞

i.e. it exists.

1.3.1 Repetitorium 1

U unitary:

D(U)

= R(U)

= H∥∥∥Uψ∥∥∥ =‖ψ‖ ∀ψ ∈ H

Ω isometric:

D(Ω)

= H∥∥∥Ωψ∥∥∥ =‖ψ‖


1.4. SCATTERING

but in general:

R (Ω) 6= H

‖ψ‖ =√〈ψ|ψ〉 =

√ψ,ψ

|ψ〉 = limt→∞|ψt〉

if limt→∞‖ψ − ψt‖ = 0

⇔ limt→∞,t′→∞

‖ψt − ψt′‖ = 0 Cauchy

Special:

|ψt〉 =t∫

0

dτ |ϕτ 〉

then:∞∫0

dτ‖ϕτ‖ <∞

⇔ limt→∞,t′→∞

t′∫t

dτ‖ϕτ‖ = 0

⇒ limt→∞,t′→∞

∥∥∥∥∥∥∥t′∫t

dτ |ϕτ 〉

∥∥∥∥∥∥∥ = 0

Remark:

|ψt〉 −−−→t→∞

|ψ〉

⇒ 〈ϕ|ψt〉 −−−→t→∞

〈ϕ|ψ〉 ∀ϕ ∈ H

But the converse is not true:let e.g. |ψt〉 be a wave packet ‖ψt‖ = 1 ∀t.Nevertheless, the wave packet will spread thus:

limt→∞〈ϕ|ψt〉 = 0 ∀ϕ ∈ H

1.3.2 Operator limit

At has a limit A if

At |ψ〉 −−−→t→∞

|ϕ〉 =: A |ψ〉 ∀ψ ∈ H

Notation:

limt→∞

At = A

At −−−→t→∞

A



1.4 Scattering

IMAGE3Note:Potentialmust fallof fasterthanCoulomb-potentialand nothave asingularity,so e.g.Yukawa-potentialis okay

1.4.1 Quantum scattering

t.d.S.E. (time-dependent Schroedinger equation):

i~∂

∂t|ψt〉 = H |ψt〉

H = H0 + V (time independent)

free motion:

∣∣∣~p2∣∣∣

2m=− ~2

2m∆

~p 7→ −i~~∇Formal solution t.d.S.E.

|ψt〉 = U (t) |ψ0〉|ψ0〉 ∈ H ⇒ |ψt〉 ∈ H ∀t

H = L2(R3,C

)ψ (~x) = 〈~x|ψ〉 wave function

For any |ψ〉 ∈ H we shall call U(t) |ψ〉 the orbit (specified by ψ).Now we let the orbit U(t) |ψ〉 be the evolution of some scattering process.then for t→ −∞ it behaves (almost) as a free wave packet. So we expect (U0 is time operatorfor free movement, U(t) = exp

(−i/~t · H

)):

U (t) |ψ〉 −−−−→t→−∞

U0(t) |ψin〉

Uo(t) = exp(− i~tH0

)Likewise:

U(t) |ψ〉 −−−−→t→+∞

U0(t) |ψout〉

Assumptions:

V (~x) = V(|~x|)

= V (r)

Definition: “V (r) = O (rp)” if ∃c ∈ R∣∣∣∣∣V (r)

∣∣ ≤ c · rp1. V (r) = O

(r−3−ε

)r →∞ (ε > 0)

2. V (r) = O(r−

3/2+ε)r → 0

3. V (r) is finite, piecewise continuous, 0 < r <∞


1.4. SCATTERING

Asymptotic condition:

∀ |ψin〉 ∈ H∃ |ψ〉 ∈ H∣∣∣U(t) |ψ〉 − U0(t) |ψin〉

−−−−→t→−∞

0

This is similar for ψout and +∞.Argument:

to show: |ψ〉 − U †(t)U0(t) |ψin〉 −−−−→t→−∞

0 The substracted term has a limit.

now: calculateddt U

†(t)U0(t) = i

~ei/~tH

(H − H0

)e−i/~tH0

= i

~U †(t)V U0(t)

⇒ U †(t)U0(t) |ψin〉+ i

~

t∫0

dτU(τ)†V U0 (τ) |ψin〉

This integral-term should have a limit for t→ −∞.It is sufficient that:

0∫−∞

dτ∥∥∥U †V U0(t)ψin

∥∥∥ <∞⇔

0∫−∞

dτ∥∥∥V U0(τ)ψin

∥∥∥ <∞ U(t) unitary


t.d.S.E.:

i~∂

∂t|ψt〉 = H |ψt〉

H = H0 + V

solved by:|ψt〉 = U(t) |ψ〉|ψ〉 = |ψ0〉

L ∼ − ~2

2m∆

U(t)exp(− i~tH

)U0 = exp

(− i~tH0

)U(t) |ψ〉 is called orbit specified by |ψ〉 ∈ H = L2

(R3,C

)“asymptotes”:

U(t) |ψ〉 −−−−→t→−∞

U0(t) |ψin〉

U(t) |ψ〉 −−−−→t→+∞

U0(t) |ψout〉



asymptotic condition:

∀ |ψin, out〉 ∈ H∃ |ψ〉 ∈ HU(t) |ψ〉 − U0(t) |ψin, out〉

0 for t→ ∓∞in fact:|ψ〉 = Ω± |ψin, out〉

with:Ω± := lim

t→∓∞U †(t)U0(t) Møller–Operators

Ω± are isometric (norm preserving) but not unitary

Now: Continuing the last lecture. To show:0∫

−∞

dτ∥∥∥V U0(t)ψin

∥∥∥ <∞idea: special case:

ψin (~x) = 〈~x|ψin〉 = 1(b√π)3/2

e−|~x− ~a|2/2b2

Now we calculate and find:∣∣〈~x|U0(τ)|ψin〉∣∣2 = 1

b3π3/2

1(1 + ~2τ2

m2b4

)3/2e−|~x− ~a|

2/(b2 + τ2~2/m2b2

)∣∣ψin (τ, ~x)

∣∣2 = above

∥∥∥V U0(τ)ψin∥∥∥2

=∫

d3x∣∣V (~x)

∣∣2 1b2π3/2

1(1 + ~2τ2

m2b4

)3/2e−|~x− ~a|

2/(b2 + τ2~2/m2b2

)

≤(∫

d3x∣∣V (~x)

∣∣2) 1(1 + ~2τ2

m2b4

)3/2

1b2π3/2

⇒0∫

−∞

dτ∥∥∥V U0(τ)ψin

∥∥∥ ≤ √∫ d3x∣∣V (~x)

∣∣2︸︷︷︸<∞

1b2π3/2

0∫−∞

dτ 1(1 + ~2τ2

m2b4

)3/4

︸︷︷︸<∞

<∞

The statement “Any ψ ∈ L2 (R,C) can be suitably approximated by a finite sum of Gaussians”then completes the argument.So we found:“Any |ψin〉 ∈ H is the “in”–asymptote of some U(t) |ψ〉”In fact, the actual state |ψ〉 at (e.g.) t = 0 is given by:

|ψ〉 = limt→−∞

U †(t)U0(t) |ψin〉 =: Ω+ |ψin〉

also:

|ψ〉 = limt→+∞

U †(t)U0(t) |ψout〉 =: Ω− |ψout〉


1.4. SCATTERING

with:

Ω± := limt→∓∞

U †(t)U0(t)

the Møller(Wave)Operators.IMAGE4let

|ψin〉 = |ϕ〉 |ψin〉 7→Ω+|ψ〉 ←[Ω− |ψout〉 |ψin〉 =: |ϕ+〉

|ψout〉 = |χ〉 |ϕ〉 7→ |ϕ+〉 |ψout〉 =: |χ−〉|χ−〉 ←[ |χ〉

1.4.3 Orthogonality and Completeness

Question: Does every |φ〉 in H define an orbit U(t) |ψ〉 that has asymptotes? No!Orthogonality: “all bound states” B ∈ H ⊥ “all states with asymptotes”States with asymptotes:

R+ =|ψ〉 ∈ H

∣∣∣ |ψ〉 = Ω+ |ψin〉

R− =|ψ〉 ∈ H

∣∣∣ |ψ〉 = Ω− |ψout〉

now R± ⊂ H and B⊥R±Argument: let |ψ〉 ∈ R+ i.e. U(t) |ψ〉 −−−−→

t→−∞U0(t) |ψin〉

or |ψ〉 = Ω+ |ψin〉let |ϕ〉 ∈ B: H |ϕ〉 = E |ϕ〉then:

〈ϕ|ψ〉 = 〈 ϕ|U †(t)︸︷︷︸“close to the potential origin”

U(t)|ψ︸︷︷︸move away

〉 independent of t

but also t→ −∞, ⇒ 〈ϕ|ψ〉 = 0.Also:

〈ϕ|ψ〉 = eλ/~tE 〈ϕ|U(t)|ψ〉 = limt→−∞

eiE/~t 〈ϕ|U0(t)|ψin〉 = 0

because of “spreading”.

⇒ B⊥R±

Asymptotic completeness statesR+ = R− = R and H = B ⊕Rwithout proof!

1.4.4 Summary

|ψ〉 ∈ R

U |ψ〉 7→t→−∞ U0(t) |ψin〉U |ψ〉 7→t→+∞ U0(t) |ψout〉

or: |ψ〉 = Ω+ |ψin〉 |ψ〉 = Ω− |ψout〉



every ψin, out labels a unique actual orbit U(t) |ψ〉.Ω± are isometric (but not unitary).IMAGE5

1.4.5 Scattering Operator

Ω− is isometric, Ω− has an inverse.

|ψ〉 = Ω− |ψout〉

⇒ |ψout〉 = Ω†− |ψ〉|ψout〉 = Ω†−Ω+︸︷︷︸

=:S

|ψin〉

where S is the scattering operator.|ψin〉 = |ϕ〉 prepared by an accelerator|ψout〉 = |χ〉 measured by a detectorLemma:

HΩ± = Ω±H0 intertwining relations

argument: consider (τ fixed):

ei/~τHΩ± = ei/~τH lim(

ei/~tHe−i/~tH0

)= lim

(ei/~(t+τ)He−i/~tH0

)=(

lim ei/~(t+τ)He−i/~(t+τ)H0

)e−i/~τH0

= Ω±ei/~τH0(ddτ

(ei/~τHΩ±

))τ=0

=(

ddτ Ω±ei/~τH0

)τ=0

⇒ HΩ± = Ω±H

we have

Ω†±Ω± = 1

HΩ± = Ω±H0

Thus follows:

Ω†±HΩ± = Ω†±Ω±H0 = H0

now:

SH0 = Ω†−Ω+H0 = Ω†−HΩ+ =(HΩ−

)†Ω+

=(Ω−H0

)†Ω+

= H0Ω†−Ω+ = H0S

⇒[S, H0

]−

= 0


1.4. SCATTERING

We draw from that:

⇒ 〈ψin|H0|ψin〉 = 〈ψin|S+H0S|ψin〉〈ψout|H0|ψout〉

We used here:

SS+ = S+S = 1[H0, S

]−

= 0

So we find that the “ingoing energy” (expectation value) is equal to the “outgoing energy”(expectation value)(H is time independent) Energy conservation

We can always write:

|ψin〉 ∈ H = L2(R3,C

)ψin (~x) = 〈~x|ψin〉 =

∫d3kain

(~k)ϕ~k (~x)

remember: ϕ~k (~x) = 〈~x|~k〉 = 1(2π)3/2

ei(~k·~x)

H0ϕ~k (~x) =~2∣∣∣~k∣∣∣2

2m ϕ~k (~x)

ϕ~k /∈ H

nevertheless:

a ∈ L2(R3,C

)⇒ ψ ∈ L2

(R3,C

)ϕ~k → 〈ϕ~k|ϕ~q〉 = δ(3)

(~q − ~k

)1.4.6 Repetitorium 3

|ψ〉 = Ω± |ψin,out〉 = limt∓∞

U †(t)U0(t) |ψin,out〉

Møller–OperatorsR (Ω+) = R (Ω−) = R

R+B = HR⊥B isometric

Scattering operator: S := Ω†−Ω+

S : H → H unitaryIntertwining relations

HΩ± = Ω±H[S, H0

]−

= 0

⇒ Energy Conservation

IMAGE6



for |ψin〉 ∈ H

ψin (~x) = 〈~x|ψin〉 =∫

d3kain(~k)ψ~k (vecx)

where ϕ~k (~x) = 〈~x|~k〉 = 1(2π)3/2

ei(~k·~x)

H0ϕ~k (~x) =~2∣∣∣~k∣∣∣2

2m ϕ~k (~x)a ∈ H ψin ∈ H

but ϕ~k ∈ L2(R3,C

)we have 〈ϕ~k|ϕ~q〉 = δ(3)

(~k − ~q

)ψout (~x) =

∫d3k aout

(~k)ϕ~k (~x)

with

ain,out(~k)

= 〈ϕk|ψin〉 = 1(2π)3/2

∫d3x e−i

(~k·~x)ψin,out (~x)

aout(~k)

= 1(2π)3/2

∫d3xe−i(veck·~x)S

∫d3q ei(~q·~x)ain (~q)︸︷︷︸

ψin(~x)

=∫

d3q ain (~q) 〈ϕ~k|S|ϕ~q〉

〈ϕ~k|S|ϕ~q〉 is the probability amplitude to go to from ~q → ~k.Now we have

0 = 〈ϕ~k|[H0, S

]|ϕ~q〉

= 〈ϕ~k|H0S − SH0|ϕ~q〉

=

~2∣∣∣~k∣∣∣2

2m − ~2|~q|2

2m

〈ϕ~k|S|ϕ~q〉⇒ 〈ϕ~k|S|ϕ~q〉 will vanish unless:

E~k =~2∣∣∣~k∣∣∣2

2m = E~q = ~2|~q|2

2m

We expect 〈ϕ~k|S|ϕ~q〉 ∝ δ(Ek − Eq

)We write:

S = 1 + R

The 1 means no scattering, the R is the scattering operator.We write:

〈~k|R|~q〉 = −2π · i · δ(Ek − Eq

)t(~k ← ~q

)or 〈~k|S|~q〉 = δ(3)

(~q − ~k

)− 2π · i · δ

(Ek − Eq

)t(~k ← ~q

)Here, t

(~k ← ~q

)is the transition amplitude, which is only defined for Ek −Eq. δ

(Ek − Eq

)means scattering took place, δ(3)

(~q − ~k

)is the part for no scattering.


1.4. SCATTERING

Define the scattering amplitude:

f(~k, ~q

):= − (2π)2m

~2 t(~k ← ~q

)⇒ 〈~k|S|~q〉 = δ(3)

(~q − ~k

)+ i~2

2π ·mδ(Ek − Eq

)f(~k, ~q

)1.4.7 Quantum cross–section

Starting point: Probability that for some |ψin〉 we will scatter into a cone with direction ~k andsolid angle Ω.IMAGE7

W (Ω← ϕin) = Ω∞∫0

dk k2∣∣∣∣ϕout

(~k)∣∣∣∣2

k =∣∣∣~k∣∣∣

ϕin(~k)

= Φ(~k)

= e−i(~b·~k)Φ(~k)

assume: Φ(~k)

is centeredaround ~k0 = ~k~e3

Assume: That the displacement ~b is random uniform in plane ⊥~k0. b =∣∣∣~b∣∣∣ is called the impact

parameter.Then:

NSc =(Ω,~k

)=∑i

W (Ω← ϕi)

'∫

plane ⊥~k0

d2b nincW(Ω← Φ~b

)

= ninc

∫⊥~k0

d2bW(Ω← Φ~b

)

Here, NSc is the number of scatterings into the cone and ninc is the density, that is, particlesper area.This last integral is the cross-section (area)!Cross–section:

σ (Ω← Φb) =∫⊥~k0

d2bW (Ω← Φb)

=∫

d2bΩ∞∫0

dk k2|Φout|2


uniform distribution (ninc) of wave packets ϕb(~k)

= e−(~b·~k)ϕ(~k), in which ϕb are the Four-

ier–components



IMAGE7

NSC (Ωk) = ninc

∫d2bW (Ωk ← ϕb)︸︷︷︸

(effective) cross section σ(Ωk←~k0

)(area)

W (Ωk ← ϕb) = Ωk

∞∫0

dk k2∣∣∣∣ψout

(~k)∣∣∣∣2

S Matrix element

〈~k|S|~q〉 = δ(3)(~k − ~q

)−2πiδ

(Ek − Eq

)t(~k ← ~q

)scattering amplitude

f(~k, ~q

)= −(2π)2m

~2 t(~k ← ~q

) ∣∣∣∣Ek−Eq

H0 |~k〉 =~2∣∣∣~k∣∣∣2

2m |~k〉

Let’s continue:

ϕout(~k)

=∫

d3q 〈~k|S|~q〉ϕin (~q)

= ϕin(~k)

+ i~2

2πm

∫d3q δ

(Ek − Eq

)f(~k, ~q

)ϕin (~q)

ϕin(~k)

= e−i(~k·~b)ϕ(~k)

with ‖ϕ‖ = 1

ϕ(~k)≈ 0 if ~k 6= ~k0

(no forward scattering)

≈ i~2

2πm

∫d3q δ

(Eq − Ek

)f(~k, ~q

)e−i(~q·~b)ϕ (~q) for ~k 6= ~k0

σ

Ωk ← ϕ︸︷︷︸≈~k0

= Ωk~4

(2π)2m2

∫d2b

∞∫0

dk k2

We now need the expressions:∫

d3q δ(Ek − Eq

)f(~k, ~q

)e−i(~q·~b)ϕ (~q)∫

d3q′ δ(Ek − Eq′

)f(~k, ~q′

)∗ei(~q′·~b)ϕ∗ (~q)

Notation:

IMAGE8Now: ∫

d2b ei((~q′−~q)·~b

)= (2π)2 δ(2)

(~q′⊥ − ~q⊥

)


1.4. SCATTERING

Now it follows:

⇒∫

d2bei((~q′−~q)·~b

)δ(Ek − Eq′

)δ(Ek − Eq

)= (2π)2 m

~2qqδ(Eq − Ek

) (δ(3)

(~q′ − ~q

)+ δ

(~q′ + ~q

))

σ(Ωk ← ~k0

)= Ωk ·

~2

m

∞∫0

dk k2∫

d3qδ(Eq − Ek

)qq

·

∣∣∣∣f (~k, ~q)∣∣∣∣2∣∣ϕ (~q)

∣∣2 + f(~k, ~q

)f∗(~k,−~q

)ϕ (~q)ϕ∗

(~−q)

︸︷︷︸ϕ(~k)=0 for ~k 6=~k0⇒0

σ(Ωk ← ~k0

)= Ωk

∫d3q

k

qq

∣∣∣∣f (~k, ~q)∣∣∣∣2∣∣ϕ (~q)∣∣2 ∣∣∣∣|~q|=∣∣~k∣∣=:k

Again ϕ (~q) = 0 if ~q 6= ~k0 ⇒ qq = k0 = k

σ(Ωk ← ~k0

)= Ωk

∣∣∣∣f (~k,~k0)∣∣∣∣2 ∫ d3q

∣∣ϕ (~q)∣∣2︸︷︷︸

=1

= Ωk

∣∣∣∣f (~k,~k0)∣∣∣∣2∣∣~k∣∣=∣∣~k0

∣∣Write:

σ(Ωk ← ~k0

)= Ωk

dσdΩk

⇒ dσdΩk

(~k ← ~k0

)=∣∣∣∣f (~k,~k0

)∣∣∣∣2∣∣~k∣∣=∣∣~k0∣∣

1.4.9 Rotational Invariance (central Potentials only)

Rotation in R3 (~ω rotation axis)~ω/|~ω| direction of rotation axis|~ω| rotation angleWe write:

g~ω ∈ SO(3)

g~ω · ~x =(exp

(A (~ω)

))~x

where A (~ω)~h =[~ω × ~h

]Rotation of a wave funtion ψ (~x), ψ ∈ L2

(R3,C

):

(D (g~ω)ψ

)(~x) := ψ

(g−1ω ~x

)=

exp(− i~

(~ω · ~L

))ψ

(~x)

Now, rotational invariance means: [D (g~ω) , H

]−

= 0

⇔[Lk, H

]−

= 0



D (gω) is unitary.now:

g ∈ SO(3)D(g)Ω± = D(g) lim

t→∓∞ei/~tHe−i/~tH0

= Ω±D(g)⇒[D(g), S

]−

=[D(g), Ω†−Ω+

]−

= 0

so: S = D†(g)SD(g)

scattering operator S commutes with rotations.[Lk, S

]= 0⇐ D

(g(~ω)

)k = 1, 2, 3

therefore:

〈~k′|S|~k〉 = 〈g~k′|S|g~k〉

since

〈~x|~k〉 = ϕ~k (~x) = 1(2π)3/2

exp(i ·(~k · ~x

))(Dgϕ~k

)(~x) = ϕ~k

(g−1ω ~x

)= 1

(2π)3/2ei(~k·g−1~x

)= 1

(2π)3/2ei(g~k·~x

)= ϕ

g~k(~x) = 〈~x|g~k〉

also:

f(~k′,~k

)= f

(g~k′, g~k

)although f

(~k′,~k

)depends in general on 5 variables

Ek = ~k2

2m = ~2k′2

2m = Ek′

for central potentials it only depends on 5− 3 = 2 variables, i.e.

Ek = ~2k2

2mand θ = ]

(~k,~k′

)Common Eigenfunctions of H0,

∣∣∣∣~L∣∣∣∣2 , L3

Starting point:

ψ (~x) = 〈~x|ψ〉 =∫

d3k 〈~x|~k〉︸︷︷︸= 1

(2π)3/2ei(~k·~x)

〈~k|ψ〉︸︷︷︸(F−1ψ)

(~k)= 1

(2π)3/2e−i(~k·~y)ψ(~y)

where ϕ~k = 〈~x|~k〉 common eigenfunctions (/∈ H = L2(R3mC

)) of p1, p2, p3 with eigenvalues

~k1, ~k2, ~k3.Also:

H0ϕ~k (~x) =~2∣∣∣~k∣∣∣2

2m ϕ~k (~x)


1.4. SCATTERING

Alternatively: Common eigenfunctions of H0,

∣∣∣∣~L∣∣∣∣2 , L3 (regular at r = 0)spherical coordinates r, θ, ϕ

ϕklµ (r, θ, ϕ) = N · jl (kr)Ylµ (θ, ϕ)

with: jl(x) := (−x)l(

1x

ddx

)l (sin(x)x

)l ∈ N0

Note that ϕklm (r, θ, ϕ) /∈ H. jl(x) are the spherical Bessel–Functions.

H0ϕklµ = ~2k2

2m ϕklµ

In fact:

ei(~k·~x)

= 4π∞∑l=0

iljl

(∣∣∣~k∣∣∣|~x|) l∑µ=−l

Y ∗lµ

(~k0)Ylµ (~x0)

~k0 =~k∣∣∣~k∣∣∣

~x0 = ~x

|~x|

We shall use:

ΦElµ (~x) =

√2mπ~2

√2mE~

iljl

(√2mE~|~x|)Ylµ (~x0)

=: 〈x|Elµ〉

Remember: Ek =∣∣~k∣∣2~2

2mthen:

〈E′l′µ′|Elµ〉 = δ(E′ − E

)δl′lδµ′µ

follows from:∞∫0

dz z2jn (αz) jn (βz) = π

2α2 δ (α− β)

for the Fourier–Transform we find(F−1ϕElµ

) (~k)

= 〈~k|Elµ〉

〈~k|Elµ〉 = ~√m∣∣∣~k∣∣∣δ

E − ~2∣∣∣~k∣∣∣2

2m

Ylm (~k0)

Completeness:

1 =∫

dE∑l,µ

|Elµ〉〈Elµ|



Consistency check:

〈~x|~k〉 =∫

dE∑l,µ

〈~x|Elµ〉〈Elµ|~k〉

=∫

dE∑l,µ

~√m∣∣∣~k∣∣∣δ

E − ~∣∣∣~k∣∣∣22m

Y ∗lµ (~k0)·

√2mπ~2

√2mE~

iljl

√2mE~|~x|

Ylµ (~x0)

= . . .

= 1(2π)2/3

4π∑l,µ

iljl

(∣∣∣~k∣∣∣ ·|~x|)Y ∗lmu~kYlµ~x0 = ei~k·~x 1(2π)3/2


dσdΩ

(~k′ ← ~k

)=∣∣∣∣f (~k′,~k)∣∣∣∣2∣∣~k′∣∣=∣∣~k∣∣

differential cross section scattering amplitude

for central Potentials V (|~x|)[~L, S

]−

= 0[~H0, S

]−

= 0

“Basis”

ϕ~k (~x) = 〈~x|~k〉 = 1(2π)3/2

ei(~k·~x)

“Alternate Basis”

ϕElµ (~x) = 〈~x|Elµ〉 =

√m

π~2

√2mE~

iljl

(√2mE~|~x|)Ylµ (~x0)

~x0 = ~x

|~x|

common eigenfunction of

pα = −i~ ∂

∂xα

H0 =

∣∣∣~p∣∣∣22m

with eigenvalues

~kα

~2∣∣∣~k∣∣∣2

2m

〈~k|S|~q〉 = δ(3)

(~k − ~q

)+ i

2πm~2δ(Ek − Eq

)f(~k, ~q

)⇒∣∣∣~k∣∣∣ = |~q|


1.4. SCATTERING

〈E′l′µ′|Elµ〉 = δ(E′ − E

)δl′lδµ′µ

1 =∫

dE∑l,µ

|Elµ〉〈Elµ|

〈~k|Elµ〉 = ~√m∣∣∣~k∣∣∣δ

E − ~2∣∣∣~k∣∣∣2

2m

Ylµ (~k0)

~k0 =~k∣∣∣~k∣∣∣

〈~k|~q〉 = δ(3)(~k − ~q

)∫

d3k |~k〉〈~k| = 1

〈~x|~k〉 = 1(2π)3/2

ei(~k·~x)

Let us continue:

〈~k′|S − 1|~k〉 = i~2

2πmδ (Ek′ − Ek) f(~k′,~k

)=∫

dE′∑l′µ′

∫dE

∑lµ

〈~k′|E′l′µ′〉〈E′l′µ′|(S − 1

)|Elµ〉〈Elµ|~k〉

We have

〈E′l′µ′|[S, H0

]|Elµ〉 = 0 and also

[Lk, S

]−

= 0

⇒ 〈E′l′µ′|S|Elµ〉[∣∣∣∣~L∣∣∣∣ , S

]−

= 0

= δ(E′ − E

)δl′lδµ′µSlµ(E)

We have:

L± = L1 ± iL2

L−SL+ = SL−L+ = S(∣∣∣∣~L∣∣∣∣2 − ~L3 − L2

3

)⇒ 〈Elµ|L−SL+|Elµ〉

=(~√l(l + 1)− µ (µ+ 1)

)2〈Elµ+ 1|S|Elµ+ 1〉

= 〈Elµ|S|Elµ〉 ~2(l (l + 1)− µ− µ2

)= Slµ(E) does not depend on µ: write Sl(E)



i~2

2πmδ (Ek′ − Ek) f(~k′,~k

)=∫

dE∑l,µ

~√m∣∣∣~k′∣∣∣δ (E − Ek′)Ylµ

(~k′0

)

Ek′ =~∣∣∣~k′∣∣∣22m[

Sl(E)− 1] ~√

m∣∣∣~k∣∣∣δ (E − Ek)Y ∗lµ

(~k0)

~2∣∣∣~k∣∣∣2

2m ⇒ δ (Ek′ − Ek) δ (E − Ek)

⇒ f(~k′,~k

)= 2πi∣∣∣~k∣∣∣∑l,µ

Ylµ(~k′0

) (Sel(E)− 1

)Y ∗lµ

(~k0)

Ek =~∣∣∣~k∣∣∣22m =

~∣∣∣~k′∣∣∣22m

f(g~k′, g~k

), g ∈ SO(3)

= f(~k′,~k

)We conclude: Take ~k0 = ~e3. Then:

Ylµ (~e3) =√

2l + 14π δµ,0

with ~k′0 = (θ, ϕ)

Yl0 (θ, ϕ) =√

2l + 14π Pl (cos θ) Legendre–Polynomial

⇒ f (Ek, θ) = f

(~k′,∣∣∣~k∣∣∣~e3

)= 2πi∣∣∣~k∣∣∣

∞∑l=0

2l + 14π=2

(Sl (Ek)− 1

)Pl (cos θ)

This is called the partial wave expansion of the scattering amplitude.Remember: The eigenvalue of an unitary operator has complex modulus 1 (i.e. is a phasefactor).We write (because S is unitary):

Sl(E) = e2iδl(E)

⇔∣∣Sl(E)

∣∣ = 1

δl(E) is called the scattering phase. We can also define:

fl(E) := Sl(E)− 12i∣∣∣~k∣∣∣


1.4. SCATTERING

partial wave scattering amplitude.

= e2iδl(E) − 12i∣∣∣~k∣∣∣ = eiδl(E) sin

(δl(E)

)∣∣∣~k∣∣∣f (E, θ) =

∞∑l=0

(2l + 1) fl(E)Pl (cos θ)

We have:

dσdΩ =

∣∣f (E, θ)∣∣2

Total elastic (angular integrated) cross section:

σ(E) =∫

dΩ∣∣f (E, θ)

∣∣2=

2π∫0

dϕπ∫

0

dθ sin θ∣∣f (E, θ)

∣∣2. . .

σ(E) =∞∑l=0

σl(E)

where

σl(E) = 4π (2l + 1) sin2 (δl(E))∣∣∣~k∣∣∣2

≤ 4π2l + 1∣∣∣~k∣∣∣2 unitary bound

E =~2∣∣∣~k∣∣∣2

2m

with σl being the partial wave cross section.


Central potential scattering

f (E, θ) =∞∑l=0

(2l + 1)fl(E)Pl(cos (θ)

)= f

(~k′, k ·~e3

)Here Pl are the Legendre-Polynomials.

E = ~2k2

2m∣∣∣~k′∣∣∣ =∣∣∣~k∣∣∣ = k

θ = ](~k′,~k

)



particle scattering amplitude:

fl(E) = sl(E)− 12ik = e2iδl(E) − 1

2ik = eiδl(E) sin(δl(E)

)k

particle scattering phase (shift) δl(E)

dσdΩ(E, θ) =

∣∣f(E, θ)∣∣2

σ(E) =∫

dΩ∣∣f(E, θ)

∣∣2 =∞∑l=0

σl(E)

σl(E) = 4π (2l + 1) sin2 (δl(E))

k2

Let’s continue.

1.4.12 Green’s Operator (Resolvent) and Transition Operator

Green’s Operator (resolvent) to self-adjoint operators

H0 =

∣∣∣~p∣∣∣22m

H = H0 + V

Definition:

G0(z) :=(z · 1− H0

)−1∀z ∈ C /∈ Spec

(H0)

G(z) :=(z · 1− H

)−1∀z ∈ C /∈ Spec

(H)

Certainly exist for =[z] 6= 0, then:(z · 1− H0

)G0(z) = 1

(z · 1− H

)G(z) = 1

We have

〈~x|H0|ψ〉︸︷︷︸(H0ψ

)(~x)

= − ~2

2m∆ 〈~x|ψ〉︸︷︷︸ψ(~x)

⇒(

~2

2m∆ + z

)〈~x|G0(z)|~y〉 = δ(3) (~x− ~y)

So 〈~x|G0(z)|~y〉 is the coordinate space matrix element of G0(z) and is the Green’s function

to the operator(~2/2m∆ + z · 1

)Likewise, the c.s.m.e. of G(z) is the Green’s function of the operator(

~2

2m∆− V (~x) + z · 1)

Inspect:

Let |n〉 be an eigenvector of H with eigenvalue En

⇒(En · 1− H

)︸︷︷︸this has no inverse

|n〉 = 0


1.4. SCATTERING

We conclude that G(z) is undefined for z = En ∈ Spec(H)

Example:

Suppose H has a discrete Spectrum only, |n〉 with eigenvalues En.∑n

|n〉〈n| = 1

G(z) =[z − H

]−1· 1∑n

|n〉〈n|

=∑n

1z − En

|n〉〈n|

Definition:

G(z) is called an analytic (operator) function of the complex variable z if 〈χ|G(z)|ψ〉 is ananalytic function of z for all |χ〉 , |ψ〉 ∈ H.Example:

〈χ|G(z)|ψ〉 =∑n

〈χ|n〉〈n|χ〉z − En

analytic z /∈ Enat z = En this has simple poles with residues: ∝ 〈χ|n〉〈n|ψ〉⇔ G(z) . . . |n〉〈n| Projection on the n-th eigenstate.

Example 2:

H has only a continuous spectrum: → H0 (e.g.), then

G0(z) =∞∫0

dE∑l,µ

|Elµ〉〈Elµ|z − E

However, this is only ok if z /∈ R+.IMAGE9Inspect:

limε↓0〈χ|G0 (E0 + iε) |ψ〉 − 〈χ|G0 (E0 − iε) |ψ〉

for E > 0

= limε↓0

∞∫0

dE( 1E + iε− E

− 1E − iε− E

)︸︷︷︸

exercise:−2πiδ(E0−E)

∑l,µ

〈χ|Elµ〉〈Elµ|ψ〉

limε↓0〈χ|G0 (E0 + iε) |ψ〉 − 〈χ|G0 (E0 − iε) |ψ〉

= −2πi∑l,µ

〈χ|Elµ〉〈Elµ|ψ〉

Knowledge on G(z) ⇔ knowledge on eigenvalues and eigenfunctions of H.IMAGE10Relation between G(z), G0(z)is given by the Lippmann-Schwinger–Equation:

G(z) = G0(z) + G0(z) · V · G(z)



Transition operator: T (z)

T (z) = V + V G (z) V⇒ L-S–Eq.

T (z) = V G0 (z) T (z)

Let V be weak:

T (z) = V + V G0(z)V + V G0V G0V + . . .


Green–Operator G(z) (Resolvent) and transition operator T (z) (H = H0 + V )

G(z) =(z1− H

)−1∀z ∈ C,∀z /∈ spec

(H)

G0(z) =(z1− H0

)−1∀z /∈ spec

(H0)

T (z) = V + V G (z) V

Lippmann-Schwinger–Eq.

G(z) = G0(z) + G0(z)V G(z)T (z) = V + V G0(z)T (z)

IMAGE10Born–Series

T (z) = V∞∑k=0

(G0(z)V

)k= V + V G0(z)V + V G0V G0(z)V + . . .

1.4.14 Relation to the Møller-Operators

Remember:

if |ψin〉 = |ϕ〉 ∈ H then at t = 0 : |ψ〉 = Ω+ |ψin〉 = Ω+ |ϕ〉 =: |ϕ+〉if |ψout〉 = |ϕ〉 ∈ H then at t = 0 : |ψ〉 = Ω− |ϕ〉 =: |ϕ−〉

Ω± = limt→∓∞

U †(t)U0(t)

we also know:

ddt U

†(t)U0(t) = i

~U †(t)U0(t)

|ϕ−〉 = Ω− |ϕ〉 = limt→+∞

U †(t)U0(t) |ϕ〉

|ϕ−〉 = |ϕ〉+ i

~

∞∫0

dτ U †(τ)V U0(τ) |ϕ〉


1.4. SCATTERING

(this is absolutely convergent (exist. Møller-Operators))

= |ϕ〉+ limε↓0

∞∫0

dτ e−ετ U †(τ)V U0(τ) |ϕ〉 1. exercise

Likewise:

|ϕ+〉 = |ϕ〉+ i

~limε↓0

−∞∫0

dτ eετ U †(τ)V U0(τ) |ϕ〉

To evaluate U0(τ) |ϕ〉, we insert∫

d3k |~k〉〈~k| = 1:

|ϕ−〉 = |ϕ〉+ i

~limε↓0

∞∫0

dτ∫

d3k[e−ετU †(τ)V U0(τ) |~k〉〈~k| |ϕ〉

]

we use H0 |~k〉 =~2∣∣∣~k2∣∣∣

2m︸︷︷︸Ek

⇒ U0(τ) |~k〉 = e−i/~Ekτ |~k〉

Inspect: [e−ετU †(τ)V U0(τ)

]→[e−ετei/~τHe−i/~Ekτ V

]= exp

(− i~τ(Ek1−i (ε~)1− H

))V

Now:∞∫0

dτ exp(Ek − iε− H

)V

= −(i

~Ek − iε− H

)−1exp

(− i~τ(Ek − iε− H

))︸︷︷︸

(0−1)

∣∣∣∣∞0

= −i~G (Ek − iε) · V

⇒ |ϕ−〉 = Ω− |ϕ〉 = |ϕ〉+ limε↓0

∫d3k G (Ek − iε) V |~k〉〈~k|ϕ〉

also:

|ϕ+〉 = Ω |ϕ〉 = |ϕ〉+ limε↓0

∫d3k G (Ek + iε) V |~k〉〈~k|ϕ〉

These expressions give Ω±in terms of the operators:

G (Ek ± iε)V︸︷︷︸=G0(E±iε)T (E0±iε)

Now the scattering operator:

S := Ω†−Ω+

〈χ|S|ϕ〉 = 〈χ|Ω†−Ω+|ϕ〉 = limt→∞t′→−∞

〈χ|U †0 U(t)U †(t′)U0(t′)|ϕ〉



Take t′ = −t, same trick . . . :

〈χ|S|ϕ〉 = 〈χ|ϕ〉 − i

~limε↓0

∞∫0

dt e−εt 〈χ|U †0(t)V U(2t)U †0(t) + U †0(t)U(2t)V U †0(t)|ϕ〉

We used here:

U †(−t) = exp(− i~

(−t) H)†

= U(t)

So we calculate:

⇒ 〈~k′|S|~k〉 = δ(3)(~k′ − ~k

)− i

~limε↓0

∞∫0

dt 〈~k′|V ei/~(Ek′+Ek+iε−2H

)t + ei/~

(Ek′+Ek+iε−2H

)tV |~k〉

= . . .

= δ(3)(~k′ − ~k

)+ 1

2 limε↓0〈~k′|V G

(Ek′ + Ek

2 + iε

)+G

(Ek′ + Ek

2 + iε

)V |~k〉

with V G(z) = T (z)G0(z)and G(z)V = T (z)G0(z)

= δ(3)(~k′ − ~k

)+ 1

2 limε↓0

1Ek′+Ek

2 − Ek + iε+ 1

Ek′+Ek2 − Ek′ + iε

〈~k′|T(Ek + Ek′

2 + iε

)|~k〉

= δ(3)(~k′ − ~k

)+ lim

ε↓0

1

Ek′ − Ek + iε+ 1Ek − Ek′ + iε

︸︷︷︸

=−2πiδ(Ek−Ek′)

〈~k′|T(Ek + Ek′

2 + iε

)|~k〉

〈~k′|S|~k〉 = δ(3)(~k′ − ~k

)− 2πiδ (Ek − Ek′) lim

ε↓0〈~k′|T (Ek + iε) |~k〉

with:

Ek =~2∣∣∣~k∣∣∣2

2m

Ek′ =~2∣∣∣~k′∣∣∣22m

Where 〈~k′|T (z)|~k〉 is the “off-shell” T -matrix, which fulfills the Lippmann-Schwinger–Eq.,i.e.

〈~k|T (z)|~k〉 = 〈~k′|V |~k〉+∫

d3q〈~k′|V |~q〉z − Eq

〈~q|T (z)|~k〉

Therefore:

f(~k′,~k

)= − (2π)2m

~2 limε↓0

limε↓0〈~k′|T (Ek + iε) |~k〉

∣∣∣∣Ek=Ek′=

~2|~k|22m


1.4. SCATTERING

In Born–approximation:

〈~k′|T (Ek + iε) |~k〉 ≈ 〈~k′|V |~k〉∣∣∣∣Ek=Ek′

= 1(2π)3

∫d3x e

i

((~k−~k′

))·~xV (~x)

~~q := ~~k′ − ~~k momentum transfer1

(2π)3/2

(F−1V

)(~q)

Local potential:

δ (~x− ~y)V (~x)

〈~k′|V |~k〉 =∫

d3x d3y 〈~k′|~x〉〈~x|V |~y〉〈~y|~k〉

1.4.15 Stationary scattering states

For an orbit with an in-asymptote |φ〉 we have:

|φ〉 =∫

d3k |~k〉φ(~k)

Also at t = 0:

|φ+〉 := Ω+ |φ〉 =∫

d3kφ(~k)

Ω+ |~k〉︸︷︷︸|~k+〉

|~k±〉 = Ω± |~k〉

Intertwining relations:

HΩ± = Ω±H0

Thus:

H |~k±〉 = HΩ± |~k〉 = Ω±H0 |~k〉 = Ω±~2∣∣∣~k∣∣∣2

2m |~k〉

= Ω±Ek |~k〉 = Ek(Ω± |~k〉

)= Ek |~k±〉

The states |k±〉 fulfill the stationary Schrödinger-Equation H |ψ〉 = E |ψ〉again: |~k±〉 /∈ H. Nevertheless:

〈~k′ + |~k−〉 = 〈~k′|Ω†+Ω+|~k〉 = 〈~k′|~k〉 = δ(3)(~k′ − ~k

)We had:

|φ±〉 = |φ〉+ limε↓0

∫d3k G (Ek ± iε) V |~k〉〈k|φ〉

⇒∫

d3k φ(~k)|~k±〉 = lim

ε↓0

∫d3k φ

(~k) (|~k〉+ G (Ek ± iε) V |~k〉

)∀φ ∈ H

|~k±〉 = |~k〉+ limε↓0

G (Ek ± iε) V |~k〉



Some relations:

T (Ek ± iε) |~k〉 =[V + V G (Ek ± iε) V

]| |~k〉 lim

ε↓0

= V(1 + G (Ek ± iε) V

)|~k〉 = V |k±〉

Transition amplitude:

t(~k′ ← ~k

)= 〈~k′|V |~k±〉

|k±〉 = |k〉+ limε↓0

G (Ek ± iε) V |~k〉 = |~k〉+ limε↓0

G0 (Ek ± iε) T (Ek ± iε)

|~k±〉 = |~k〉+ limε↓0

G0 (Ek ± iε)V |k±〉


f(~k′,~k

)= − (2π)2m

~limε↓0〈~k〉′|T (Ek + iε) |~k︸︷︷︸t

(~k′←~k

)∣∣∣∣Ek=

~2|~k|22m =

~2|~k′|22m

where:

〈~k′|T (z)|~k〉 = 〈~k′|V |~k〉+∫

d3q 〈~k′|V |~k〉 1z − Eq

〈q|T (z)|~k〉

|~q〉 G0(z) 〈~q| Lippmann-Schwinger Equation for the transition operator T .

|ϕ±〉 := Ω± |ϕ〉 = |ϕ〉+ limε↓0

∫d3k G (Ek ± iε) V︸︷︷︸

G0(Ek±iε)T (Ek±iε)

|~k〉〈~k|ϕ〉

Ek =~2∣∣∣~k∣∣∣2

2m|ϕ〉 =

∫d3k |~k〉〈~k|ϕ〉

|ϕ±〉 =∫

d3k |~k±〉〈~k|ϕ〉

|k±〉 := Ω± |~k〉H |~k±〉 = Ek |~k±〉

|k±〉 = |~k〉 limε↓0

G0 (Ek ± iε) V |~k±〉

Lippmann-Schwinger Equation for stationary scattering states.Let’s continue:


1.4. SCATTERING

We know (Home exercise):

limε↓0〈~x|G0 (E ± iε) |~y〉 = − m

(2π) ~2|~x− ~y|exp

(±i|~x− ~y| k

)E = ~2k2

2mk =

∣∣∣~k∣∣∣⇒ 〈~x|~k±〉 = 〈~x|~k〉+

∫d3y 〈~x|G0 (E ± iε) |~y〉V (~y) 〈~y|~k±〉

= 〈~x|~k〉 − m

(2π) ~2

∫d3y

e±ik|~x−~y||~x− ~y|

V (~y) 〈~y|~k±〉

|y| |x|

expand |~x− ~y| = |x|

√√√√1− 2 (~x · ~y)|~x|2

+ |~y|2

|~x|2'|~x|

(1− (~x · ~y)

|~x|2+O

(|~y|)|~x|)

~x0 := ~x

|~x|

⇒ 〈~x|~k±〉 ' 〈~x|~k〉︸︷︷︸1

(2π)3/2·ei(~k·~x)

−me±ik|~x|2π~2|~x|

∫d3y e∓ik(~x0·~y)︸︷︷︸

(2π)3/2〈±k~x0|~y〉

V (~y) 〈~y|~k±〉

' 1(2π)3/2

·

ei(~k·~x)− (2π)2m

~2 〈±k~x0|V |~k±〉︸︷︷︸=f(±k~x0,~k

)e±ik|~x||~x|

For |~x| “range” R of the Potential.Since:

V |~k±〉 = limε↓0

T (E ± iε) |~k〉

In particular:

〈~x|~k+〉 −−−−→|~x|→∞

1(2π)3/2

(ei(~k·~x)

+ f(k~x0,~k

) eik|~x||~x|

)

Sommerfelds-Radiation condition (“Abstrahlbedingung”).

1.4.17 Partial wave stationary scattering states

(central potential V(|~x|))

Define:

|Elµ±〉 := Ω± |Elµ〉H |Elµ+〉 = E |Elµ+〉∣∣∣∣~L∣∣∣∣2 |Elµ+〉 = ~2l(l + 1) |Elµ+〉

L3 |Elµ+〉 = ~µ |Elµ+〉



We had:

〈~x|Elµ〉 = ϕElµ (~x) =

√2mπ~2

√2m~

iljl

(√2mE~|~x|)Ylµ (~x0)

H0ϕElµ (~x) = EϕElµ (~x)

similarly:

〈~x|Elµ+〉 =

√2mπ~2

√2mE~

il1∣∣∣~k∣∣∣Rlk

(|~x|)Ylµ (~x0)

where: − ~2

2m

(d2

dr2 + 2r

ddr −

l(l + 1)r2

)+ V (r)− E

Rlk(r) = 0

E = ~2k2

2m


〈~x|~k+〉 −−−−→|~x|→∞

1(2π)3/2

(ei(~k·~x)

+ f(k~x0,~k

) eik|~x||~x|

)

Ek = ~2k2

2mk =

∣∣∣~k∣∣∣~x0 = ~x

|~x|i.e. if |ψin〉 =

∫d3k ϕ

(~k)|~k〉

then |ψ〉 =∫

d3kϕ(~k)|~k+〉

is the actual scattering state at t = 0.Central potentials: V

(|~x|)

|Elµ+〉 := Ω+ |Elµ〉H |Elµ+〉 = E |Elµ+〉

Wave function:

〈~x|Elµ+〉 =

√2mπ~2

√2mE~

il1kRlk

(|~x|)Ylµ (~x0)

Ek = ~2k2

2m

Fulfills: − ~2

2m

(d2

dr2 + 2r− l(l + 1)

r2

)+ V (r)− E

Rlk(r) = 0


1.4. SCATTERING

Let’s continue.Choose solutions which are regular at r = 0, i.e. r ·Rlk(r)

∣∣∣∣r=0

= 0.also: one requires:

〈E′l′µ′ + |Elµ+〉 = δl′lδµ′µδ(E′ − E

)⇒∞∫0

dr r2R∗lk′(r)Rlk(r) = π

2 δ(k′ − k

)fixes the “normalisation”.Starting point:

〈~x|~k〉 =∫

dE∑l,µ

〈~x|Elµ〉〈Elµ|~k〉

=∫

dE∞∑l=0

l∑µ=−l

~√m∣∣∣~k∣∣∣δ

E − ~2∣∣∣~k∣∣∣2

2m

Y ∗lµ (~k0 〈~x〉 |Elµ)

also: 〈~x|~k+〉 = 〈~x|Ω+|~k〉

=∫

dE∑l,µ

〈~x|Elµ+〉︸︷︷︸=〈~x|Ω+|Elµ〉

~2√m∣∣∣~k∣∣∣δ

E − ~2∣∣∣~k∣∣∣2

2m

Y ∗lµ (~k0)

=∫

dE∑l,µ

√2mπ~2

√2mE~

ilRlk(|~x|) 1k

~√mk

δ

(E − ~2k2

2m

)Y ∗lµ

(~k0)Ylµ (~x0)

We use:l∑

µ=−lY ∗lµ (~x0)Ylµ

(~k0)

= 2l + 14π Pl

(cos

(](~x0,~k0

)))

and thus get:

=∞∑l=0

2l + 14π Pl

(~x0 · ~k0

)√ 2mπ~2k

~√mk

ilRlk(|~x|) 1k

= 1(2π)3/2

∞∑l=0

(2l + 1) · ilRlk(|~x|)Pl(~x0 · ~k0

) 1k

expansion of the stationary scattering states in terms of Rlk (~x).We know:

〈~x|~k+〉 −−−−→|~x|→∞

1(2π)3/2

ei

(~k·~x)

+ f(k~x0,~k

) eik|~x||~x|

also:

f(k~x0,~k

)=∞∑l=0

(2l + 1) fl(∣∣∣~k∣∣∣)Pl (~x0 · ~k0

)〈~x|~k+〉 −−−→

~x→∞

1(2π)3/2

∞∑l=0

[iljl

(∣∣∣~k∣∣∣|~x|)+ fl

(∣∣∣~k∣∣∣) eik|~x||~x|

]Pl(~x0 · ~k0

)



it follows (r = |~x|):

ilRlk (r)∣∣∣~k∣∣∣ −−−→r→∞

iljl (kr) + fl(k)eikrr

r ·Rlk(r) −−−→r→∞

kr · jl (kr) + kfl(k)eikr e−iπ/2·l︸︷︷︸=i−l

−−−→r→∞

sin(kr − lπ2

)+ eiδl(k) sin

(δl(k)

)︸︷︷︸e2iδl(k)−1

2i

ei(kr)−lπ/2

r ·Rlk(r) −−−→r→∞

= ei(kr−lπ/2) − e−i(kr−lπ/2)2i + e2iδl(k) − 1

2i · ei(kr−lπ/2)

= . . .

= eiδl(k) sin(kr − lπ2 + δl(k)

)Thus for r →∞ the radial part of the stationary partial wave scattering state has the form ofa free stationary wave, but with a phase-shift δl(k).Suppose we know R<lk(r) which is regular at the origin (e.g. by numerical integration of thestationary Schrödinger–Equation). The normalisation is arbitrary. Then build:

αl(k) := R ·[

ddr log

(R<lk(r)

)]r=R

where V (r) ≈ 0 for r ≥ R.IMAGE11for r > R we neglect the Potential and the solution to the stationary Schrödinger–Equationcan be written as a superposition of the solutions of the free Schrödinger–Equation.

jl(kr) and ηl(kr)

for r > R.Proposition: In fact:

R>lk(r) = k · eiδl(k)

cos(δl(k)

)jl(kr)− sin

(δl(k)

)ηl(kr)

indeed has the asymptotic form given by our previous approach to Rlk(r).Argument:

R>lk(r) = k

eiδl(k) cos(δl(k)

)jl(kr)− e−iδl(k) sin

(δl(k)

)ηl(kr)

Using:

sin(δl (k)

)= eδl(k) − e−δl(k)

2i

we can rewrite to:

= kjl (kr) + ke2iδl(k) − 1

2i

eiπ/2︸︷︷︸i

jl(kr)− ηl(kr)


1.4. SCATTERING

We also know:

ηl(kr) −−−−→kr→∞

−cos

(kr − lπ2

)kr

jl(kr) −−−−→kr→∞

−sin(kR− lπ2

)kR

Continuity of the logarithmic deivative at r = R then yields:

αl(k) = R ·ddrR

<lk(r)

R<lk(r)

∣∣∣∣r=R

!= R ·ddrR

>lk(r)

R>lk(r)

∣∣∣∣r=R

αl(k) is known (analytically or numerically)

⇒ αl(k) = k ·R · cos(δl(k)

)j′l(kR)− sin

(δl(k)

)η′l(kR)

cos(δl(k)

)jl(kR)− sin

(δl(k)

)ηl(kR)

= . . .

⇒ tan(δl(k)

)= (kR) j′l(kR)− αl(k)jl(kR)

(kR) η′l(kR)− αl(k)ηl(kR)

Example: hard sphere potentialIMAGE12

tan(δl(k)

)= jl(kR)ηl(kR)

δ0(k) = −kR

low energy scattering ~2k2

2m |V0|IMAGE13

ρ = kr

jl(ρ) ' ρl

(2l + 1)!!

ηl(ρ) ' −(2l − 1)!!ρl+1

ρ→ 0k → 0

tan(δl(k)

)=

ρ · l·ρl−1

(2l+1)!! − αl(k) ρl(2l+1)!!

ρ (2l − 1)!! (l + 1) 1ρl+2 + αl(k) (2l−1)!!

ρl+1

tan(δl(k)

)−−−→k→0

ρ2l+1 l − αl(k)(2l + 1)!! (2l − 1)!!

(l + 1 + αl(k)

)for low energies only partial waves with small l contribute.Special cases: if l + 1 + αl(k) = 0, i.e. αl (kR) ≈ −(l + 1)then tan

(δl (kR)

)=∞, i.e. δl (kR) ≈

(n+ 1

2

)π, n ∈ Z

then σl (kR) = 4πk2R

(2l + 1) sin2 (δl (kR))

= 4π(kR)2 (2l + 1) unitary bound.



Let’s expand around ER = ~2k2R

2m , αl (kR) = −(l + 1).

αl(E) = αl (ER) + (E − ER)α′l (ER) + . . .

= − (l + 1) + (E − ER)α′l (ER) + . . .

tan(δl(E)

)≈ (2l + 1) ρ2l+1

(l −

(l (l + 1) + (E − ER)α′l (ER)

))(2l + 1)!! (E − ER)α′l (EK)


Let R<lk (regular at r = 0) solution to− ~2

2m

(d2

dr2 + 2r

ddr −

l(l + 1)r2

)+ V (r)− ~2k2

2m︸︷︷︸=E

Rlk(r) = 0

for R with V (r) = 0 (r ≥ R) build αl(k) = R

[ddr log

(R<lk(r)

)]r=R

then:

tan(δl(k)

)= (kR)j′l(kR)− αl(k)jl(kR)kR · n′l(kR)− αl(k)nl(kR)

kR1≈ (kR)2l+1 l − αl(k)(2l + 1)!!(2l − 1)!!(l + 1 + αl(k))

Let’s continue:

let: αl (kR) = −l + 1 ER = ~2k2R

2mexpand: αl(E) = αl (ER) + α′l (ER) (E − ER) +O (E − ER)2

→ tan(δl(E)

)≈ (2l + 1)ρ2l+1

(l − (−l − 1) + (E − ER)α′l (ER)

)(2l + 1)!! (E − ER)α′l (ER)

E ≈ ER ρ = kr

= ρ2l+1 (2l + 1)2

(2l + 1)!!α′l (ER)1

E − ER− (2l + 1) ρ2l+1

(2l + 1)!! +O (E − ER)2

The first term is the dominant one and thus called the resonance term, the other terms aresmall regular terms.

tan(δl(E)

) E≈ER≈ − γρ2l+1

E − ER+ smaller regular terms . . .

γ = − 1αl (ER)

(2l + 1)2

(2l + 1)!!scattering amplitude:

fl(k) = e2iδl(k) − 12i · k

...= tan δl(k)k(1− i · tan

(δl(k)

))σl = 4π(2lm)

k2tan2 (δl(k)

)1 + tan2 (δl(k)

) E≈ER≈ 4πk2 (2l + 1)

(γ(kR)2l+1

)2

(E − ER)2 +(γ(kR)2l+1)2

This is the Breit-Wigner formular for the partial cross section.

Remember:

fl(k) = e2iδl(k) − 12ik

⇒ kfl(k)


1.4. SCATTERING

This is called the Argand amplitude (see exercises).

= 12

(i+ ei(2δl(k)−π/2)

)If E → Ek δl(k)→ π

2 , and kfl(k)→ i.

fl(k) E≈ER≈ 1k

−γ(kR)2l+1

(E − ER) + iγ(kR)2l+1

IMAGE14As a complex function of complex energies E has a pole of E = ER − iγ(kR)2l+1.Now: PDF-presentation (see web page).


Low energy scattering

Ek V

ρ = kR 1

tan δl(k) = (kR)2l+1(l − αl(k)

)(2l + 1)(

(2l + 1)!!)2 (

l + 1 + αl(k))

αl log derivative at r = R where V (r) = 0, r > R

if αl (Ek) = −l − 1 ⇒ δl (kR) =(n+ 1

2

)π

σl(k) maximal ↔ resonancethen

tan(δl(E)

)= −γ (kR)2l+1

E − ER+ regular terms

γ = − 1α′l (Ek)

(2l + 1)(2l + 1)!!

E ≈ ER

fl(E) = −1k

−γ(kR)2l+1

E − ER + iγ(kR)2l+1 Breit-Wigner Formula

scattering amplitude has a pole at

E = ER − iγ(kR)2l+1

fl(k) = e2iδl(k) − 12ik

kfl(k) = 12

(i+ ei(2δl(k)−π/2)

)IMAGE14Compare

fl(E) = −1k

A(E)E −

(ER − iΓ(k)/2

)Γ (kR) = 2γ (kRR)2l+1

is called width of the resonance.



Let’s continue...We consider l = 0 for very low energies kR 1IMAGE15

tan δ0(k) = kR · 0− α0(k)1 + α0(k)

⇒ cot δ0 = −1 + α0α0

1kR

⇔ k · cot δ0 = − 1α0R

− 1R

= − 1a0

+O(k2)

︸︷︷︸12 r0k

2

a0 is called the scattering length and r0 is the effective range (perhaps later).

⇒ − 1R

1 + α0(0)α0(0) = − 1

a0

⇔ α0(0)1 + α0(0) = a0

R

kR 1 in fact k → 0

sin2 δ0 ≈ tan2 δ0 ≈ δ20 = (kR)2 α0(0)2(

1 + α0(0))2 = k2R2 a

20R2 = k2a2

0

then σ0 ≈ 4πa20 partial wave cross section for l = 0.

→ scattering amplitude:

f0 = e2iδ0 − 12ik = 2i

cot δ0 − i· 12ik

= 1−1/a0 − ik

= a0−1− ika0

⇒ f0 ≈ −a0 if ka0 1

We can inspect f0 and find:

f0(k) = a0−1− ika0

has a pole at k = iκ with κ = 1/a0

This corresponds to an energy E = ~2k2/2m = −~2κ2/2m < 0Radial Schrödinger-Equation (l = 0):− ~2

2m

(d2

dr2 + 2r

ddr

)+ V (r)− E

Rk(r) = 0

for r > R where V (r) = 0

uk(r) := r ·Rk(r)⇔ u′′(r) = κ2u(r)

where indeed: E = −~2k2

2m < 0

⇒ u′′(r) ∝ e−κr


1.4. SCATTERING

is the regular solution for r →∞ with eigenvalue E = − ~2k2/2mSo for a0 > 0 then this corresponds to a bound state with Energy −κ2~2/2m and a bound statewave function

∝ e−κrr

= e−r/a0

r

Note:

|κR| = |kR| 1

We conclude: The Scattering length a0 is then determined by the properties of a very weaklybound state with wave function e−r/a0/rThen we find:

σ0 = 4π a20

1 + k2a20

= 4π~2

2m~22m

1a2

0+ ~2k2

2m=

2π ~2

m

Ek − Eb

Let’s inspect the stationary scattering state

r > R R>ok(r) = eiδ0(j0(kr) cos δ0 − η0(kR) sin δ0

)(l = 0) = eiδ0

kr

(sin(kr) cos δ0 + cos(kr) sin δ0

)kR 1 ≈ eiδ0

kr

(kr · cos (δ0) ·

(

1− 12 (kr)2

)sin δ0

)

= eiδ0kr

sin δ0 (1 + kr · cot δ0)

= eiδ0kr

(1− kr

ka0

)sin δ0

= eiδ0 sin δ0kr

(1− r

a0

)r smallscattering wave function crosses the r-axis at r = a0. For a0 < 0, there are no bound states.IMAGE16

2. remarks: IMAGE17Application:

n-p:

mpc2 ' mnc

2 ≈ 1 GeV a(

3S1)

total spin S = 1 l = 0

with J = L+ S and 2S + 1 = 3experimental number: 5.4 fmEstimate for bound state energy of(

3S1)

= ~2/2 · µ1a2

0

= (~c)2

2µc2 · a20

= (0.2 GeV fm)2

1 GeV · (5.4)2 fm2 = 1.3 MeV



µ is the reduced mass

µ = mp ·mn

mp +mn≈ 1

2mp = 12mn

experimentally the binding energy of the deuteron (mp in 3S1)

Eexpb ≈ 2.2 MeV

1.4.21 A small excursion on Jost-function

Special case l = 0:

H = − ~2

2m∆ + V (r)l=0⇔ Φ′′k(r) +

(k2 − U(r)

)ϕk(r) = 0

U(r) := 2mV (r)~2

Φk(r) = r ·RK(r)

Consider solutions with special boundary conditions (r = 0), i.e. Φk(0) = 0, Φ′k(r)∣∣∣∣r=0

= 1U(r) = 0 for r > RStatements:

1. There are functions f(k), g(k) with

Φk(r) = 12ik

[f(k)eikr − g(k)e−ikr

]Argument: of course: for r →∞:

Φ′′k(r) = −k2Φk(r)

tho boundary conditions Φk(0) = 0, Φ′k(r) = 1 then fixes f(k), g(k)

2. it follows:

g(k) = f(−k) for real kf∗(−k∗

)= f(k) for complex k

Argument: inspect

Φ−k(r) = − 12ik

(f(−k)e−ikr − g(−k)eikr

)at least for r > R

Φ′′−k(r) = −k2 · Φ−k(r)

uniqueness of the solution (fixed by boundary conditions) then requires: f(−k) = g(k)for real k. Therefore:

Φk(r) = 12ik

[f(k)eikr − f(−k)e−ikr

]


1.4. SCATTERING

is called the Jost-solution.Consider

(Φk(r)

)∗ and Φk∗(r). . .φk∗(r) =

(φk(r)

)∗. . .⇒ f∗(−k) = g∗(k) = f (k∗)⇒ f(−k) =

(f (k∗)

)∗And f(k) is called the Jost-function.

3. inspectuk(r)r

:= Ω+j0 (kr) uk(r) = r ·Rk(r)

rRk(r) −−−→r→∞

r · j0(kr) + r · S0(k)− 12ik

eikrr

= eikr − e−ikr2ik + S0(k)− 1

2ik eikr

= − 12ik

(−S0(k)eikr + e−ikr

)⇒ φk(r)

f(−k) = − 12ik

(− f(k)f(−k)eikr + e−ikr

)

⇒ φk(r)f(−k) = uk(r)

⇒ S0(k) = f(k)f(−k)

We already had Rk(r).


low energy scattering (l = 0)

− 1a0

= limk→0

k cot(δ0(k)

)scattering length

f0(k) ' a0−1− ika0

(l = 0) scattering amplitude

σo ' 4π a20

1 + k2a20

partial wave cross-section

∃ bound state w.f. u(v) ∝ e−r/a0

with

Eb = ~k2

2m = − ~2

2ma20

σ0(E) ≈ 2π~2/m

E − Ebstationary scattering wave for r > R, kR 1, u>(r) ∝

(1− r/a0

)IMAGE16IMAGE18

Jost-function: rad. Schrödinger-equation (l = 0,Φ(r) = r ·R(r), U(r) = 2mV/(r)~2)

Φ′′k(r) +(k2 − U(r)

)Φk(r) = 0

b.c. Φk(r) = 0 Φ′k(0) = 1

⇒ Φk(r) = 12ki

[f(k)eikr − f(−k)e−ikr

]



Properties of the Jost-function f(k):

f(−k) =(f(k∗))∗

S-matrix S0(k) = f(k)f(−k)

uk(r) = φk(r)f(−k)

for:

uk(r)r

:= Ω+j0(kr)

S0(k) = e2iδ0(k)

Let’s continue. . .Proposition:

f(k) =∣∣f(k)

∣∣ · eiδ0(k)

Arguments:

let f(k) =∣∣f(k)

∣∣ · eiφ(k)

Inspect real k:

f(−k) =(f(k)

)∗ =∣∣f(k)

∣∣ e−iφ(k)

k ∈ R

⇒ f(k)f(−k) = e2iφ(k) = s0(k) = e2iδl(k)

⇒ φ(k) = δ0(k)

Zeroes of f(−k) ↔ poles of the s0-matrix.

1. Suppose k ∈ R. If f(−k) = 0 ⇒ f∗(k) = 0, because f(−k) = f(k)∗ if k ∈ R, ⇔ f(k) = 0⇒ φk ≡ 0, but this contradicts Φ′k(0) = 1.⇒ “Jost-Function does not have zeroes for real k.

2. Let k = iκ, κ ∈ R, κ > 0

φ(ik)(r) = − 12κ[f(ik)e−κr + f(−iκ)eκr

]Now, if f(−iκ) = 0, then:

φ(ik) = − 12κf(ik)e−κr

This is a square integrable solution of the (l = 0) Schrödinger-Equation with eigenvaluek2 = −κ2 < 0, so this corresponds to a bound state.

3. k = −iκ, k ∈ R, κ > 0 ⇒ no square integrable solution of the Schrödinger-Equation(anti-bound states).

4. k = κ + iλ ∈ C, k 6= 0, λ > 0, so k is in the upper complex half plane. For f(−k) = 0,we then have φk ∝ e−λreikr which is square-integrable. However, the eigenvalue is k2 =κ2 − λ2 + 2iκλ.


1.4. SCATTERING

5. Let k = κ− iλ, k 6= 0, λ > 0then

ϕk = 12ik

(f(k)eλreiκr − f(−k)e−λre−iκr

)suppose f(−k) = 0, we have f(k)eλreiκr, this is not square-integrable, but will correspondto a stationary scattering solution, in fact to a resonance.IMAGE19

f(−α− iβ) = 0⇒ f(α− iβ) = 0

fm f(−k) = f∗(k∗)

1. example: Suppose that f(k) can (for some k) be approximated by:

f(k) = (k − k0)(k + k∗0

)where

k0 = κ+ iλ |λ| |κ|

Check:

f(−k) = (−k − k0)(−k + k∗0

)= (k + k0)

(k − k∗0

)f(k∗)

=(k∗ − k0

) (k∗ + k∗0

)→(f(k∗))∗

=(k − k∗0

)(k + k0)

So this is okay as an “Ansatz”.Calculate s-wave scattering amplitude

f0(k) = s0(k)− 12ik

. . .

f0(k) = −2λk2 −

(k2 + λ2)+ 2ikλ

σ0λκ≈ 4πλ2(

k2 − κ2)+ 4κ2λ2


Jost–solution:

Φk(r) = 12ki

f(k) · eikr − f(−k) · e−ikr

of Φ′′k(r) +

(k2 − U(r)

)· Φk(r) = 0

Φk(0) = 0Φ′k(0) = 1

We have: f(−k) =(f(k∗))∗

s0(k) = f(k)f(−k)

f(k) =∣∣f(k)

∣∣ eiδ0(k)

Singularities of s0(k)⇔ zeroes of f(−k)



1. Ansatz: f(k) = (k − k0)(k + k∗0

), k0 = κ+ iλ, λ κ

σ0(k) = 4πλ2(k2 − κ2)2 + 4k2λ2

Let’s continue. . .Example 2:

f(k) = k + iβ

k − iαα, β ∈ R+

f(−k) = −k + iβ

−k − iα= k − iβk + iα

f(k∗)

= k∗ + iβ

k∗ − iα

→(f(k∗))∗

= k − iβk + iα

= f(−k) okay!

Ansatz:

f(−k)⇔ −k + iβ = 0 ⇒ bound state

k = iβ Eb = −~2β2

2mlimk→∞

f(k) = 1

f(k) =√k2 + β2

k2 + α2 eiδ0(k)

f(k) = k + iβ

k − iα= (k + iβ) (k + iα)

k2 + α2

f(k) = k2 − αβ + i (α+ β) k(k2 + α2)

⇒ tan δ0(k) = (α+ β) kk2 − αβ

cot δ0(k) = k2 − αβ(α+ β) k

⇒ k · cot δ0(k) = k2 − αβα+ β

= − αβ

α+ β+ 1α+ β

· k2

For k very small, compare with effective range expansion:

= − 1a0

+ 12r0k

2

With a0 the scattering length and r0 the effective range. So we get for the scattering length:

a0 = 1β

+ 1α

⇒ 1a0

= αβ

α+ β= β (α+ β)

α+ β− β2

α+ β= β − 1

2r0β2

Eb = −~2β2

2m


1.4. SCATTERING

And for the effective range:

r0 = 2α+ β

Example: n–p scattering: 3S1–channel, i.e. S = 1, L = 0, → J = 1Experimental numbers:

r0 = 1.72 fma0 = 5.40 fm

E3S1b = 2.225 MeV

⇒ 1β

=√

~2c2|Eb|2µc2 ≈ 4.32 fm

µ ≈ 12 · 1

GeV/c2 reduced mass

1(5.40)

?= 1(4.32) −

12

(1.72)(4.32)2

⇔ 0.185 ≈ 0.183


2 Relativistic Wave equations

Intro:

non-rel. Schrödinger-Equation:

i~∂

∂tψ (t, ~x) =

[− ~2

2m∆ + V (~x)]ψ (t, ~x)

If rotational invariance:

V (~x) = V(|~x|)[

H,Dg

]−

= 0

where:(Dgψ

)(t, ~x) := ψ

(t, g−1~x

)g ∈ SO (3)

Dg is unitary, in face:

Dg(~ω) = e−i/~

(~ω·~L)

g : R3 → R3(~ω · ~L

):= ω1L1 + ω2L2 + ω3L3

Li components of angular momentum operator

Consequences: If ψ (t, ~x) is a solution to the Schrödinger-Equation(V (~x) = V

(|~x|))

thenalso Dgψ (t, ~x) is also a solution of the Schrödinger-Equation.“Solutions are representatives of the rotation group in L2

(R3,C

)”

Now we shall turn this around: We want to have an equation of which the solutions arerepresentations of a symmetry group.We shall investigate the Poincaré–Group.This group acts in R4 with Minkowski–Metric.

x, y ∈ R4 g(x, y

):= gµνx

µyν

This is called Einstein’s summation convention.R4, g (_,_)


is called Minkowski–Space,

M = R1,3

x ∈ R4 x =(x0, ~x

)x0 = ct ~x ∈ R3

g (x, x) =(x0)2−(x1)2−(x2)2−(x3)2

=(x0)2−|~x|2 = gµνx

µxν

g00 = 1 gii = −1, i = 1, 2, 3gµν = 0 otherwise

Subgroup of the Poincaré–Group: Lorentz–Group L

Λ ∈ L

⇔ Λ : M→M∣∣∣∣∀x ∈M : g (Λx,Λx) = g (x, x)

⇔ f(Λx,Λy

)= g

(x, y

)∀x, y ∈M

i.e. the Lorentz–transformations Λ are isometries of the metric g (_,_)Poincaré–Group P are all Lorentz–transformations plus space and time translations, i.e.

A ∈ P : ∃a ∈M,∃Λ ∈ LAx = Λx+ a(a0~a

)= a

Statement:

Λ is Lorentz-Transformation

∀Λ∣∣∣∣g (Λx,Λx) = g (x, x) ∀x ∃ν1, ν2 ∈ 0, 1

∃ν ∈ R∃O1,O2 ∈ SO (3)

such that:

Λ = O1Λ0O2Pν1T ν2

O1 ⇔

1 0 0 000 O10

O1 : R3 → R3

O1 : M→M

with:

P

(x0

~x

)=(x0

−~x

)space reflection

T

(x0

~x

)=(−x0

~x

)time inversal


CHAPTER 2. RELATIVISTIC WAVE EQUATIONS

Λ0 is a boost in the 1–direction. Its matrix form is:

Λ0 (v) :

γ γβ 0 0γβ γ 0 00 0 1 00 0 0 1

where:

β := v

cγ := 1√

1− β2

We now define a representation of the Poincaré–group in the space of complex functions:

Φ : M→ C

The idea is that this should give an equation of motion. Let indeed:

Φ : M→ Cx 7→ Φ (x)

with:

Ax = Λx+ a = A (Λ, a)x

then define:(ρ (A) Φ

)(x) = Φ

(A−1x

)indeed:

A−1x = Λ−1 (x− a)

Check:

A(A−1x

)= Λ

(A−1x

)+ a

= Λ(Λ−1 (x− a)

)+ a

= x− a+ a = x

In order that ρ(A) is a representation: one should have ρ (1)M→M = 1 (multiplied by 1)and also:

ρ (A1) ρ (A2) = ρ (A1A2)

with:

A1A2 = (Λ1Λ2, A1a2 + a1) exerciseA1 (Λ1, a1)A2 (Λ2, a2)


We are looking for an operator K (invariant) (for free motion), such that

Kφ (x) = 0 ⇒(K(ρ(A)φ

))(x) = 0

Now:

φ : M→ C

Known operators:

M→M

which can be used to construct:

(M→ C)→ (M→ C)

1. 1 → 1

2. xµ, so inspect: (xµφ

)(x) = xµ · φ (x)

multiply by xµ, µ = 0, 1, 2, 3

3.

pµ := i~∂

∂xµ= i~∂µ (derivative)

p0 = i~∂0 = i~∂

∂x0 = i~1c

∂

∂t

pk = i~∂k = i~∂

∂xk=(i~~∇

)k

i = 1, 2, 3

pk = i~∂

∂xk= −i~∇k k = 1, 2, 3

Now:

1. 1 is Poincaré–invariant:

ρ (A)1 = 1 · ρ (A)

candidate for a part of K

2.

x2 := xµxµ

this however is not Poincaré–invariant




Wanted: Operator K on φ : R1,3 → C with(Kφ

)(x) = 0 ⇒

(K(ρ(A)

)φ)

(x) = 0

∀A ∈ PA = A (Λ, a)Ax = Λx+ a

Λ ∈ L

i.e.(Λx · Λy

)=(x · y

)=(x0y0

)− (~x · ~y)

(ρ(A)φ

)(x) := φ

(A−1x

)⇒

ρ (A1) ρ (A2) = ρ (A1A2)ρ (1) = 1

linear operators:

1,xµ

pµ = i~∂

∂xµ= i~δµ

p0 = p0 = i~c

∂

∂t

pi = i~∂

∂xi

pi = −i~ ∂

∂xi

i = 1, 2, 3

Let’s continue. . .Remark:

if[K, ρ (A)

]−

= 0 then indeed:

(K(ρ(A)φ

))(x) =

ρ(A) · Kφ︸︷︷︸0

(x) = 0

Candidates for K:

1. K = 1, 1ρ(A) = ρ(A)1

2. x2 = (x · x) = xµxµ is indeed Lorentz-invariant but not Poincaré-invariant.Argument: take e.g.

Ax = x+ a Λ = 1

ρ(A)(x2φ (x)

)= (x · a)2 φ (x− a)

=(x2 − 2 · (x · a) + a2

)φ (x− a)

6= x2ρ(A)φ (x)φ(x−a)

if a = 0


3. also something like xµpµ is not Poincaré-invariant, thus no candidate for K

4. but pµpµ is indeed Poincaré-invariant. (exercise!)

So the simplest operator is:

K = pµpµ −m2c21

with m being the mass of the particle and c the velocity of light.

= −~2

c2∂2

∂t2+ ~2︸︷︷︸

=:−~2

∆−m2c2

D’Alembert-Operator: := 1c2∂2

∂t2−∆

So the equation of motion for a relativistic particle with mass m (Spin 0):[+ m2c2

~2

]φ (x)︸︷︷︸

=φ(t,~x),x0=ct

= 0

This is the free Klein–Gordon equation.Goal: On the space of solutions to the Klein–Gordon Equation we want to define a scalar-product in order to construct a Hilbert-space. Then ρ(A) should be unitary with respect tothis scalar-product.NRQM:

ρ (t, ~x) = ψ∗ (t, ~x)ψ (t, ~x)~ (t, ~x) = ~2mi

[ψ∗ (t, ~x) ~∇ψ (t, ~x)− ψ (t, ~x) ~∇ψ∗ (t, ~x)

]⇒ ρ (t, ~x) + div~ (t, ~x) = 0

This is the continuity equation (compare E-Dyn.).

if: i~∂tψ (t, ~x) = Hψ (t, ~x)

So we are looking for a 4-vector field:

jµ (x) =(cρ (t, ~x) ,~ (t, ~x)

)j0 (x) = cρ (t, ~x)ji (x) = ji (t, ~x)

with the properties:

1.

()µ (x) = jµ[φ]

(x)

= Λµν jν(Λ−1x

)for φ (x) = φ

(Λ−1x

)



2. (∂µj

µ [φ])

(x) = 0 continuity equation

if[+ m2c2

~2 φ (x)]

= 0

Then indeed:

〈φ, φ〉 =‖φ‖2 :=∫t=0

d3x j0 [φ] (t = 0, ~x)

should define a norm, which is Poincaré-invariant.


Poincaré-invariant operator for scalar functions:

φ : R1,3 → C

K = pµpµ −m2c2 · 1 ⇔ Klein-Gordon Equation

:= 1c2∂2

∂t2−∆(

+ m2c2

~2

)φ (x) = 0

Wanted: “scalar” product, norm → Hilbert-space ⇒

current j [φ]j0 (x) = c · ρ (t, ~x)j (x) =

(c · ρ (t, ~x) ,~ (t, ~x)

)such that for

φ (x) = φ(Λ−1 (x− a)

)we have:

µ (x) = jµ[φ]

(x) = Λµν jν(Λ−1x

)a = 0

with µ (x) being a 4-vector-field. Furthermore:(+ m2c2

~2

)φ (x) = 0 ⇒ ∂µj

µ [φ] (x) = 0

This is the continuity equation. Then:

‖φ‖2C = 〈φ, φ〉C =∫

d3x j0 [φ] (t, ~x)

∣∣∣∣∣∣t=0


should define a Poincaré-invariant norm.Let’s continue . . .Inspect a hypersurface (3-dimensional) ⊂ R1,3, call this σ. σ can be parametrised by paramet-ers: u, v, w, so:

σ = xµ | xµ (u, v, w)

We can then define an infinitesimal surface / volume element:

dσµ (x) := εµνρτ∂xν

∂u

∂xρ

∂v

∂xτ

∂wdudv dw

with εµνρτ being the Levi-Civita tensor and:

ε0123 = +1

It is antisymmetric in all indices. Now built:

∗j (x) = jµ (x) dσµ (x)

How does this transform under Lorentz-transformations? Suppose j is a 4-vector-field:

µ (x) = Λµν jν(Λ−1x

)then:

∗ (x) := µ (x) dσµ (x) = Λµκjκ(Λ−1x

)︸︷︷︸

=:y

εµνρτ∂xν

∂u

∂xρ

∂v

∂xτ

∂wdudv dw

x = Λy

= jκ(y)

ΛµκΛµκΛναΛρβΛτγεµνρτ︸︷︷︸|detΛ|︸︷︷︸

=1

εκαβγ

∂yα

∂u

∂yβ

∂v

∂yγ

∂wdu dv dw

= jκ(y)

dσκ(y)

= ∗j(y)

y = Λ−1x

Proposition: ∫σ

∗ (x) =∫σ

(∗j) (x)

if

∂µjµ (x) = 0

Argument: ∫σ

∗ (x) =∫σ

jµ(y)

dσµ(y)

y = Λ−1x σ = Λ−1σ

=∫σ

∗j(y)



Inspect:

∆Q :=∫σ

jµ(y)

dσµ(y)−∫σ

jµ (x) dσµ (x)

First consider a finite hypersurface σ:IMAGE20

∆Q =∫∂V4

jµ (x) dσµ (x)−∫

∂V4\σ,σ

jµ (x) dσµ (x)

(orientation!)

=∫V4

∂µjµ (x)︸︷︷︸

=0

d4x−∫

∂V4\σ,σ

jµ (x) dσµ (x)

The second part goes to zero, too, if the borders of σ (σ)→∞ where we suppose:

limx→∞

j (x) = 0

So ∆Q = 0:

⇒∫σ

µ (x) dσµ (x) =∫σ

jµdσµ (x)

Independent of the parametrisation.

Now take:

t = 0 u = x1, v = x2, w = x3

then:

dσµ (x) = ε0123︸︷︷︸=1

dx1 dx2 dx3︸︷︷︸d3x

Then:

jµ (x) dσµ (x) = j0 (t = 0, ~x) 1 d3x

⇒∫0 (t = 0, ~x) d3x =

∫j0 (t = 0, ~x) d3x

We propose:

jµ [φ] (x) = i~2mc φ

∗ (x) (∂µφ) (x)−(∂µφ∗

)(x)φ (x)

For solutions of:(+ m2c2

~2

)φ (x) = 0


Then indeed:

∂µjµ (x) = 0 (exercise!)

and also:

µ (x) =(j[φ])µ

(x) = Λµν jν(Λ−1x

)φ (x) = φ

(Λ−1x

)i.e. j is a 4-vector field (exercise!).And finally (already proved): ∫

t=0

d3x j0 [φ] (x)

is indeed invariant.However, first inspect the solutions to the free Klein-Gordon Equation:(

+ m2c2

~2

)φ (x) = 0

Ansatz:

φ (x) =∫

d4k f (k) · e−i(k·x)

φ (x) =∫

d4k f (k) (−ikµ)(−ikµ

)e−i(k·x) = −

∫d4k f (k) · k2e−i(k·x)

k2 = (k · k)(+ m2c2

~2

)φ (x) =

∫d4kf (k)

(m2c2

~2 − k2)

e−i(k·x) != 0

⇒(m2c2

~2 − k2)f (k) = 0

⇒if k2 6= m2c2

~2 then f (k) = 0

So f (k) 6= 0 only if (k0)2−∣∣∣~k∣∣∣2 = m2c2/~2

⇔ k0 = ±

√m2c2

~2 +∣∣∣~k∣∣∣2

Define:

Ωk =

√m2c4

~2 +∣∣∣~k∣∣∣2 c2

thus: k0 = ±Ωk

C

IMAGE21which is the support of the solutions to the free Klein-Gordon equation.



We thus write:

f (k) = g (k) δ(k2 − m2c2

~2

)

accordingly:

φ (x) =∫

d4k δ

(k2 − m2c2

~2

)g (k) e−i(k·x)

Now for a Lorentz-transformation:

Λ ∈ L ⇒ k2 =2

δ

(k2 − m2c2

~2

)invariantk = Λk

d4k = |detΛ|︸︷︷︸=1

d4k = d4k

we could propose a norm:

‖φ‖ =∫

d4k δ

(k2 − m2c2

~2

)∣∣g (k)∣∣2

Now two questions arise:

1. Is this indeed Poincaré-invariant? Yes! (exercise!)

2. Is this norm consistent with the current norm ‖φ‖C we defined with j0? No!

Remarks on item 2:We had:

jµ (x) = i~2mc φ

∗ (x) (∂µφ) (x)−(∂µφ∗

)(x)φ (x) (2.1)

We take the form:

φ (x) =∫

d4k δ

(k2 − m2c2

~2

)g (k) e−i(k·x) (2.2)

=∫

d3k

∫dk0 δ

k20 −

(∣∣∣~k∣∣∣2 + m2c2

~2

)︸︷︷︸

Ω2k/c2

g(k0,~k

)e−i(k0x0−

(~k·~k))

(2.3)


We remember the mass shell and find:

Ek = ~Ωk =√√√√√m2c4 +

∣∣∣~~k∣∣∣2︸︷︷︸~p

·c2 (2.4)

δ(f(x)

)=∑i

δ (x− xi)∣∣f ′ (xi)∣∣ (2.5)

for f (xi) = 0 f ′ (xi) 6= 0(2.6)

=∫

d3k

∫dk0 C

2Ωk δ(k0 − Ωk

c

)+ δ

(k0 + Ωk

c

) g(k0,~k

)e−i(k0·x0−~k·~x

)(2.7)

x0 = ct (2.8)

=∫

d3kC

2Ωk g(

Ωk

c,~k

)︸︷︷︸

=:g+(~k)

e−i(

Ωkt−(~k·~x))

+ g

(−Ωk

c,~k

)︸︷︷︸

=:g−(~k)

e−i(−Ωkt−

(~k·~x)) (2.9)

we have:

φ (x) = φ (t, ~x) =∫

d3kC

2Ωkg+(~k)

ei

((~k·~x)−Ωkt

)+∫

d3kC

2Ωkg−(~k)

ei

((~k·~x)+Ωkt

)(2.10)

= φ+ (t, ~x) + φ− (t, ~x) (2.11)

Linear combination of positive / negative frequency solutions.Now with equation (2.11) calculate:

‖φ‖C =∫

d3x j0 [φ] (t = 0, ~x)

. . .

(2π)3 ~2m

∫ d3k

2Ωk∣∣∣∣g+

(~k)∣∣∣∣2 −∣∣∣∣g− (~k)∣∣∣∣2 =‖φ‖C

So this is unfortunately indefinite!

. . .however:

‖φ‖2 =∫

d3k δ

(k2 − m2c2

~2

)∣∣g (k)∣∣2

. . . = c

∫ d3k

2Ωk∣∣g+ (k)

∣∣2 +∣∣g− (k)

∣∣2 2.0.26 Repetitorium 16

Current:

hµ (x) = i~2mc ψ

∗ (x) (∂µφ) (x)−(∂µφ∗

)(x)φ (x)



Klein-Gordon Equation:(+ m2c2

~2

)φ (x) = 0

⇒ ∂µj [φ]µ (x) = 0(ρ (A)φ

)(x) = φ (x) := φ

(Λ−1 (x− a)

)→ µ (x) =

(j[φ]µ

(x))

= Λµν jν(Λ−1 (x− a)

)4-vector field

⇒∫

d3x j0 (t, ~x)

∣∣∣∣∣∣t=0

=‖φ‖2c

Poincaré-invariant

φ (x) =∫

d4k δ

(k2 − m2c2

~2

)e−i(k·x) · g (k)

=∫

d3kc

2Ωkg+(~k)

ei

((~k·~x)−Ωkt

)︸︷︷︸

=:φ+(t,~x)

+∫

d3kc

2Ωkg−(~k)

ei

((~k·~x)+Ωkt

)︸︷︷︸

=:φ−(t,~x)

Ωk :=

√|k|2 c2 +

(mc2)2~2

g±(~k)

= g

(±Ωk

c,~k

)

‖φ‖2c = (2π)3 · ~2m ·

∫ d3k

2Ωk

∣∣∣∣g+(~k)∣∣∣∣2 −∣∣∣∣g− (~k)∣∣∣∣2

indefinite

Let’s continue. . .We now define a sesqui-linear form (“scalar product”)

〈φ1, φ2〉c := i~2mc

∫d3x

[φ∗1 (x)

(∂0φ2

)(x)− φ2 (x)

(∂0φ∗1

)(x)]∣∣∣∣∣∣x0=ct=0

Which is (anti-)linear in the first and second argument.Proposition:

〈φ+, φ−〉c = 0


Argument:

〈φ+, φ−〉c = i~2mc

∫d3x

[φ∗+ (x)

(∂0φ−

)(x)−

(∂0φ∗+

)(x)φ− (x)

]∣∣∣∣∣∣x0=0

⇒ 〈φ+, φ−〉c = i~2m

∫d3x

∫d3k

∫d3k′

g∗+

(~k′)

e−i(~k′·~x)

(iΩk) g−(~k)

ei(~k′·~x)

− (iΩk′) g∗+(~k′)

e−i(~k′·~x)

(iΩk) ·1

2Ωk· 1

2Ωk′g−(~k)

ei(~k′·~x)

∫d3x e

i

((~k−~k′

)·~x)

= (2π)3 δ(3)(~k − ~k′

)= − ~

2m12

∫ d3k

2Ωk

(g∗+

(~k)g−(~k)− g∗+

(~k)g−(~k))

= 0In fact one can write the norms ‖φ±‖C in an explicit Poincaré-invariant form (apart fromtime reflections).

‖φ+‖C =∫

d4k δ

(k2 − m2c2

~2

)Θ(k0)∣∣∣∣g (~k)∣∣∣∣2 ~

2mc (2π)3

‖φ−‖C = −∫

d4k δ

(k2 − m2c2

~2

)Θ(−k0

)∣∣∣∣g (~k)∣∣∣∣2 ~2mc (2π)3

Θ(x) =

1 for x > 0,0 for x < 0.

Thus:

H = H+ ⊕H−

with:

H± :=

φ± : R1,3 → C∣∣∣ (+ m2c2

~2

)φ± = 0;‖φ±‖2C ≷ 0; g∓ = 0

H :=

φ : R1,3 → C∣∣∣ (+ m2c2

~2

)φ = 0

With H+⊥H−. Moreover H± are invariant subspaces:

ρ(A)H± ⊂ H±A representation of P on H thus decomposes in irreducible representations on H± with ‖φ±‖2Cpositive / negative definite.Both irreducible representations describe particles with spin 0 and mass m with a frequency:

ωk = Ωk =

√∣∣∣~k∣∣∣2 c2 +(mc2)2~2

and:

E~p =√|~p|2 c2 +

(mc2)2 for φ+

and frequency:

ωk = −Ωk for φ−




NRQM: Spin 1/2 particles (Pauli-Equation)for rotations (∈ SO(3)) representations of SU(2):

ψ ∈ L2(R3,C2

)(D (g)

)(~x) = gψ

(ρ (g)−1 ~x

)g ∈ SU(2)g = ρ (g) ∈ SO(3)

e.g. if:

g~ω = e−i/2σ(~ω)

σ (~ω) = ω1σ1 + ω2σ2 + ω3σ3

then:

g = ρ (g) = eA(~ω)

with:

A (~ω) ~x = [~ω × ~x]

and we found:

D (g~ω) = e−i/~

(~ω· ~J)

~Jk = Lk ⊗ 1C2 + 1⊗ SkC2

Sk := 12~σk

SU(2) covering group of SO(3):ρ : SU(2)→ SO(3)

surjective only:

1SU(2) → 1SO(3)

−1SU(2) → 1SO(3)

ϑ = traceless, herm. 2× 2 matrices, basis σk

gσ(h)g† = σ(ρ(ρ (g)

)~h)∀~h ∈ R3

g ∈ SU(2)

This is a basic property of ρ (g)

ϑ σ(~h)

σ−1

g // gσ (h) g†

σ−1

ϑ

R3 ~h

σ

LL

ρ(g) // ρ (g)~h

σ

KK

R3


Let’s continue. . .We also want to construct a covering group for L↑+ (group of all orthochronous Lorentz-trafoswith detΛ = +1)We start with the Poincaré-Group P

A ∈ PA (Λ, a)

Ax = Λx+ a

Λ ∈ La ∈ R1,3 = M

A1A2 = (Λ1, a1) · (Λ2, a2) = (Λ1Λ2, a1 + Λ1a2)

This is called the semi-direct product and one writes P = LoMfor the covering group of P, denoted by P = LoMSo to construct:

ρ :SL (2,C)→ L↑+

SL (2,C) =α∣∣∣α ∈ End C2 : detα = ±1

α : C2 → C2

Note: SU(2) ⊂ SL (2,C)now put:

σ (h) := h012 +3∑i=1

hiσi σ : R1,3 → ϑ′

ϑ′ = hermitean 2× 2 matrices

and then for g ∈ SL (2,C)

gσ (h) g† = σ(ρ (g)h

)ρ : SL (2,C)→ L↑+

Properties:

1.

ρ (g1g2) = ρ (g1) ρ (g2) group homomorphism

ρ(1SL(2,C)

)= 1L↑+

2. indeed for g ∈ SL (2,C) then ρ (g) ∈ L↑+this follows from:

(h · h) = det[σ (h)

]σ (h) =

(h0 + h3 −ih2 + h1

ih2 + h1 h0 − h3

)det

[σ(ρ(g)h

)]= det

[σ (h)

]⇒(ρ(g)h · ρ(g)h

)= (h · h) ∀h ∈M



3.

ker [ρ] = 1,−1

indeed:

g ∈ ker [ρ] ⇔ ρ(g) = 1

⇒ g(σ (h)

)g† = σ (h) ∀h ∈M

Take h0 6= 0, hi = 0, i = 1, 2, 3

⇒ gg† = 1 g unitary

Here, the Kernel ker means all elements mapped to the unity matrix. For unitary matrices,we already know ker 1,−1Again we have a covering only, the correspondence is 2 to 1

4.

σ(p)σ (h)σ

(p)

= σ(Gph

)Gp := ΛpΛe0

with Λph = h− 2 ·(p · h

)p

with(p · p

)= 1

5.

∃p|(p · p

)= 1 p0 > 0

g(σ (h)

)g† = σ(p)σ(h)σ(p)

g =σ(p) = σ

GpO︸︷︷︸∗

h

= σ (Λh)

O =

1 ∅

∅ O ∈ SO(3)

∗: most general orthochronous, proper Lorentz-trafo. So indeed:

Range [g] = L↑+

Now define: representations of spin-1/2 particles.

g ∈ SL (2,C)ρ(g) ∈ L↑+ψ ∈ L2

(M,C2

)(D (g, a)ψ

)(x) = g ψ

(Λ−1 (x− a)

)(D(g, a

)ψ

)(x) = g ψ

(Λ−1 (x− a)

)g :=

(g†)−1


D is called the adjoint representation. In general for g ∈ SL (2,C):(g†)−16= g

If however:

g ∈ SU(2) ⊂ SL (2,C)

g =(g†)−1

= g

Indeed, these are representations:(D (g1, a1)D (g2, a2ψ)

)(x) def= g1

(D (g2, a2)ψ

) (Λ−1

1 (x− a1))

= g1g2ψ

(Λ−1

2

(Λ−1

1 (x− a1)− a2))

= g1g2ψ

Λ−12 Λ−1

1︸︷︷︸=(Λ1Λ2)−1

(x− a1 − Λ1a2)

=(D((g1, a1) (g2, a2)

)ψ)

(x)

also:(D (g1, a1) D (g2, a2)ψ

)(x) = g1g2ψ

(Λ−1

2

(Λ−1

1 (x− a1)− a2

))now:

g1g2 =(g†1

)−1 (g†2

)−1=(g†2g†1

)−1=((g1g2)†

)−1= ˆg1g2

= ˆg1g2ψ((Λ1Λ2)−1 (x− a1 − Λ1a2)

)=(D((g1, a1) (g2a2)

)ψ)

(x)

Remark:

g =(g†)−1

⇒ g−1 = g†

⇒ gg† = 1(g†)†g† = 1

⇒ gg† = 1

⇒ g−1 = g†

⇒ g =(g−1

)†Define the so-called Weyl-Operators: W, W : C2 → C2.

W = σµ∂

∂xµ= 1

∂

c∂t− σ

(~∇) (

= σµ∂µ)

(use summation convention). And:

W = −σµ∂

∂xµ= −1 ∂

c∂t− σ

(~∇)



where:

σ0 := 1 = σ0

σi := −σi

. . . One can show:

W D = DWWD = DW

Such relations are called intertwining relations. So D and W do not really commute: Therepresentation goes over into the adjoint representation.Nevertheless, the following statements hold:

1. if ψ (x) solves Wψ = 0 then also Dψ solves this.

2. if ψ (x) solves Wψ = 0 then also Dψ solves this.

Argument:

1.

W Dψ = DWψ︸︷︷︸=0

= 0

2.

WDψ = D Wψ︸︷︷︸=0

= 0

The Weyl-equations:

Wψ = 0Wψ = 0

Calculate:

WW = −σµ∂

∂xµσν

∂

∂xν= 1

2

(σµσν

∂2

∂xν∂xµ+ σν σµ

∂2

∂xµ∂xν

)

= −12(σµσν + σν σµ

) ∂2

∂xµ∂xν

= −gµν∂2

∂xµ∂xν

= − = WW

Thus:

Wψ = 0 ⇒ 0 = WWψ = −ψWψ = 0 ⇒ 0 = WWψ = −ψ

ψ and ψ solve the Klein-Gordon equation but with m = 0.So the solutions of the Weyl equations are spin 1

2 particles with m = 0.



Representations for spin 1/2 particles:

D (g, a)D (g, a)

H = L2(R1,3,C2

)→ H

g ∈ SL (2,C)a ∈ R1,3

Λ = ρ(g) ∈ L↑+ρ : SL (2,C)→ L↑+(

D (g, a)ψ)

(x) = gψ(Λ−1 (x− a)

)(D (g, a) ψ

)(x) = gψ

(Λ−1 (x− a)

)g :=

(g†)−1

=(g−1

)†(adjoint representation)

Weyl-operators:

W := σµ∂

∂xµ= σµ∂

µ = 1∂

∂ct− σ

(~∇)

W := −σµ∂

∂xµ= −σµ∂µ = −1 ∂

∂ct− σ

(~∇)

σ(~∇)

=3∑i=1

σi∂

∂xi

σ0 = σ0 = 12

σk = −σkk = 1, 2, 3

Intertwining relations:

W D (g, a) = D (g, a)WWD (g, a) = D (g, a) W

Weyl-equation:

Wψ = 0 ⇒W Dψ = DWψ = 0Wψ = 0 ⇒ WDψ = DWψ = 0

WW = WW = −Wψ = 0Wψ = 0⇒ ψ = 0⇒ ψ = 0

Klein-Gordon equation mass m = 0 particles. Let’s continue. . .



In order ~ describe particles (spin 1/2) with a finite mass m, we try:(W + mc

~12

)ψ (x) = 0

⇒ D(W + mc

~12

)ψ = 0

⇒ DWψ = −mc~Dψ

calculate: (W + mc

~12

)Dψ (x) = W Dψ (x) + mc

~Dψ (x)

= DWψ − DWψ =(D − D

)︸︷︷︸6=0

Wψ 6= 0

We observe: ψ (x) 7→ Dψ (x)→ WDψ (x) = DWψ (x)Wψ transforms like ψ.So we will use coupled equations:

Wψ + mc

~ψ = 0

Wψ + mc

~ψ = 0

We thus take as an equation of motion:(0 W

W 0

)(ψ

ψ

)+ mc

~

(ψ

ψ

)= 0

We check:

W(Dψ

)+ mc

~(Dψ) = DWψ + mc

~Dψ

= D

Wψ + mc

~ψ︸︷︷︸

=0

= 0 ok!

W (Dψ) + mc

~

(Dψ

)= . . . = 0 ok!

We can now define:

D :=(

0 W

W 0

)D : C4 7→ C4

D2 =(WW 0

0 WW

)=(− 00 −

)


Then indeed:

0 =(D + mc

~14

) (ψ

ψ

)︸︷︷︸

R1,3 7→C4

ψ, ψ : R1,3 7→ C2

⇒(D − mc

~14

)(D + mc

~14

)(ψ

ψ

)

=(D2 − m2c2

~2 14

)(ψ

ψ

)= 0

⇒

−− m2c2

~2 00 −− m2c2

~2

(ψψ

)= 0

All four components of (ψ

ψ

)

fulfil the free Klein-Gordon equation with mass m!⇒ (free) Dirac-equation:(

D + mc

~14

)(ψ

ψ

)= 0(

ψ

ψ

)=ψD

Dirac-Spinor:

ψD : R1,3 7→ C4

D =(

0 W

W 0

)=(

0 σµ∂µ

−σµ∂µ 0

)=(

0 σµσµ 0

)⊗ ∂µ

define:

−iγµ :=(

0 σµ−σµ 0

)

⇔ γµ =(

0 iσµ−iσµ 0

)

specifically:

γ0 := γ0 =(

02 i12−i12 02

)

γk := −γk =(

02 −iσk−iσk 02

)k = 1, 2, 3



Then: (−iγµ∂µ + mc

~14

)ψ0 (x) = 0

⇔(iγµ∂µ −

mc

~14

)ψ0 (x) = 0

Properties of the γ-matrices:

γ0γ0 =(γ0)2

= 14

γ0γk =(σk 00 −σk

)

γkγ0 =(−σk 0

0 σk

)⇒[γ0, γk

]+

= 0

⇒[γ0, γ0

]+

= 214

k 6= l[γk, γl+

]= −2δkl14

k, l ∈ 1, 2, 3

In short:

[γµ, γν ]+ = 2gµν14

Take:

U = 1√2

(12 i · 1212 −i · 12

)∈ U(4)

then:

γ0D = Uγ0U † =

(12 O2O2 −12

)γkD = UγkU †

=(

02 σk−σk 02

)

This is called the Dirac-representation of the γ matrices.



Dirac-Equation:(0 W

W 0

)(ψ

ψ

)+ mc

~

(ψ

ψ

)= 0

(D (g, a)ψ

)(x) = gψ

(Λ−1 (x− a)

)(D (g, a) ψ

)(x) = gψ

(Λ−1 (x− a)

)=(g−1

)†=(g†)−1

W = σµ∂µ = 1∂

∂(ct) − σ(~∇)

W = −σµ∂µ = −1 ∂

∂(ct) − σ(~∇)

σ0 = σ0 = 12

σk = −σkk = 1, 2, 3g ∈ SL (2,C)

W D = DWWD = DW

also: (iγµ∂µ −

mc

~14

)ψD (x) = 0

ψD =(ψ

ψ

)

γ0 =(

0 i12−i12 0

)

γk =(

0 −iσk−iσk 0

)[γµ, γν ]+ = 2gµν14

in Weyl-representation.

γµD := UγµU †

U = 1√2

(12 i1212 −i12

)U † = U−1

U ∈ U(4)

→ Dirac-representation

γ0D =

(12 00 −12

)

γkD =(

0 σk−σk 0

)k = 1, 2, 3



Let’s continue. . .Poincaré-invariance of D + mc

~ 14

D =(

0 W

W 0

)

Let:

A = (g, a)g ∈ SL (2,C)a ∈ R1,3

A ∈ SL (2,C) oR1,3

We had:

D(A)ψ (x) = gψ

(A−1x

)D(A)ψ (x) = g︸︷︷︸

(g−1)†=(g†)−1

ψ(A−1x

)

A = (Λ, a) ∈ L↑+ oR1,3

A−1x = Λ−1 (x− a)Λ = ρ(g)

ρ : SL (2,C)→ L↑+

Now declare:

DD(AψD

)(x) = S(g)ψD

((ρ(g)

)−1 (x− a))

with

S(g) =(g 00 g

)S(g) : C4 7→ C4

ψD =(ψ

ψ

)

Proposition:

DD(A)D = DDD

(A)

D =(

0 W

W 0

)

or

DD(A)DDD

(A)−1

= D


Argument:

DD(A)DDD

(A)−1

=

D(A)

00 D

(A)( 0 W

W 0

)DD

(A)−1

=

0 D(A)W

D(A)W 0

DD (A)−1

=

0 W D(A)

WD(A)

0

DD (A)−1

=(

0 W

W 0

)D(A)

00 D

(A)

︸︷︷︸DD(A)

DD(A)−1

︸︷︷︸14

= D

The identity part:mc

~14

is trivial, so we have that the representation[DD

(A), D

]−

= 0

commutes.equivalent forms: (

iγµ∂µ −mc

~

)ψD (x) = 0

U ∈ GL (4,C)

⇒ U

(iγµ∂µ −

mc

~

)U−1UψD (x) = 0

=

i UγµU−1︸︷︷︸=:γµ

∂µ −mc

~14

(UψD)︸︷︷︸=:ψD

= 0

We had:

γµγν + γνγµ = 2gµν14 = [γµ, γν ]+⇒ [γµ, γν ]+ = 2gµν14

We had in Weyl-representation:(γ0)†

= γ0

and (γk)†

= −γk k = 1, 2, 3



We require:(γ0)† != γ0

⇔(Uγ0U−1

)†= Uγ0U−1(

U−1)† (

γ0)†U † = Uγ0U−1(

U−1) (U−1

)†γ0U †U = γ0

⇒ U †U = 14

⇒ U † = U−1

U ∈ U(4)

Then also: (γk)†

=(UγkU−1

)†= U

(γk)†U † = −UγkU †

= −γk

γ0 is hermitean, γk antihermitian in all representations.

U † = U−1

γµ = UγµU †

Proposition:

S−1(g)γµS(g) = Λµνγν

Argument:

S(g) =(g 00 g

)

S−1(g)γµS(g) =(g−1 00 g−1

)(0 iσµ

−iσµ 0

)(g 00 g

)σ0 = σ0 σk = −σkσ0 = σ0 σk = −σk

=(

0 ig−1σµg−ig−1σµg 0

)

=

0 ig−1σµ(g−1

)†g−1 (−iσµ)

(g−1

)†0

=(

0 iΛµνσν−iΛµν σν 0

)= Λµνγν

So:



then also: (USU †

)−1 (UγµU †

) (USU †

)= US−1 U †U︸︷︷︸

=1

γµ U †U︸︷︷︸=1

SU † = US−1γµSU †

= ΛµνUγνU † = Λµν γν

So all basic properties remain the same after applying UAU † = A.We had:


also with:

S(g) = US(g)U−1

γµ = UγµU−1

⇒ S−1(g)γµS(g) = γνΛµν

Notation: a, b ∈ C4

〈a, b〉C4 =4∑

k=1a∗kbk

Define current:

jµ [ψD] (x) = 〈ψD (x) , γ0γµψD (x)〉C4 = ψD (x) γ0γµψD (x)∂µj

µ (x) = 〈(∂µψD

)(x) , γ0γµψD (x)〉C4 + 〈ψD (x) , γ0γµ

(∂µψD

)(x)〉C4

⇒ ∂µjµ = 〈(γµ)†

(γ0)† (

∂µψD)

(x) , ψD (x)〉C4

+ 〈ψD (x) γ0γµ(∂µψD

)(x)〉C4

now:(γ0)†

= γ0(γk)†

= −γk(γ0)† (

γ0)†

= γ0γ0 = 14(γk)† (

γ0)†

= −γkγ0 = γ0γk

∂µjµ = 〈γ0γµ

(∂µψD

)(x) , ψD (x)〉C4 + 〈ψD (x) , γ0γµ

(∂µψD

)(x)〉C4

use γµ∂µψD (x) = −imc~ψD (x) Dirac-equation

∂µjµ = 〈γ0

(−imc

~ψD (x)

), ψD (x)〉

C4+ 〈ψD (x) , γ0 − imc

~ψD (x)〉

C4= 0

Using anti-linearity.Is jµ [ψD] (x) a 4-vector field? Check this for A = (g, 0) = gCompute:

〈DD (g)ψD (x) , γ0γµDD (g)ψD (x)〉C4 = 〈S(g)ψD(ρ(g)−1x

), γ0γµS(g)ψD

(ρ(g)−1x

)〉C4

= 〈ψD(ρ(g)−1x

), S(g)†γ0γµS(g)ψD

(ρ(g)−1x

)〉C4

= 〈ψD(ρ(g)−1x

), γ0

(γ0S†(g)γ0

)γµS(g)ψD

(ρ(g)−1x

)〉C4



We just inserted γ0γ0 here.interlude:

γ0S(g)†γ0 =(

0 i12−i12 0

)(g† 00 g†

)(0 i12−i12 0

)(

0 ig†

−ig† 0

)(o i12−i12 0

)=(g† 00 g†

)

=(g−1 00 (g)−1

)= S(g)−1

= 〈ψD(ρ−1(g)x

), γ0 S−1(g)γµS(g)︸︷︷︸ψD (ρ−1(g)x

)〉C4

= Λµν 〈ψD(ρ−1x

), γ0γνψD

(ρ−1(g)x

)〉C4

= Λµν jν(ρ−1(g)x

)jµ [ψ] indeed transforms as a 4-vector field. Then we shall have:

‖ψD‖2 = 〈ψD1 , ψD2〉 :=∫

(t=0)

d3(x) j0 [ψD] (x)

=∫t=0

d3x 〈ψD (x) , γ0γ0︸︷︷︸14

ψD (x)〉C4

=∫t=0

d3xψ† (0, ~x)ψ (0, ~x) ≥ 0

This defines a positive semi-definite norm and this is Poincaré invariant.Other covariants: Define:

ψ (x) = ψ† (x) γ0

adjoint spinor.Define: scalar field:

s [ψ] (x) = ψ (x)ψ (fvx) = ψ† (x) γ0ψ (x)

Define: pseudoscalar field:

s [ψ] (x) = iψ (x) γsψ (fvx)

with γ5 = iγ0γ1γ2γ3

vector field:

jµ [ψ] (x) = ψ (x) γµψ (x)

pseudovector field:

µ [ψ] (x) = ψ (x) γ5γµψ (x)

antisymmetric tensor field:

tµν (x) = ψ (x) γµγνψ (x) µ > ν



free Dirac-equation:(iγµ∂µ −

mc

~14

)ψ (x) = 0

in the Weyl-representation:(DD

(A)ψ

)(x) = S(g)ψ

(ρ(g)−1 (x− a)

)S(g) =

g 00(g−1

)†

g :=(g−1

)†=(g†)−1

A = (g, a) ∈ SL (2,C) oR1,3

ρ(g) ∈ L↑+

other representation:

γµ = UγµU †

U ∈ U(4)ψ = Uψ(

DD(A)ψ

)(x) = US(g)U−1ψ

(ρ(g)−1 (x− a)

)jµ [ψ0] (x) := 〈ψD (x) , γ0γµψD (x)〉C4(

iγµ∂µ −mc

~14

)ψD = 0

⇒ ∂µjµ (x) = 0

ψ := ψ (x)† γ0 bilinearsγ5 = iγ0γ1γ2γ3

S [ψ] (x) = ψ (x)ψ (x) scalarS [ψ] (x) = ψγ5 (x)ψ (x) pseudoscalarjµ [ψ] (x) = iψγµ (x)ψ (x) vector [ψ] (x) = ψγ5γµ (x)ψ (x) axial vector

tµν [ψ] (x) = ψγµγν (x)ψ (x) tensor (µ > ν)

Dirac-representation:

γ0 =(1 00 1

)

γk =(

0 σk−σk 0

)

γ5 =(

0 1

1 0

)Let’s continue. . .



2.1 Solutions of the free Dirac-equation

Define:

/a = aµγµ = aµγµ

⇒[/a, /b

]+

=[aµγ

µ, bνγν]

+

= aµbνgµν · 2 · 14 = (a · b) · 2 · 14

Dirac-equation: (iγµ∂µ −

mc

~

)ψ = 0(

i/∂ − mc

~

)ψ (x) = 0

Ansatz:

ψ (x) = w (k) e−i(k·x)

w (k) ∈ C4

We know:

ψ (x)− m2c2

~2 ψ (x) = 0

k =(k0,~k

)⇒ k2

0 −∣∣∣~k∣∣∣2 = m2c2

~2

So we have:

k0 = ±

√m2c2

~2 +∣∣∣~k∣∣∣2 = ±Ωk

c

Solutions:

w

(Ωk

c,~k

)e−i(

Ωkt−(~k·~x))

and:

w

(−Ωk

c,~k

)e−i(−Ωkt−

(~k·~x))∼ ei/~Et

This is the positive / negative frequency solution. We now introduce:

Ωk := Ωk

~k = −~k


2.1. SOLUTIONS OF THE FREE DIRAC-EQUATION

The second solution can then also be written as:

w

(− Ωk

c,−~k

)ei

(Ωkt−

(~k·~x))

So we shall write:

u (k) e−i(k·x)

v (k) ei(k·x)

with k0 = Ωk

c> 0

We still have: (/k − mc

~

)u (k) = 0(

/k + mc

~

)v (k) = 0 (2.12)

Special case: resting particle: ~k = 0

k0 = mc

~~k = 0

⇒(γ0mc

~− mc

~14

)u(k0,~0

)= 0(

γ0 − 14)u(k0,~0

)= 0(

γ0 + 14)v(k0,~0

)= 0

in Dirac-representation: (1 00 −1

)

⇒(

0 00 −2 · 1(2)

)u(k0,~0

)︸︷︷︸

u0

= 0

(21 00 0

)v(k0,~0

)︸︷︷︸

v0

= 0

Solutions for a particle at rest:

u(1)0 =

1000

u(2)0 =

0100

u

(r)0 =

(χ

(r)s

0

)



where:

χ(r)s ∈ C4

χ(1)s =

(10

)

χ(2)s =

(01

)V

(1)0 =

0010

V(2)

0 =

0001

V

(r)0 =

(0χ

(r)s

)

For ~k 6= ~0 now

u(r) (k) = N

(mc

~1 + /k

)u

(r)0

and

v(r) (k) = N

(mc

~1− /k

)u

(r)0

Will fulfil equations (2.12) since:(/k − mc

~

)(/k + mc

~

)= k2 − m2c2

~2 = 0

Normalisation: calculate:

u(r) (k)u(s) (k) := 〈u(r) (k) , γ0u(s) (k)〉C4



Now:

u(r) (k) = N∗u(r)†0

(mc

~+ /k

)†γ0†

γ0 = γ0†

. . . = N∗u(r)†0 γ0

(mc

~+ /k

)γk = −

(γk)†

k = 1, 2, 3

u(r) (fvk)u(s) (fvk) = |N |2 u(r)0

(mc

~+ /k

)(mc

~+ /k

)u

(s)0(

mc

~+ /k

)(mc

~+ /k

)= m2c2

~2 + 2mc~/k + k2︸︷︷︸

m2c2~2

= 2mc~

(mc

~+ /k

)

u(r) (fvk)u(s) (fvk) = 2|N |2 mc~u

(r)0

(mc

~+ /k

)u

(s)0

= 2|N |2 mc~

(χ(r)† 0

)(1 00 −1

)mc~ + Ωk

c −σ(~k)

σ(~k)

mc~ −

Ωkc

(χ(s)

0

)

= 2|N |2 mc~

(mc

~+ Ωk

c

)δrs

ψ (x) =

u (k) e−i(k·x)

v (k) e+i(k·x)(/k − mc

~

)u (k) = 0(

/k + mc

~

)v (k) = 0

Normalised solutions:

u(r) (k) =

√

Ek+mc22mc2 χ(r)

~cσ(~k)√

2mc2(Ek+mc2)χ(r)

v(r) (k) =

~cσ(~k)√

2mc2(Ek+mc2)χ(r)√

Ek+mc22mc2 χ(r)

χ(1) =

(10

)

χ(2) =(

01

)



These are the Pauli-Spinors

Ek =√m2c4 + ~2

∣∣∣~k∣∣∣2 c2 = ~Ωk

u(r) (k)u(s) (k) = δrs

u(r) (k) v(s) (k) = 0v(r) (k)u(s) (k) = 0v(r) (k) v(s) (k) = −δrs


plane wave solutions to free Dirac equation(i/∂ − mc

~

)ψ (x) = 0

/a := aµγµ

⇔(iγµ∂µ −

mc

~

)ψ (x) = 0

Ansatz:

ψ (x) =

u (k) e−i(k·x);(/k − mc

~

)u (k) = 0

v (k) ei(k·x);(/k + mc

~

)v (k) = 0

“Normalised” solutions:

u(r) (k) =

√

Ek+mc22mc2 χ(r)

~cσ(~k)√

2mc2(Ek+mc2)χ(r)

v(r) (k) =

~cσ(~k)√

2mc2(Ek+mc2)χ(r)√

Ek+mc22mc2 χ(r)

χ(1) =

(10

)

χ(2) =(

01

)Pauli–Spinors. Then:

u(r) (k)u(s) (k) = δrs u(r) (k) v(s) (k) = 0v(r) (k)u(s) (k) = 0 v(r) (k) v(s) (k) = −δrs

u = u†γ0

v = v†γ0

with:

Ek =√m2c4 + ~2

∣∣∣~k∣∣∣2 c2 = ~Ωk

k0 = +Ωk

c

σ(~k)

= k1σ1 + k2σ2 + k3σ3



Let’s continue. . .The positive definite density was given by

ψ† (x)ψ (x) = ψ (x) γ0ψ (x) = j0 (x)

So let’s calculate u† (x)u (k). We know: (/k − mc

~

)u (k) = 0

⇒ u† (k)(/k − mc

~

)†= 0

⇒ u† (k)(/k − mc

~

)†γ0 = 0

⇔ u† (k)(γ0k0 − γlkl −

mc

~14

)γ0 = 0

⇒ u† (k) γ0(γ0k0 + γlkl −

mc

~14

)= 0

⇒ u (k)(/k − mc

~14

)= 0

Then follows:

⇒ u(r) (k)[/k − mc

~, γµ

]+u(s) (k) = 0

⇒ u(r) (k)[/k, γµ

]+ u

(s) (k) = 2mc~u(r) (k) γµu(s) (k)

Now we also have:

[kνγν , γµ]+ = 2gνµkν14 = 2kµ14

⇒ u(r) (k) γµu(s) (k) = ~kµ

mcu(r) (k)u(s) (k)

µ=0−−→ u(r)† (k)u(s) (k) = u(r) (k) γ0u(s) (k) = ~k0

mcu(r) (k)u(s) (k) = Ek

mc2 δrs > 0

indeed:

u(r)† (k)u(s) (k) = Ekmc2︸︷︷︸

note this factor

δrs > 0

Then also for:

ψ(+)(r)k (x) = u(r) (k) e−i(k·x)

we have: (ψ

(+)(r)k (x)

)†ψ

(+)(s)k (x) = Ek

mc2 δrs

also:

v(r) (k)[/k + mc

~, γµ

]+v(s) (k) = 0

⇒ v(r) (k) γµv(s) (k) = −~kµ

mcv(r) (k) v(s) (k)

⇒ v(r) (k)† v(s) (k) = −~k0

mc· (δrs) = Ek

mc2 δrs > 0



So this is also positive!Remark:

γ5γ0γk =(

0 −11 0

)(0 σk−σk 0

)=(σk 00 σk

)=: Σk

Spin–Operator:

Sk : C4 → C4

Sk = 12~Σk

⇒ u, v are eigenfunctions of S3 with

S3u(1/2)0 e−i

mc~ ct =

(+/−

) 12~u

(1/2)0 e−i

mc2~ t

S3v(1/2)0 ei

mc~ ct =

(+/−

) 12~v

(1/2)0 ei

mc2~ t

Here,(

1/2)means 1 or 2, like

(+/−

)means + or −. So these solutions describe spin 1/2-particles

(now, the fraction is meant).Furthermore: For

ψ(+)(r)k (x) = e

−i(k0x0−

(~k·~x))u(r) (k) k =

(k0,~k

)ψ

(−)(s)k (x) = e

+i(k0x0−

(~k·~x))v(s) (k) k =

(k0,−~k

)(ψ

(−)(s)k

(x))†ψ

(+)(r)k (x) = 0 prove yourself!

Thus states of positive and negative energy are naturally orthogonal if they have opposite en-ergies, but the same three-momentum ~k.IMAGE22

Dirac–equations:

iγµ∂µ = iγ0 ∂

∂ (ct) + i3∑

k=1γk

∂

∂xk= i

cγ0 ∂

∂t+ i

(~γ · ~∇

)(iγµ∂µ −

mc

~

)ψ (x) = 0

define:

p0 = i~∂

∂x0 = i~c

∂

∂t

pk = i~∂

∂xk= i~∇k

k = 1, 2, 3

⇔(γµpµ~− mc

~

)ψ (x) = 0

⇔(γµpµ −mc14

)ψ (x) = 0

⇔(/p−mc14

)ψ (x) = 0



“minimal coupling”:

pµ 7→ pµ −q

cAµ (x)

Aµ (x) 4-vector potential. Dirac equation for a particle of mass m, charge q in an electromag-netic field specified by Aµ (x) µ = 0, 1, 2, 3(

γµpµ −q

cγµAµ (x)−mc14

)ψ (x) = 0

electromagnetic field:

Fµν (x) = ∂µAν (x)− ∂νAµ (x)

We now inspect (proposition):[γµ(i~∂µ + q

cAµ (x)

)−mc

](Cψ) (x) = 0

where

(Cψ) (x) := γ5γ2ψ∗ (x)

If ψ fulfils our former Dirac equation. In e.g. Weyl-representation:

γ5γ2 = −(

0 iσ2−iσ2 0

)Argument:[

γµ(i~∂µ + q

cAµ (x)

)−mc

](Cψ) (x) = γ5

[γµ(−i~∂µ −

q

cAµ (x)

)−mc

]γ2ψ∗ (x)

= γ5γ2

γ0(i~∂0 + q

cA0

)+ γ1

(i~

∂

∂x1 + q

cA1

)− γ2

(i~

∂

∂x1 + q

cA2

)+ γ3

(i~

∂

∂x1 + q

cA3

)−mc

ψ∗ (x)

Weyl-representation:

iγ0 =(

0 −11 0

)

iγk =(

0 σkσk 0

)k = 1, 2, 3

σ2 =(

0 −ii 0

)

⇒ σ∗2 =(

0 i−i 0

)= −σ2

So we can write:

= γ5γ2

[i~γµ

∂

∂xµ− q

cγµAµ −mc

]ψ (x)

∗

= 0

Because of the Dirac equation.



2.1.2 Wave packets

Let’s start with linear combinations of positive energy solutions only:

ψ(+) (x) =∫ d3k

(2π)3mc2

Ek

∑r=1,2

b(~k, r

)u(r) (k) e−i(k·x)

b ∈ C

Normalisation:∫d3x j(+)0 (t, ~x) =

∫d3x

∫ d3k′

(2π)3m2c4

EkEk′

∑r,r′

b∗(~k, r

)b(~k′, r′

)u(r)† (k)u(r′) (k′) e

+i(Ωk−Ωk′)t−i((~k−~k′

))·~x

=∑r

∫ d3k

(2π)3

∣∣∣∣b (~k, r)∣∣∣∣2 mc2

Ek= 1

If J (+) is the total current, we can do (with l = 1, 2, 3):

J (+)l (t, ~x) =∫

d3x j(+)l (t, ~x) =∫ d3k

(2π)3m2c4

E2k

∑r,r′

b∗(~k, r

)b(~k′, r′

)u(r)† (k) γ0γl︸︷︷︸

=:αl

u(r′) (k)

= 〈plc

Ek〉

(which is still to show).


Wave packet (pos. energy comp. only)

ψ+ (x) =∫ d3k

(2π)3mc2

Ek

∑r=1,2

b(~k, r

)u(r) (k) e−i(k·x)

Norm:∫d3x j(+)0 (t, ~x) =

∑r

∫ d3k

(2π)3

∣∣∣∣b (~k, r)∣∣∣∣2 mcEk︸︷︷︸probability density

= 1

spatial component of current:

j(+)l(t) =∫

d3x j(+)l (t, ~x) =∫ d3k

(2π)3m2c4

E2k

∑r,r′

b∗(~k, r

)b(~k, r′

)u(r)† (k) γ0γlu(r′) (k)

Let’s continue. . .Gordon-identity:

Let u be a positive energy solution of the Dirac–equation:Then:

u(r) (k) γµu(s)(q)

= ~2meu

(r) (k)[(k + q)µ + iσµν (k − q)ν

]us(q)

σµν := i

2 [γµ, γν ]−



follows from:

0 = u(r) (k)[/a

(/q −

mc

~

)+(/k − mc

~

)/a

]u(s)

(q)

/a = aµγµ

We find:

u(r) (k) γ0︸︷︷︸u(r)(k)

γlu(r)′ (k) = ~2mcu

(r) (k) 2 · klu(r′) (k)

=

(~kl)c

mc2 δrr′

⇒ j(+)l =∑r

∫ d3k

(2π)3mc2

Ek

∣∣∣∣b (~k, r)∣∣∣∣2︸︷︷︸probability density

plc

Ek

The term plc/Ek is the group velocity of the wave-packet.Let’s inspect a Gaussian wave packet.

ψ (0, ~x) = 1(πd2)3/4

e−12|~x|2

d2 w

w is a fixed spinor

w ∈ C4

w =(φ0

)φ ∈ C2

In general:

ψ (t, ~x) =∫ d3k

(2π)3mc2

Ek

∑r

[b(~k, r

)ur(k)e−i(k·x) + d∗

(~k, r

)· V (r) (k) ei(k·x)

]

We have:∫d3x e−

12|~x|2

d2−i(~k·~x)

=(4πd2

)3/2e−

12

∣∣~k∣∣2d2

. . . by comparison:

(4πd2

)3/4e−

12

∣∣~k∣∣2d2· w = mc2

Ek

∑r

[b(k, r)u(r) (k) + d∗

(k, r

)v(r)

(k)]

k =(k0,−~k

). . .⇒ b

(~k, r

)=(4πd2

)3/4e−

12

∣∣~k∣∣2d2 (u(r)† (k)w

)d∗(~k, r

)=(4πd2

)3/4e−

12

∣∣~k∣∣2d2 (v(r)† (k)w

)



If w =(φ0

)then

∣∣∣∣∣d∗b∣∣∣∣∣ ∝

∣∣∣~~kc∣∣∣(Ek +mc2)

thus∣∣a∗∣∣ ∼|b| if |~p| c ≈ mc2

if d ~cmc2 = ~

mc

then components with typical momenta of pc ∼ mc2 ~c/d are suppressed.Then negative energy solutions are unimportant, but if the packet is smaller than ~/mc (=Comptonwavelength of a particle of mass m), negative energy components are important.


General wave packet:

ψ (t, ~x) =∫ d3k

(2π)3mc2

Ek

∑r

[b (k, r)u(r) (k) e−i(k·x) + d∗ (k, r) v(r) (k) ei(k·x)

]Norm:∫ d3k

(2π)3mc2

Ek

∑r

(∣∣b (k, r)∣∣2 +

∣∣d (k, r)∣∣2) = 1

new:

J l(t) =∫ d3k

(2π)3mc2

Ek

plc

Ek

∑r

(∣∣b (k, r)∣∣2 +

∣∣d (k, r)∣∣2)

+ i∑r,r′

(b∗(k, r

)d∗(k, r′

)e2 i~Ektu(r)

(k)σlov(r′) (k)

)

−b(k, r

)d(k, r′

)e−2 i~Ektv(r′) (k)σlou(r)

(k)

For further information see: Schwabl 10.1.2

σµν = i

2 [γµ, γν ]

l = 1, 2, 32Ek~∼ 2mc2

~∼ 2 · 1021 Hz

ampl. ∼ ~cmc2 ∼ 4 · 10−13 m electrons

These small rapid oscillations are called “Zitterbewegung”.Let’s continue. . .up to now: free motion: (

iγµ∂

∂xµ− mc

~14

)ψ (x) = 0

ψ (x) ∈ C4

electric field: 4-vector potential:

Aµ (x)µ = 0, 1, 2, 3

A0 (t, ~x) = ψ (t, ~x)Ai (t, ~x) is a (3-vector) field



minimal coupling / minimal substitution:

pµ 7→ pµ − q

cAµ (t, ~x)

p0 = i~1c

∂

∂t

pk = i~∂

∂xk= −i~ ∂

∂xk

Dirac equation in an electromagnetic field:

i~∂

∂tψ (t, ~x)− cq

cA0 (t, ~x)ψ (t, ~x) =

c · 3∑k=1

αk

(pk − q

cAk (t, ~x)

)+ βmc2

ψ (t, ~x)

β = γ0

αk = γ0γk

k = 1, 2, 3

⇒ i~∂

∂tψ (t, ~x) =

cα

(−i~~∇− q

c~A (t, ~x)

)+ βmc2 + qΨ (t, ~x)

ψ (t, ~x)

α (~p) =3∑

k=1pkαk =

3∑k=1

pkγ0γk

α (~p) : C4 → C4

Standard/Dirac representation:

β =(12 0202 −12

)

αk =(

02 σkσk 02

)k = 1, 2, 3

We now write:

ψ =(φχ

)φ, χ ∈ C2

act. φ, χ : R3 × R→ C2

i~∂

∂t

(φχ

)= c

02 σ(−i~~∇− q

cA)

σ(i~~∇− q

c~A)

02

(φχ

)+mc2

(1 0202 12

)(φχ

)+ qΦ

(φχ

)

i~∂

∂t

(φχ

)= c

σ(Π)χ

σ(Π)φ

+mc2(φ−χ

)qΦ (t, ~x)

(φχ

)

Π := −i~~∇− q

cA (t, ~x)

σ (~a) =3∑

k=1akσk



Introduce:(φχ

)= e−

i~mc

2t

(φχ

)

i~∂

∂t

(φχ

)= mc2

(φχ

)+ e−i

i~mc

2t · i~ ∂∂t

(φχ

)⇒ i~

∂

∂t

(φχ

)= c

σ(Π)φ

σ(Π)χ

+ qΦ(φχ

)− 2mc2

(0χ

)

Approximation:

qΦ (t, ~x) mc2

i~∂

∂tχ (t, ~x) 2mc2 · χ (t, ~x)

0 ≈ c · σ(Π)φ− 2mc2 · χ

⇒ χ 'σ(Π)

2mc · φ

Approximately:

i~∂

∂tφ (t, ~x) =

σ(Π)σ(Π)

2m φ (t, ~x) + qΦ (t, ~x)φ (t, ~x)

We know:

σ (~a)σ(~b)

=(~a ·~b

)12 + iσ

([~a×~b

])⇒ σ (~a)σ

(~b)

=∣∣∣Π∣∣∣2 + iσ

([Π× Π

])Indeed:

[Π× Π

]k= εklm

(−i~ ∂

∂xl− q

cAl (t, ~x)

)(−i~ ∂

∂xm− q

cAm (t, ~x)

). . .

= i~q

cBk

~B =[~∇× ~A

]

i~∂

∂tφ (t, ~x) =

∣∣∣−i~~∇− q

c~A (t, ~x)

∣∣∣22m − q~

2mc~σ(~B)

+ qΦ (t, ~x)

φ (t, ~x)

which is the Pauli-equation for the upper components φ of the Dirac-spinor. And:

χ ≈σ(Π)

2mc φ

Special case:

Φ (t, ~x) = 0

A (~x) = 12[~B × ~x

]



time independent constant magnetic field ~B, indeed:

~B = (curl A) (~x)(div ~A

)(~x) = 0

Coulomb-gauge.Neglect the

∣∣∣ ~A∣∣∣2-terms:

i~∂

∂tφ (t, ~x) = − ~2

2m∆− q

2mc~B ·

~L+ 2︸︷︷︸g-factor

S

φ (t, ~x)

(~a · A

):=

3∑k=1

akAk

Lk = −i~εklmxl∂

∂xmangular momentum operator

where Sk = 12~σk

σ(~B)

= 112~

3∑k=1

BkSk

The g-factor is the g-factor of the electron (Spin 1/2).Dirac-equation in a central electric field:

~A (t, ~x) = 0qΦ (t, ~x) = V

(|~x|)

Now, put:

ψ (t, ~x) = e−i~E·tψ0 (~x)

HDψ0 (~x) :=

−i~ 3∑k=1

αk∂

∂xk+mc2 · β + V

(|~x|)14

ψ0 (~x) = E · ψ0 (~x)

Remarks:

Parity: (Pψ0) (~x) = γ0ψ0 (P~x) = γ0ψ0 (−~x)⇒ [P, HD]− = 0

Define

Jk := Lk · 14 + ~2Σk

Σk := γ5αk = γ5γ0γk

k = 1, 2, 3

Dirac-representation:

Σk =(σk 00 σk

)[Jk, HD

]−

= 0

Try to find common eigenfunctions (R3 7→ C4) (of HD) and ˆ|J |2, J3, P.



We had already for eigenfunctions (R3 7→ C2) ˆ|J |2, J3, Px: C2 7→ C2.

ˆ∣∣∣ ~ ∣∣∣2Jχljmj (~x0) = ~2j(j + 1)χljmj (~x0)

χljmj : S2 7→ C2

~x0 ∈ S2

~x0 = ~x

|~x|~x ∈ R3

J3χljmj (~x0) = ~mj · χljmj (~x0)P(χ)χljmj (~x0) = (−1)l χljmj (−~x0)

spherical (Pauli)-spinors:

χljmj (~x0) =∑ml,ms

ml+ms=mj

〈lml12ms|jmj〉Ylml (~x0)χms

l = j ± 12

ms = ±12

This motivates the “ansatz”:

ψ0 (~x) =

χl=j∓ 12 ,j,mj

(~x0)F(|~x|)

χl′=j± 12 ,j,mj

(~x0)G(|~x|)

In fact:

χl′=j± 12 ,j,mj

(~x0) = σ (~x0)χl=j∓ 12 ,j,mj

(~x0)

Also:

Kχljmj (~x0) = κ · χljmj (~x0)

where K = 12 + 1~σ(~L)

κ =

l + 1 if l = j − 12

−l if l = j + 12

⇒ κ2 =(j + 1

2

)2

Through a lot of calculation, we may arrive at:

F (r) = f(r)G(r) = ig(r)

~c(f ′(r) + 1− κ

rf(r)

)+(E +mc2 − V (r)

)g(r) = 0

~c(g′(r) + 1 + κ

rg(r)

)−(E −mc2 − V (r)

)f(r) = 0




Dirac-equation in a central electric field

HDψ0 (~x) :=

−i~ 3∑k=1

αk∂

∂xk+mc2β + V

(|~x|)14

ψ0 (~x) = Eψ0 (~x)

ψ0 : R3 → C4

Ansatz:

ψ0 (~x) =

χl=j∓ 12 ,j,mj

(~x0)F(|~x|)

χl′=j± 12 ,j,mj

(~x0)G(|~x|)

~x0 = ~x

|~x|

χljmj (~x0) =∑ml,ms

ml+ms=mj

〈lml12ms|jmj〉Ylml (~x0)χms︸︷︷︸

∈C4

Common eigenfunction of∣∣∣∣ ~J ∣∣∣∣2, JS , P, Pψ0 (~x) = γ0ψ0 (−~x)

χl′=j± 12 ,j,mj

(~x0) = σ (~x0)χl=j∓ 12 ,j,mj

(~x0) Kχljmj (~x0) = κ · χljmj (~x0)

where K = 12 + 1~σ(~L)

κ =

l + 1 if l = j − 12

−l if l = j + 12

⇒ κ2 =(j + 1

2

)2

~c(f ′(r) + 1− κ

rf(r)

)+(E +mc2 − V (r)

)g(r) = 0

~c(g′(r) + 1 + κ

rg(r)

)−(E −mc2 − V (r)

)f(r) = 0

V (r) = −Ze2

rα = e2

~cr →∞ f(r), g(r) ∼ e−γr, γ = 1

~c√m2c4 − E2

r → 0 f(r), g(r) ∼ rγ , γ = 1 +√κ2 − (Zα)2

Let’s continue. . .Maybe important for the exam:

ρ = √ε+ε− · r

ε± = 1~c

(mc2 ± E

)f ′ (ρ) + 1− κ

ρf (ρ) +

√ε+ε−

+ Zα

ρ

g (ρ) = 0

g′ (ρ) + 1 + κ

ρg (ρ) +

√ε−ε+− Zα

ρ

g (ρ) = 0



Ansatz:

f(ρ) = ργ

∑k

akρk

e−ρ

g(ρ) = ργ

∑k

bkρk

e−ρ

a0 6= 0b0 6= 0

Coefficients of ρq+γ−1e−ρ:

(q + γ + 1− κ) aq − aq−1 + ε+ε−bq−1 + (Zα) bq = 0

(q + γ + 1 + κ) bq − bq−1 + ε−ε+aq−1 + (Zα) aq = 0

With some calculation, we arrive at:

bq = (Zα)− ε (q + γ + 1− κ)ε (Zα) + q + γ + 1 + κ

aq

ε :=√ε−ε+

. . .

aqaq−1

= ε (Zα) + q + γ + 1 + κ

ε (Zα) + q + γ + κ·

2ε (q + γ) + (Zα)(ε2 − 1

)ε[(Zα)2 + (q + γ + 1)2 − κ

]Inspect for q →∞:

aqaq−1

≈ 2q

compare with: e2ρ =∞∑q=0

1q! (2ρ)q

CqCq−1

= 2q

⇒ series must terminate:



⇒ ∃N∣∣∣∣aN+1 = 0

⇒ bN+1 = 0

⇒ 2ε (N + 1 + γ) + (Zα)(ε2 − 1

) != 0

→quantisation of E

(. . .)n2

E = mc2√√√√1 + (Zα)2(N+√k2+(Zα)2

)2

N = 0, 1, 2, . . .N ∈ N0

κ2 =(j + 1

2

)2

Remark:

bN =√ε−ε+aN

Inspect:

f(x) = 1√1 + x

(N+√κ2−x2)2

f(0) = 1

f ′ (0) = −11

1(N +|κ|

)2f ′′ (0) = 3

41(

N +|κ|)4 − 1(

N +|k|)3|k|

E = mc2

1− 12

(Zα)2(N +|κ|

)2 + 12 (Zα)4

34

1(N +|κ|

)4 − 1(N +|k|

)3|k| + . . .

N +|κ| = n

n ∈ N

= mc2

1− 12

(Zα)2

n2 − 12

(Zα)4

n3

1j + 1

2︸︷︷︸

free structure

+ 34

1n. . .

bound States notation:



n = N +|κ| N κ(= j + 1/2

)j l spectroscopic notation Energy/mc2

1 0 1 1/2 0 1S1/2

√1− (Zα)2

2 1 +1 1/2 0 2S1/2√

1+√

1−(Zα)2

2−1 1/2 1 2P1/2

0 2 3/2 1 2P3/212

√4− (Zα)2

Table 2.1: bound states notation

IMAGE23


Radial equation central field problem:

~c(f ′ (r) + 1− κ

rf(r)

)+(E +mc2 − V (r)

)g(r) = 0

~c(g′ (r) + 1 + κ

rg(r)

)+(E −mc2 − V (r)

)f(r) = 0

ψ0 (r, θ, φ) =(f(r)ξl=j∓1/2lmj (θ, φ)ig(r)ξl′=j±1/2lmj (θ, φ)

)

V (r) = −~cZαr

α = e2

~c≈ 1

137

f(ρ) = ργ(∑

κ

aκρκ

)e−ρ

g(ρ) = ργ(∑

κ

bκρκ

)e−ρ

ρ = √ε+ε−r

ε± = 1~c

(mc2 ± E

)γ = −1 +

√κ2 − (Zα)2

∃N∣∣∣∣aN+1 = bN+1 = 0

n = N +|κ|

|κ| = j + 12

⇒ E = mc2√√√√1 + (Zα)2(N+√κ2−(Zα)2

)2

≈ mc2

1− 12 (Zα)2 1

n2 −12

(Zα)4

n3

(1

j + 12− 3

41n

)+ . . .

Let’s continue. . .IMAGE23Experiment:



IMAGE24wave functions for N = 0(√

κ2 − (Zα)2 − κ)a0 + (Zα) b0 = 0(√

κ2 − (Zα)2 + κ

)b0 + (Zα) a0 = 0

⇒ b0 =κ−

√κ2 − (Zα)2

Zα︸︷︷︸>0

a0

also:

b0 =√ε−ε+︸︷︷︸>0

a0

⇒ κ > 0

lowest state we have:

κ = 1 ⇒ j = 12

N = 0 n = 1

Upper component spinor:

ξ0 12m

⇒ ψn=1, s︸︷︷︸l=0

,j=1/2 (~x) wave function

Scale in the wave functions is√ε+ε− = 1

~c√m2c4 − E2

We have:

E21s1/2 = m2c4

(1− (Zα)2

)Scale for the 1s1/2 function:√

ε1s1/2+ ε

1s1/2− = mc2

~c(Zα) = Z

aB

aB = ~mcα

is the Bohr–Radius

f1s1/2(r) ∝(Zr

aB

)√1−(Zα)2−1

e−(ZraB

)

g1s1/2 =√ε−ε+f1s1/2(r) =

1−√

1− (Zα)2

Zαf1s1/2(r)

spherical Spinor:

ξ01/2mj (~x0) =∑mlms

=0

〈0012mj |

12mj〉︸︷︷︸

=1

Y00 (~x0)︸︷︷︸1√4π

ξmj



thus:

ψ1s1/2m (r, θ, φ) ∝ e−ZraBZr

aB

√1−(Zα)2−1

ξm

i1−√

1−(Zα)2

Zα σ (~x0) ξm

~x0 = ~x

|~x|

r → 0 has a mild singularity, because:

(Zr

aB

)√1−(Zα)2−1

≈ exp

(−12 (Zα)2

)log

(Zr

aB

)


3 Fundamentals of many-body problems

N -body system:

Φ : R3 × . . .× R3︸︷︷︸N -times

7→ C

with spin:

Ψ : R3 × . . .× R3︸︷︷︸N -times

7→ C2 ⊗ . . .⊗ C2 =N⊗ C2

Let ξ ± 1/2 is a basis of C2:

ξ+ 12

=(

10

)

ξ− 12

=(

01

)

Basis ofN⊗ C2:

ξα1 ⊗ ξα2 ⊗ . . .⊗ ξαN

with:

〈ξα1 ⊗ . . .⊗ ξαN , ξβ1 ⊗ . . .⊗ ξβN 〉 =N∏k=1

δαkβk

N -particle Hilbert-space is given by:Ψ∣∣∣∣‖Ψ‖2 = 〈ψ,ψ〉 <∞

Ψ : R3 × . . .× R3︸︷︷︸N -times

7→N⊗ C2S+1

for spin s-particles. Schrödinger equation:

i~∂

∂tΨ = HΨ

New symmetry:Let σ be a permutation for 1, . . . , NThis acts on ψ : as

(Pσψ) (~x1, . . . , ~xN ) = Φα1...αN

(xσ(1), . . . , xσ(N)

)ξαα(1) ⊗ . . .⊗ ξαα(N)

=(φασ(1)...ασ(N) (~x1, . . . , ~xN ) ξαα(1) ⊗ . . .⊗ ξαα(N)

)


CHAPTER 3. FUNDAMENTALS OF MANY-BODY PROBLEMS

Building a basis for L2(R3Nm

(C2s+1

)N).

Let H be a 1-particle Hilbert-space H = L2(R3,C2s+1

)with a scalar product 〈 , 〉 and ONB

φαie.g. and oscillator-basis αi = ni, li,mli ,msi

Permutation group SN acts onN⊗ H (i.e. N -particle Hilbert space)

Pσ(φα1 ⊗ φα2 ⊗ . . .⊗ φαN

)= φασ(1) ⊗ φασ(2) ⊗ . . .⊗ φασ(N)

Let

AN (H) =ψ ∈

N⊗ H

∣∣∣∣Pσψ = ε(σ)ψ,∀σ ∈ SN

be the space of antisymmetric N -particle states (here ε(σ) signum of σ).

SN (H) =ψ ∈

N⊗ H

∣∣∣∣Pσψ = ψ,∀σ ∈ SN

be the space of symmetric N -particle states.Define: For ω1 ∈ AN (H), ω2 ∈ AM (H) define a skew-symmetric (= anti-symmetric) product:

ω1 ∧ ω2 ∈ AN+M (H) := 1N !M !

∑σ∈SN+M

ε (σ)Pσ (ω1 ⊗ ω2)

Likewise for ω1 ∈ SN , ω2 ∈ SM we define symmetric product:

ω1 ∨ ω2 := 1N !M !

∑σ∈SN+M

ε (σ)Pσ (ω1 ⊗ ω2)

Properties:

1.

ωk ∈ ANkω1 ∧ (ω2 ∧ ω3) = (ω1 ∧ ω2) ∧ ω3 = ω1 ∧ ω2 ∧ ω3

ω1 ∨ (ω2 ∨ ω3) = (ω1 ∨ ω2) ∨ ω3 = ω1 ∨ ω2 ∨ ω3

2.

ω1 ∈ AN ω2 ∈ AMω2 ∧ ω1 = (−1)NM ω1 ∧ ω2

ω1 ∈ SN ω2 ∈ SMω2 ∨ ω1 = ω1 ∨ ω2

3. for φα1 ⊗ φα2 ⊗ . . .⊗ φαN an ON basis ofN⊗ H:

⇒ 〈φα1 ⊗ φα2 ⊗ . . .⊗ φαN , φβ1 ⊗ φβ2 ⊗ . . .⊗ φβN 〉 =N∏k=1

δαkβk

4.

〈φα1 ∧ . . . ∧ φαN , φβ1 ∧ . . . ∧ φβN 〉 = N !N∏k=1

δαkβk

⇒ φα1 ∧ φα2 ∧ . . . ∧ φαN = 1√N !

∑σ∈SN

ε(σ)Pσ(φα1 ⊗ φα2 ⊗ . . .⊗ φαN

)


“determinant”is an ONB of AN . Likewise:

1√N !n1!n2! . . . nk!

φ∨n1α1 ∨ φ

∨n2α2 ∨ . . . ∨ φ

∨nkαk

with N =k∑i=1

ni

φ∨nα1 = φαk ∨ φαk ∨ . . . ∨ φαk︸︷︷︸n times

“permanent”is orthonormal Basis of SN

5. let φ ∈ H:

a† :

AN 7→ AN+1 for ω ∈ ANSN 7→ SN+1 for ω ∈ SN

a† (φ)ω := φ ∧ ω 1√N + 1

a† (φ)ω := φ ∨ ω 1√N + 1

a† creation operator, creates 1 extra particle in the state φ.adjoint of a† is defined by:

〈ωN+1, a†(φ)ωN 〉 =: 〈aωN+1, (φ)ωN 〉

Then for AN

λ ∈ C

a(λφk

)φβ1 ∧ . . . ∧ φβj ∧ . . . ∧ φβN+1 =

0 if 6 ∃i∣∣∣∣βi = k,

λ∗√N + 1 (−1)j−1 δβj ,kφβ1 ∧ . . . ∧

φβj ∧ . . . ∧ φβN+1

Likewise for SN+1

a(λφk

)φβ1 ∨ . . . ∨ φβj ∨ . . . ∨ φβN+1 =

0 if ∃i∣∣∣∣βi = k,

if k = β1 λ∗√N + 1 (n1 + 1) δβj ,kφβ1 ∨ . . . ∨ φβN+1

let’s say k = β1.

Definition: Fock–Space

A =∞⊕N=0AN

S =∞⊕N=0SN

A0 = S0 = C|0〉 ∈ A0

|0〉 ∈ S0

vacuum.




N -particle Hilbert-space:

HN =N⊗ H

φαiONB H

then: 1√N !φα1 ∧ . . . ∧ φαk ONB AN Fermions

1√N !n1! . . . nk!

φvn1α1 ∧ . . . ∧ φ

nkαk

ONB AN Bosons

when:

ω1 ∈ AN1

ω2 ∈ AN2

ω1 ∧ ω2 = 1N1!N2!

∑σ∈SN1+N2

ε (σ)Pσ (ω1 ⊗ ω2) = (−1)N1N2 ω2 ∧ ω1

ω1 ∈ SN1

ω2 ∈ SN2

ω1 ∨ ω2 = 1N1!N2!

∑σ∈SN1+N2

Pσ (ω1 ⊗ ω2) = ω2 ∨ ω1

φvnα = φα ∨ φα ∨ . . . ∨ φα︸︷︷︸n-times

PG(ψα1 ⊗ . . .⊗ ψαN

)= ψασ(1) ⊗ . . .⊗ ψασ(N)

Building up:

a† : AN → AN+1

a†(ψ)ω = ψ ∧ ω 1√N + 1

a† : SN → SN+1

a†(ψ)ω = ψ ∨ ω 1√N + 1

ω ∈ AN ∈ SN creation operator〈aψ(N+1), ψ(N)〉 := 〈ψN+1, a†ψ(N)〉 ∀ψ(N+1), ψ(N) annihilation operator

a : AN+1 → AN

a : SN+1 → SN

in fact:

a (λφk)φβ1 ∧ . . . ∧ φβN+1 =

0 if 6 ∃i∣∣∣∣βi = k,

λ∗√N + 1 (−1)j−1 δβj ,kφβ1 ∧ . . . ∧

φβj ∧ . . . ∧ φβN+1

a(λφk

)φβ1 ∨ . . . ∨ φβj ∨ . . . ∨ φβN+1 =

0 if ∃i∣∣∣∣βi = k,

if k = β1 λ∗√N + 1 (n1 + 1) δβj ,kφβ1 ∨ . . . ∨ φβN+1∑

l

nl = N + 1


Let’s continue. . .

(FermionicBosonic

)Fock-space:

A = ⊕∞N=0AN

S = ⊕∞N=0 SN

A0,S0=C

→contain 1-element: |0〉 vacuum〈0|0〉 = 1

defined by: a (ψ) |0〉 = 0

Basis AN :1√N !φα1 ∧ . . . ∧ φαN =a†α1 . . . a

†αN|0〉

Remark:

a† (φα)=a†α

Basis SN : (α†α1

)n1. . .(α†αk

)nk |0〉= 1√N !φvn1α1 ∨ . . . ∨ φ

vnkαk

Still has a (normalisation)2 = n1! . . . n2k

For AN : (a (φk) a (φl) + a (φl) a (φk)

)φα1 ∧ . . . ∧ φ αi︸︷︷︸

=k

∧ . . . ∧ φ αj︸︷︷︸=l

∧ . . . ∧ φαN

∝

(−1)j−1 (−1)i−1 + (−1)i−1 (−1)j−2︸︷︷︸=0

φα1 ∧ . . . ∧

φ αi︸︷︷︸

=k

∧ . . . ∧

φ αj︸︷︷︸=l

∧ . . . ∧ φαN

⇒[α (φk) , α (φl)

]+ = 0

take adjoint−−−−−−−→[α† (φk) , α† (φl)

]+

= 0

For SN :

⇒[α (φk) , α (φl)

]− = 0

take adjoint−−−−−−−→[α† (φk) , α† (φl)

]−

= 0

For AN :(a (φk) a† (φl) + a† (φl) a (φk)

)φα1 ∧ . . . ∧ φαN

k = l:

φk ∈φαi

0 + 1√N + 1

√N + 1 = 1

φk 6∈φαi 1√

N + 1√N + 1 + 0 = 1

. . .



k 6= l

→ = 0

⇒[a (φk) , a† (φl)

]+

= δkl · 1

For SN : [a (φk) , a† (φl)

]−

= δkl · 1

For Fermions:

a (φk)φα1 ∧ . . . ∧ φαN =

0 if k 6∈ αi√N + 1 (−1)j−1 φα1 ∧ . . . ∧

φαj ∧ . . . ∧ φαN if k = αj

a† (φk) a (φk)φα1 ∧ . . . ∧ φαN =

0 if k 6∈ αi√N + 1 (−1)j−1 1√

N+1φαj ∧ φα1 ∧ . . . ∧φαj ∧ . . . ∧ φαN if k = αj

= φα1 ∧ . . . ∧ φαj ∧ . . . ∧ φαNω ∈ AN∑

k

a† (φk) a (φk) = N · ω

So this counts the total number of particles.Bosons, also ∑

p

a†(φp)a(φp)ω = N · ω

ω ∈ SN

Generalisation for arbitrary states φ, ψ (upper sign valid for A, lower for S):[a (φ) , a† (ψ)

]±

= 〈φ, ψ〉 · 1[a (φ) , a (ψ)

]± = 0[

a† (φ) , a† (ψ)]±

= 0

Number operator N(φ):

N(φ) = a† (φ) a (φ)

1.

N(φ) |0〉 = a† (φ) a (φ) |0〉 = 0

2. Bosons:

〈φ, φ〉 = 1[N(φ), a(φ)

]− =

[a†(φ)a(φ), a(φ)

]−

= a†(φ)[a(φ), a(φ)

]−︸︷︷︸

=0

+[a†(φ), a(φ)

]−︸︷︷︸

−1·1

a(φ)

= −a(φ)[N(φ), a(φ)

]− = N(φ)a(φ)− a(φ)N(φ)

→ Annihilation operator decreases the number of particles by 1



+ fermions− bosons[

a(φ), a†(φ)]±

= 〈φ, ψ〉 · 1[a†(ψ), a†(φ)

]±

= 0[a(ψ), a(φ)

]± = 0

(occupation) number operator

N(φ) := a†(φ)a(φ)

|0〉 ∈

A0

S0 N(φ) |0〉 = 0

Bosons:[N(φ), a(φ)

]− = −a(φ)[

N(φ), a†(φ)]−

= a†(φ)

a†(φ) =

AN → AN+1

SN → SN+1

a(φ) =

AN → AN−1

SN → SN−1

Let’s continue. . .Remarks: (Fermions upper, Bosons lower row)

N(φ)a†(φ) |0〉 = a†a(φ)a†(φ) |0〉

= a†(φ)(1∓ a†(φ)a(φ)

)|0〉 = a†(φ) |0〉

⇒ eigenvalue of N(φ) on a†(φ) |0〉 = +1also: for Fermions:

N(φ)2 = a†(φ)a(φ)a†(φ)a(φ)

= a†(φ)(1− a†(φ)a(φ)

)a(φ)

= a†(φ)a(φ) = N(φ)

⇒ eigenvalue of N(φ) can be 0 or 1 only.Operators in this occupation number formalism:

Typical Hamilton operator (N -particle system):

H =

sum of single particle operators︷︸︸︷N∑k=1

U(k) +N∑

i<j=1V (i, j)

︸︷︷︸sum of all pair interactions



First consider ∑Nk=1 U(k) and matrix-element (Fermions):

〈φα′1...α′N ,N∑k=1

U(k)φα1...αN 〉 = 〈φα′1...α′N ,N∑k=1

U(k)∑σ∈SN

ε(σ)φασ(1) ⊗ . . .⊗ φασ(N)〉1√N !

Suppose that the single particle basisϕαi

is a ONB with:

U(i)ϕαi = Uαiϕαi

with Uαi =∫

d3xϕ∗αi (~x) U (~x)ϕαi (~x)

i.e. U is diagonal inϕαi

.

. . . = 〈φα′1...α′N ,∑σ

ε(σ)N∑k=1

Uασ(k)︸︷︷︸=∑

kUαk

φασ(1) ⊗ . . .⊗ φασ(N)〉1√N !

= 〈φα′1...α′N ,∑α

N (ϕα)Uαφα1...αN 〉

because: ˆN (ϕα)ϕα1 . . . ϕαN =

1 if α ∈ αi0 if α 6∈ αi

thus: U =N∑k=1

U(k) =∑α

N (ϕα)Uα

=∑α

〈ϕα, Uϕα〉︸︷︷︸Uα

a† (ϕα) a (ϕα)

U =∑α

〈ϕα, Uϕα〉 a† (ϕα) a (ϕα)

This is true in general, i.e. for fermions and bosons.Now consider a basis trafo to an other ONB

ψβ:

ψβ =∑α

〈ϕα, ψβ〉ϕα

define:

a†(ψβ)

=∑α

〈ϕα, ψβ〉 a† (ϕα)

a(ψβ)

=∑α

〈ψβ, ϕα〉︸︷︷︸=〈ϕα,ψβ〉∗

a (ϕα)

⇒[a(ψβ′), a†

(ψβ)]±

= δβ′β1 (show yourself)


U =∑α

〈ϕα, Uϕα〉 a† (ϕα) a (ϕα)

=∑α

〈ϕα, Uϕα〉∑β,γ

〈ψβ, ϕα〉 a†(ψβ)〈ϕα, ψγ〉 a

(ψγ)

U =∑β,γ

〈ψβ, Uψγ〉 a†(ψβ)a(ψγ)

where: 〈ψβ, Uψγ〉 =∫

d3xψ∗β (~x) U (~x)ψγ (~x)

⇒[U , N

]−

= 0

N =∑α

a† (ϕα) a (ϕα)

=∑α

N (ϕα)

Now the 2-body term:(Fermions)

〈φα′1...α′N ,N∑

i<j=1V (i, j)φα1...αN 〉 = 〈φα′1...α′N ,

N∑i<j=1

V (i, j)∑σ

ε(σ)ϕασ(1) ⊗ . . .⊗ ϕασ(N)〉1N !

Again, take a 2-particle basis, such that V is diagonal:

V (~x, ~y)ϕα (~x)⊗ ϕβ (~y) = Vαβϕα (~x)⊗ ϕβ (~y)

e.g. suppose:

V (~x, ~y) =∑µ

Cµθµ (~x) 2ωµ (~y)

e.g.:

1|~x− ~y|

=∞∑l=0

l∑m=−l

4π2l + 1

rl<rl+1>

Y ∗lm

(Θ′, ϕ′

)Ylm (Θ, ϕ)

r< = min(r, r′

)r> = max

(r, r′

)~x(r′,Θ′, ϕ′

)~y(r′,Θ′, ϕ′

)

= 〈φα′1...α′N ,∑σ∈SN

ε(σ)12

N∑i 6=j=1

V (i, j)φασ(1) ⊗ . . .⊗ φασ(i) ⊗ . . .⊗ φασ(j) ⊗ . . .⊗ φασ(N)〉1√N !

= 〈φα′1...α′N ,∑σ∈SN

ε(σ)12

N∑i 6=j=1

Vασ(i)ασ(j)φασ(1) ⊗ . . .⊗ φασ(N)〉1√N !

= 〈φα′1...α′N ,∑α,β

VαβPϕαϕαφα1...αN 〉



where:

Pϕαϕβ should count number of pairs:

if ϕα 6∈

ϕαi

or ϕβ 6∈

ϕαi

0

if α 6= β nαnβ

if α = β nα (nα − 1)

Now define:

Nα = N (ϕα)a† (ϕα) = a†α

Pαβ := NαNβ − δαβNα = a†αaαa†βaβ − δαβa

†αaα

= a†α

([aα, a

†β

]±∓ a†βaα

)aβ − δαβa†αaα

= a†αδαβ ∓ a†αa

†β aαaβ︸︷︷︸=∓aβaα

−δαβa

†αaα

= a†αa†βaβaα

Both for fermions and bosons:

V =N∑

i<j=1V (i, j) =1

2∑αβ

〈ϕαϕβ|V |ϕαϕβ〉 a†ϕαa†ϕβaϕβaϕα

with:

〈ϕαϕβ|V |ϕαϕβ〉 =∫

d3x

∫d3y ϕ∗α (~x)ϕ∗β (~y)V (~x, ~y)ϕα (~x)ϕβ (~y)

Trafo to a general ONBψβ:

V =∑

α,β,γ,δ

〈ψαψβ|V |ψγψδ〉 a† (ψα) a† (ψα) a (ψδ) a(ψγ)

where:

〈ψαψβ|V |ψγψδ〉 =∫

d3x

∫d3y ψ∗α (~x)ψ∗β (~y)V (~x, ~y)ψγ (~x)ψδ (~y)

H =∑α,β

〈ϕα, Uϕβ〉 a†αaβ + 12∑

α,β,γ,δ

〈ϕαϕβ|V |ϕγϕδ〉 a†αa†βaδaγ[

H, N]−

= 0

N :=∑α

a†αaα

The whole Fock-space is block-diagonal for the interactions we know up to now:IMAGE25

a†(ψβ)

=∑α

〈ϕα, ψβ〉 a† (ϕα)


We can also define:

ψ† (~x) :=∑α

〈ϕα,~x〉︷︸︸︷ϕ∗α (~x) a† (ϕα)

This defines the field creation operator.

ψ† (~x) :=∑α

ϕα (~x) a (ϕα)

This defines the field annihilation operator.[ψ (~x) , ψ† (~y)

]±

=∑α,β

ϕβ (~x)ϕ∗α (~y)[a(ϕβ), a† (ϕα)

]±

=∑α

ϕα (~x)ϕ∗α (~y) = δ(3) (~x− ~y) · 1

completeness of the basis

H =∫

d3x ψ† (~x) U (~x)ψ (~x) + 12

∫d3x d3y ψ† (~x) ψ† (~y)V (~x, ~y) ψ† (~y) ψ† (~x)︸︷︷︸

Order!

Particle density operator:

n (~x) =N∑i=1

δ(3) (~x− ~xi) =∑α,β

a†αaβ

∫d3y ϕ∗α (~y) δ(3) (~x.~y)ϕβ (~y)

=∑α,β

a†αaβϕ∗α (~x)ϕβ

(~β)ψ† (~x) ψ (~x)

Number operator:

N :=∫

dd3x n (~x) =∫

d3x ψ† (~x) ψ (~x) =∑α

a†αaα

Current-density operator:

~ (~x) := ~2im

[ψ† (~x)

(~∇ψ)

(~x)−(~∇ψ†

)(~x) ψ (~x)

]


[aα, aβ

]±

=[a†α, a

†β

]±

= 0[aα, a

†β

]±

= δαβ



Fermions up and bosons down. Occupation number operator:

Nα = a†αaα

N =∑α

Nα

∑α

a†αaα

a†α := a (ϕa)ϕαi

ONB

H =∑α,β

〈ϕα, Uϕβ〉 a†αaβ + 12∑

α,β,γ,δ

〈ϕαϕβ|V |ϕγϕδ〉 a†αa†βaδaγ

H : F 7→ Fwith: ψ† (~x) =

∑α

ϕ∗αa†α

ψ (~x) =∑β

ϕβ (~x) aβ

⇒[ψ (~x) , ψ (~y)

]±

=[ψ† (~x) , ψ† (~y)

]±

= 0[ψ (~x) , ψ† (~y)

]±

= δ(3) (~x− ~y)

H =∫

d3x ψ† (~x) U (~x)ψ (~x) + 12

∫d3x d3y ψ† (~x) ψ† (~y)V (~x, ~y) ψ† (~y) ψ† (~x)︸︷︷︸

Order!

[H,N ]− = 0

Let’s continue. . .Lagrange–density for Schrödinger field

x =(x0, ~x

)= (ct, ~x) = (t, ~x)

L[ψ,ψ∗,

(∂µψ

),(∂νψ

∗)] (x) :=[

12 i~c

(ψ∗ (x)

)(∂0ψ) (x)−

(∂0ψ (x)

)∗ψ (x)− ~2

2m(~∇ψ∗ (x) · ~∇ψ (x)

)]

action:

S :=∫

d4xL [. . .] (x)

δS != 0

Euler-Langrange equation:

∂

∂xµ∂L

∂(∂µψ

)∗ − ∂L∂ψ∗

= 0

∂0

(−1

2 i~cψ (x))− ~2

2m∆ψ (x)− 12 i~ (∂0ψ) (x) = 0

⇒ i~∂

∂tψ (t, ~x) = − ~2

2m∆ψ (t, ~x)

define conjugate momenta:

Π∗ (x) := ∂L∂ (∂0ψ) = i

12~cψ

∗ (x)

Ψ (x) := ∂L∂ (∂0ψ)∗ = −i12~cψ (x)


Hamiltonian–density:

H = Π∗ (x) ∂0ψ (x) +(∂0ψ

∗) (x) Π (x)− L

Hamiltonian:

H =∫

d3xH = ~2

2m

∫d3x

((~∇ψ)∗

(t, ~x) · ~∇ψ (t, ~x))

P.I.=∫

d3xψ∗ (t, ~x)(− ~2

2m∆ψ)

(t, ~x)

Eigenstates in a finite volume: V = L3

periodic boundary conditions for the fields:

− ~2

2m∆ψ(†)n (~x) = E(0)

n ψ(†)n (~x)

ψ(†)n (t, ~x) = 1

L3/2ei(~kn·~x−iω0

nt)

= 1V 1/2

e−i(k·x)

with:

kn = 2πL~n

~n ∈ N3

ω0n = E

(0)n

~=

~∣∣∣~kn∣∣∣22m

⇒∫

d3xψ(†)∗~n (t, ~x)ψ~n (t, ~x) = δ

~n, ~n′

Now expand:

ψ (x) =∑n

anψ(†)n (x)

ψ† (x) =∑n

a†nψ(†)∗n (x)

consider an, a†n to be annihilation / creation operators.Now indeed with

[an, a

†n′

]±

= δnn′ :

H =∫V

d3x

(∑n

a†nψ(†)∗ (x)

)− ~2

2m ∆∑n′

an′ψ(†)n′︸︷︷︸

=En′ψ†n(x)

(x)

=∑n

E(0)n a†nan



Klein–Gordon field:

L[φ, φ∗,

(∂µφ

),(∂µφ

∗)]~ = c = 1(

∂µφ∗) (∂µφ)−m2φ∗φ

Euler–Lagrange:

∂µ∂µφ+m2φ = 0

→ H→ H = −∫

d3xφ∗ (x) ~∂t∂φ

∂t(x)

ψ~∂tφ = ψ (∂tφ)− (∂tψ)φ

The ~∂t has an arrow in both directions, to show it is applied in both directions (not a vector).Expand:

φ (x) =∑n

[anφ

(†)n (x) + c†nφ

(−)−~n (x)

]φ (x) =

∑n

[a†nφ

(†)∗n (x) + cnφ

(−)∗−~n (x)

]φ

(+)~n = 1√

2EnL3 e±i(~kn·~n−Ent

)En =

√m2 +

∣∣∣~kn∣∣∣2K.G.:

±δnn′ = i

∫d3xφ(±)

n (x) ~∂tφ(±)n′ (x)

0 = i

∫d3xφ(+)

n (x) ~∂tφ(−)n′ (x)

. . . . . .

H =∑n

Ena†nan + cnc

†n

now: cnc†n = ∓c†ncn + 1

Here: fermions above, bosons below

H =∑n

En(a†nan ∓ c†ncn

)+

∑n

En︸︷︷︸〈0|H|0〉→∞

Energy is positive definite only if[cn, c

†n

]−

= δnn′ , so for bosons only.


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