SCPY152 General Physics II 19 Quantum Wells in Three...

SCPY152 General Physics II

19 Quantum Wells in Three Dimensions

Udom Robkob

Department of Physics, Faculty of Science, Mahidol UniversityOffice: SC4-202, Science 4 Bld., Salaya

E-mail: [email protected]

February 21, 2017

Udom Robkob SCPY152 General Physics II 19 Quantum Wells in Three Dimensions

Today Topics

Cartesian coordinates system

Schrodinger equation in 3D

Infinite potential box

3D harmonic potential well

Central potential problems

Spherical coordinate system

Radial and angular equations

Free angular solutions



One first reviews our mathematical formulation.

For position three dimensional Euclidean space, wedenoted it as ~r . In Cartesian coordinate system, one hasthree coordinates: ~r = x i + y j + zk ≡ (x , y , z), wherethe system of basis vectors (i , j , k) is understood.



In 3D Cartesian system, one define ”gradient” as adirectional derivative as

∇ = ∂x i + ∂y j + ∂z k = (∂x , ∂y , ∂z) (1)

with ∂x = ∂/∂x , and so on. It is used to apply on scalarfunction, f (x , y , z), to be vector function, i.e.,

∇f (x , y , z) = Fx i + fy j + Fz k ≡ ~g(fx , fy , fz) (2)

where fx = ∂f /∂x , and so on.The ”divergence” is the application on vector function toget scalar function, i.e.,

∇ · ~g = ∂x fx + ∂y fy + ∂z fz = (∂2x + ∂2

y + ∂2z )f (3)

The ”Laplacian” is defined to be

∇2 = ∇ · ∇ = ∂2x + ∂2

y + ∂2z (4)


Schrodinger equation in Three Dimensions

On can write Schrodinger equation in 3D Euclideanspace, using Cartesian system, by first applying aparticle-wave correspondence:

1D 3D

x ↔ x , ~r ↔ ~r (5)

px ↔ −i~d

dx, ~p ↔ −i~∇ (6)

Then one has(− ~2

2m∇2 + V (~r)

)ϕ(~r) = Eϕ(~r) (7)

Or in a more suitable form:

∇2ϕ(~r) +2m

~2(E − V (~r))ϕ(~r) = 0 (8)



Equation (8) is partial differential equation. It is hard tofind solution.

We are looking at one special case in which the potentialfunction is ”separable”;

V (~r) ≡ V (x , y , z) = V (x) + V (y) + V (z) (9)

Since the Laplacian is already separated, i.e.∇2 = ∂2

x + ∂2y + ∂2

z , then one can apply the separation ofthe particle wave function in the form

ϕ(~r) ≡ ϕ(x , y , z) = ϕx(x)ϕy (y)ϕz(z) (10)



With the separated energy

E = Ex + Ey + Ez , (11)

equation (8) will be separated into three equation as(1

ϕx(x)ϕ′′x(x) +

2m

~2(Ex − V (x))

)+(

1

ϕy (y)ϕ′′y (y) +

2m

~2(Ey − V (y))

)+(

1

ϕz(z)ϕ′′z (z) +

2m

~2(Ez − V (z))

)= 0 (12)

Since (x , y , z) are independent variables, then theseequations must be separately equal to zero.



Then one has three independent Schrodinger equations ineach direction as

ϕ′′x(x) +

2m

~2(Ex − V (x))ϕx(x) = 0 (13)

ϕ′′y (y) +

2m

~2(Ey − V (y))ϕy (y) = 0 (14)

ϕ′′z (z) +

2m

~2(Ez − V (z))ϕz(z) = 0 (15)

So that quantum physics in three dimensions is justsolving three Schrodinger equations. After success, onejust recombine them as in equations (10,11).



The first simplest case is the problem of infinite potentialbox of sizes (a, b, c), i.e., the potential function is

V (x , y , z) =

∞, x < 0, y < 0, z < 00, 0 < x < a, 0 < y < b, 0 < z < c∞, x > a, y > b, z > c

(16)



There are no waves outside the box, but only standingwaves inside the box.

The Schrodinger equation is simply separated with zeropotential function inside the box, i.e.,

ϕ′′x + k2

xϕx = 0, ϕ′′y + k2

yϕy = 0, ϕ′′z + k2

zϕz = 0 (17)

where ϕx = ϕx(x), k2x = 2mEx/~2, and so on.

As we know, standing waves in three directions are

ϕnx (x) =

√2

asin(nxπx/a), nx = 1, 2, 3, ... (18)



Cont.

ϕny (y) =

√2

bsin(nyπy/b), ny = 1, 2, 3, ... (19)

ϕnz (z) =

√2

csin(nzπz/c), nz = 1, 2, 3, ... (20)

The quantum energies of ap article is written in terms ofthree quantum numbers nx , ny , nz in the form

Enx ,ny ,nz = Enx +Eny +Enz =~2π2

2m

(n2x

a2+

n2y

b2+

n2z

c2

)(21)



In case of potential cube, a = b = c , one has

Enx ,ny ,nz =~2π2

2ma2(n2

x + n2y + n2

z) (22)

Since a cube has four-fold symmetry in each direction,then one observe a numbers of ”degeneracy” of energylevels of the system. (Degeneracy means a number ofdifferent quantum states, specified with a different sets ofquantum numbers, that have the same quantum energy.)

Look at the following table:



Degenerated states and degeneracy of a potential cube:

nx ny nz Enx ,ny ,nz ( ~2π2

2ma2 ) Degeneracy1 1 1 3 12 1 1 61 2 1 6 31 1 2 62 2 1 92 1 2 9 31 2 2 93 1 1 111 3 1 11 31 1 3 112 2 2 12 1



The three dimensional isotropic harmonic oscillatorpotential is

V (x , y , z) =1

2mω2(x2 + y 2 + z2) (23)

which appears in separable form.

The quantum energy of this system can be written in theform

Enx ,ny ,nz = ~ω(nx + ny + nz +

3

2

)(24)

with nx , ny , nz = 0, 1, 2, ....

Exercise Look for the degenerated states and count itsdegeneracy of this system.



This system contains more degenerated states:

nx ny nz Enx ,ny ,nz (~ω) Degeneracy0 0 0 3/2 11 0 0 5/20 1 0 5/2 30 0 1 5/21 1 0 7/21 0 1 7/20 1 1 7/2 62 0 0 7/20 2 0 7/20 0 2 7/2

It grows up like: 1, 3, 6, 9, 12, ..., with increasing energystep of ~ω.


Central potential problems

In case of the potential function appear as a function ofradial distance r =

√x2 + y 2 + z2, i.e.,

V = V (r). (25)

On one hand, one says that the system has ”sphericalsymmetry”, i.e., have free angular motion.

On the other hand, one knows that Schrodinger equationof this system is not separable in Cartesian coordinatesystem.

But it is separable in ”spherical coordinate system”.


Spherical coordinates system

One can change from Cartesian to spherical coordinatesas:

x = r sin θ cosφ, y = r sin θ sinφ, z = r cos θ (26)



Mathematically, one says thatx = x(r , θ, φ), y = y(r , θ, φ) and z = z(r , θ).

One looks for the inversion: r = r(x , y , z), θ = θ(x , y , z)and φ = φ(x , y , z), by first apply the variations:

dx =∂x

∂rdr +

∂x

∂θdθ +

∂x

∂φdφ (27)

dy =∂y

∂rdr +

∂y

∂θdθ +

∂y

∂φdφ (28)

dz =∂z

∂rdr +

∂z

∂θdθ (29)

One can see the coefficients: ∂x/∂r , ... . Then oneevaluate their inversions: ∂r/∂x ,..., by inverse matrixmethod. (See details in the lecture notes.)



Next one try to write

∂

∂x=

∂r

∂x

∂

∂r+∂θ

∂x

∂

∂θ+∂φ

∂x

∂

∂φ(30)

∂

∂y=

∂r

∂y

∂

∂r+∂θ

∂y

∂

∂θ+∂φ

∂y

∂

∂φ(31)

∂

∂z=

∂r

∂z

∂

∂r+∂θ

∂z

∂

∂θ(32)

From this one has

∂

∂x= sin θ cosφ

∂

∂r+

cos θ cosφ

r

∂

∂θ− sinφ

r sin θ

∂

∂φ(33)

∂

∂y= sin θ sinφ

∂

∂r+

cos θ sinφ

r

∂

∂θ+

cosφ

r sin θ

∂

∂φ(34)

∂

∂z= cos θ

∂

∂r− sin θ

r

∂

∂θ(35)



And one write

∇2 =∂2

∂x2+

∂2

∂y 2+

∂2

∂z2

=1

r 2

∂

∂r

(r 2 ∂

∂r

)+

1

r 2

[1

sin θ

∂

∂θ

(sin θ

∂

∂θ

)+

1

sin2 θ

∂2

∂φ2

](36)

Apply this into Schrodinger equation, and separate theparticle wave function into the form

ϕ(x , y , z)→ ϕ(r , θ, φ) = R(r)A(θ, φ) (37)


Radial and angular equations

The Schrodinger equation will be separated into radialequation

d2R(r)

dr 2+

2

r

dR(r)

dr+

[2m

~2(E − V (r))− α

r 2

]R(r) = 0

(38)and the angular equation

1

sin θ

∂

∂θ

(sin θ

∂A

∂θ

)+

1

sin2 θ

∂2A

∂φ2+ αA = 0 (39)

where α is a constant of independent.


Free angular solution

For free angular equation

1

sin θ

∂

∂θ

(sin θ

∂A

∂θ

)+

1

sin2 θ

∂2A

∂φ2+ αA = 0 (40)

It is known in the name of ”spherical harmonic equation”(Reference to:http://mathworld.wolfram.com/SphericalHarmonic.html)

Its solution exists when α = l(l + 1), with l = 0, 1, 2, ...,and additional index m, withm = −l ,−(l − 1),−(l − 2), ..., (l − 2), (l − 1), l .

The solution get the same name, spherical harmonic:Y ml (θ, φ).



References to:http://mathworld.wolfram.com/SphericalHarmonic.htmland https://en.wikipedia.org/wiki/Table-of-spherical-harmonics



and More



The angular distribution of particle wave function appearsas in the following figure:


Radial solution

Radial solution relies on the potential function.

The simplest radial problem is a problem of free particle,i.e., V (r) = 0. The radial equation becomes

r 2R ′′ + 2rR ′ +(k2r 2 − l(l + 1)

)R = 0 (41)

After one has multiplied through with r 2 and definedk2 = 2mE/~2.

Let one define ρ = kr , then one get R = R(ρ) and

ρ2R ′′ + 2ρR ′ + (ρ2 − l(l + 1))R = 0 (42)

This equation is known in the name of ”spherical Besselequation”. (http://mathworld.wolfram.com/SphericalBesselDifferentialEquation.html)


Radial solution

Its solution exists in the form of ”spherical Besselfunction”

R(ρ) = C1jl(ρ) + C2nl(ρ) (43)

where jl(ρ) is called ”spherical Bessel function of the firstkind”


Radial solution

While nl(ρ) is known in the name of ”spherical Besselfunction of the second kind”’:

Since particle wave appear everywhere include origin, butnl(kr) not finite at r = 0, then one has to ignore it bychoosing C2 = 0.


Radial solution

Our solution is then

R(r) = Cl jl(kr) (44)

See some examples


Infinite potential sphere

The potential function of infinite potential sphere is

V (r) =

{0, 0 ≤ r < a∞, r ≥ a

(45)

The particle wave exists only inside the sphere and itswave function vanish at the spherical boundary.

From equation (41), one says that, for the l = 0 state

R0(a) = 0 = C0j0(k0a)→ k0a = nπ (46)

Or k0n = nπ/a. Then one has discrete energies

E0n =~2k2

0n

2m=

~2π2n2

2ma2, n = 1, 2, 3, ... (47)


Infinite potential sphere

The other zeros of jl(kla) can be calculated athttp://keisan.casio.com/has10/SpecExec.cgi.

One can collect in a table form as

Let make a list of increasing energy levels from the lowestenergy state, (l = 0, n = 1), or ground state.


Summary

Discrete energies of infinite potential box

Discrete energies of infinite potential cube, withdegeneracy

Discrete energies of isotropic harmonic oscillator, withdegeneracy

Particle wave function of free particle in sphericalcoordinates

Discrete energies of infinite potential sphere


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