Physics 2D Lecture SlidesDec 1
Vivek SharmaUCSD Physics
QM in 3 Dimensions • Learn to extend S. Eq and its
solutions from “toy” examples in 1-Dimension (x) → three orthogonal dimensions (r ≡x,y,z)
• Then transform the systems – Particle in 1D rigid box 3D
rigid box– 1D Harmonic Oscillator 3D
Harmonic Oscillator • Keep an eye on the number
of different integers needed to specify system 1 3 (corresponding to 3 available degrees of freedom x,y,z)
y
z
x
ˆˆ ˆr ix jy kz= + +
Quantum Mechanics In 3D: Particle in 3D BoxExtension of a Particle In a Box with rigid walls
1D → 3D⇒ Box with Rigid Walls (U=∞) in
X,Y,Z dimensions
yy=0
y=L
z=L
z
Ask same questions:• Location of particle in 3d Box• Momentum • Kinetic Energy, Total Energy• Expectation values in 3D
To find the Wavefunction and various expectation values, we must first set up the appropriate TDSE & TISE
U(r)=0 for (0<x,y,z,<L)
x
The Schrodinger Equation in 3 Dimensions: Cartesian Coordinates
2 2 22
2 2
2 2 2 2 2
22
2
22
2
2 2
Time Dependent Schrodinger Eqn:( , , , )( , , , ) ( , , ) ( , ) .....In 3D
2
2 22
2
x y z tx y
x y
z t U x y z x t im t
m x m y m
z
mSo
∂ ∂ ∂∇
∂Ψ− ∇ Ψ + Ψ =
∂
− ∇ = +⎛ ⎞ ⎛ ⎞∂ ∂ ∂− − + −⎜ ⎟ ⎜ ⎟∂ ∂ ∂⎝ ⎠ ⎝
+∂
⎠
= +∂ ∂
x
2
x x [K ] + [K ] + [K ] [ ] ( , ) [ ] ( , ) is still the Energy Conservation Eq
Stationary states are those for which all proba
[ ]
=
bilities
so H x t E
K
x t
z⎛ ⎞
=
Ψ
⎟⎠
=
⎜⎝
Ψ
-i t
are and are given by the solution of the TDSE in seperable form: = (r)eThis statement is simply an ext
constant in time
(ension of what we
, derive
, , ) ( , )d in case of
x y z t r t ωψΨ = Ψ1D
time-independent potential
y
z
x
Particle in 3D Rigid Box : Separation of Orthogonal Spatial (x,y,z) Variables
1 2 3
1
2
2 3
2
in 3D:
x,y,z independent of each ( , , ) ( ) ( ) ( )and substitute in the master TISE, after dividing thruout by = ( ) ( ) (
- ( , , ) ( ,
other , wr
, ) ( , ,
)and
) ( ,
ite
, )
n
2mx y z
TISE x y z U x y z x y z E x y
x y zx y
z
zψ ψ ψ ψ
ψ ψ ψ
ψ
ψ
ψ ψ∇
=
+ =
221
21
22222
232
3
21
2
2
( )12 ( )
This can only be true if each term is c
oting that U(r)=0 fo
onstant for all x,y,z
( )12 ( )
(2
r (0<x,y,z,<L)
( )12 (
)
z E Constm z z
ym
xm x x
xm
y yψ
ψψ
ψ
ψ
ψψ⎛ ⎞∂
− +⎜ ⎟⎛ ⎞∂
+ − =
⇒
⎛ ⎞∂−⎜ ⎟∂⎝ ⎠∂⎝ ⎠
=
⇒
−
⎜ ⎟⎝ ⎠
∂
∂
223
3 32
222
2 21 12 2
1 2 3
) ( ) ;
(Total Energy of 3D system)
Each term looks like
( ) ( ) ;2
With E
particle in
E E E=Constan
1D box (just a different dimension)
( ) ( )2
So wavefunctions
t
z E zm z
yy
E x Ex
ym
ψ ψ ψ ψ ψ∂− =
∂−
∂
=
==∂
+ +
∂
3 31 2 21must be like ,( ) sin x ,( ) s ) s nin ( iyy kx k z k zψψ ψ∝ ∝∝
Particle in 3D Rigid Box : Separation of Orthogonal Variables
1 1 2 2 3 3
i
Wavefunctions are like , ( ) sin
Continuity Conditions for and its fi
( ) sin y
Leads to usual Quantization of Linear Momentum p= k .....in 3D
rst spatial derivative
( )
s
sin x ,
x
i i
z k z
n k
x
L
yk
p
k ψ
ψ π
ψ
π
ψ∝ ∝
⇒
=
∝
=
1 2 3
2
2
1 3 1
2 2 22
2
2 3 ; ;
Note: by usual Uncertainty Principle argumen
(n ,n ,n 1,2,3,.. )
t neither of n , n ,n 0! ( ?)
1Particle Energy E = K+U = K +0 = )2
(m 2
(
zy
x y z
n
why
p nL
nmL
p nL L
p p p
π
π
π⎛ ⎞ ⎛ ⎞= = ∞⎜ ⎟ ⎜⎛ ⎞= ⎜ ⎟⎝ ⎠
⎟⎝ ⎠ ⎝
=
⎠
+ + = 2 2 21 2 3
2
1 2 3
2
1
Ei
3
-
3
1
)
Energy is again quantized and brought to you by integers (independent)and (r)=A sin (A = Overall Normalization Cosin y
(r)
nstant)
(r,t)= e [ si
n ,n , nsin x
sin x ysn in ]t
k
n n
k
A k k
k
k
z
z
ψ
ψ
+ +
=ΨE-i
et
Particle in 3D Box :Wave function Normalization Condition
3
*
1
1
21
x,y,
E E-i -i
2
E Ei i*2
2 22
2
3
*3
z
2
(r) e [ sin y e
(r) e [ s
(r,t)= sin ]
(r,t)= sin ]
(r,t)
sin x
sin x
sin x
in y e
[ si
Normalization Co
(r,t)= sin ]
ndition : 1 = P(r)dx
n y
dyd
1
z
t t
t t
k z
k
k
k
A k
A k
A k zk k
A
z
ψ
ψ
Ψ
Ψ
Ψ
⇒
Ψ
=
=
=
∫∫∫L L L
2
3 3E
2 2 21 2 3
x=0 y=
2 2 -
1
z
3
i
0
2
=0
sin x dx
s
sin y dy sin z dz =
(
2 2 2
2 2 an r,t)=d [ s sinii ex yn ] nt
L
k
L LA
A kL
k k k
k zL
⎛ ⎞⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠⎝ ⎠
⎡ ⎤ ⎡ ⎤⇒ = ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦Ψ
∫ ∫ ∫
Particle in 3D Box : Energy Spectrum & Degeneracy
1 2 3
2 22 2 2
n ,n ,n 1 2 3 i
2 2
111 2
2 2
211 121 112 2
2
3Ground State Energy E2
6Next level 3 Ex
E ( ); n 1, 2, 3... , 02
s
cited states E = E E2
configurations of (r)= (x,y,z) have Different ame energy d
i
mL
mL
n n n nmL
π
π
ψ ψ
π= + + = ∞ ≠
=
⇒ = =
⇒ egeneracy
yy=L
z=Lz
xx=L
2 2
211 121 112 2
Degenerate States6E
= E E2mLπ
= =
x
y
z
E211 E121 E112ψ
E111
x
y
z
ψ
Ground State
Probability Density Functions for Particle in 3D Box
Same Energy Degenerate StatesCant tell by measuring energy if particle is in
211, 121, 112 quantum State
Source of Degeneracy: How to “Lift” Degeneracy • Degeneracy came from the
threefold symmetry of a CUBICAL Box (Lx= Ly= Lz=L)
• To Lift (remove) degeneracy change each dimension such that CUBICAL box Rectangular Box
• (Lx≠ Ly ≠ Lz)• Then
2 22 2 2 231 2
2 2 22 2 2x y z
nn nEmL mL mL
ππ π⎛ ⎞⎛ ⎞ ⎛ ⎞= + +⎜ ⎟⎜ ⎟ ⎜ ⎟⎜ ⎟ ⎝ ⎠⎝ ⎠ ⎝ ⎠
Ener
gy
The Hydrogen Atom In Its Full Quantum Mechanical Glory
2 2 2
By example of particle in 3D box, need to use seperation of variables(x,y,z) to derive 3 in
1 1
This approach willdependent d
( ) M
iffer
ore compli
ential. eq
cated form of U than bo
get
x
ns.
U rr x y z
∝ = ⇒+ +
2 22
2
To simplify the situation, use appropriate variablesIndependent Cartesian (x,y,z) Inde. Spherical Polar (r, ,
very ugly since we have a "conjoined triplet"
Instead of writing Laplacian
)
x y
θ φ
∂ ∂∇ = +
∂ ∂
→
2
2
2 2
2
2 2 2
2
2
2
22
2 2 2
22
2
, write
1sin
TISE for (x,y,z)= (r, , ) become
1r
1 (r, , ) (r, , )r
s
1 2m+
1= s
(E-U(r))
insi
si
n
1 sins
(r, , )n
in
r
r
rr
z
r
r
rr r
r θθ
θ θ θ
θθ θ
φ
ψ ψ
θψ θ φ ψ θ φ
ψ θ φ
θ φ
θ φ
∂ ∂⎛ ⎞⎜ ⎟∂ ∂⎝ ⎠
∂ ∂⎛
∂ ∂⎛
⎞⎜ ⎟∂ ∂
∂+
∂
∂∂
∂∂
⎞∇ + +⎜ ⎟∂ ∂⎝ ⎠
∂ ∂⎛ ⎞+ +⎜ ∂ ⎠⎝⎟∂ ⎝⎠ !!!! fun!!!
(r, , ) =0 ψ θ φ
r
2
( ) kZeU rr
=
Spherical Polar Coordinate System
2
( sin )Vol
( )( ) = r si
ume Element dV
ndV r d rd dr
drd dθ φ θ
θ θ φ
=
Don
’t P
anic
: Its
sim
pler
than
you
thin
k ! 2
2
2 2 22
2
2 2
2 2
2
1 2m+ (E-U (r))sin
T ry to free up las
1 (r, , ) =0 r
all except
T his requires m ulti
t term fro
plying thruout by sin
si
1 sinsi
si si
m
n
n
n
rr
r r
r
r
rr r
ψ ψ ψ ψ θ φ
φ
θ
ψ
θθ θ θ
θθ
θ
θ
φ∂ ∂⎛ ⎞
⎜ ⎟∂ ∂⎝ ⎠
⇒
∂ ∂⎛ ⎞⎜ ⎟∂ ∂⎝ ⎠
∂ ∂⎛ ⎞+ +⎜ ⎟∂ ∂⎝ ⎠
∂+
∂
∂∂
2
2
2 2 2
2
2m ke+ (E+ )r
(r, , )=R (r). ( ). ( ) P lug it in to the T ISE above & divide thruout by (r, , )=R (r). ( ).
sin =0
For Seperation of V ariables, W rite
( , , )r
N ote tha
(
t :
)
n r
r
φψ θ φ θ φ
ψ ψ θ ψ
ψ θ φ
φ
θ
θ
θ
φ
θ∂⎛ ⎞ +⎜ ⎟∂⎝ ⎠
∂
∂∂
Θ
Ψ
Θ ΦΦ
∂
2 2 22 2
2 22
( ). ( )
( , , ) ( ) ( )
( , , ) ( ) ( )
s
R (r) r
( ) w hen substitu ted in T I
in sin =0
R earrange by ta
sin
king the
sin
SE
( )
1 2m ke+ (E+
)r
r R r
r R r
R rrR r r
θ φ
θ φ φθθ φ θ
θ
θ
θφ
φ
θ θφ
θθ θ
θ
= Θ Φ
∂Ψ= Φ
∂∂Ψ
= Θ∂
∂ ∂⎛ ∂ ∂Θ⎛ ⎞+ +⎜ ⎟Θ⎞
⎜ ∂ ∂⎝⎟∂ ∂
∂∂
∂Θ⇒
∂∂Φ
∂
∂∂⎝ Φ⎠⎠
Φ
2 2
2
2 2 2
22
2
2m ke 1+ (E+ )r
LH S is fn . of r, & R H S is fn of only , for equality to be true for all r, ,
LH S= constant = R H
term on R H S
sin s
S =
sin sin
m
in =-
l
R rrR r r φ
θ φ θ φ
θ θ
φ
θθ θ
θ∂ ∂⎛ ⎞⎜ ⎟
∂ ΦΦ ∂
⇒
∂ ∂Θ⎛ ⎞+ ⎜ ⎟Θ ∂ ∂⎝⎝ ⎠ ⎠∂ ∂
2 2 22 2
2
2
2
2
sin sin =m
Divide Thruout by sin and arrange all terms with r aw
Now go break up LHS to seperate the terms...r .. 2m keLHS: + (E+ )
a
& sin si
y fromr
1
n lR r
r
rr
r
R r
R
θ θ
θ
θ θθ θ
θ
θ
∂ ∂Θ⎛ ⎞+ ⎜ ⎟Θ ∂ ∂⎝ ⎠∂ ∂⎛
∂
⎞
∂∂
⎜ ⎟∂ ∂⎝ ⎠⇒
2
2
2 2
2
m 1 sinsin sin
Same argument : LHS is fn of r, RHS is fn of , for them to be equal for a
LHS = const =
2m ke(E+ )=r
What do we have after shuffl
ll r,
= ( in1) g RHS
l
l l
R rr
θθ θ θ θ
θ θ
∂ ∂Θ⎛ ⎞− ⎜ ⎟⎛ ⎞ +
Θ ∂ ∂⎝ ⎠
⇒ +
⎜ ⎟∂⎝ ⎠
22
2
2
2
2 22
2 2 2
!
m1 sin ( 1) ( ) 0.....(2)sin sin
...............
d ..(1)
1 2m ke ( 1)(E+ )- ( ) 0....(
m 0.
3
.
)r
T
l
l
d R r l lr R rr dr
d d ld
d
r
ld
r
θ
φ
θθ θ θ θ
⎡ ⎤Θ⎛ ⎞ + + − Θ
⎡ ⎤∂ +⎛ ⎞ + =⎜ ⎟ ⎢ ⎥∂⎝ ⎠ ⎣ ⎦
=⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦
Φ+ Φ =
hese 3 "simple" diff. eqn describe the physics of the Hydrogen atom.All we need to do now is guess the solutions of the diff. equationsEach of them, clearly, has a different functional form
Solutions of The S. Eq for Hydrogen Atom2
22
dThe Azimuthal Diff. Equation : m 0
Solution : ( ) = A e but need to check "Good Wavefunction Condition"Wave Function must be Single Valued for all ( )= ( 2 )
( ) = A e
l
l
l
im
im
dφ
φ
φ
φφ φ φ π
φ
Φ+ Φ =
Φ⇒ Φ Φ +
⇒ Φ ( 2 )
2
2
A e 0, 1, 2, 3....( )
The Polar Diff. Eq:
Solutions : go by the name of "Associated Legendre Functions"
1 msin
( 1) ( ) 0sin si
Quantum #
n
l
l
iml
d d
Magnetic
ld
m
ld
φ π
θ θθ θ θ θ
+
⎡ ⎤Θ⎛ ⎞ + + − Θ =⎜ ⎟ ⎢ ⎥⎝ ⎠
=
⎣ ⎦
⇒ = ± ± ±
only exist when the integers and are related as follows 0, 1, 2, 3....
: Orbital Q
;
ua
p
nt
osit
um N
ive numb
umber
er
1For 0, =0 ( ) = ; 2
For
l
l
l
l ml l
l m
l
m
θ
= ± ± ± ± =
= ⇒ Θ
2
1, =0, 1 Three Possibilities for the Orbital part of wavefunction
6 3[ 1, 0] ( ) = cos [ 1, 1] ( ) = sin 2 2
10[ 2, 0] ( ) = (3cos 1).... so on and so forth (see book) 4
l
l l
l
l m
l m l m
l m and
θ θ θ θ
θ θ
= ± ⇒
= = ⇒ Θ = = ± ⇒ Θ
= = ⇒ Θ −
Φ
Solutions of The S. Eq for Hydrogen Atom2 2
22 2 2
2
20
1 2m ke ( 1)The Radial Diff. Eqn: (E+ )- ( ) 0r
: Associated Laguerre Functions R(r), Solutions exist only if:
1. E>0 or has negtive values given by ke 1E=-2a
d R r l lr R rr dr r r
Solutions
n
⎡ ⎤∂ +⎛ ⎞ + =⎜ ⎟ ⎢ ⎥∂⎝ ⎠ ⎦
⎝
⎣
⎛ 2
0 2
2. And when n = integer such that 0,1,2,3,4,, , ( 1)
n = principal Quantum # or the "big daddy" qunatuTo
; Bohr
Summa
m # : The hy
Rad
r drogeize n
ius
atom
a
l
mke
n= −
⎞ = =⎜ ⎟⎠
n = 1,2,3,4,5,....0,1,2,3,,4....( 1)
m 0, 1, 2, 3,.Quantum # appear only in Trappe
is brought to you
d systems
The Spati
by the letters
al Wave Function o
f the
Hydrogen Atom
..
l
l nl
∞= −
= ± ± ± ±
lm( , , ) ( ) . ( ) . ( ) Y (Spherical Harmonics)
l
l
mnl lm nl lr R r Rθ φ θ φΨ = Θ Φ =
Radial Wave Functions & Radial Prob Distributions
0
0
-r/a3/20
r-2
a3/2
00
23
23/20 00
R(r)= 2 e
a
1 r(2
n
1 0 0
2 0 0
3 0 0
- )e a2 2a
2 r(27 18 2 )a81 3a
l
rar e
a
l m
−
− +
n=1 K shelln=2 L Shell n=3 M shelln=4 N Shell……
l=0 s(harp) sub shelll=1 p(rincipal) sub shelll=2 d(iffuse) sub shelll=3 f(undamental) ssl=4 g sub shell……..
Symbolic Notation of Atomic States in Hydrogen
2 2
4 4
2
( 0) ( 1) ( 2) ( 3) ( 4
3 3 3
) .....
1
4
3
1s p
s l p l d l f l g l
s
l
sd
n
ps p
= = =→ = =
↓
5 5 5 5 4
5 4
5s p d f gd f
Note that: •n =1 non-degenerate system •n1>1 are all degenerate in l and ml.
All states have same energyBut different spatial configuration
2
20
ke 1E=-2a n
⎛ ⎞⎜ ⎟⎝ ⎠
Facts About Ground State of H Atom -r/a
3/20
-r/a100
0
2 1 1( ) e ; ( ) ; ( )a 2 2
1 ( , , ) e ......look at it caref
1. Spherically s
1, 0,
ymmetric no , dependence (structure)
2. Probab
0
ully
i
ln l r
ra
m R θ φπ
θ φπ
θ φ
⇒ = Θ = Φ =
Ψ =
⇒
= = =
22
100 30
Likelihood of finding the electron is same at all , and depends only on the radial seperation (r) between elect
1lity Per Unit Volume : ( ,
ron & the nucleus.
3 Energy
,
of Ground ta
)
S
rar e
aθ
πθ φ
φ−
Ψ =
2
0
kete =- 13.62a
Overall The Ground state wavefunction of the hydrogen atom is quite Not much chemistry or Biology could develop if there was only the ground state of the Hydrogen Atom!
We nee
boring
eV= −
d structure, we need variety, we need some curves!
Interpreting Orbital Quantum Number (l)
2
RADIAL ORBITAL
RADIAL ORBI
2 22
2 2 2
22
TAL2 2
K
1 2m ke ( 1)Radial part of S.Eq
K ;substitute this form for
n: (E+ )- ( ) 0r
For H Atom: E = E
K K
K + U =
1 2m ( 1)-2m
d dR r l lr R rr dr dr r
d dR l lrr dr dr r
ker
⎡ ⎤+⎛ ⎞ + =⎜ ⎟ ⎢ ⎥⎝ ⎠ ⎣ ⎦
+⎛ ⎞ +⎜
+
⎟
−
+⎝ ⎠
[ ]
ORBITAL
RADIAL
OR
2
2
22
B
2
2
( ) 0
( 1)Examine the equation, if we set get a diff. eq. in r2m
1 2m ( ) 0 which
Further, we also kno
depends only on radi
K
K us
w
r of orb
K tha
i
t
t
R r
l l thenr
d dRr R rr dr dr
⎡ ⎤=⎢ ⎥
⎣ ⎦
+
⎛ ⎞ + =⎜ ⎟⎝ ⎠
=
22
ITAL ORBITorb
2
AL 2
2
ORBITAL 22
L= r p ; |L| =mv r
( 1
1 ; K2 2
Putting it all togather: K magnitude of) | | ( 1)2m
Since integer=0,1,2,3...(
A
n-1) angular mome
ng.2
nt
Mom
orbit
l l L l lr
l
Lmvmr
L
posi
mr
tive
× ⇒
+= = +
=
=
= ⇒
=
⇒
um| | ( 1)
| | ( 1) : QUANTIZATION OF Electron's Angular Momentum
L l l discrete values
L l l
= + =
= +
pr
L
Magnetic Quantum Number ml (Right Hand Rule)
QM: Can/Does L have a definite direction
Classically, direction & Magnitud
? Proof by NegatˆSuppose L was precisely known/defined (L || z)
e of L
S
always well defi
n
ed
:
n
io
L r p= ×
2
z
z z
Electron MUST be in x-y orbit plane
z = 0 ;
, in Hydrogen atom, L can not have precise measurable
ince
Uncertainty Principle & An
p p ; !!!
gular Momentum
value
: L
2
pz E
L r p
Som
φ
= × ⇒
⇒ ∆ ∆ ∆ ⇒ ∆ ∞
∆
∞
∆
=∼ ∼ ∼
∼
Z
Z
The Z compo
Arbitararily
nent of
L vector spins around Z axis (precesses).
|L | ; 1, 2, 3...
( 1)
It
picking Z axis as a referen
L
:| L |
ca
| | (always)
s
n
in
ce directi
c
o
:
e
n
l l
l
m
Note L
m l
m l l
<
= = ± ± ± ±
< +
Z never be that |L | ( 1) (breaks Uncertainty Principle)So you see, the dance has begun !
lm l l= = +
L=2, ml=0,±1, ± 2 : Pictorially
Sweeps Conical paths of different ϑ: Cos ϑ = LZ/L and average
<LX> = 0 <LY> = 0
What’s So “Magnetic” ?
Precessing electron Current in loop Magnetic Dipole moment µ
More in this in Tomorrow’s lecture when we look at Energy States
Radial Probability Densities l
2* 2 2
m
2
( , , ) ( ) . ( ) . ( )
( , , ) |
Y
Probability Density Function in 3D:
P(r, , ) = =| | Y |
: 3D Volume element dV= r .sin . . .
Prob. of finding parti
| | .
cle in a ti
n
l
l
l
nlnl lml
n
m
mll
Note d
r R r
r
r d
R
d
R
θ φ θ φ
θθ φ
φ
φ
θ θ
Ψ = Θ Φ =
ΨΨ Ψ =
l
l
2 2
22
m0
2
2
0
2 2
m
y volume dV isP.dV = | Y | .r .sin . . .The Radial part of Prob. distribution: P(r)dr
P(r)dr= | ( ) |
When
| ( ) |
( ) & ( ) are auto-normalized then
P(r)dr
| | .
| |
=
.
|
l
l
l
lm
l
m
m
n
nl
n
llR
R r d
R
dr
d
d
r
d
dπ π
θ
θ φ
φ φ
θ φ
θ
θ
Θ
Θ
Φ
Φ
∫ ∫
2 2
2 2nl
2 2
0
0
in other words
Normalization Condition: 1 = r |R | dr
Expectation Values
P(r)=r |
<f(
| |. . ;
r)>= f(r).P(r)dr
nll r r Rd
∞
∞
∫
∫
Ground State: Radial Probability Density
0
0
0
0
2 2
22
3
22
30
2
2
0
0
0
( ) | ( ) | .4
4( )
Probability of finding Electron for r>a
To solve, employ change of variable
2rDefine z= ; limits of integra
4
12
tiona
ra
r aa
r a
ra
r e dr
P r dr r r dr
P r dr r ea
change
P a
P z
ψ π
−∞
>
∞
>
−
=
⇒ =
⎡ ⎤⎢ ⎥
=
=
⎣ ⎦
∫
2 22
(such integrals called Error. Fn)
1 66.7% =- [ 2 2] | 5 0.6672
!!
z
z
e
z e
z
e
d
z −
−
∞+ + = = ⇒
∫