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Physics 2D Lecture SlidesMar 10
Vivek SharmaUCSD Physics
• 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)
QM in 3 Dimensions
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
x
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)
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
y z [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 2 22 2 2 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
( ) kZeU rr
=
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
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
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