schrodinger’s equation for three dimensions. qm in three dimensions the one dimensional case was...

Schrodinger’s Equation for Three Dimensions

QM in Three Dimensions

The one dimensional case was good for illustrating basic features such as quantization of energy.

QM in Three Dimensions

The one dimensional case was good for illustrating basic features such as quantization of energy.

However 3-dimensions is needed for application to atomic physics, nuclear physics and other areas.

Schrödinger's Equa 3Dimensions

For 3-dimensions Schrödinger's equation becomes,

t

irUm

2

2

2



Where the Laplacian is

t

irUm

2

2

2

2

2

2

2

2

22

zyx



Where the Laplacian is

and

t

irUm

2

2

2

2

2

2

2

2

22

zyx

zyxUrU ,,


The stationary states are solutions to Schrödinger's equation in separable form, tiertr

,


The stationary states are solutions to Schrödinger's equation in separable form,

The TISE for a particle whose energy is sharp at is,

rErrUrm

22

2

tiertr ,

E

Particle in a 3 Dimensional Box


The simplest case is a particle confined to a cube of edge length L.



The potential energy function is

for That is, the particle is free within the box.

0,, zyxU Lzyx ,,0



The potential energy function is

for That is, the particle is free within the box. otherwise.

0,, zyxU Lzyx ,,0

zyxU ,,


Note: If we consider one coordinate the solution will be the same as the 1-D box.



The spatial waveform is separable (ie. can be written in product form):

r

zyxzyxr ,,



The spatial waveform is separable (ie. can be written in product form):

Substituting into the TISE and dividing by

we get,

r

zyxzyxr ,,

r


The independent variables are isolated. Each of the terms reduces to a constant:

Ezmymxm

23

2

3

2

22

2

2

2

21

2

1

2

222

121

2

1

2

2E

xm

222

2

2

2

2E

ym

323

2

3

2

2E

zm

1... 2...

3...


Clearly EEEE 321


Clearly The solution to equations 1,2, 3 are of the

form where

EEEE 321

kxsin kxcos, 22 mEk



form where Applying boundary conditions we find,

EEEE 321

kxsin kxcos, 22 mEk

zkykxkAzyx 321 sinsinsin,,




where

EEEE 321

kxsin kxcos, 22 mEk


11 nLk 22, nLk 33, nLk




where Therefore,

EEEE 321

kxsin kxcos, 22 mEk


11 nLk 22, nLk 33, nLk

2322

21

2

2kkk

mE

m

pppE zyx

2

222


with and so forth.1kpx L

n 1


with and so forth. Using restrictions on the wave numbers

and boundary conditions we obtain,

1kpx L

n 1




2322

212

22

321 2nnn

mLEEEE

1kpx L

n 1




Thus confining a particle to a box acts to quantize its momentum and energy.

2322

212

22

321 2nnn

mLEEEE

1kpx L

n 1


Note that three quantum numbers are required to describe the quantum state of the system.



These correspond to the three independent degrees of freedom for a particle.



These correspond to the three independent degrees of freedom for a particle.

The quantum numbers specify values taken by the sharp observables.


The total energy will be quoted in the form

2322

212

22

,, 2321nnn

mLE nnn


The ground state ( ) has energy

1321 nnn

2

22

1,1,1 2

3

mLE


Degeneracy


Degeneracy: quantum levels (different quantum numbers) having the same energy.



Degeneracy is a natural phenomena which occurs because of the same in the system described (cubic box).



Degeneracy is a natural phenomena which occurs because of the same in the system described (cubic box).

For excited states we have degeneracy.


There are three 1st excited states having the same energy. They correspond to combinations of the quantum numbers whose squares sum to 6.


There are three 1st excited states having the same energy. They correspond to combinations of the quantum numbers whose squares sum to 6.

That is

2

22

2,1,11,2,11,1,2 2

6

mLEEE


The 1st five energy levels for a cubic box.

n2 Degeneracy

12 none

11 3

9 3

6 3

3 none

4E0

11/3E0

2E0

3E0

E0


The formulation in cartesian coordinates is a natural generalization from one to higher dimensions.


The formulation in cartesian coordinates is a natural generalization from one to higher dimensions.

However it not often best suited to a given problem. Thus it may be necessary to convert to another coordinate system.


Consider an electron orbiting a central nucleus.

Consider a particle in a two-dimensional (infinite) well, with Lx = Ly.

1. Compare the energies of the (2,2), (1,3), and (3,1) states? Explain your answer?

a. E(2,2) > E(1,3) = E(3,1)

b. E(2,2) = E(1,3) = E(3,1)

c. E(1,3) = E(3,1) > E(2,2)

2. If we squeeze the box in the x-direction (i.e., Lx < Ly) compare E(1,3) with E(3,1): Explain your answer?

a. E(1,3) < E(3,1)

b. E(1,3) = E(3,1)

c. E(1,3) > E(3,1)

Example 1

42

Consider a particle in a two-dimensional (infinite) well, with Lx = Ly.

1. Compare the energies of the (2,2), (1,3), and (3,1) states?

a. E(2,2) > E(1,3) = E(3,1)

b. E(2,2) = E(1,3) = E(3,1)

c. E(1,3) = E(3,1) > E(2,2)

2. If we squeeze the box in the x-direction (i.e., Lx < Ly) compare E(1,3) with E(3,1):

a. E(1,3) < E(3,1)

b. E(1,3) = E(3,1)

c. E(1,3) > E(3,1)

Example 1

E(1,3) = E(1,3) = E0 (12 + 32) = 10 E0

E(2,2) = E0 (22 + 22) = 8 E0

Example 2: Energy levels (1)• Now back to a 3D cubic box:

Show energies and label (nx,ny,nz) for the first 11 states of the particle in the 3D box, and write the degeneracy D for each allowed energy.

Use Eo= h2/8mL2.

z

x

yL

L

L

E

44

z

x

yL

L

L

D=1

6Eo (2,1,1) (1,2,1) (1,1,2)D=3

E

(1,1,1)3Eo

(nx,ny,nz)

222

2

2

8 zyxnnn nnnmL

hE

zyx

nx,ny,nz = 1,2,3,...

Example 2: Energy levels (1)• Now back to a 3D cubic box:Show energies and label (nx,ny,nz) for the first 11 states of the particle in the 3D box, and write the degeneracy D for each allowed energy.

Use Eo= h2/8mL2.

E

3Eo

6Eo

9Eo

11Eo

(nx,ny,nz)

z

x

y

L1

L2 > L1

L1

Example 3: Energy levels (2)• Now consider a non-cubic box:

22

2

222

2

1

2

88 yzxnnnn

mL

hnn

mL

hE

zyx

Assume that the box is stretched only along the y-direction. What do you think will happen to the cube’s energy levels below?

(1) The symmetry of U is “broken” for y, so the “three-fold” degeneracy is lowered…a ”two-fold” degeneracy remains due to 2 remaining equivalent directions, x and z.

(1,1,1)D=1

(1,2,1)D=1D=2(2,1,1) (1,1,2)

(2) There is an overall lowering of energies due to decreased confinement along y.

E

3Eo

6Eo

9Eo

11Eo

(nx,ny,nz)

Example 3: Energy levels (2)• Now consider a non-cubic box:

Assume that the box is stretched only along the y-direction. What do you think will happen to the cube’s energy levels below?

z

x

y

L1

L2 > L1

L1

222

222

21

2

88 yzxnnn nmL

hnn

mL

hE

zyx

schrodinger’s equation for three dimensions. qm in three dimensions the one dimensional case was...

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