2D square box – separation of variables

ny=1,2,3,….nx=1,2,3,….

Ground state of 1D infinite square well.

ny=1,2,3,….

nx=1,2,3,….

2D square box

ny=1,2,3,….nx=1,2,3,….

What is the energy of the ground state of the 2D infinite square well?

A)0B)E0

C)2E0

D)4E0

E)5E0

nx=1, ny=1

E=E0(12+12)

2D square box

ny=1,2,3,….nx=1,2,3,….

What is the energy of the first excited state of the 2D infinite sqaure well?

A)2E0

B)3E0

C)4E0

D)5E0 E)8E0

nx=1, ny=2

E=E0(12+22)

Or nx=1, ny=2

2D square box – probability densities

ny=1,2,3,….nx=1,2,3,….

Contour maps of |(x,y)|2 Ground state (1,1)

Q. The potential seen by the electron in a hydrogen atom is…

A. Independent of distance

B. Spherically symmetric

C.An example of a central force potential

D.Constant

E. More than one of the above

The potential seen by the electron is

rkerV

2)(

Spherically symmetric (doesn’t depend on direction). It depends only on distance from proton so it is a central force potential.

)(rVr

rkerV

2)(

On your own – no discussion

3-D central force problemsThe hydrogen atom is an example of a 3D central force problem.

The potential energy depends only on the distance from a point (spherically symmetric)

xy

z

r

Spherical coordinates is the natural coordinate system for this problem.

What is the x-coordinate of our point (r,)?

A)B)C)

3-D central force problemsThe hydrogen atom is an example of a 3D central force problem.

The potential energy depends only on the distance from a point (spherically symmetric)

xy

z

r

cossinsincossin

rzryrx

Spherical coordinates is the natural coordinate system for this problem.

Engineering & math types sometimes swap and .General potential: V(r,,)

Central force potential: V(r)

ErVrrr

rrrme

)(

sin1sin

sin11

2 2

2

2222

2

2

The Time Independent Schrödinger Equation (TISE) becomes:

We can use separation of variables so )()()( ),,( rRr

3-D central force problemsThe hydrogen atom is an example of a 3D central force problem.

The potential energy depends only on the distance from a point (spherically symmetric)

xy

z

r

cossinsincossin

rzryrx

Spherical coordinates is the natural coordinate system for this problem.

Engineering & math types sometimes swap and .General potential: V(r,,)

Central force potential: V(r)

ErVrrr

rrrme

)(

sin1sin

sin11

2 2

2

2222

2

2

The Time Independent Schrödinger Equation (TISE) becomes:

We can use separation of variables so )()()( ),,( rRr

The partThe variable only appears in the TISE as

E

2

2

So we should not be surprised that the solution is ime )(Note that m is a separation variable and not the electron mass.

We use me for the electron mass.x

y

z

r

cossinsincossin

rzryrx

Are there any constraints on m?

What can we say about and ?)( )2( They have to be the same! )2()(

mimeim sincos)( Since cosine and sine have periods of , as long as m is an integer (positive, negative, or 0) the constraint is satisfied.

Angular momentum about the z-axis is quantized:

Note is similar towhich is the solution to the free particle with

Angular momentum quantization about z-axis ime )(

xy

z

r

cossinsincossin

rzryrx

ikxe

kp

As k gives the momentum in the x direction, m gives the momentum in the direction (angular momentum).

mLz

There is nothing truly special about the z-axis.

We can point the z-axis anywhere we want to.

It is just the nature of the coordinate system that treats the z-axis different than the x and y axes.

The part

xy

z

r

cossinsincossin

rzryrx

The solution to the part is

more complicated so we skip the details of the solution.

The end result is that there is another quantum variable ℓ which must be a non-negative integer and ℓ ≥ |m|.

The ℓ variable quantizes the total angular momentum: )1( L

Note that LZ cannot be larger than the total L.

Total angular momentum is )1( L ℓ can be 0, 1, 2, 3, …

The z-component of the angular momentum is mLz where m can be 0, ±1, ±2, … ±ℓ