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Angular momentum operators in spherical coordinates

Note: variables with a hat are operators (such as ).

 (14) (15) (16)

So,

 (17) (18) (19)

We can expand the term :,

 (20)

We can write out the entire operator, and use the fact that , , and are orthogonal to compute :
 (21) (22)

The key point here is that

 for (23)

So, and commute. If two operators commute, then you can measure the physical quantities associated with those simultaneously.

If , then when you measure energy (putting it in a well-defined state), then angular momentum is put into an unknown state of energy - even if you had previously put angular momentum into a well-defined state by measuring it.

In any physical system, it is useful to find the maximal set of operators which commute with each other.

In order to compute the commutator, the first part of only has radial terms, so it will commute. The computation breaks down to: Does ?

Recall the following relation between the components of angular momentum:

 (24)

where

To illustrate cyclic/anti-cyclic order:

We are ready to perform the commutator computation:

 (25)

Consider:

 (26)

(as because ). Then,
 (27) (28) (29) (30)

From (27):

 (31)

We now consider . We can follow the same derivation process to get:
 (32) (33)

Plugging (34) and (36) into (29), we get

 (34)

Similarly,

 (35)

Also, . If was a function of and not , then these would not commute.

We know that the eigenstates of the angular momentum operators are

and

Comment: The maximal set of commuting operators is , , and . We could have chosen or instead of ; we choose because it has the simplest form in spherical coordinates.

Because of this, if we make measurements of and , then we collapse the wave function entirely.

In describing atoms with one electron, the interaction with the nucleus only depends on the Coulumb potential, which is spherical symmetrical. Without loss of generality, the wave function for an electron in a potential may be written

 (36)

The set of all these states is complete" in the sense that any wave function can be written in the above basis.

Hence, implies:

does not depend on .

Friday, 29 August 2008

Possible times for makeup classes:

• Tuesday 1:30 - 2:30 PM
• Thursdays 5 - 6 PM or 1 - 2 PM
• Fridays 3 - 4 PM

Discussing HW problem: The time-dependent Schrödinger equation is:

If , then is a stationary state - Why?

We say it is a stationary state because:

Hence, the probability of finding the particle at a place is independent of time.

We know solutions of

are complete. Meaning that any function, in the domain can be expanded in the . So,

From the sum, the probability that we find energy state where is .

Hermitian operators have eigenfunctions such that Here, the inner product is:

Thus,

What is the wave function for this problem for ? Thus,

The linear momentum is

The expected value is

For the last problem of the homework,

Note about the book: Equation 7-28 is not quite correct. The only comes in because the book has not normalized the angular part of the wave function. Book uses and instead of .

Schrödinger equation for an energy eigenstate of a particle in a central potential , satisfies:

Thus may be written

 (37)

We know

where is a spherical harmonic and

So write and substitute into (40), then

 (38)

where labels the energy eigenstate and eigenvalue.

Normalization of an energy eigenstate :

 (39) (40)

Therefore, since

we have

 (41)

The radial probability is then .

Monday, 1 September 2008 Labor Day, no class.

Wednesday, 3 September 2008

Next: General Central Potential Problem Up: Central Field Problems Previous: Quantum Treatment   Contents
Brian Bockelman 2008-09-11