Note: variables with a hat are operators (such as ).
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:
We are ready to perform the commutator computation:
Plugging (34) and (36) into (29), we get
We know that the eigenstates of the angular momentum operators are
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
The set of all these states is ``complete" in the sense that any wave function can be written in the above basis.
Friday, 29 August 2008
Possible times for makeup classes:
Discussing HW problem: The time-dependent Schrödinger equation is:
We say it is a stationary state because:
We know solutions of
From the sum, the probability that we find energy state where is .
Hermitian operators have eigenfunctions such that Here, the inner product is:
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:
Normalization of an energy eigenstate :
The radial probability is then .
Monday, 1 September 2008 Labor Day, no class.
Wednesday, 3 September 2008