Recent content by borgwal

What this example also shows is that in *finite* dimensions you cannot have two hermitian operators A and B satisfying [A,B]=cI

For j=1/2 there are two states: m=+/- 1/2, and for j=3/2 there are 4 different m-states. The "weak" Zeeman effect just refers to a situation where the energy shift due to the magnetic field is small and can be treated with perturbation theory: the unperturbed Hamiltonian has split the l=1 level...

It's because the interaction that splits the energies of the state is the spin-orbit coupling, proportional to \vec{L}\cdot\vec{S}, which can be rewritten as being proportional to the difference \vec{J}^2-\vec{L}^2-\vec{S}^2, which is dependent only on the quantum numbers j and l (s=1/2 in...

Once you have answered part (b), you will see that the relation that you're supposed to prove in part (a) cannot be quite correct!

Yes, 3g/2L it is.

You clearly did not state the full problem so I have to keep guessing: were you supposed to diagonalize the Hamiltonian and find U such that [tex] H=\sum_k E(k) b^+_kb_k [/itex]?

If the b's are fermionic annihilation operators, then that *means* they satisfy the anticommutation relations that, as you figured out, are equivalent to U being unitary. Done.

You need more information to prove any of those relations. You must have been given some info about what the b's are supposed to be, for instance. I assumed that you had been told that the b's are fermionic annihilation operators.

First, is the vector \vec{r} in E meant to be the *unit* vector in the r-direction, or not?

It's because a photon carries angular momentum, but in 1-D the concept of angular momentum is not defined.

So far so good! Now the other thing you have been given is that delta v_x is 1% of v_x.

You just took the square of the wavefunction, instead of the absolute value squared.

Yeah.

There are (at least) two sorts of average you can take: If you want to calculate a force averaged over *distance*, you can use the change in kinetic energy, divided by distance: Work =\delta E=F_average * d If you want to calculate an average over *time*, then that would be given by the change...

No, you have to prove U is unitary. Edit: you already seem to know that U being unitary is equivalent to the b's satisfying the same anticommutation relations as the c's. But that's all there is to it...