Trying to understand Selection Rules in Cohen-Tannoudji

[kq6up]

On the bottom half of page 249 Cohen-Tannoudji it talks about selection rules in terms of off diagonal elements of the matrix generated by $\langle \phi _{ n^{ \prime },\tau ^{ \prime } } \mid \hat { B } \mid \phi _{ n,\tau } \rangle$. I thought all off diagonal matrix elements would be zero due to the $\delta_{n^{\prime},n}$ nature of these state vectors? Is this because it is not just a double sum with a single good quantum number? I am confused. Could you shed some light?

Thanks,
KQ6UP

 

[kith]

On the bottom half of page 249 Cohen-Tannoudji it talks about selection rules in terms of off diagonal elements of the matrix generated by $\langle \phi _{ n^{ \prime },\tau ^{ \prime } } \mid \hat { B } \mid \phi _{ n,\tau } \rangle$. I thought all off diagonal matrix elements would be zero due to the $\delta_{n^{\prime},n}$ nature of these state vectors?

I don't have the book handy, but by "$\delta_{n^{\prime},n}$ nature of these state vectors" you probably mean that $\langle \phi _{ n^{ \prime },\tau ^{ \prime } } \mid \phi _{ n,\tau } \rangle \propto \delta_{n^{\prime},n}$ right? If yes, why do you expect something similar to hold if you include the $\hat { B }$ operator? Acting with it on the state to the right may give you nonzero coefficients for states where the principal quantum number is equal to $n'$.

 

[kq6up]

I don't have the book handy, but by "$\delta_{n^{\prime},n}$ nature of these state vectors" you probably mean that $\langle \phi _{ n^{ \prime },\tau ^{ \prime } } \mid \phi _{ n,\tau } \rangle \propto \delta_{n^{\prime},n}$ right? If yes, why do you expect something similar to hold if you include the $\hat { B }$ operator? Acting with it on the state to the right may give you nonzero coefficients for states with $n'$.

Is that because some operators can change the state (like ladder
operators)? I always view those as in a class of their own because
they are not part of a normal eigen function / eigen value equal where
the function remains unchanged (e.g. A f(x)=a f(x). It seems like off
diagonal elements of a regular operator like momentum would all have
to be zero. Am I understanding you clearly, or no?

Thanks,
KQ6UP

 

[kith]

Is that because some operators can change the state (like ladder
operators)?

Yes, if you apply an operator to a state it will change the state in most cases. Only the eigenstates of the operator remain unchanged.

It seems like off diagonal elements of a regular operator like momentum would all have
to be zero.

The off-diagonal elements of the matrix representation of an operator are only zero if you chose a basis of eigenstates of said operator for the matrix representation. Are you familiar with the Pauli matrices? They represent the spin-1/2 operators $S_x$, $S_y$ and $S_z$ in a basis of eigenstates of $S_z$. As you see, only the $z$-matrix is diagonal, the others are not.

 

[kq6up]

That makes perfect sense thank you.

Chris