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Trying to understand Selection Rules in Cohen-Tannoudji

  1. Feb 24, 2016 #1
    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
     
  2. jcsd
  3. Feb 24, 2016 #2

    kith

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    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'##.
     
    Last edited: Feb 24, 2016
  4. Feb 24, 2016 #3
    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
     
  5. Feb 24, 2016 #4

    kith

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    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.

    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.
     
  6. Feb 24, 2016 #5
    That makes perfect sense thank you.

    Chris
     
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