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Dirac Hydrogen atom

  1. Jun 4, 2015 #1

    ChrisVer

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    Hey I was reading through a text and came across:
    I can understand the second statement from the Pauli matrices... However I think that I don't understand the 1st statement as it is... why would the diagonal elements of an odd-operator be zero if parity is definite?
     
    Last edited: Jun 4, 2015
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  3. Jun 4, 2015 #2

    blue_leaf77

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    I think what they mean as "fixed" is actually "definite".
     
  4. Jun 4, 2015 #3

    ChrisVer

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    Yup understand that... (a flaw in my translation :biggrin:)...I don't understand why the diagonal elements are zero.
     
  5. Jun 4, 2015 #4

    blue_leaf77

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    If ##A## is an odd operator and ##\pi## is a parity operator, then we have ##\pi^\dagger A \pi = -A##. If furthermore a state ##|\psi\rangle## which has a definite parity sandwhiches the expression ##\pi^\dagger A \pi = -A## from left and right in each side of the equation, you will get ##\langle \psi| A |\psi \rangle = -\langle \psi|A|\psi \rangle## because ##\langle \psi|\pi^\dagger A \pi|\psi \rangle = \langle \psi| A |\psi \rangle##. You see then we must require ##\langle \psi| A |\psi \rangle = 0## so that the previous relation can hold.
     
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