Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Momentum operator and parity

  1. Jan 30, 2015 #1

    dyn

    User Avatar

    What does it mean when it is said that the momentum operator flips the parity of the function on which it operates ?
     
  2. jcsd
  3. Jan 30, 2015 #2
    It means the following. Suppose you have a state, ##|\psi\rangle## with parity ##P=\pm1##. Then you apply the momentum operator to it, i.e. ##\hat p|\psi\rangle##. Then the parity of the new state is going to be, ##\hat P\left(\hat p|\psi\rangle\right)=\mp\hat p|\psi\rangle##.
     
  4. Jan 30, 2015 #3

    dyn

    User Avatar

    Thanks for that. In terms of parity what does the del operator do acting on its own ?
     
  5. Jan 31, 2015 #4
    What do you mean by 'on its own'? An operator is always defined through its action. However (I don't know if this is what you were looking for) when you perform a parity transformation, ##x\to-x## (let's stay in 1 dimension for simplicity) you have that ##\frac{d}{dx}\to-\frac{d}{dx}##. Is this what you want to know?
     
  6. Jan 31, 2015 #5

    dyn

    User Avatar

    Does this mean if I diiferentiate an even function I get an odd function and vice versa ? If the function is neither odd or even it remains that way after differentiation ?
     
  7. Jan 31, 2015 #6
    Yes for the first question. If the function doesn't have a well defined parity then you can't really say anything about its derivative.
     
  8. Jan 31, 2015 #7

    dyn

    User Avatar

    Thanks. My original question was about the momentum operator flipping the parity of functions. It seems as though this is solely due to the differential part of the momentum operator ? The -ih(bar) part of the operator doesn't affect the parity ?
     
  9. Jan 31, 2015 #8
    Well, that's a constant, which by definition is an even function under parity.
     
  10. Jan 31, 2015 #9

    dyn

    User Avatar

    Thanks for your help
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Momentum operator and parity
  1. Operators - Parity (Replies: 2)

  2. Parity Operator (Replies: 4)

Loading...