Momentum operator and parity

  • Thread starter dyn
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  • #1
dyn
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What does it mean when it is said that the momentum operator flips the parity of the function on which it operates ?
 

Answers and Replies

  • #2
470
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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##.
 
  • #3
dyn
716
49
Thanks for that. In terms of parity what does the del operator do acting on its own ?
 
  • #4
470
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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?
 
  • #5
dyn
716
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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 ?
 
  • #6
470
58
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.
 
  • #7
dyn
716
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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 ?
 
  • #8
470
58
Well, that's a constant, which by definition is an even function under parity.
 
  • #9
dyn
716
49
Thanks for your help
 

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