# Momentum operator and parity

1. Jan 30, 2015

### dyn

What does it mean when it is said that the momentum operator flips the parity of the function on which it operates ?

2. Jan 30, 2015

### Einj

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. Jan 30, 2015

### dyn

Thanks for that. In terms of parity what does the del operator do acting on its own ?

4. Jan 31, 2015

### Einj

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. Jan 31, 2015

### dyn

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. Jan 31, 2015

### Einj

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. Jan 31, 2015

### dyn

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. Jan 31, 2015

### Einj

Well, that's a constant, which by definition is an even function under parity.

9. Jan 31, 2015