Anti-commutation of parity operator

[sydfloyd]

Homework Statement

The parity operator is defined as $$P \psi (x) = \psi (-x)$$. Show that $$P$$ and $$p_x$$ anti-commute, that is, $$\{ P,p_x \} = Pp_x + p_xP = 0$$.

Homework Equations

$$P \psi (x) = \psi (-x)$$
$$p_x = - i \hbar \frac{\partial}{\partial x}$$

The Attempt at a Solution

$$\{ P,p_x \} \psi(x) = ( Pp_x + p_xP ) \psi(x) = -i \hbar \left[ P \frac{\partial}{\partial x} \psi (x) + \frac{\partial}{\partial x} [ P \psi (x) ]\right] = -i \hbar \left[ \frac{\partial}{\partial (-x)} \psi (-x) + \frac{\partial}{\partial x} \psi (-x) \right] = -i \hbar \left[ - \frac{\partial}{\partial x} \psi (-x) + \frac{\partial}{\partial x} \psi (-x) \right] = 0$$

Is it valid to say that $$P \frac{\partial}{\partial x} \psi (x) = \frac{\partial}{\partial (-x)} \psi (-x)$$ ?

 

[thebigstar25]

for the part where you found (pxP) this one is correct ..

as for your question, I won't say that I am 100% sure that it is valid .. but I would say that I am 80% agree with what you suggested in order to get your answer ..

what I know about the parity operator is that when you apply it on psi(x) you get psi(-x), so whenever you have x you simply change it to -x and vice versa (that would be applied to other things such as velocity and force) ..

 

[sydfloyd]

I'm under the impression that the parity operator transforms $$x \rightarrow -x$$.

Let's say that $$f(x) = \frac{\partial \psi (x) }{\partial x}$$ .

Then $$P f(x) = f(-x) = \frac{\partial \psi (-x) }{\partial (-x)} = - \frac{\partial \psi (-x) }{\partial x}$$ , right?

I feel that something is wrong here.

 

[thebigstar25]

I'm under the impression that the parity operator transforms $$x \rightarrow -x$$.

Let's say that $$f(x) = \frac{\partial \psi (x) }{\partial x}$$ .

Then $$P f(x) = f(-x) = \frac{\partial \psi (-x) }{\partial (-x)} = - \frac{\partial \psi (-x) }{\partial x}$$ , right?

I feel that something is wrong here.

I told you I am not 100% sure .. but i still think it is right the way you did ..

hopefully there would be someone else to confirm what I think ..

 

[kuruman]

I'm under the impression that the parity operator transforms $$x \rightarrow -x$$.

Let's say that $$f(x) = \frac{\partial \psi (x) }{\partial x}$$ .

Then $$P f(x) = f(-x) = \frac{\partial \psi (-x) }{\partial (-x)} = - \frac{\partial \psi (-x) }{\partial x}$$ , right?

Right. I don't see anything wrong. Use a trial function of x, like a few terms of a polynomial, that doesn't have definite parity and see how it works.