Anti-commutation of parity operator

[sydfloyd]

Homework Statement

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

Homework Equations

[tex]P \psi (x) = \psi (-x)[/tex]
[tex]p_x = - i \hbar \frac{\partial}{\partial x}[/tex]

The Attempt at a Solution

[tex] \{ 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 [/tex]

Is it valid to say that [tex] P \frac{\partial}{\partial x} \psi (x) = \frac{\partial}{\partial (-x)} \psi (-x) [/tex] ?

 

[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 [tex]x \rightarrow -x[/tex].

Let's say that [tex] f(x) = \frac{\partial \psi (x) }{\partial x}[/tex] .

Then [tex]P f(x) = f(-x) = \frac{\partial \psi (-x) }{\partial (-x)} = - \frac{\partial \psi (-x) }{\partial x}[/tex] , right?

I feel that something is wrong here.

 

[thebigstar25]

I'm under the impression that the parity operator transforms [tex]x \rightarrow -x[/tex].

Let's say that [tex] f(x) = \frac{\partial \psi (x) }{\partial x}[/tex] .

Then [tex]P f(x) = f(-x) = \frac{\partial \psi (-x) }{\partial (-x)} = - \frac{\partial \psi (-x) }{\partial x}[/tex] , 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 [tex]x \rightarrow -x[/tex].

Let's say that [tex] f(x) = \frac{\partial \psi (x) }{\partial x}[/tex] .

Then [tex]P f(x) = f(-x) = \frac{\partial \psi (-x) }{\partial (-x)} = - \frac{\partial \psi (-x) }{\partial x}[/tex] , 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.