Commutation Relation: Hi Parity Operator?

[fatema]

hi, do the translation operator commute with parity operator?

 

[blue_leaf77]

Parity operator $\hat\pi$ is defined such that when acting on a position eigenvector $|x\rangle$ to be $\hat\pi|x\rangle = |-x\rangle$. Start from
$$\hat x \hat\pi = \int dx' \ \hat x \hat\pi |x'\rangle \langle x'|$$
and with the help of the eigenvalue relation $\hat x|x'\rangle = x'|x'\rangle$.

 

[hilbert2]

It's enough to find a single example of a translation $T\psi (x) = \psi (x+\Delta x)$ and a function $\psi (x)$ for which $TP\psi (x)$ and $PT\psi (x)$ don't have the same value at some point $x$.

 

[blue_leaf77]

It's enough to find a single example of a translation $T\psi (x) = \psi (x+\Delta x)$ and a function $\psi (x)$ for which $TP\psi (x)$ and $PT\psi (x)$ don't have the same value at some point $x$.

I was thinking since it's easy to show that the two operators do not commute, why not push it a little further to know the exact relation between $TP$ and $PT$.

 

[samalkhaiat]

hi, do the translation operator commute with parity operator?

On coordinate basis $|x \rangle$, the action of translation operator $T_{a} = e^{- i a p}$ is given by $$T_{a} | x \rangle = | x + a \rangle \ .$$ And in the same basis, the parity operator is given by $$\pi = \int dy \ |-y \rangle \langle y | \ .$$ Now it is an easy exercise to show that $$T_{a} \ \pi = \int dy \ |y \rangle \langle - y + a | \ ,$$ $$\pi \ T_{a} = \int dy \ |y \rangle \langle - y - a |\ .$$ So, in general they do not commute. This becomes clear if you use the above two equations to evaluate the action on the wave function $$\left( T_{a} \ \pi \Psi \right) ( - x) = \Psi (x + a) \ ,$$ $$\left( \pi \ T_{a} \Psi \right) ( - x) = \Psi ( x - a) \ .$$