# Prove that the parity operator is Hermitian

We know that an operator is Hermitian when:

$$\langle f|\hat{O}g\rangle$$ = $$\langle \hat{O} f|g\rangle$$

Parity operator in 1D is simply defined as:

$$\hat{\Pi} f(x) = f(-x)$$

I don't know anything about the eigenvalues of parity operator (that is asked in the next problem).

How can I show it is Hermitian?

$$\langle f(x)|\hat{\Pi}|g(x)\rangle$$ = $$\langle f(x)|g(-x)\rangle$$

$$\langle \hat{\Pi} f(x)|g(x)\rangle$$ = $$\langle f(-x)|g(x)\rangle$$

These last integrals are not equal, unless both functions are symmetric.

How can I prove it?

• The eigenfunctions of the parity operator are those that are symmetric, as you say, $f(-x)=f(x)$ but also those that are anti-symmetric, $f(-x)=-f(x)$. Does that help at all? This also provides a strong hint as to what the eigenvalues of such an operator might be. Oct 18, 2020 at 18:20
• Take the variable substitution x -> -x Oct 18, 2020 at 18:24
• @daydreamer if I do that (y=-x) in second integral I will have $- \int f(y) g(-y) dy$, which is equal to $- \int f(y) \hat{\Pi} g(y) dy$. There is an extra minus sign.
– AA10
Oct 18, 2020 at 18:58
• If you're integrating over the entire real line or any other symmetric interval, you get the same result... Oct 18, 2020 at 19:48
• @daydreamer Ok, I understood now. Thanks.
– AA10
Oct 18, 2020 at 22:01

Set $$x=-\xi$$ in $$\int_{-\infty}^{\infty} f(x)g(-x)\,dx$$ to get $$\int_{-\infty}^{\infty} f(x)g(-x)\,dx=\int_{+\infty}^{-\infty}f(-\xi)g(\xi)d(-\xi)\\ = - \int_{+\infty}^{-\infty}f(-\xi)g(\xi)d\xi= \int_{-\infty}^{+\infty}f(-\xi)g(\xi)d\xi$$ so $$\langle Pf,g\rangle = \langle f,Pg\rangle$$
• So, can it be used this way: $\langle \hat{\Pi} \hat{\Pi} f(x)|g(x)\rangle$ = $\langle f(x)| \hat{\Pi} \hat{\Pi}g(x)\rangle$ because $\hat{\Pi} \hat{\Pi} \psi(x) = \psi(x)$ ?