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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?

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    $\begingroup$ 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. $\endgroup$
    – Charlie
    Oct 18, 2020 at 18:20
  • $\begingroup$ Take the variable substitution x -> -x $\endgroup$
    – daydreamer
    Oct 18, 2020 at 18:24
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    $\begingroup$ @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. $\endgroup$
    – AA10
    Oct 18, 2020 at 18:58
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    $\begingroup$ If you're integrating over the entire real line or any other symmetric interval, you get the same result... $\endgroup$
    – daydreamer
    Oct 18, 2020 at 19:48
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    $\begingroup$ @daydreamer Ok, I understood now. Thanks. $\endgroup$
    – AA10
    Oct 18, 2020 at 22:01

2 Answers 2

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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 $\langle Pf,g\rangle = \langle f,Pg\rangle$

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Regarding eigenvalues, notice that the parity operator is an involution, in the present context means it is it's own inverse. Next, use that every function can be expressed as the sum of its symmetric and antisymmetric part. Think that it does the job.

Please note that we assume the operator is hermitian with respect to some integration interval. For an accessible discussion on this, check Shankar's Principles of Quantum Mechanics chapter 2 or 3 I think

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  • $\begingroup$ 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)$ ? $\endgroup$
    – AA10
    Oct 18, 2020 at 22:08
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    $\begingroup$ Well it sure can, but then it its a mere triviality (redundant equivalence). What I'm saying is that this implies the square of the parity operator eigenvalues are 1. Please upvote if my answer was helpful. $\endgroup$
    – daydreamer
    Oct 18, 2020 at 22:12

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