- #1
Sunnyocean
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If the parity operator ##\hat{P}## is hermitian, then:
##\langle \phi | \hat{P} | \psi \rangle = (\langle \psi | \hat{P} | \phi \rangle)^*##
Let us see if the above equation is true.
The left hand side of the above equation is:
## \langle \phi | \hat{P} | \psi \rangle = \int_{-\infty}^{\infty} \phi (x)^* \hat{P} \psi (x)\,dx ##
The right hand side of the above equation is:
## (\langle \psi | \hat{P} | \phi \rangle)^* = (\int_{-\infty}^{\infty} \psi (x)^* \hat{P} \phi (x)\,dx)^* = (\int_{-\infty}^{\infty} \psi (x)^* \phi (-x)\,dx)^* = \int_{-\infty}^{\infty} \phi (-x)^* \psi (x)\,dx ##
Now, changing the variable of integration from x to -t, we obtain:
## \int_{\infty}^{-\infty} \phi (t)^* \psi (-t)\,dt = \int_{\infty}^{-\infty} \phi (t)^* \hat{P} \psi (t)\,dt = - \int_{-\infty}^{\infty} \phi (t)^* \hat{P} \psi (t)\,dt = - \langle \phi | \hat{P} | \psi \rangle ##
(The minus in front of the integral appears when we switch the limits of integration)
So it would appear that P (the parity operator) is ANTI Hermitian (i.e. it is equal to the negative of its Hermitian conjugate), not Hermitian.
The “best trick” I’ve seen so far is that those who “prove” that the parity operator is Hermitian simply ignore the limits of integration. In other words, they do the calculations above for the *indefinite* integral. In this case, of course, you don’t get the minus from switching the sign of the integral above, so it appears that the parity operator is indeed Hermitian.
But it is very odd, to say the least, that an operator is hermitian when used with indefinite integrals and anti hermitian when used with definite integrals. After all, it is the same operator.
Can anyone see any mistake in my calculations above? If yes, what is it?
If what I wrote above is wrong, than can anyone show me the proof that the parity operator is hermitian?
*Please note: I am aware that this question has been asked before on Physics Forums and I *did* read the answers, however I did not find them to be conclusive. So please don’t jut copy past answers.
##\langle \phi | \hat{P} | \psi \rangle = (\langle \psi | \hat{P} | \phi \rangle)^*##
Let us see if the above equation is true.
The left hand side of the above equation is:
## \langle \phi | \hat{P} | \psi \rangle = \int_{-\infty}^{\infty} \phi (x)^* \hat{P} \psi (x)\,dx ##
The right hand side of the above equation is:
## (\langle \psi | \hat{P} | \phi \rangle)^* = (\int_{-\infty}^{\infty} \psi (x)^* \hat{P} \phi (x)\,dx)^* = (\int_{-\infty}^{\infty} \psi (x)^* \phi (-x)\,dx)^* = \int_{-\infty}^{\infty} \phi (-x)^* \psi (x)\,dx ##
Now, changing the variable of integration from x to -t, we obtain:
## \int_{\infty}^{-\infty} \phi (t)^* \psi (-t)\,dt = \int_{\infty}^{-\infty} \phi (t)^* \hat{P} \psi (t)\,dt = - \int_{-\infty}^{\infty} \phi (t)^* \hat{P} \psi (t)\,dt = - \langle \phi | \hat{P} | \psi \rangle ##
(The minus in front of the integral appears when we switch the limits of integration)
So it would appear that P (the parity operator) is ANTI Hermitian (i.e. it is equal to the negative of its Hermitian conjugate), not Hermitian.
The “best trick” I’ve seen so far is that those who “prove” that the parity operator is Hermitian simply ignore the limits of integration. In other words, they do the calculations above for the *indefinite* integral. In this case, of course, you don’t get the minus from switching the sign of the integral above, so it appears that the parity operator is indeed Hermitian.
But it is very odd, to say the least, that an operator is hermitian when used with indefinite integrals and anti hermitian when used with definite integrals. After all, it is the same operator.
Can anyone see any mistake in my calculations above? If yes, what is it?
If what I wrote above is wrong, than can anyone show me the proof that the parity operator is hermitian?
*Please note: I am aware that this question has been asked before on Physics Forums and I *did* read the answers, however I did not find them to be conclusive. So please don’t jut copy past answers.
