Proof that the parity operator is hermitian

[Brad Barker]

hey, it's good to be back at pf.

anyway, today i had an exam in my honors modern course, and one of the questions was a proof that the parity operator is hermitian. i don't think i got it right. :/

here's what i did:

1:
$$\int(P_(op) \psi_2(x))^* \psi_1(x) dx<br /> <br /> = \int \psi_2^*(-x) \psi_1(x) dx.$$

and
2:

$$\int \psi_2^*(x) P_(op) \psi_1(x) dx<br /> <br /> = \int \psi_2^*(x) \psi_1(-x)dx.$$But... that doesn't really get me anywhere. if (1) equaled (2), then i'd be satisfied, but... it doesn't appear that this is the case.so how would you have gone about this?

 

[topsquark]

hey, it's good to be back at pf.

anyway, today i had an exam in my honors modern course, and one of the questions was a proof that the parity operator is hermitian. i don't think i got it right. :/

here's what i did:

1:
$$\int(P_(op) \psi_2(x))^* \psi_1(x) dx<br /> <br /> = \int \psi_2^*(-x) \psi_1(x) dx.$$

and
2:

$$\int \psi_2^*(x) P_(op) \psi_1(x) dx<br /> <br /> = \int \psi_2^*(x) \psi_1(-x)dx.$$


But... that doesn't really get me anywhere. if (1) equaled (2), then i'd be satisfied, but... it doesn't appear that this is the case.


so how would you have gone about this?

What do you know about the eigenvalues of a Hermitian operator?

-Dan

 

[Brad Barker]

they're real. (that was the next question on the exam!)

but psi1 and psi2 needn't be eigenfunctions of the parity operator.

 

[Physics Monkey]

Hi Brad,

The point is that your choice of psi1 and psi2 have nothing to do with it. The easiest approach is just to show directly that the eigenvalues are real. Hint: think about what $$P^2$$ is. Another approach is to look at things like $$P | x \rangle$$ where $$| x \rangle$$ are the position eigenstates.

 

[Brad Barker]

ah, i forgot that eigenvalues real => operator hermitian, and not just vice-versa. sucks that i forgot that direction; i showed my professor a proof of it after class on tuesday!

oh well.

 

[Galileo]

Is that true? What about the matrix:
$$\left( \begin{array}{cc}1 & 4\\1 & 1\end{array}\right)$$.
It's not Hermitian, but it has real eigenvalues 3 and -1.

 

[Brad Barker]

Is that true? What about the matrix:
$$\left( \begin{array}{cc}1 & 4\\1 & 1\end{array}\right)$$.
It's not Hermitian, but it has real eigenvalues 3 and -1.

:/

well, i know there's a reason why i found it tough to swallow. but that fact(?) is in my prof's lecture notes. i'll email him.

 

[topsquark]

:/

well, i know there's a reason why i found it tough to swallow. but that fact(?) is in my prof's lecture notes. i'll email him.

I would have to look it up in my notes as well, but I suspect that the actual theorem is that any Unitary Hermitian operator must have real eigenvalues and that if you have only real eigenvalues you may construct a Unitary Hermitian matrix from them.

-Dan

 

[Galileo]

Ah, but if you can show P is unitary, then the result follows immediately from P^2=1.

 

[Meir Achuz]

hey, it's good to be back at pf.

anyway, today i had an exam in my honors modern course, and one of the questions was a proof that the parity operator is hermitian. i don't think i got it right. :/

here's what i did:

1:
$$\int(P_(op) \psi_2(x))^* \psi_1(x) dx<br /> <br /> = \int \psi_2^*(-x) \psi_1(x) dx.$$

and
2:

$$\int \psi_2^*(x) P_(op) \psi_1(x) dx<br /> <br /> = \int \psi_2^*(x) \psi_1(-x)dx.$$


But... that doesn't really get me anywhere. if (1) equaled (2), then i'd be satisfied, but... it doesn't appear that this is the case.


so how would you have gone about this?

Just make the variable substitution x'=-x.

 

[Brad Barker]

Just make the variable substitution x'=-x.

yep, that's the way to do it, it turns out!


also, the theorem is that operator is hermitian <=> expectation values are real.

this, i guess, is not quite the same as "eigenvalues are real."

 

[Meir Achuz]

[QUOTEalso, the theorem is that operator is hermitian <=> expectation values are real.
This, I guess, is not quite the same as "eigenvalues are real."[/QUOTE]
Proving the evs are real is a bit longer, but still simple. An expectation value can be real even if the evs are not all real.

 

[Poop-Loops]

yep, that's the way to do it, it turns out!

I have to do this as a homework problem now and I'm just not seeing it. I can get <Pf(x)|g(x)> = <f(x')|Pg(x')> where x' = -x, but I don't see how that's a proof, since it's comparing two different variables.

 

[Poop-Loops]

I still haven't figured this one out. I tried <Pf(x)|g(x')> = <f(x')|Pg(x)>, but that's still not the same thing has having f(x) and g(x), so I don't know what else to do.

 

[Galileo]

That P is unitary and P^2=1 is all you need. so just prove those properties.

 

[Poop-Loops]

P^2 = 1 was given to me.

I don't understand how I'd prove that it's unitary, though.

Is it that since <Pf(x)|g(x)> makes <f(x)| -> <f(-x)| and the same for |g(x)> -> |g(-x)>, so I can multiply P* and P and get 1?

 

[Brad Barker]

P^2 = 1 was given to me.

I don't understand how I'd prove that it's unitary, though.

Is it that since <Pf(x)|g(x)> makes <f(x)| -> <f(-x)| and the same for |g(x)> -> |g(-x)>, so I can multiply P* and P and get 1?

well, i think it would be most illustrative to write the brackets out as integrals in position space. then you can see how the variable substitution will get you what you need.

 

[Poop-Loops]

I did, got what you got in the first post, and no amount of switching around variables has helped me see the answer.

 

[Brad Barker]

I did, got what you got in the first post, and no amount of switching around variables has helped me see the answer.

ok, my tex'ing is kind of rusty...

$$\int_{-\infty}^\infty(P\psi(x))^*\psi(x)dx = \int_{-\infty}^\infty\psi^*(-x)\psi(x)dx.$$

let y = -x.

then the above becomes...

$$-\int_{\infty}^{-\infty}\psi^*(y)\psi(-y)dy = \int_{-\infty}^\infty\psi^*(y)P\psi(y)dy.$$

note that the variable of integration is a dummy variable. you can return it all to the dirac bra-ket notation, if you would like. in any case, it's clear now that we proved what we wanted.

 

[Poop-Loops]

Okay, but now you have a function of y's instead of x's. If I were to take it back into x's, I'd get something totally different. How can you justify setting x = y if you state that x = -y before that?

 

[Brad Barker]

Okay, but now you have a function of y's instead of x's. If I were to take it back into x's, I'd get something totally different. How can you justify setting x = y if you state that x = -y before that?

it's not a function of x or y! they're just a complete set of states. you can "remove" them entirely, in a sense, and go back to bra-ket notation, where you'll clearly see that we've solved the problem.

if you'd like, you can let y --> x in that last integral. it doesn't matter. it's all mathematically equivalent.

if you've ever solved an integral using "u-substitution" you should realize that substitutions don't change the integral itself but instead put it in a more useful form.


just yesterday, in fact, in one of my classes, we performed at least seven substitutions for a single problem (computing the green function for a certain case of helmholtz' equation). the answer to the problem is the answer is the answer is the answer.

 

[Poop-Loops]

it's not a function of x or y! they're just a complete set of states. you can "remove" them entirely, in a sense, and go back to bra-ket notation, where you'll clearly see that we've solved the problem.

if you'd like, you can let y --> x in that last integral. it doesn't matter. it's all mathematically equivalent.

if you've ever solved an integral using "u-substitution" you should realize that substitutions don't change the integral itself but instead put it in a more useful form.

Yes, but if you make a substitution and go back, you should end up with the initial state you had. If I substitute y for -x and then go back at the end, I get two functions of -x instead of x. I just don't see how you can say that a function of -x = a function of x unless the entire function is odd.

 

[Brad Barker]

Yes, but if you make a substitution and go back, you should end up with the initial state you had. If I substitute y for -x and then go back at the end, I get two functions of -x instead of x. I just don't see how you can say that a function of -x = a function of x unless the entire function is odd.

why would you want to undo the transformation? the "x" in the integral is a dummy variable, just as "y" is.

let's start and end even further:

$$<br /> <P\psi|\psi> = \int_{-\infty}^\infty<P\psi|x><x|\psi>dx =<br /> \int_{-\infty}^\infty(P\psi(x))^*\psi(x)dx = \int_{-\infty}^\infty\psi^*(-x)\psi(x)dx.<br />$$

with y = -x...

$$-\int_{\infty}^{-\infty}\psi^*(y)\psi(-y)dy = \int_{-\infty}^\infty\psi^*(y)P\psi(y)dy =<br /> \int_{-\infty}^\infty<\psi|y>P<y|\psi>dy = <\psi|P|\psi>.$$

recall that

$$\int |x><x| dx = 1 = \int |y><y| dy$$.