Is \( i\hat{p} \) Hermitian?

[PineApple2]

Hi. In a question I needed to figure out whether $-\frac{i\hbar}{m} \hat{p}$ is hermitian or not. Since the constant doesn't matter this is similar to whether $i \hat{p}$ is hermitian or not. I thought that since $\hat{p}$ is hermitian, then $i$ times it would not be, since it would not "transfer" properly to the other side of the bra-ket expression. But the solutions say it is hermitian. Can anyone explain?

 

[voko]

Recall the definition of "Hermitian".

 

[PineApple2]

Recall the definition of "Hermitian".

voko - thanks for the excellent answer - you really took the time to explain everything... with much elaboration. Next time - if you don't want to help, please don't reply...

For others - I thought about the following: since we know
$$\langle pA|B \rangle = \langle A|pB \rangle$$

we therefore know that

$$\langle ipA|B \rangle = \langle A|-ipB \rangle$$

is that true

 

[voko]

voko - thanks for the excellent answer - you really took the time to explain everything... with much elaboration. Next time - if you don't want to help, please don't reply...

You have read the rules to this forum, haven't you? You must have noticed that we are required not to give blow-by-blow explanations to homework and coursework problems.

I gave you a hint. The problem asks "is this Hermitian"? To answer it, you need to recall the definition. Then it will surely be trivial.

 

[PineApple2]

You have read the rules to this forum, haven't you? You must have noticed that we are required not to give blow-by-blow explanations to homework and coursework problems.

I gave you a hint. The problem asks "is this Hermitian"? To answer it, you need to recall the definition. Then it will surely be trivial.

I already know the solution as I mentioned, and it is not homework, but an exercise I am doing for practice. I have seen helpful answers in the forum so I know there are people that are willing to help. Your answer doesn't give any information, I can say "recall the definition" about anything.

 

[voko]

I already know the solution as I mentioned, and it is not homework, but an exercise I am doing for practice. I have seen helpful answers in the forum so I know there are people that are willing to help. Your answer doesn't give any information, I can say "recall the definition" about anything.

If this is not an assignment for you, you should really post elsewhere. This section of the forum is for assistance on homework and coursework assignments. We can bent rules here and there, but still.

A Hermitian operator has to satisfy this: <Ax|y> = <x|Ay>. If A = cB, where c is any complex constant, and B is a Hermitian operator, then <Ax|y> = <cBx|y> = c*<Bx|y> = c*<x|By> = <x|c*By> ≠ <x|Ay>. That's all there is to it.

 

[PineApple2]

If this is not an assignment for you, you should really post elsewhere. This section of the forum is for assistance on homework and coursework assignments. We can bent rules here and there, but still.

A Hermitian operator has to satisfy this: <Ax|y> = <x|Ay>. If A = cB, where c is any complex constant, and B is a Hermitian operator, then <Ax|y> = <cBx|y> = c<Bx|y> = c<x|By> = <x|cBy> = <x|Ay>. That's all there is to it.

I did not see other appropriate place for posting exercises, would be happy to hear.
Thanks for the answer, I thought that the constant also has to be complex-conjugated. Does that happen only in the position-representation, $\langle x|\psi \rangle$ ?

 

[dextercioby]

It's trivial to show that, if A is symmetric, then iA is antisymmetric, because the scalar product on a complex Hilbert space is sesquilinear and not bilinear.

For $\psi,\phi \in D(A) \subseteq D\left(A^{\dagger}\right)$ A being symmetric means that

$$\langle \psi, (iA)\phi\rangle = \langle (iA)^{\dagger} \psi, \phi\rangle = i \langle \psi, A\phi\rangle = \langle - i\psi, A\phi\rangle = \langle -iA^{\dagger}\psi, \phi\rangle = ...$$

 

[voko]

II thought that the constant also has to be complex-conjugated.

Sorry, I was doing copy and paste, then I got distracted and posted unfinished stuff. Multiplication by an arbitrary complex constant does not generally preserve Hermiticity, multiplication by i in particular turns a Hermitian operator into anti-Hermitian and vice versa. Another way to see that is by looking at their eigenvalue expansion, then multiplication by i will convert real eigenvalues into imaginary and vice versa.

 

[PineApple2]

Sorry, I was doing copy and paste, then I got distracted and posted unfinished stuff. Multiplication by an arbitrary complex constant does not generally preserve Hermiticity, multiplication by i in particular turns a Hermitian operator into anti-Hermitian and vice versa. Another way to see that is by looking at their eigenvalue expansion, then multiplication by i will convert real eigenvalues into imaginary and vice versa.

So you're saying that the professor's answer that it's Hermitian is wrong?

 

[Dick]

So you're saying that the professor's answer that it's Hermitian is wrong?

p is hermitian. ip is not hermitian. So yes.