Linear and Hermitian Operator on a 2-variable Hilbert Space

[CalcYouLater]

Homework Statement

I am given that an operator is defined by the following:

$$\hat\pi\psi(x,y)=\psi(y,x)$$

in a 2 variable Hilbert space of functions $$\psi(x,y)$$

I have to show that the operator is linear and Hermitian

Homework Equations

An operator is self adjoint if:

$$\left\langle\psi\right|\hat{A}^{\dagger}\left|\right\phi\rangle=\left\langle\phi\right|\hat{A}\left|\right\psi\rangle^{*}$$

The Attempt at a Solution

With the information given, how can I apply the above condition? Should I put x one side and y on the other? Or should I use psi(x,y) on each side. If so, does the fact that one of the psi's is a bra and the other is a ket change anything?

To show that it is linear, I need to show that it is distributive. Is it as simple as saying:

$$\hat{\pi}(A\psi(x,y)+B\psi(x,y))=\hat{\pi}(A\psi(x,y))+\hat{\pi}(B\psi(x,y))$$

If I remember correctly, my professor said that since the space is linear then the operators are distributive. That doesn't make sense to me. If an operator can be anything, then couldn't it be something that isn't linear as well?

 

[Dick]

What's your definition of the inner product? It's an integral, right? Showing it's linear isn't at all hard. And showing it's hermitian I think is just a changes of dummy variables, isn't it? An operator can be anything, but you are asked to show this one is linear. I'm not sure what your last question is about.

 

[CalcYouLater]

Hi. Thanks for responding (and for the help in a different thread).

What's your definition of the inner product? It's an integral, right? Showing it's linear isn't at all hard. And showing it's hermitian I think is just a changes of dummy variables, isn't it? An operator can be anything, but you are asked to show this one is linear. I'm not sure what your last question is about.

Inner product:

$$\left\langle\psi\right|\hat{A}\left|\right\psi\rangle=\int_{-\infty}^{\infty}\psi^{*}\hat{\pi}\psi{dx}$$

Hmm, dummy variables. Would that mean that the above integral changes from integrating over dx to over dy?

$$\int_{-\infty}^{\infty}\psi^{*}\hat{\pi}\psi{dy}$$

 

[Dick]

Hi. Thanks for responding (and for the help in a different thread).

Inner product:

$$\left\langle\psi\right|\hat{A}\left|\right\psi\rangle=\int_{-\infty}^{\infty}\psi^{*}\hat{\pi}\psi{dx}$$

Hmm, dummy variables. Would that mean that the above integral changes from integrating over dx to over dy?

$$\int_{-\infty}^{\infty}\psi^{*}\hat{\pi}\psi{dy}$$

They are functions of two variables, so it's probably a double integral dx*dy, isn't it? Applying pi just flips x and y, doesn't it? Does that affect the integral?

 

[CalcYouLater]

They are functions of two variables, so it's probably a double integral dx*dy, isn't it? Applying pi just flips x and y, doesn't it? Does that affect the integral?

Yes, a double integral does make sense. I think that flipping x & y does not change the integral because it doesn't matter in which order we integrate because the limits are the same for both x and y.

But, does it change the complex conjugation? I say no because I do not see any evidence based off of the information given that complex numbers are affected by this permutation.

 

[Dick]

Yes, a double integral does make sense. I think that flipping x & y does not change the integral because it doesn't matter in which order we integrate because the limits are the same for both x and y.

But, does it change the complex conjugation? I say no because I do not see any evidence based off of the information given that complex numbers are affected by this permutation.

Right. The limits for x and y are the same. Write out the integral expressions for <psi,pi(phi)> and <pi(psi),phi> and try to convince yourself interchanging the labels 'x' and 'y' makes no difference.

 

[CalcYouLater]

If I say:

$$\left\langle\psi\right|\hat{\pi}\left|\right\phi\rangle=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\psi^{*}\hat{\pi}\phi{dx}{dy}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\psi^{*}{dx}{dy}\hat{\pi}\phi=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\psi^{*}\hat{\pi}\phi{dx}{dy}$$

and likewise:$$\left\langle\hat{\pi}\psi\right|\left\right\phi\rangle=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\hat{\pi}\psi^{*}\phi{dx}{dy}=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\hat{\pi}\psi^{*}{dx}{dy}\phi=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\phi\hat{\pi}\psi^{*}{dx}{dy}$$

Isn't that assuming the operator is linear? I must be missing something here. I was thinking that showing that moving around the integrand does not change the integration proves they're hermitian, but now I think that is based off the idea that (pi*psi*phi)=(psi*pi*phi).

 

[Dick]

integral conjugate(psi(y,x))*phi(x,y)*dx*dy=integral conjugate(psi(x,y))*phi(y,x)*dx*dy. True or false? They only differ by an interchange of x and y, don't they? Do you see what I'm saying?

 

[CalcYouLater]

Ahh. You are good. Does that mean I get a two-fer? Is the fact that the integral is unchanged enough (I think sufficient is the word) to show the operator is linear as well. Or is there a condition for linearity that I am not considering?

 

[Dick]

Ahh. You are good. Does that mean I get a two-fer? Is the fact that the integral is unchanged enough (I think sufficient is the word) to show the operator is linear as well. Or is there a condition for linearity that I am not considering?

No, linearity is separate. You want to show stuff like pi(A(x,y)+B(x,y))=pi(A(x,y))+pi(B(x,y)). It's really pretty direct if you think about what pi is doing. It's just switching x and y around.