Parity operator and a free particle on a circle

[dyn]

Hi.
I have just looked at a question concerning a free particle on a circle with ψ(0) = ψ(L). The question asks to find a self-adjoint operator that commutes with H but not p.

Because H commutes with p , i assumed there was no such operator.
The answer given , was the parity operator. It acts as Pψ_N = ψ_-N.The answer then states it can be seen to be self-adjoint by writing it as a matrix in this basis.
I do not know how to write the parity operator as a matrix in this basis. For starters is the basis not infinite-dimensional ?
Any help would be appreciated. Thanks

 

[PeterDonis]

I have just looked at a question

Can you give a specific reference?

 

[dyn]

It is an old exam question I found. It is not online.

 

[Nugatory]

Because H commutes with p , i assumed there was no such operator.

You've probably already figured out that that assumption is not valid - even if A and B commute with C, it does not follow that A and B commute with one another. A more familiar example will be angular momentum: $L_x$, $L_y$, and $L_z$ all commute with $L^2$ but not with one another.

I do not know how to write the parity operator as a matrix in this basis. For starters is the basis not infinite-dimensional ?

It is, but that doesn't have to stop you. There's a (rather straightforward) closed-form expression for the elements of the matrix $P_{ij}$, and you can use that to check for self-adjointness.

 

[dyn]

You've probably already figured out that that assumption is not valid - even if A and B commute with C, it does not follow that A and B commute with one another. A more familiar example will be angular momentum: $L_x$, $L_y$, and $L_z$ all commute with $L^2$ but not with one another.

Thank you that is a very helpful example.

It is, but that doesn't have to stop you. There's a (rather straightforward) closed-form expression for the elements of the matrix $P_{ij}$, and you can use that to check for self-adjointness.

The closed form expression for the matrix might be straightforward but I have no idea what it is ?

 

[dyn]

I found on the net that the parity operator matrix was diag( -1 , -1 , -1 , ….) but that doesn't change e^ikx into e^-ikx. So can anybody tell me what the parity operator matrix is ?
Thanks

 

[Nugatory]

Your boundary condition constrains the possible values of $k$.

 

[dyn]

I appreciate your reply but it doesn't make the picture any clearer for me

 

[PeterDonis]

I found on the net that the parity operator matrix was diag( -1 , -1 , -1 , ….) but that doesn't change e^ikx into e^-ikx.

Why not?

Also, when you say you "found on the net", where? Please give a specific reference.

 

[dyn]

Why not?
.

Because (-1)(e^ikx) does not equal e^-ikx

 

[PeterDonis]

Because (-1)(eikx) does not equal e-ikx

But what if the $-1$ acts on $x$ instead of $e^{ikx}$?

 

[dyn]

Then it would equal e^-ikx but surely the (-1) acts on the wavefunction as in Pψ(x) = ψ(-x)

 

[PeterDonis]

surely the (-1) acts on the wavefunction as in Pψ(x) = ψ(-x)

Look at what you just wrote. If $\psi(x) = e^{ikx}$, then what is $\psi(-x)$? And if $P\psi(x) = \psi(-x)$, what does that tell you?

 

[dyn]

If ψ(x) = e^ikx then ψ(-x) = e^-ikx but I still don't know what the parity operator is ?

 

[PeterDonis]

I still don't know what the parity operator is ?

What operator did you write down in post #12?

 

[dyn]

What operator did you write down in post #12?

I wrote the parity operator as -1 or diag(-1 , -1 , -1 , ….) but I don't see how (-1) changes e^ikx into e^-ikx.
It just changes e^ikx into -e^ikx

 

[PeterDonis]

I wrote the parity operator as -1 or diag(-1 , -1 , -1 , ….)

Not in post #12 you didn't. In that post you wrote down an equation involving $P \psi(x)$. What was it?

 

[dyn]

Not in post #12 you didn't. In that post you wrote down an equation involving $P \psi(x)$. What was it?

I wrote Pψ(x) = ψ(-x)

 

[PeterDonis]

I wrote Pψ(x) = ψ(-x)

Yes. What is $P$ in that equation?

 

[dyn]

the parity operator either in matrix form or acting on individual wavefunctions

 

[PeterDonis]

the parity operator

Right. And that equation in post #12 appeared as part of this:

surely the (-1) acts on the wavefunction as in Pψ(x) = ψ(-x)

So you have a $-1$ acting on the wave function, giving the result you're looking for. So what, exactly, is the problem?

 

[PeterDonis]

the parity operator either in matrix form or acting on individual wavefunctions

Perhaps this is part of your confusion: the parity operator "in matrix form" and "acting on individual wavefunctions" are not two different things. They're the same thing, just written in different forms. A wave function is just a continuous form of a state vector, and a state vector is just a discrete form of a wave function.

So maybe you should start with this: if the Hilbert space is finite dimensional, can you write down the parity operator? And show how it acts on a state vector and why it's self-adjoint?

 

[dyn]

because -1 acting on e^ikx gives -e^ikx and this is not the same as e^-ikx which is what you should get after the partity operator acts on e^ikx

 

[PeterDonis]

because -1 acting on eikx gives -eikx

But the equation you wrote down in post #12 doesn't have the $-1$ acting on $\psi(x)$. It has the $-1$ acting on $x$. Look at what you wrote. You wrote $P \psi(x) = \psi(-x)$. What is being multiplied by $-1$ in that expression?

 

[dyn]

But the equation you wrote down in post #12 doesn't have the $-1$ acting on $\psi(x)$. It has the $-1$ acting on $x$. Look at what you wrote. You wrote $P \psi(x) = \psi(-x)$. What is being multiplied by $-1$ in that expression?

I appear to be going round in circles here. If P is -1 then it is acting on the wavefunction but that gives -e^ikx and not e^-ikx which is what it should be

 

[PeterDonis]

I appear to be going round in circles here.

Yes, because you're not responding to the questions I'm asking, you just keep repeating yourself.

Can you answer the question I asked in post #24, just as I asked it?

 

[dyn]

But the equation you wrote down in post #12 doesn't have the $-1$ acting on $\psi(x)$. It has the $-1$ acting on $x$. Look at what you wrote. You wrote $P \psi(x) = \psi(-x)$. What is being multiplied by $-1$ in that expression?

The wavefunction ψ(x) is being multiplied by -1

 

[PeterDonis]

Another suggestion if the line I've been trying to get you to follow just won't work:

If you are going to represent operators by matrices, you have to represent states by state vectors, not wave functions. So you need to think of a state vector $| \psi \rangle$ instead of a wave function $\psi(x)$. But you can also write down a correspondence between state vectors and wave functions: if we have a state vector $| \psi \rangle$, then the wave function is given by $\psi(x) = \langle x | \psi \rangle$. Similarly, the wave function $P \psi(x)$ that you get by acting on the state with an operator $P$ is written in matrix/state vector form as $P \psi(x) = \langle x | P | \psi \rangle$.

Two things you might consider at this point:

First, how would you translate $P \psi(x) = \psi(-x)$ into the matrix/state vector form given the above?

Second, in the expression $\langle x | P | \psi \rangle$, you can view $P$ as acting on $| \psi \rangle$ from the left, or as acting on $\langle x |$ from the right. What does that suggest in the light of the equation $P \psi(x) = \psi(-x)$?

 

[PeterDonis]

The wavefunction ψ(x) is being multiplied by -1

No. You're not even reading what you wrote. I'll try once more: we have the equation $P \psi(x) = \psi(-x)$. What gets multiplied by $-1$ going from the LHS to the RHS, just looking at the equation as it's written, forgetting everything you think you know about the operator $P$? "The wave function" $\psi$ is not the right answer.

 

[dyn]

No. You're not even reading what you wrote. I'll try once more: we have the equation $P \psi(x) = \psi(-x)$. What gets multiplied by $-1$ going from the LHS to the RHS, just looking at the equation as it's written, forgetting everything you think you know about the operator $P$? "The wave function" $\psi$ is not the right answer.

Going from the RHS to the LHS , x is multiplied by -1. But that doesn't tell me what P is.
Its very late here. I need to get some sleep. Thank you for your time

 

[PeterDonis]

I found on the net that the parity operator matrix was diag( -1 , -1 , -1 , ….)

One more comment: I asked you for a specific reference for this, but you never gave one. You should consider the possibility that either the reference you got this from is wrong, or you are misinterpreting what it says. Giving a specific reference would help others to help you figure out whether one of those things is going on.

 

[PeterDonis]

Going from the RHS to the LHS , x is multiplied by -1

Exactly. So your thinking that "$P$ multiplies the wave function by $-1$", which is what was suggested by the thing you "found on the net" (see the other post I made just now), is evidently not correct. In fact, if $P$ "multiplies" anything by $-1$, it's $x$, not $\psi(x)$.

that doesn't tell me what P is.

Not by itself, no. But it should tell you something. See above.

 

[dyn]

I know P multiplies x by -1 to give -x but I don't know what P is that multiplies e^ikx to give ^-ikx

 

[Avodyne]

On a circle of circumference $L$, the eigenvalues of the momentum operator are $2\pi k/L$, where $k$ is an integer (positive, negative, or zero). The eigenfunction of the momentum operator corresponding to the eigenvalue $2\pi k/L$ is $\psi_k(x) = (1/\sqrt{L})\exp(ikx/L)$. These form an orthonormal basis, $$\int_0^L dx\,\psi_k(x)^*\psi_n(x)=\delta_{kn}.$$
The matrix elements of any operator $A$ in this basis are given by
$$A_{kn} =\int_0^L dx\,\psi_k(x)^*A\psi_n(x).$$
So now we can think about the parity operator $P$, which, acting on any wave function $\psi(x)$, yields $\psi(-x)$:
$$P\psi(x)=\psi(-x).$$
From this, you should be able to compute the matrix elements of the parity operator, $P_{kn}$.

To get an idea of what it looks like in this basis, write out $P$ as a matrix for $k=-2,-1,0,1,2$ and $n=-2,-1,0,1,2$.

Is this matrix self-adjoint?

 

[dyn]

On a circle of circumference $L$, the eigenvalues of the momentum operator are $2\pi k/L$, where $k$ is an integer (positive, negative, or zero). The eigenfunction of the momentum operator corresponding to the eigenvalue $2\pi k/L$ is $\psi_k(x) = (1/\sqrt{L})\exp(ikx/L)$. These form an orthonormal basis, $$\int_0^L dx\,\psi_k(x)^*\psi_n(x)=\delta_{kn}.$$
The matrix elements of any operator $A$ in this basis are given by
$$A_{kn} =\int_0^L dx\,\psi_k(x)^*A\psi_n(x).$$
So now we can think about the parity operator $P$, which, acting on any wave function $\psi(x)$, yields $\psi(-x)$:
$$P\psi(x)=\psi(-x).$$
From this, you should be able to compute the matrix elements of the parity operator, $P_{kn}$.

To get an idea of what it looks like in this basis, write out $P$ as a matrix for $k=-2,-1,0,1,2$ and $n=-2,-1,0,1,2$.

Is this matrix self-adjoint?

So I get the following matrix
$\begin{pmatrix} 0&0&0&0&1\\ 0&0&0&1&0\\ 0&0&1&0&0\\ 0&1&0&0&0\\ 1&0&0&0&0 \end{pmatrix}$

I hope this is right ? It converts ψ₁into ψ_-1 etc , so it seems to work and it is self-adjoint
I still have 2 questions -
1- if I ask what is P in the case Pe^ikx = e^-ikx , is that a valid question ? Because its not related to the matrix above and its actually e^-2ikx
2 - is the matrix form of P dependent on the form of the wavefunction(although always being self-adjoint) ?
Because using the integrals above for the even functions of a infinite symmetric well I would get P as diag(1 ,1 , 1, ….) and for the odd functions I would get P as diag(-1 ,-1 , -1 , …).
All different matrices but all self-adjoint

Thanks