Undergrad 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