Parity operator and a free particle on a circle

[PeterDonis]

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

That's what I get too.

f I ask what is P in the case Peikx = e-ikx , is that a valid question ? Because its not related to the matrix above and its actually e-2ikx

It's a valid question, but you seem to be confused about what the question is actually asking. I was trying to make that point earlier but I didn't do a good job; let me try again from a different starting point.

First, we have to be clear about what we're talking about. We are talking about a Hilbert space, which is a space of thingies usually called "vectors" that have certain properties, and another kind of thingies called "operators" that act on the vectors in certain ways and have certain other properties. But there are (at least) two very different ways of representing the two different kinds of thingies, which could be anything for all we know, in terms of mathematical objects we are familiar with.

Consider the equation $P \psi(x) = \psi(-x)$. What does it mean? The first thing it means is that we are interpreting the thingies in the Hilbert space as functions, in this case functions from the complex numbers to the complex numbers. More precisely, as square integrable functions from the complex numbers to the complex numbers. And this works because it turns out that the space of square integrable functions from the complex numbers to the complex numbers satisfies all of the properties that define a Hilbert space.

Under this interpretation, an operator like $P$ is a map that takes functions into functions. So the equation $P \psi(x) = \psi(-x)$ is saying that $P$ maps the function $\psi(x)$ into another function that has the same value at $-x$ as $\psi$ has at $x$, for all $x$. So if we know that the function $\psi(x)$ is $e^{ikx}$, then obviously the function $P \psi(x)$ is $e^{-ikx}$. But we didn't figure that out by multiplying $\psi(x)$ by anything, still less by multiplying it by some element of a matrix, or multiplying it by a function that made it into $\psi(-x)$ (which is how you got $e^{-2ikx}$). Nothing we've done so far has anything to do with multiplying or matrices. We just asked ourselves, what function will have the same value at $-x$ that $e^{ikx}$ has at $x$, and the answer popped out by inspection. And it's pretty much that easy for any function that you have a formula for: just plug in $-x$ wherever $x$ appears.

What is not so easy, using this representation, is to figure out if an operator like $P$ is self-adjoint. It can be done, but it can't be done by inspection the way figuring out $P \psi(x)$ from the formula for $x$ can. So if we're interested in the self-adjointness of an operator (or in many other properties of operators), we would like to find a different kind of representation. And there is one: we can represent vectors in the Hilbert space as column vectors or row vectors, and operators as matrices.

With this representation, the Dirac bra-ket notation is much more commonly used, so we would write our operator equation as $P | x \rangle = | - x \rangle$. (Actually, I have just pulled a bit of a fast one here, as we will see below; but it will do for now.) And what this is saying is that the operator $P$ corresponds to the matrix that, if we the column vector $| x \rangle$ by it, gives us the column vector $| - x \rangle$. So, for example, if we take the column vector

$$
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}
$$

which describes $| x = 2 \rangle$ (at least in the labeling convention I have just implicitly adopted), and multiply it by the matrix you obtained for $P$, we get the column vector

$$
\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}
$$

which describes $|x = -2 \rangle$.

And of course, as you saw, once we have a matrix representation for $P$, it's simple to show that $P$ is self-adjoint; you just do it by inspection, by looking at the matrix and seeing that it's the same as its conjugate transpose. That's why the matrix representation is preferred for questions like that.

But what happened to $e^{ikx}$, you say? [Actually, this doesn't quite apply here, because you're actually working in the momentum basis, not the position basis -- see post #40. But all this is still worth keeping in mind since the position basis is another valid basis that is natural to use.] Well, remember that, if we go back to the wave function representation, the function $\psi(x) = e^{ikx}$ describes a state of momentum $k$; but what we just did in the matrix representation was to apply the operator $P$ to a state with position $x$! So $e^{ikx}$ disappeared because we weren't looking at that state in the matrix representation; we were looking at a different state, whose wave function representation would be $\delta(x)$--and we would then have $P \delta(x) = \delta(-x)$ as the direct translation into the wave function representation of what we wrote above in the matrix representation. And the reason I picked the state $| x \rangle$, or $\delta(x)$, to start with in the matrix representation is that it's the easiest kind of state to write down in that representation, especially if we are going to try to write the result of applying $P$ to it. In fact, in the matrix representation, there isn't even a way to write down the exact state that is $e^{ikx}$ in the wave function representation. [Edit: not in the position basis, but there is in the momentum basis -- see post #40.]

So the answer to your question is that, if we are using the wave function representation, $P$ is just a map from functions to functions, and you figure out what functions it maps to what other functions by inspection--just look at the formula and plug in $-x$ wherever $x$ appears. It has nothing whatever to do with multiplying or matrices. We only drag in matrices and multiplication (by matrices) because it makes it much easier to see that $P$ is self-adjoint.

 

[PeterDonis]

is the matrix form of P dependent on the form of the wavefunction(although always being self-adjoint) ?

No, it's not. It's determined by the properties of the operator $P$. In the matrix representation, as we saw, those properties are most easily used to determine the matrix $P$ by looking at the action it must have on the basis vectors. But once you know the matrix $P$, you can use it to act on any state vector whatever, and its components will be the same.

 

[PeterDonis]

once you know the matrix $P$, you can use it to act on any state vector whatever, and its components will be the same.

More precisely, they will be the same as long as you stay in the same basis you used to figure out the matrix. The version you first wrote down (and which I quoted two posts ago) was determined using the position [Edit: momentum -- see post #40] basis. But the version you get by doing the integrals with the even and odd functions is in the eigenbasis of the operator, which is different, because the eigenstates of $P$ are not position [Edit: or momentum] states, they are the even and odd states (which in the wave function representation are just even and odd functions of $x$).

As an exercise, you might try figuring out what the even and odd states look like in the state vector/matrix representation that you used. Even states will stay the same when acted on by $P$, while odd states will flip sign. Then, figure out what the 5 x 5 matrix representation of $P$ looks like in this basis. One way to do that is to figure out the unitary transformation matrix $U$ that rotates the position states into the even and odd states, and then take the matrix representation $P$ that you figured out in the position basis and compute $U P U^{-1}$.

 

[PeterDonis]

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 , …).

One other note, to add on to the exercise I suggested in my last post: neither of these give you the full matrix of $P$, since to get all the matrix elements you need to work with a complete basis of states. The position states $\delta(x)$ form a complete basis (if you include all possible values of $x$), and the momentum states $e^{ikx}$ do as well (if you include all possible values of $k$), but the even functions by themselves, or the odd functions by themselves, do not. So you need to figure out what complete set of basis states includes the even and the odd functions, or at least enough even and odd functions to allow you to write any other even or odd function in terms of them. (Hint: there is at least one well-known set of even and odd functions that has this property, which can be constructed from the momentum states.) Then you need to compute the matrix elements of each possible pair of basis functions (which would at least have to include three possibilities: two even functions, two odd functions, and one even/one odd).

 

[PeterDonis]

The version you first wrote down (and which I quoted two posts ago) was determined using the position basis. But the version you get by doing the integrals with the even and odd functions is in the eigenbasis of the operator

Actually, looking back at @Avodyne's post that you responded to, I got the first part of this wrong. He was asking you to look at the matrix elements between momentum states, not position states. So actually you were computing the matrix $P$ in the momentum basis, in which each basis vector labels a different value of $k$, not $x$. I.e,. the column vector

$$
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}
$$

is the state $| k = 2 \rangle$, not $| x = 2 \rangle$. But the reasoning still goes through the same way, because, of course, the state you get by operating on $| k = 2 \rangle$ with $P$ is the state $| k = -2 \rangle$.

This all sheds even more light on the "what happened to $e^{ikx}$" question. After all: what is the state with momentum $k$ in the wave function representation? It's $e^{ikx}$! And if you take a state with momentum $k$ and operate on it with $P$, you get the state with momentum $-k$, which is? It's $e^{-ikx}$! In other words, $P$ doesn't "care" whether you interpret the minus sign as going from $\psi(x)$ to $\psi(-x)$, or going from a state with momentum $k$ to a state with momentum $-k$. It's the same either way.

 

[dyn]

Thank you for all your replies.I guess I was looking for what is P in Pψ(x) = ψ(-x) in a similar way as if P was the momentum operator it would be -iħd/dx in 1-D. But it seems that it should just be taken as an instruction to take the wavefunction and change every x to -x.

If I consider any basis of 5 wavefunctions are you saying that the Parity operator matrix is always

$\begin{pmatrix} 0&0&0&0&1\\ 0&0&0&1&0\\ 0&0&1&0&0\\ 0&1&0&0&0\\ 1&0&0&0&0 \end{pmatrix}$ ?

Because I can see that for the particle on a ring this matrix converts ψ₁ into ψ_-1 which is the correct operation. But for the odd and even states of an infinite well it doesn't work but diag(1,1,1,1,1) would convert ψ₁ into ψ₁ which would be correct for even functions and diag(-1,-1.-1,-1,-1) would convert ψ₁ into -ψ₁ which would be correct for odd functions

 

[PeterDonis]

If I consider any basis of 5 wavefunctions are you saying that the Parity operator matrix is always

No. It just is in the momentum basis (and the position basis).

for the odd and even states of an infinite well it doesn't work but diag(1,1,1,1,1) would convert ψ1 into ψ1 which would be correct for even functions and diag(-1,-1.-1,-1,-1) would convert ψ1 into -ψ1 which would be correct for odd functions

As I said before, to get the correct matrix elements for any basis, you have to have a complete basis. Neither the even functions by themselves nor the odd functions by themselves form a complete basis. But there is a complete basis that consists of a family of even functions and a family of odd functions. The matrix for P is diagonal in that basis, but it's not a diagonal of all 1's or all -1's; it's a mixture of the two.

I again suggest trying to work out what this complete basis is and what the matrix for P looks like in that basis. You can do it for the same 5-dimensional subspace we have been working with: as I said before, the even and odd functions in the complete basis I just referred to, for this subspace, can be formed from linear combinations of the momentum basis states, i.e., the $k = -2, -1, 0, 1, 2$ states. So, as I said before, all you need to do is find the unitary transformation matrix $U$ that rotates the momentum basis states into the new even/odd basis states, and then take the matrix $P$ in the momentum basis, which you know, and compute $U P U^{-1}$. That will give you the matrix $P$ in the new basis.

 

[PeterDonis]

diag(1,1,1,1,1) would convert ψ1 into ψ1

By the way, there is another point to be made here. When you say this, you are implicitly redefining what function $\psi_1$ refers to. In other words, you are using the symbol $\psi_1$ to mean the column vector

$$
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}
$$

In the momentum basis, this column vector denotes the function $e^{i k_1 x}$, where $k_1$ is whichever value of $k$ appears first in our list of $k$ values. But in the basis you implicitly adopted in the above quote, the same column vector now denotes a different function, namely whatever even function appears first in our list of even functions.

There is another way to approach this, however. Suppose you were to find a column vector that was left invariant by the matrix

$$
P = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{bmatrix}
$$

(I can think of three such column vectors.) Such a column vector, interpreted in the momentum basis (which is the basis to interpret it in since we are using the $P$ matrix that we found for that basis), would denote a function in the subspace of functions that could be formed by taking linear combinations of functions of the form $e^{ikx}$, with $k$ ranging over our list of 5 $k$ values. The components of the column vector would give the coefficients in the linear combination. And that function would be an even function.

Similarly, if you were to find a column vector that got taken to minus itself by the above matrix $P$, that column vector would denote an odd function.

And if you could find a set of 5 mutually orthogonal functions in this way (some even, some odd), those functions could form a different complete basis of the 5-dimensional subspace of functions that we've been looking at. And their matrix elements, computed using the same integrals as before, would then tell you the form of $P$ in that basis.

 

[dyn]

I can't believe that the parity operator is so complicated !

No. It just is in the momentum basis (and the position basis).
As I said before, to get the correct matrix elements for any basis, you have to have a complete basis. Neither the even functions by themselves nor the odd functions by themselves form a complete basis. But there is a complete basis that consists of a family of even functions and a family of odd functions. The matrix for P is diagonal in that basis, but it's not a diagonal of all 1's or all -1's; it's a mixture of the two.

I again suggest trying to work out what this complete basis is and what the matrix for P looks like in that basis. You can do it for the same 5-dimensional subspace we have been working with: as I said before, the even and odd functions in the complete basis I just referred to, for this subspace, can be formed from linear combinations of the momentum basis states, i.e., the $k = -2, -1, 0, 1, 2$ states. So, as I said before, all you need to do is find the unitary transformation matrix $U$ that rotates the momentum basis states into the new even/odd basis states, and then take the matrix $P$ in the momentum basis, which you know, and compute $U P U^{-1}$. That will give you the matrix $P$ in the new basis.

You state that the matrix
$$
P = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{bmatrix}
$$
is the parity operator in the momentum or position basis with 5 basis vectors. But then you state that with the infinite well the matrix is diagonal with a mixture of 1's and -1's but surely this is still in the position basis ; so why is it diagonal and not of the form shown ?

 

[PeterDonis]

But then you state that with the infinite well the matrix is diagonal with a mixture of 1's and -1's but surely this is still in the position basis ;

No, it isn't; it's in the basis formed by a particular set of even and odd functions.

 

[dyn]

I thank you for your patience with this but it's giving me brain ache again and its late so I will return to this tomorrow. Thanks again

 

[PeterDonis]

it's giving me brain ache again

I sympathize. My brain has ached plenty of times when I've tried to work through the gory details of something that looks simple on the surface. There's nothing wrong with taking your time.

 

[dyn]

If I go back to the particle on a circle ; I solve Hψ=Eψ and I get functions of the form e^ikx where takes discrete values. I assume these are basis vectors (infinite dimensional) in position space labelled by integer values of k. So the parity operator takes the form

$$
P = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{bmatrix}
$$

If I now go the symmetric infinite well I again solve Hψ=Eψ and this time I get odd functions of the form sin kx and even functions of the form cos kx.
I assume these are the (infinite dimensional) basis vectors in position space and so should also give the parity operator matrix form shown above. As far as I have always been aware the set of cos kx and sin kx functions forms a basis in position space

 

[PeterDonis]

I assume these are basis vectors (infinite dimensional) in position space labelled by integer values of k

No, they're basis vectors in momentum space. Integer values of $k$ label momentum values, not position values. Position basis vectors would be delta functions $\delta(x)$.

So the parity operator takes the form

This is correct whether you use the position basis or the momentum basis.

If I now go the symmetric infinite well I again solve Hψ=Eψ and this time I get odd functions of the form sin kx and even functions of the form cos kx.

Yes.

I assume these are the (infinite dimensional) basis vectors in position space

No, they are, again, basis vectors in momentum space. Position space basis vectors would still be delta functions.

Also, you seem not to have noticed that $\cos kx$ and $\sin kx$ are linear combinations of $e^{ikx}$ and $e^{-ikx}$. And since the energy eigenvalue $E$ is the same for $k$ and $-k$, the functions $e^{ikx}$ and $e^{-ikx}$ are also solutions of $H \psi = E \psi$ for the infinite well (and, conversely, cosines and sines are also solutions for the particle on the circle). Alternatively, you could just express $e^{ikx}$ and $e^{-ikx}$ as linear combinations of $\cos kx$ and $\sin kx$, since any linear combination of solutions for the same eigenvalue is also a solution for that eigenvalue.

As far as I have always been aware the set of cos kx and sin kx functions forms a basis in position space

I don't know where you got that idea from, but it's wrong. Every QM textbook I've ever seen that discusses the position basis makes it clear that the position basis functions are delta functions.

 

[PeterDonis]

$\cos kx$ and $\sin kx$ are linear combinations of $e^{ikx}$ and $e^{-ikx}$

This was what I was getting at before when I talked about looking for a column vector that was left invariant (up to a possible minus sign) by the matrix $P$. In the basis $e^{ikx}$, where $k = 2, 1, 0, -1, -2$, we would write, for example, $\cos kx$ as the column vector

$$
\begin{bmatrix} 0 \\ \frac{1}{2} \\ 0 \\ \frac{1}{2} \\ 0 \end{bmatrix}
$$

and $\sin kx$ (or more precisely $i \sin kx$) as the column vector

$$
\begin{bmatrix} 0 \\ \frac{1}{2} \\ 0 \\ - \frac{1}{2} \\ 0 \end{bmatrix}
$$

How does the matrix $P$ that you wrote down (most recently in post #48) act on these column vectors?

 

[dyn]

So particle on a circle basis vectors in momentum space are e^ikx leading to

$$
P = \begin{bmatrix} 0 & 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0 & 0 \end{bmatrix}
$$

but then you state that for the symmetric potential well the basis vectors in momentum space are
cos(kx) and sin (kx) so why do we not have the parity operator matrix form shown above in this situation ? In both cases we have basis vectors in momentum space.

 

[PeterDonis]

but then you state that for the symmetric potential well the basis vectors in momentum space are
cos(kx) and sin (kx)

I said that they are "basis vectors", yes. I did not say they are "the basis vectors".

There is no unique set of basis vectors on a Hilbert space, any more than there are in ordinary Euclidean space. You can always do the equivalent of rotating the coordinates to make a new set of vectors the basis vectors. One way of doing that in the case under discussion would be to "rotate" from momentum space to position space. But another way is to "rotate" between different "orientations of axes" in momentum space. We can do that because, as I said, cosines and sines can be expressed as linear combinations of complex exponentials, and vice versa.

The basis in which the parity matrix $P$ is what you wrote in post #48 is the basis of complex exponentials in momentum space. In that basis, the column vector

$$
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}
$$

with the values of $k$ we chose, denotes the function $e^{2ix}$, i.e., the complex exponential with $k = 2$. The column vectors I wrote in post #50 are how you would write $\cos x$ and $i \sin x$, i.e., the cosine and sine for $k = 1$, in that same basis.

But you could rotate to a different basis, a basis in which $\cos x$ and $i \sin x$ were in the set of basis vectors. In that basis, the column vector

$$
\begin{bmatrix} 1 \\ 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}
$$

would denote whichever one of the cosine or sine functions appeared first in the list of basis vectors. And in that basis, the matrix $P$ would have a different form. In fact, as I said in a previous post, it would be diagonal, with some of the diagonal entries being $1$ and some being $-1$. The exercise I suggested previously is to figure out, first, what cosine and sine functions are in the list of basis vectors in this basis, and second, what is the exact form of the parity matrix $P$ in this basis.

 

[PeterDonis]

particle on a circle basis vectors in momentum space

for the symmetric potential well the basis vectors in momentum space

Everything I said in post #52 (and indeed everything I've said in pretty much all of this thread) applies equally well to the particle on the circle and the symmetric potential well. The term "free particle on a circle" is a bit misleading, because putting the particle on the circle instead of the real line makes it behave, as far as most QM properties are concerned, like a bound particle instead of a free particle. In particular, there are even and odd states (cosines and sines) for the particle on a circle, just as there are for the symmetric well. They're basically the same scenario.

 

[dyn]

So there is no unique matrix representation of the parity operator

 

[PeterDonis]

So there is no unique matrix representation of the parity operator

I believe I pointed out many posts ago that the components of the matrix representation of an operator (or a state vector, for that matter) depend on the basis. That's true for any operator (or state vector).

 

[dyn]

Can I ask one more question - if I see a wavefunction written as ψ( x ) then unless its the delta function the ψ(x) is in momentum space even though its a function of position. Is that correct ?

 

[PeterDonis]

if I see a wavefunction written as ψ( x ) then unless its the delta function the ψ(x) is in momentum space even though its a function of position. Is that correct ?

No.

First of all, a wave function isn't "in momentum space" or "in position space". It's in the Hilbert space of the quantum system being studied. "Momentum space" and "position space" are just another way of referring to the momentum basis and the position basis of that Hilbert space.

As far as function notation is concerned, the strictly correct rules are: if you write $\psi(x)$, you are using the position basis. If you want to use function notation in the momentum basis, you write $\psi(p)$ (or $\psi(k)$ if you prefer the notation $k$ for momentum).

We have been sloppy in this thread by writing all of our $\psi$ functions in function notation as functions of $x$, even when we were talking about the momentum basis. That isn't completely wrong; after all, basis functions are functions in the Hilbert space, and you can write any function in any basis. And many people are much more familiar with thinking of functions of position than they are with thinking of functions of momentum. But if we're going to do that, we need to carefully keep in mind the distinction between the function notation we're using--what we're thinking of the functions as functions of--and the basis we're talking about. This becomes particularly important when we're dealing with matrix notation and we have to keep clear about which functions correspond to which column vectors.

 

[dyn]

No.

First of all, a wave function isn't "in momentum space" or "in position space". It's in the Hilbert space of the quantum system being studied. "Momentum space" and "position space" are just another way of referring to the momentum basis and the position basis of that Hilbert space.

As far as function notation is concerned, the strictly correct rules are: if you write $\psi(x)$, you are using the position basis. If you want to use function notation in the momentum basis, you write $\psi(p)$ (or $\psi(k)$ if you prefer the notation $k$ for momentum).
.

When I solved Hψ=Eψ for particle on circle I get ψ(x) = e^ikx. I know its a function of x because it was found by solving a differential equation in x. So according to the above its in the position basis but earlier (#49) you said it was in the momentum basis