I Parity operator and a free particle on a circle

[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