# The Position Operator

1. May 19, 2009

### Selectron

Usually in QM we say that a wavefunction psi is an eigenfunction of some operator if that operator acting on psi gives eigenvalue * psi.

The position operator is just "multiply by x". So any psi would seem to fit the above description of an eigenfunction of the position operator with eigenvalue x. But obviously this doesn't make sense because in general wavefunctions do not have definite positions. So where does the above reasoning break down? Is it to do with x being a continuous variable?

2. May 19, 2009

### Demystifier

The eigenfunction of the position operator is the Dirac delta function. It even has the the unit norm if you choose the correct normalization, namely if you divide it by the square root of the space volume.

3. May 19, 2009

### Hurkyl

Staff Emeritus
(Except, of course, that the dirac delta isn't a function, it doesn't make sense to ask for its norm, and the space-volume is already an ill-defined notion, let alone its square root)

4. May 19, 2009

### tirrel

you see that a funtion $$f(x)$$ seen as a vector in $$L^2(R)$$ is not proportional to the vector $$xf(x)$$

two vector are proportional if $$w=\alpha v$$, with $$\alpha$$ belonging to the field where the vector space has its scalars, that is C in this case... x is not a fixed complex number so $$xf(x)$$ is not proportional to $$f(x)$$...

eigenvectors of the position operator are the delta distributions, in fact formally (which means: in some sense not well specified) $$x\delta(x-a)=a\delta(x-a)$$ ... they are not member of $$L^2(R)$$ so they put some difficulty in defining the space where to work... unluckily I don't remember well how the situation is usually mathematically established (maybe you have to pass to dual spaces and reinterpret all the equations)...

Last edited: May 19, 2009
5. May 19, 2009

### Hurkyl

Staff Emeritus
Yes. In infinite dimensional linear algebra*, we use the notion of a "spectrum" -- the spectrum of T consists of those $\lambda$ for which $T - \lambda \mathbf{1}$ is not invertible. (Note this is the same as eigenvalue in finite dimensions, or when T has a discrete spectrum) Or at least it's something like that -- I'm more used to nicer cases, and the description at wikipedia includes other constraints.

Instead of eigenvectors, we can invoke the "spectral theorem".

But that can be somewhat awkward if you're not used to it**. It's more common to study an extension of your Hilbert space to a larger vector space -- and extend our operators to act on these new vectors -- such that this new vector space really and truly does have eigenvectors for each element of the spectrum.

The catch is that this larger vector space is not as nicely behaved as our original vector space. In particular, the inner product is not defined on them -- if $| \psi \rangle$ and $| \phi \rangle$ are two of these generalized kets, then $\langle \phi | \psi \rangle$ is nonsense. In fact, if $| \theta \rangle$ is an ordinary ket, then $\langle \theta | \psi \rangle$ is still nonsense. In order to have an inner product, $| \theta \rangle$ usually has to be a very nice ket; in the case you're looking at, it's usually required to be a "test function".

If you really want to know the details of this larger space, wikipedia gives a concise description. Also, look at this page, specifically the section on Tempered distributions, and the "problem of multiplication" section.

*: or at least, in "nice" infinite-dimensional linear algebra
**: e.g. I find it awkward.

Last edited: May 19, 2009
6. May 19, 2009

### Demystifier

Of course!
But it does not prevents us from pretending that such formal problems do not exist and to use such ill defined entities to obtain physically sensible results that agree with experiments.
This is what I call the difference between mathematics and physics!

And of course, if you prefer to be rigorous, you can allways work with a finite discretized space and obtain the same results by taking the continuous limit at the END of calculation.
But a good physicist intuitively bypasses such technicalities.

7. May 19, 2009

### Hurkyl

Staff Emeritus
Nor does it prevent us from pretending that such formal problems do not exist and to use such ill defined entities to obtain physically nonsensical results. :tongue:

And besides, I don't know of any use for "normalizing" a delta function.

If that actually worked, then there wouldn't be a problem!

8. May 19, 2009

### Demystifier

Here is an example where such a normalization can be useful (though not necessary). Consider a particle scattering in QFT. Due to conservation of 4-momentum, the scaterring amplitude is proportional to the delta function in the 4-momentum space. But you are interested in probability, which is proportional to the amplitude squared, which in turn is proportional to the squared delta function. But squared delta function is ill defined. Then you make the trick and divide it by the 4-volume of spacetime and say that it is transition probability per unit time and unit 3-volume of space, which leaves only the first power of the delta function. But if your initial delta function in spacetime was normalized correctly (i.e., divided by the square root of the spacetime volume), then you would not need to artificially divide the squared amplitude by the 4-volume of spacetime - the correct factor would already be there.

9. May 19, 2009

### Selectron

Thanks. I knew that the eigenfunction of position is a delta-function, but I just wanted to understand why the logic I outlined at the start is wrong. I think I understand that from tirrel's post as the position operator means multiply by the variable x, not some constant value, so it doesn't work. Thanks for all the other interesting posts as well.

10. May 19, 2009

### Hurkyl

Staff Emeritus
Could you be more explicit? The thing it sounds like you're describing is a calculation that very much requires the delta not to have a scale factor put in front of it -- but I can't be sure without knowing the details of the calculation.

11. May 20, 2009

### kexue

You mean the inner product of the eigenvectors of the position operator is the Dirac delta? Or what do you mean?

12. May 20, 2009

### Demystifier

Hurkyl and kexue,
OK, I have to admit that my verbal explanation above was a mess. I was not concentrated while writing it. Please, ignore it! I will write it more carefully later.

13. May 20, 2009

### kexue

Note when we calculate stationary states ( when the Hamiltonian is time independent), we solve the eigenvectors ( the stationary states) and the eigenvalues of the Hamiltonian: H|phi>=E|phi>. In position space, <x|H|phi>=E<x|phi>, this gives the time-independent Schrödinger equation. Note |phi> are the stationary states, the eigenvectors of the time independent Schrödinger equation.

To get |psi>, the solution to the original, time dependent Schrödinger equation, you build the propagator U(t,t_0) from the stationary states and eigenvalues of the Hamiltonian.

Now to your question. If you have a system, of course, you can not take any |phi>. Only the ones that you calculate from the Hamiltonian as just described.

When you ask for the position of the particle, you do <x|X|phi>=<x|phi>. You do not calculate for eigenvectors or eigenvalues of X here! Actually, you just project the |phi>, the eigenvectors of the stationary Hamiltonian in position space.