Can q Have Eigenvectors Other Than Zero in Standard Quantization Procedure?

[kvl]

In standard quantization procedure we should apply commutation rules $$[p,q]=i$$. But let's do a simple calculation:

$$i \langle q_0 | q_0 \rangle = \langle q_0 | [p,q] | q_0 \rangle = \langle q_0 | pq | q_0 \rangle - \langle q_0 | qp | q_0 \rangle = q_0 \langle q_0 | p | q_0 \rangle - q_0 \langle q_0 | p | q_0 \rangle) = 0$$

which means that in Hilbert space q can have no eigenvectors other then zero (the same holds for p). But then, what does completeness of q means ? How can q be an observable ?

 

[Galileo]

It means you cannot have the commutation rule [p,q]=i in finite dimensions.
Your calculation is basically part of taking the trace of the commutator:
0=tr([p,q])=iN, where N is the dimension of your Hilbert space.

In infinite dimensions the orthogonality and completeness relations are given by:
$$\langle q|q' \rangle=\delta(q-q')$$
$$\int dq |q\rangle \langle q|=I$$

That's how it works in practice anyway, even though it has some mathematical 'issues' which someone else can undoubtedly clarify.

 

[kvl]

It means you cannot have the commutation rule [p,q]=i in finite dimensions.
Your calculation is basically part of taking the trace of the commutator:
0=tr([p,q])=iN, where N is the dimension of your Hilbert space.

AFAIK the trace of an operator is given by
$$\mbox{tr}(\hat{f}) = \int dq \langle q | \hat{f} | q \rangle$$
and can be undefined in infinite-dimension Hilbet space. But $$\langle q_0 | [p,q] | q_0 \rangle$$ is just a scalar product of two vectors which should have finite value even in infinite-dimension case by definition of Hilbert space. Am I wrong ?

 

[Galileo]

AFAIK the trace of an operator is given by
$$\mbox{tr}(\hat{f}) = \int dq \langle q | \hat{f} | q \rangle$$
and can be undefined in infinite-dimension Hilbet space.

But that definition already assumes infinite dimensions. For finite dimensions it is just:
$$tr(\hat f)=\sum_{i} \langle q_i | \hat{f} | q_i \rangle$$
where the sum is finite and the |q_i>'s form a orthonormal basis.

But $$\langle q_0 | [p,q] | q_0 \rangle$$ is just a scalar product of two vectors which should have finite value even in infinite-dimension case by definition of Hilbert space. Am I wrong ?

It does when |q_0> is just a normal eigenvector, but you can't have a normalizable eigenvector of the operators q and p given the commutation relation [p,q]=i.
The inner product of two of these eigenstates is given by a delta function (like in the previous post.)

You know that [X,P]=ih where x and p are the position and momentum operators in 1-D.
Given its interpretation, an eigenstate of X , say , |x>, would mean a well defined position of the particle. So it's probability density should be an infinitely high and narrow spike at the position x, (so <x'|x>=d(x'-x)).
An eigenstate of the momentum operator P is a plane wave, which has the same modulus squared everywhere and is not normalizable.

These properties hold generally for observables whose commutator is some imaginary number.

 

[Demystifier]

For more examples of such impossible identities in quantum mechanics, as well as methods for correct treatment of such expressions, see
http://lanl.arxiv.org/abs/quant-ph/9907069 [Rep.Prog.Phys. 63 (2000) 1893]

 

[George Jones]

The position and momentum operators usually do not have eigenvectors that live in a Hilbert space, physicists just pretend (usefully) that such things exist.

Workarounds:

1) Since, e.g., infinitely sharp position measurements are impossible, use wavepackets wihout ever leaving the comfy confines of (separable) Hilbert space.

2) use rigged Hilbert space, in which weak eigenvectors live as distributions (e.g., the Dirac delat function).

Ses posts #13 and #15 by me in this thread.

Ses also this puzzle.

Some historical speculations by me.

 

[javierR]

We should be careful with our dimensionality in general: a Hilbert space is finite dimensional or countably infinite if there is a discrete spectrum of states, e.g. a particle in a 1D potential well, in which case *sums* over states may appear; it is continuous (uncountably infinite) if there's a continuous spectrum, like e.g. the free particle, in which case integrals over state labels may appear. In both of those cases, calculations involving position space wavefunctions are done with integrals over x.
With regard to the original question, it doesn't matter what the dimension of the Hilbert space is; if we are dealing with systems that can have a continuous range of expectation values for x and p, the expectation value of [x,p] should be non-zero. But as was pointed out somewhat, writing the operator equation x|x0>=x0|x0> means you are calling |x> position eigenstates with eigenvalue x0; therefore what is the action of the operator acting on such a state p|x0>...? If you cannot define this properly, how can you compute expectation values and add and subtract them? So your problem is not having a well-defined operator equation. (As was pointed out, if you write the problem in terms of position space wavefunctions, the wavefunction you are claiming is a spatial delta function; if a definite momentum state, then a spatial plane wave).

 

[George Jones]

it is continuous (uncountably infinite) if there's a continuous spectrum, like e.g. the free particle,

As far as I am aware, uncountably infinite-dimensional (i.e., topologically non-separable) Hilbert spaces almost never make appearances in quantum theory (Sakurai notwithstanding). For example, the state space for a free particle is the Hilbert space (completion) of Lebesque square-integrable functions, and this Hilbert space has a countably infinite orthonormal basis.

 

[kvl]

For more examples of such impossible identities in quantum mechanics, as well as methods for correct treatment of such expressions, see
http://lanl.arxiv.org/abs/quant-ph/9907069 [Rep.Prog.Phys. 63 (2000) 1893]

Thanks, its very interesting reading !

The position and momentum operators usually do not have eigenvectors that live in a Hilbert space, physicists just pretend (usefully) that such things exist.

Workarounds:

1) Since, e.g., infinitely sharp position measurements are impossible, use wavepackets wihout ever leaving the comfy confines of (separable) Hilbert space.

2) use rigged Hilbert space, in which weak eigenvectors live as distributions (e.g., the Dirac delat function).

Ses posts #13 and #15 by me in this thread.

Ses also this puzzle.

Some historical speculations by me.

Thanks, for the links.

 

[Bork]

Ok, unfortunately at this point I have not yet learned Latex. I'm enthusiastic about high tech in a lot of ways but pretty old-fashioned when it comes to calculating stuff. So instead of writing an ugly string of code, I will try and explain what I think is the flaw in your reasoning. Bear in mind, I'm kinda rusty with the more advanced results for Hilbert Spaces, so if this has nothing to do with the usual position and momentum operators then I apologize.

You have to remember that the operator p when represented in position space becomes i times a derivative on the whole ket/bra product with respect to q0. So p sandwiched between the q0's becomes the derivative of 1, so you get 0 for that part. And p sandwiched between <q0| on the left and q0|q0> on the right gives you i times the derivative of q0 w.r.t. q0 which is just =i, so your final answer is i=i+0=i, and it works as it's supposed to.

Hope this helps.

 

[Bork]

Oh silly me, my apologies. I didn't realize you bracketing the operators between 2 q0's is a different situation than usual. Since <q0|q0> is not normalized to 1 (it's the infinite part of a delta spike), everything I just said is irrelevant, and the other posters should have your answers. Plus my math was wrong upstairs, but that doesn't matter given the fact it's entirely irrelevant.