Understanding the Dimensionality of Probability in Quantum Mechanics

[alexepascual]

About the "rigged" Hilber space

I am not the best person to answer this. I guess one of the more mathematically inclined guys in this thread may answer it better than I can. But I think it has to do with the definition of a Hilbert space. These kets of definite position and definite momentum don't seem to comply with one or more of the expected mathematical properties of Hilbert space.
Thinking about it better, these states don't even represent anything physical, as you can't have a particle occupy a mathematical point. If you did, the state vector would have to be infinitelly long as it would give 1 when it's square modulus is integrated over an infinitesimal volume. Looking at it a different way, if you start shrinking the volume that the particle occupies, then the probability density will increase, as its integral over this volume will always be 1. (I am talking about a case were we know for certain that the particle is within the volume). In the limit where the volume shrinks to a mathematical point, the probability density will blow up to an infinity.
But again, this does not explain why we can't include these "infinite length" vectors in Hilbert Space. Let's see if someone else answers that.

 

[Eye_in_the_Sky]

In that case, I think the remarks on pages 46 and 47 of
http://www.math.sunysb.edu/~leontak/book.pdf
are relevant here.

Thank you very much Robphy for the reference. Although I was not able to follow it in detail, it has served to clarify for me and remind me of just how those terms ought to be used in a rigorous context.

 

[alexepascual]

Robphy:
The link you posted does talk about the "generalized eigenvectors", but it is so mathematical that it is like chinesse to me.

 

[alexepascual]

From Griffiths, pg 101-102:
"A complete inner product space is called a Hilbert space"...
On eigenvalues of operators with continuous spectra:
"Unfortunately, these eigenfunctions do not lie in Hilbert space, and hence, in the stricter sense, do not count as vectors at all. For neither of them is square integrable.."
"Nevertheless, they do satisfy a kind of orthogonality condition.."

 

[Eye_in_the_Sky]

Feynman's |mom p>

Feynman, (which I have been reading lately) does not address the problem, but when he deals with momentum, he comes up with a normalization scheme that I still don't completely understand (with respect to the units).
Feynman has in his Lectures pg.16-7 :

prob(p,dp)= | <mom p|psi> |^2 dp/ h
______

Dr. Feynman's eigenkets |mom p> are not the "normalized" eigenkets |p>.

Up to a phase factor (which itself can depend on p),

|mom p> = h^1/2|p> .

You can do the same thing to a discrete orthonormal basis |f_n>. Just write:

|g_n> = h^1/2|f_n> ,

and know that any time you want prob(n) in terms of the "ortho-non-normal" basis |g_n>, it is given by:

prob(n) = |<g_n|psi>|² (1/h) .

Of course, in the discrete case, we have no explicit "measure" (like the "dp" of the continuous case) to which we can adjoin the "weight factor" 1/h.
_______________________

P.S. Every time I try to arrive at the "ultimate" answer to the original topic of this thread, I find that I "miss the mark". Soon, however, I will reach that goal.

 

[turin]

Thinking about it better, these states don't even represent anything physical, as you can't have a particle occupy a mathematical point.
...
Looking at it a different way, if you start shrinking the volume that the particle occupies, then the probability density will increase, as its integral over this volume will always be 1.
...
In the limit where the volume shrinks to a mathematical point, the probability density will blow up to an infinity.

Not to disparage your reply, but, speaking physically, this "problem" seems quite tame compared to most of the exclusively QM notions that one encounters. If, for instance, entanglement, or wave function collapse, can be accepted dogmatically, then why should we scorn such an inocuous QM distinction as an infinite probability density? Physically, much stranger (IMO) things have already been well accepted.

If, on the other hand, you are just trying to demonstrate the mathematical (not physical) meaning of Hilbert space as a mathematical object (and that was the issue that I indeed raised), then I just realized that I don't really care that much any more. I think you are correct to say that "infinitely long" vectors cannot be members of a Hilbert space (I think this is in Shankar Ch 1). I think that there is a (confusingly) similarly named space, called a [physical/proper/something] Hilbert space mentioned that includes vectors with infinite modulus. I don't recall any of this distinction having anything to do with units, though.

 

[robphy]

Robphy:
The link you posted does talk about the "generalized eigenvectors", but it is so mathematical that it is like chinesse to me.

I think [someone correct me if I am wrong] the point is that:
although the position operator $\hat Q$ has no eigenvectors in the hilbert space,
that is, there are no square-integrable functions $\phi$ such that
$$\hat Q\phi=q_0\phi$$,

there is a generalized eigenvalue equation (with generalized eigenvector $\phi=\delta(q-q_0)$)
$$\begin{align*} \int \left[ \hat Q \phi \right] \ \tau(q) dq &=\\ \int \hat Q\ \delta(q-q_0)\ \tau(q) dq &= \int q\ \delta(q-q_0)\ \tau(q) dq \\ &=\int q_0\ \delta(q-q_0)\ \tau(q) dq \\ &=\int \left[ q_0\phi \right] \ \tau(q) dq \end{align}$$
where $\tau(q)$ is some "test function" used to evaluate the integral.

 

[alexepascual]

Turin,
I think you may have misinterpreted the intention of my posts. With respect to Hilbert space, and the exclusion from it of eigenstates of continuous observables, I was not even trying to raise an issue. To put this conversation in perspective, the original topic had to do with an apparent inconsistency in the dimensionality of some equations. Then Eye_in_the_sky talked about the need to consider the base kets of continuous observables as carying some units to obtain consistency. He referred to these as "generalized" and mentioned that they were not part of Hilbert space.
Then just as a side comment, I said to Eye, that I heard about a "rigged Hilber space" which would be a Hilber space + the "generalized" eigenvectors.
At this point you objected to this idea, which by the way was not my idea but something I read. This not being a topic with which I am very familiar, I suggested that someone else answer your question. But after that I was thinking about it and nevertheless tried to venture an explanation.
I think somewhere else I mentioned about QM being learned in a dogmatic way, and also in another post about some problems being swept under the rug. These comments were more in connection with the dimensionality problem and the learning of QM in general, and were not in connection with the issue of Hilbert space.
Later in your post you say that thinking better about it, you "don't really care that much any more" (about those vectors being outside of Hilbert space) Well, we agree in this, I don't care that much either. At least for the time being and in the context of this thread.
I would like to point out though that these issues caused a lot of grief to John Von Newman, who discussed them in his "Mathematical foundations of Quantum Mechanics" (I hope I got the title right). He didn't like that much Dirac's delta function. But this would be a subject for another thread at another time.

 

[Eye_in_the_Sky]

In Conclusion ...

... Changing the subject, I remember reading (I don't know where) about a "rigged Hilbert space" which would include the eigenvectors of continuous variables.

All I know about it is just what you say, that a "rigged" Hilbert space includes those kinds of objects among its elements. (I must add, however, that from the little I have seen, it looks to me like the rigged Hilbert space has structure which is much richer than what is necessary for the mere inclusion of those kets.)

On the other hand, in order to understand this business of "generalized" eigenvectors, we don't need to know much more than we already do.

These kets of definite position and definite momentum don't seem to comply with one or more of the expected mathematical properties of Hilbert space.

One of the features which defines a Hilbert space is that it has an "inner product". That's the bracket <f|g>. By definition, this bracket is supposed to map any pair of vectors |f>, |g> to some complex number. This implies, in particular, that for any vector |f>, <f|f> has to be a number - that is, finite. So, if we have an object |h> such that <h|h> is infinite, then by definition |h> is not a member of the Hilbert space.

And basically, that's all there is to it.

So, for example, for the eigenket |x> we get

<x|x> = Dirac_delta(0) = infinity .

Alternatively, consider |p> in position-representation:

<x|p> = h^-1/2e^ikx .

Thus,

|<x|p>|² = 1/h .

To obtain <p|p> we now need to integrate this constant 1/h over the whole real line. Once again, the result is divergent. … In conclusion, these eigenkets, and others like them, are not "vectors belonging to H". They are called "generalized" eigenvectors.

(Something of the spirit of what is "really" going on has been lost in the above simplification so as to make it appear to be an entirely trivial matter. Sorry about that.)

-------------------

I don't know much about the historical aspects of how the idea of Dirac's delta function was received by his contemporaries. But today, that object is understood to be well-defined, along with many other "generalized" functions, all put on a thoroughly rigorous foundation in a branch of mathematics called "The Theory of Distributions" or "Distribution Theory".

-------------------

Regarding the original topic of this thread, here is the answer I propose:

For an ordinary vector |phi>, the outer product |phi><phi| is a (dimensionless) projection operator, whereas, for a generalized eigenvector |s>, of an operator with a continuous spectrum, labeled by the eigenvalues s, the outer product |s><s| is a projection-operator density. It is a density with respect to the parameter s, and, therefore, has dimensions ^-1. In this way psi(s) = <s|psi> gets the dimensions ^-1/2 which it needs, in order for |psi(s)|² to have the interpretation of a probability density with respect to the parameter s ... all in a perfectly natural, self-consistent way which is true to the physical and mathematical meaning of the objects involved.

 

[alexepascual]

Robphy and Eye_in_the_Sky,
Thanks for your posts, this issue is much more clear to me now.
I may be posting some questions soon on my other thread on "momentum as the generator of translations". I hope I'll see you there.
Thanks again,
-Alex Pascual-