Understanding the Dimensionality of Probability in Quantum Mechanics

[alexepascual]

In the following equation:
<br /> \left\langle {\phi }<br /> \mathrel{\left | {\vphantom {\phi \Psi }}<br /> \right. \kern-\nulldelimiterspace}<br /> {\Psi } \right\rangle = \int {\phi ^ * } \left( x \right)\psi \left( x \right)dx<br />
my understanding would have been that the bracket represents a probability amplitude. But the dx on the integral gives it a dimension of length. OK, the square of the absolute value of the braket is not a probability but a probability density. Wouldn't then the units be probability over unit of length?
I am a little confused. Any help will be appreciated.
By the way, this equation is from Feynman Lectures, Vol III, 16-6

 

[turin]

Don't forget about normalization. To interpret it as a probability, you are already assuming that it is normalized. In other words, implicit in the interpretation is that an integral over all of space has been done and divided by.

 

[alexepascual]

If both state vectors are normalized, so would the bracket, as well as the integral on the right. Wouldn't it?. Well, I mean they would reflect the correct value which would have an absolute value typically less than one. On the other hand, doesn't normalization only affect the absolute value of the vector and not any units?. As far as I know, the state vector as well as a bracket are dimensionless. Now, it appears that the integral would give you a dimensionless quantity only if you drop the units of length from dx. If you do that, everything would be fine and the right side of the equation would be dimensionally consistent with the left.
Now, in a different situation of integration, such as W= integral F*dx , or W= integral P*dv, you do consider the dimension of the diferential. As a matter of fact, in these cases you have to consider it in order for the equation to be dimensionally consistent. So, I would assume that in physics in general, when you do an integration, if you are integrating over some variable that has units, you keep the units and use them in your integration.
In the quantum mechanics case I posted, the only apparent way to keep it dimensionally correct would be to add a factor of (1/unit of length) in front of the integral. I wonder if there is some convention where this factor is ommited even it would be needed to make the equation dimensionally consistent.

 

[tavi_boada]

Your mistake is supposing that the wave functions don't have dimensions. They do. The dimensionality is correct. Take a square well problem. The wave functions (sin(nx) and cos(nx)) must have normalization inversely proportional to the square root of a length constant. This way the product of two wave functions has dimensionality of L to the minus one, keeping the dimensionality of the scalar product correct. If you think about it you will see that normalization constants have to have dimensionality. In momentum representation it's 1 divided by the square root of momentum units and in space representation one divided by the square root of space units.keep it up.

ciao!

 

[somy]

dear tavi boada:
can you explain more.I didn't understand your explanation.
thanks in advanced.

 

[tavi_boada]

hey somy

WHen constructing the solutions to the infinite square well, we get from the time independent Schrödinger equation + boundary value considerations that the wave functions are sin(n*pie*x/L), where L is the size of the well. THe probability of finding the particle inside the well is 100%, the integral of the wave function squared must be equal to one. THus the earlier given solution is not complete, we must modify it with a factor which is the inverse square root of the value of the above integral. THus, the scalar product of the two wave funcions has the dimensions of nothing, as expected, because both wave functions are multiplied by constants with dimensionality of length to the minus one half.

 

[alexepascual]

Thanks Tavi,
Your explanation is very clear and it solves my problem.
Now I can continue reading chapters 16 and 20 from Feynman's Lectures.

 

[alexepascual]

Tavi,
I haven't had time to do much reading, but I was thinking about the following. The expression <x|psi> is equivalent to psi(x). Now, psi(x) has units, according to your post. This would make me think that <x|psi> also has units. But in that case where do the units come from?. I was not aware of bras or kets having units. I guess the ket psi by itself could not come with any units as it is a state vector that can be represented in different bases. Could the units be in the <x| bra? I tought this bra just had a label, which is the value of x, and a bare complex number without units.
Maybe you can clarify this for me. I'll appreciate it.

 

[selfAdjoint]

Tavi,
I haven't had time to do much reading, but I was thinking about the following. The expression <x|psi> is equivalent to psi(x). Now, psi(x) has units, according to your post. This would make me think that <x|psi> also has units. But in that case where do the units come from?. I was not aware of bras or kets having units. I guess the ket psi by itself could not come with any units as it is a state vector that can be represented in different bases. Could the units be in the <x| bra? I tought this bra just had a label, which is the value of x, and a bare complex number without units.
Maybe you can clarify this for me. I'll appreciate it.

Why do you say \psi(x) has units?
Here's what QM says about \psi:

- It's a square summable function forming a point in a complex Hilbert space
- It satisfies the Schroedinger equation
- \psi\bar{\psi} maps into the configuration space of an observable and gives the probability for the observable to be in the state indicated by the point on the configuration space.

None of this allows the wave function to assume the attributes of a real measurement in spacetime. It requires the action of a self-adjoint operator to create a number that can do that.

 

[alexepascual]

Self Adjoint:
Thanks for your response. If you look to the previous posts you'll see what my concern is. But in short: when integrating the wave function over all space, if properly normalized, we are supposed to get just the number one. But the integral has the differential element dx which has units of volume, ( length in one-dimensional case). So Tavi suggested that the wave function has units of length to the minus 1/2 (one dimensional case again) . I can accept that. But I had some doubts as to how this concept translates when using Dirac notation. I never suggested you could get a measurable out of a wave function without using an operator.

 

[tavi_boada]

Suppose [psi> represents the state an electron in a given potential. We know that <x[psi> times it's complex conjugate represents the probability density of the electron being at x, not the probability of finding it there. The probability of finding it there is
P(x)dx=<x[psi>·<psi[x>dx=<x[psi>^2dx.
Therefore, psi(x) must have units because a probability, by definition can't have any.
You know that a state can be expressed in terms of the eigenvectors of a hermitian operator. The position operator,x, has eigenvectors [x>, so psi(x) is the coefficient of the expansion of [psi> in terms of the proper base of x. THat's why <x[psi> squared is the probability density of measuring the position of the electron and finding it at x.

 

[alexepascual]

Tavi,
I am not questioning that \psi(x) has units. I accepted that.
(see my post before SelfAdjoints's)
My concern is, in the bracket <x|psi> , where are the units hiding?
So, see, my problem comes when expressing things in dirac notation, not when expressed as a wave function.
Perhaps I am looking at this the wrong way. I usually think of the bracket as an inner product. Maybe in this case I should look at it as just the components in x-space of |psi>, together with its units.

 

[Eye_in_the_Sky]

Solution to Problem with dimensionality

Any ket |f> belonging to the Hilbert space (regardless of whether or not |f> is normalized) is dimensionless. A "generalized" ket, such as |x>, however, is not a member the Hilbert space ... and it has dimension. If x has dimension L, then |x> has dimension L^(-1/2), and likewise for the corresponding "generalized" bra.

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

You can convince yourself of this by writing

<x|x'> = delta(x-x')

and observing that

Integral { delta(x-x') dx }

is dimensionless, implying that delta(x-x') has dimension 1/L.

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

In this way, a function f(x) = <x|f> gets the dimension it needs.

Similarly, <x|p>, where |p> is a "generalized" eigenket of the momentum operator, gets the dimension of reciprocal square-root of action.

 

[alexepascual]

Thanks a lot Eye_in_the_Sky. Your explanation is easy to understand and straight-to-the-point. I think it completely answers my question.
Thanks again,
Alex

 

[turin]

I guess I'm not appreciated at all around here.

 

[alexepascual]

Turin,
Why do you say that?
You posted the first response to this post and I answered the same day. Your post was apparently correct but it did not totally answer my question. When Tavi replied, that solved part of my puzzle, but some questions remained, which were addressed by Eye_in_the_Sky. The response by SelfAdjoint was also correct but was not answering my question.
Some times it is hard to understand what a person asking for help on a topic is really looking for. Part of this may be because the original poster has not communicated the problem in enough detail. Part might be that the person responding is looking at the subject from a different angle, which is satisfactory for him but not for the original poster.
You should not take it personally. I appreciate all responses to my posts and I remember you very well from an interesting discussion we had in another thread.
I view this forum as a collaborative effort where we are all learning from each other. In my physics classes, where instructors graded on a curve, I found that there was more competition than collaboration. Here I have enjoyed a different atmosphere, where collaboration prevails over competition.
I also think that due to the complexity of the subject and its abstract character, it is not always easy to communicate effectively, partly also due to the fact that different people have different angles of approach which others may not find satisfactory.
I, personally like to think in terms of pictures, and I like to always connect the abstract features of the theory to concrete examples. There are people who are perfectly comfortable exploring all the mathematical apparatus without making contact with "reality". That is their preference and I respect it.
As an added benefit of this forum, we get exposed to different views which we may not find useful today but may gradually start having more appeal in the future. All the disagreement and difficulties in communication may not be that counterproductive after all.
Well, Turin, you are always welcome and certainly appreciated.
-Alex Pascual-

 

[tavi_boada]

Turin,

I, personally like to think in terms of pictures, and I like to always connect the abstract features of the theory to concrete examples. There are people who are perfectly comfortable exploring all the mathematical apparatus without making contact with "reality".
-Alex Pascual-

Alex, this has nothing to do with the thread, but I find it remarcable you can think pictorialy about QM. I've tried that approach because it's always been easier for me that way in other subjects, but failed completely. I cannot attack QM prloblems without use of mathematical weapons, I envy you. In fact, come to think of it, I guess I won't be able to adress any problem in modern physics pictorialy. I guess I'm a bit depressed 'cause I can't for the life of me grasp the spirit of Misner/Thorne/Wheelers gravitation. How can they say 2-forms are like honycombs!?Any GR expert fell up to explain?

 

[robphy]

I guess I'm a bit depressed 'cause I can't for the life of me grasp the spirit of Misner/Thorne/Wheelers gravitation. How can they say 2-forms are like honycombs!?Any GR expert fell up to explain?

I posted these links on this other thread regarding visualizing field lines
http://www.ee.byu.edu/ee/forms/
http://www.lgep.supelec.fr/mse/perso/ab/IEEEJapan2.pdf [see p. 10]



In addition, you may find this useful
http://clifford-algebras.org/Beijing2000/page127-154.pdf
http://physics.syr.edu/courses/vrml/electromagnetism/references.html
I believe the field of 2-forms visualized as a honeycomb is due to Jan A Schouten.

(This is now off-topic for this thread.)

 

[alexepascual]

Tavi,
The fact that I have as a preference visualizing things, does not mean that I am being successful in doing so in quantum mechanics. But I try. Memorizing rules to manipulate symbols on a piece of paper does not give me the feeling that I am really learning about nature. On the other hand, when I have a good explanation of the math in terms of "things" in the real world, that helps. With respect to pictures, even seing the state vector as an arrow, and a basis as a set of 3-D coordinate axes gives me a more intuitive description. When you do the math, I like to know what is actually happening to the state. Is anything happening to it or are we simply changin our way to describe it?. I am not really asking here, I am saying that every time that's the kind of question I ask.
In this thread, though, my question was purelly mathematical, which means that it did not concern what is happening to the state but how we use the symbols to describe it.
In this respect I have my suspicion that the explanation given by you and Eye-in-the-Sky, (which I am sure represents common knowledge in traditional quantum mechanics) may not be the only one, or the only possible way to do things. I think this may be a matter of convention, an I am not knowlegeable enough about the subject to challenge the conventional approach, but I could venture an opinion to see what you guys think. I'll do that my next post.
With respect to the Gravitation book, I haven't read it. Probably I will within the next two years, but I think it is kind of heavy reading.

 

[alexepascual]

Robphy,
I read a little about Geometric Algebra a few weeks ago and it sounded interesting. It wonder how much it is used in quantum mechanics and what are it's advantages. It appears it may be a better way to conceptualize some operations such as products of vectors, but I am not sure about it yet. I have that in my to-do list of things to read. There is so much to learn and so little time!

 

[alexepascual]

Tavi and Eye_in_the_Sky.
Your posts have completely answered my original question. But, as I say earlier, I have my suspicion that this mathematical scheme may represent a convention and not a unique way of doing the math.
To be more explicit: In the scheme described by you, the kets corresponding to eigenvectors of a continuous variable's operator carry with them units that are the units of the variable to the -1/2. I understand that these kets are special as they represent "limits" or things not properly within the Hilbert space. Still, handling the units like this seems a bit artificial.
It may seem too audacious of me to venture opinions about these things while I am still learning and have not read everything about it. But anyway I'll say it, as you, having more knowledge about it may even mention some thinking along the same lines by other people.
I am somewhat aware that Von Newman had objections about the Dirac delta function, but I don't know the details behind his argument.
On the other hand, it appears kind of obvious that many of the infinities we get in quantum mechanics are the result of considering certain variables as continuous when they are actually discrete.
I don't know much about quantum loop gravity and superstring theory, but I think I remember reading that they address this problem.
Going back to QM. if we were to consider all variables as discrete, then the scheme you mentioned would have to be changed as it implies treating differently continuous varables from discrete ones. For instance, the spin up and spin down unit vectors, as far as I know don't carry any units, while the position eigenvectors do.
It seems to me, and here is where I may be venturing too far with my little knowledge, that a more natural way to do the math would be (taking the position eigenvectors as example) to keep them dimensionless and to add a denominator with the square root of the space units to the dx in the integration. This way, we would be doing the integration over a dimensionless domain, which would be consistent with the way we do summation for discrete variables. This scheme would also make it easier to transition to a treatment where all variables are discrete.
It seems to me that in quantum mechanics we are dealing with two big domains or languages: one being the Hilbert space, in which all the classical variables disappear and are transformed into mere lables for unit vectors, and the other domain is that of classical variables which we multiply, divide etc. and keep track of the units by dimensional analysis. The operators corresponding to measurables would allow us to "extract" the classical variable with it's units from the state vector living in Hilbert space.
If we look at it this way, when we are in Hilbert space, the numbers we manipulate should not have any units, they should only be related to the square root of a probability and a phase factor. And perhaps it should not matter wether the Hilbert space is rigged or not. Then when we "translate" to our usual "domain" or "language" of classical variables, then the units are reconstructed from the lables.
I'll appreciate your opinions about this.

 

[Eye_in_the_Sky]

From Another Angle - Seeing it Deeper

[Note: This post is intended to provide an alternative argument to that given in the post entitled Solution to Problem with dimensionality in which the Dirac delta function was used.]

Consider an observable for which the associated operator is given by

A = Sigma_n { a(n) P(n) } ,

where a(n) are the eigenvalues and P(n) are the corresponding eigenprojections. If the observable has dimensions, then the eigenvalues
a(n) will have those very same dimensions, while, on the other hand, the eigenprojections P(n) are (always) dimensionless.

For a measurement of position, the operator

X = Integral { x |x><x| dx }

is an idealization. The operator which corresponds more precisely to the actual physical measurement has a form just like A above, where

a(n) = x(n) , the "location" of each detector

and

P(n) = Integral_[region(n)] { |x><x| dx } ,

where "region(n)" is the interval over which the detector "detects".

Clearly, "|x><x| dx" must be dimensionless.

[Alex, later on today, I hope to reply to your most recent post.]

 

[turin]

tavi boada,
Regarding the honeycomb: can you picture the difference between &part;_&mu; and dx^&mu; ?

alexepascual,
Regarding units in QM: I think it is more important for the eigenvalues to have the appropriate units. Then, when your observable operates on the spin state, for instance, you get an hbar. Come to think of it, eigenvalues themselves have units, so how can that be explained with the rotation picture?
Regarding normalization: you simply divide the state vector by the square root of the self inner product. This inner product is an integration over all of the domain of interest, so it will pick up whatever units are used to describe that domain. For the spin states the units are kind of like angular values, so they should be unitless I reckon.

 

[alexepascual]

Eye_in_the_Sky:
Thanks for your post. I see you use the projector for discretized states, and an integral to obtain each of the discrete states.
I guess this approach makes sense. In the end you get to the same conclusion as in your previous post, which is nice. But I was already convinced after your post using the Dirac delta function.
I have looked for references to "generalized" kets in the books but I haven't found anything. Would they use a different terminology?. Or maybe none of the books I have from the library deal with this topic in detail. (Sakurai, Shankar, Merzbacher). I also have Feynman's Lectures and Griffiths.
You mention in your first post that a definite momentum ket would carry units of reciprocal square-root of action. I don't understand that. Being a momentum ket, shouldn't it have to have units of reciprocal of square root of momentum?. I am not really disputing what you are saying. I am sure you are right. It is just that I don't understand it.

 

[alexepascual]

Turin:
I agree with you that in general state vectors don't have units as they represent the square root of a probability + a phase (for each component).
But, as Tavi_Boada and Eye_in_the_sky have shown us, the wave function, as well as the eigenvectors of the operators of continuous variable do need some units (not the units of the variable in question) in order for the equations to be dimensionally consistent. I am not very happy with this way of doing the math, but I accept it. The books that I have seen appear to ingnore this issue.
If we limit ourselves to the case of discrete variables, where we use summations instead of integrals, then the kets are all dimensionless, like the kets for spin.
It is only when we apply a measurable's operator that we get eigenvalues together with their units (as you mention). But the units we were talking about are not these units, but the reciprocal square root of the measurable's units which are needed only for eigenkets of continuous variables.
The subject of getting units when applying the operator would fit more in my previous thread on the meaning of operators than here. But as you mention using a rotation picture used to interpret this, I'll give you my opinion. You could interpret the measurement process as a rotation of the state vector to one of the eigenvectors of the measurement operator. But I think this would have to be done "after the fact", once you know to which eigenvector it went. This picture would not give you the probability of obtaining the eigenvector, but it would give you the new eigenvector and you would not have to normalize it as it would keep it's unit modulus. In using this "rotration picture" you would stay within the realm of Hilbert space, and no measurable operators would be used (just a unitary operator to perform the rotation).
I think using measurement operators presents a different picture. You get the eigenvalues which are not in Hilbert space but are embedded in the operator. So, as I say in a previous post in this thread, I think operators provide a translation from the language of Hilbert space, where the quantities being manipulated are just complex dimensionless numbers, to the language of classical mechanics, where we use the typical dimensioned variables.
Form what I understand, using operators goes hand-in-hand with the projection picture, where besides getting the probabilities, you also get the eigenvalues which are the numbers used in classical mechanics.
I suppose there is a lot of thinking that can be done about all these things, but most people just accept them like they are presented in the books, without questioning or trying to visualize them a different way.
The way I see it, QM is usually presented as a dogma and it requires a lot of faith on your part to accept it.

 

[Eye_in_the_Sky]

I have looked for references to "generalized" kets in the books but I haven't found anything. Would they use a different terminology?

The term I can remember using was "generalized". Whether that term was applied to the word "vector" or "ket" I cannot recall. Perhaps "generalized eigenvector" is the most appropriate terminology.
_______________________________

Or maybe none of the books I have from the library deal with this topic in detail. (Sakurai, Shankar, Merzbacher). I also have Feynman's Lectures and Griffiths.

I never learned about this topic from a physics textbook.
_______________________________

You mention in your first post that a definite momentum ket would carry units of reciprocal square-root of action. I don't understand that. Being a momentum ket, shouldn't it have to have units of reciprocal of square root of momentum?

Both objects "<x|" and "|p>" are of the "generalized" variety; therefore, <x|p> has the dimensions of ([x][p])^-1/2 = (action)^-1/2

 

[Eye_in_the_Sky]

It has been suggested that each of "|x>" and "<x|" have dimension L^-1/2.

I have my suspicion that this mathematical scheme may represent a convention and not a unique way of doing the math ... I understand that these kets are special as they represent "limits" or things not properly within the Hilbert space. Still, handling the units like this seems a bit artificial.

To my eye (and I will explain why in my next post), the above scheme is the most "natural", "self-consistent", "physical" way to handle these entities and others like them. Nevertheless ...

... it implies treating [the bras and kets] differently [for operators with] continuous [spectra] ... [than for operators with] discrete [spectra].

Yes that is true. And, furthermore ...

... a more natural way to do the math would be to keep them [i.e. the bras and kets (of operators with continuous spectra)] dimensionless ...

... and use a dimensionless measure "dx" stripped of its units under the integral (or some other similar prescription).

I appreciate the care you have taken in formulating your considerations, as well as your point concerning the analogy between |x> (continuous case) and |f_n> (discrete case). The solution you offer makes perfect sense, and I suppose it would appeal to the mathematically oriented individual (for whom even trivial cases are treated in "dimensionless domains").

But the "real" analogy is between

|x><x|dx and |f_n><f_n| ... and not ... |x> and |f_n> .

I hope to convince you of this in my next post.

 

[robphy]

The term I can remember using was "generalized". Whether that term was applied to the word "vector" or "ket" I cannot recall. Perhaps "generalized eigenvector" is the most appropriate terminology.

Both objects "<x|" and "|p>" are of the "generalized" variety; therefore, <x|p> has the dimensions of ([x][p])^-1/2 = (action)^-1/2

Does "generalized" refer to something akin to "generalized function", "Dirac delta function", "distribution"?

 

[turin]

Eye,
The last thing that you mentioned has reminded me that I agree with you. Shankar talks about that in the first chapter. From what I remember, there is a stark contrast between discrete and continuous inner product spaces (and Shankar basically calls for a redefinition of the inner product itself). I guess what you have is a comparison of outer products, but I imagine that it would boil down to the same basic issue. They are just two different kinds of spaces.

Alex,
I recall that chapter 7 in Shankar (harmonic oscillator) has some sort of discussion on units. Basically, it seems that there is a "most meaningful" scale in which to solve the Schroedinger Eq., and converting to this scale involves eliminating the units. I realize that this doesn't exactly address the integration issue, but perhaps it could give some insight, and, after all, a solution to a diff. eq. is essentially the same thing as an integration (though often intrinsicly). The Hermite polynomials are normalized with a power of the variable y and then the dy also shows up, but this variable is the conveniently unitless variable that was used from the outset of the solution.

 

[Eye_in_the_Sky]

Does "generalized" refer to something akin to "generalized function", "Dirac delta function", "distribution"?

Yes, it certainly does. But exactly what the "technical definitions" of some of those terms are is not clear to me.

This much (... or should I say "little"? ), however, is clear to me:

An object like |x> or |p> is a "generalization" of an object which is called a "vector in the Hilbert space" ... the latter has finite norm, whereas |x> and |p> do not.

 

[robphy]

Yes, it certainly does. But exactly what the "technical definitions" of some of those terms are is not clear to me.

This much (... or should I say "little"? ), however, is clear to me:

An object like |x> or |p> is a "generalization" of an object which is called a "vector in the Hilbert space" ... the latter has finite norm, whereas |x> and |p> do not.

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

 

[alexepascual]

Eye:
Thanks for your explanations. I was careless when I read one of your previous posts and thought you meant the units for the ket |p> would be the reciprocal square root of action. Now I see you were talking about the bracket <x|p>, and everything is clear.
From what Turin is saying, it appears that Shankar uses in one of his chapters a dx stripped of units. So, if the dimension problem can be addressed several ways, I think most authors are just sweeping this problem under the rug and letting the reader interpret it any way they want.
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)= | <p|psi> |^2 dp/ h
(sorry for not using latex)
(Actually he has h-bar/2pi which I replaced by h.)

Now, that h in the denominator is action. If the |p> has units, then you would not need anything under dp. If on the other hand |p> is considered dimensionless, then the way to "fix it" would be to divide dp by a unit of momentum, not h. I don't get it.
Changing the subject, I remember reading (I don't know where) about a "rigged Hilbert space" which would include the eigenvectors of continuous variables.

 

[turin]

What are "eigenvectors of continuous variables?"

 

[alexepascual]

Turin,
By "eigenvectors of continuous variables" I meant:
"eigenvectors of measurement operators of continuous variables"
So these would include the base states (kets) for position and momentum.
Sorry for the sloppy terminology.

 

[turin]

Why would a Hilbert space have to be "rigged" in order to include eigenvalues? I'm probably not appreciating the meaning of "Hilbert space" in this discussion.

 

[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))
<br /> \begin{align*}<br /> \int \left[ \hat Q \phi \right] \ \tau(q) dq &amp;=\\<br /> \int \hat Q\ \delta(q-q_0)\ \tau(q) dq <br /> &amp;= \int q\ \delta(q-q_0)\ \tau(q) dq \\<br /> &amp;=\int q_0\ \delta(q-q_0)\ \tau(q) dq \\<br /> &amp;=\int \left[ q_0\phi \right] \ \tau(q) dq<br /> \end{align}<br />
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-