# Problem with dimensionality

1. Jun 20, 2004

### alexepascual

In the following equation:
$$\left\langle {\phi } \mathrel{\left | {\vphantom {\phi \Psi }} \right. \kern-\nulldelimiterspace} {\Psi } \right\rangle = \int {\phi ^ * } \left( x \right)\psi \left( x \right)dx$$
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

Last edited: Jun 20, 2004
2. Jun 21, 2004

### 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.

3. Jun 21, 2004

### 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.

4. Jun 22, 2004

Your mistake is supposing that the wave functions dont 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 lenght 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!

Last edited: Jun 22, 2004
5. Jun 22, 2004

### somy

can you explain more.I didn't understand your explanation.

6. Jun 22, 2004

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 wich 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.

7. Jun 22, 2004

### 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.

8. Jun 24, 2004

### 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.

9. Jun 25, 2004

Staff Emeritus
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.

10. Jun 25, 2004

### alexepascual

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 lenght 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.

11. Jun 25, 2004

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.

12. Jun 25, 2004

### alexepascual

Tavi,
I am not questioning that $$\psi(x)$$ has units. I accepted that.
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.

Last edited: Jun 25, 2004
13. Jun 25, 2004

### 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.

14. Jun 25, 2004

### 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

15. Jun 27, 2004

### turin

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

16. Jun 28, 2004

### 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-

17. Jun 28, 2004

Alex, this has nothing to do with the thread, but I find it remarcable you can think pictorialy about QM. I've tried that aproach 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 wont 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?

18. Jun 28, 2004

### robphy

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 [Broken] [see p. 10]

In addition, you may find this useful
http://clifford-algebras.org/Beijing2000/page127-154.pdf [Broken]
http://physics.syr.edu/courses/vrml/electromagnetism/references.html [Broken]
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.)

Last edited by a moderator: May 1, 2017
19. Jun 28, 2004

### 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.

20. Jun 28, 2004

### 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!

21. Jun 28, 2004

### 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 continous 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 dissapear 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.

22. Jun 28, 2004

### 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.]

Last edited: Jun 29, 2004
23. Jun 28, 2004

### turin

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.

24. Jun 29, 2004

### 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.

25. Jun 29, 2004

### 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.

Last edited: Jun 29, 2004