Momentum operator as the generator of translations.

1. Jun 7, 2004

alexepascual

The explanations I have found in quantum mechanics books as to why the momentum operator is considered to be the "generator of translations" are a little difficult and not very intuitive.
Could someone help me on this?
What I am looking for is some explanation in terms of pictures, or at least in terms of states and what you do to them (change them, measure them, change basis, etc.)
I'll appreciate it.
--Alex--

2. Jun 8, 2004

robphy

Try http://www.math.ohio-state.edu/~gerlach/math/BVtypset/node111.html [Broken]

Last edited by a moderator: May 1, 2017
3. Jun 8, 2004

slyboy

It might be appropriate to look into why angular momenum is the generator of rotations first, because this is slightly easier to understand. Once you have the basic idea, the connection between ordinary momentum and translations becomes easier to understand.

4. Jun 8, 2004

alexepascual

Robphy and Slyboy,
Thanks for your answers, that't get me going on this topic. I'll have to think about it, read the article, think some more...
I am sure I'll have more questions about it later. Now I'll have to do my homework (the one you guys gave me).
Thanks again,
--Alex--

Last edited: Jul 2, 2004
5. Jun 8, 2004

turin

Do you have a problem with the classical mechanics version, too?

6. Jun 9, 2004

alexepascual

Turin,
I guess I do have a problem with this concept in classical mechanics too as I don't remember studying at that time. I now looked for it in Thornton, and I don't see it.
However, I think the link posted by Robphy talks about translations of a classical funcion.
From all I have seen untill now, it appears that the clue is understanding the expression of the translation as a Taylor series, and then converting it to an exponential.
I'll keep thinking about it. If you have any clues, I'll appreciate it.

7. Jun 9, 2004

turin

I am hoping more to get a clue from your thread here than give clues.

The only things I understand about it are that the canonical momentum is the generator of generalized position translations (which covers linear as well as angular, and whatever non-physical abstractions that you decide to dream up). The fact that this is true for linear momentum and Cartesian coordinates of a free particle I believe to be the best starting point. And the issue is not trivialized by considering only one dimensional motion, so I [humbly] suggest looking into that.

The Hamiltonian is just ~p2. What the Hamiltonian means in Classical mechanics is something for which I suppose you must come to your own personal level of acceptance. I personally like the "total" energy interpretation, with the one exception that it is, in general, only the "generalized" or "canonical" energy, and may not necessarily represent the physical ability of the system to do work. It has a conjugate, the time, and the Hamiltonian is, in fact, the generator of time translations of the system.

So, anyway, the partial derivative with respect to the momentum gives you the rate of change of position. In this way I understand it as the generator of position translation. However, there is this issue of infinitesimal vs. finite, and the need to go to exponentiation to get the finite translation. I am not at all sure about this, but I think that may be one big difference between Classical and Quantum.

Last edited: Jun 9, 2004
8. Jun 9, 2004

sol2

http://focus.aps.org/stories/v10/st3/pic-v10-st3-1.jpg

Seeing double. Researchers have caught glimpses of a rare event in which a single photon splits in two. This calorimeter, which contains 400 kg of liquid krypton, detected the photon pairs.

http://focus.aps.org/story/v10/st3

To be able to create this "diversion", could you not set up spin orientations in cryptography? This woud be the jest of Penrose's quanglement issue, displayed?

Also understanding that a "medium" is essential between cryptography pairs, how could we move the understanding of the http://superstringtheory.com/forum/stringboard/messages22/66.html [Broken] ( matrices ) through feynmen toy models and understand that there must be a way in which to interpret dynamcial movement in the spacetime fabric?

Do you understand this? Any correction from others appreciated as well.

A strange thought just crossed my mind about Alice. Imagine, a http://wc0.worldcrossing.com/WebX?14@134.wWtJbPzXbda.6@.1dde7e87/2 [Broken]. Turn on your speakers if you like Don't forget to scroll down right away to the post in question.

Last edited by a moderator: May 1, 2017
9. Jun 10, 2004

slyboy

No. Unfortunately, I haven't really got a clue what you are talking about. What do you mean by cryptography requires a "medium"?

10. Jun 10, 2004

sol2

Fiber optics.

I am more interested in teleportation, and spin orientations being effected.

The quantum world is complex issue here but if we are to speak about quantum geometry and gravity, these issues must be addressed?

I believe this subject in terms of teleportation has potential. That we move from computerzation, to teleportation, raises a interesting feature of consideration to me.

If the graviton has dimensional significance, then the interpretation of that quantum world must be based on teleportation principals?

11. Jun 11, 2004

slyboy

Quantum crypto doesn't necessarily require a "medium". See the experiments on free-space quantum cryptography form example.

I don't really see what teleportation has to do with gravity.

12. Jun 11, 2004

sol2

On second part for sure.

http://www.esi-topics.com/enc/interviews/Prof-Nicolas-Gisin.jpg

I was thinking of the Professors early work with entanglement and the link he forged. His experiment with fiber optics, was over ten miles and held great interest with the telephone companies in Switzerland.

Orientations in spin can be effected by magnetic fields? Oreintations can be effected by gravitational fields? Gr on a classical level in regards to probe B, while at the quantum level how so? No?

So there is this "distance", and no medium ( this is not about the aether question so often brought up, but is about the significance of dimensions)?

Last edited: Jun 11, 2004
13. Jun 22, 2004

from generator of translation to teleportation, wow!!
Answering the original question, try working the operations out with the generator of INFINITESSIMAL translation.Express it as a taylor expansion to first order than group up the derivatives with respect to the three dimensions and you get a gradient operator, which is proportional to the momentum operator in the spacial representation.

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

alexepascual

Thank you Tavi. This week and part of last week I have been kind of busy working on other stuff. In my free time I have been doing a little more reading from Feynman. These last few days I have been a little tired at night to work on derivations. But as sson as I can I'll go back to it and I'll consider using the first two terms in a Taylor series.
Thanks again,
Alex

15. Jun 23, 2004

Hey alexpascual, I vigorously recommend Sakurai's Modern Quantum Mechanics. It is a very readable and close introduction. I haven't studied the collisions part, but the rest is great!

16. Jun 23, 2004

arivero

On the same token, I a bit puzzled about angular momentum. From Newton, it seems that any central force field will preserve it, even if not spherically symmetric. But for Noether theorem, rotational symmetry seems a prerequisite. I am missing something.

17. Jun 27, 2004

turin

Isn't a central force requirement the same thing as rotational symmetry?

18. Jul 2, 2004

alexepascual

Ok guys, I am back.
I read the link suggested by Robphy. But I still have some questions.
I will try for the moment to stay away from rotations and concentrate on translations in one spatial dimension.
It appears that the math allows you to treat translations using an operator without having to plug-in the momentum. From this, I would assume that this operator is a mathematical beast that is more general than the specific applications we may find in quantum mechanics. I am trying to narrow the field as much as possible. So I will try to understand the generator of translations as a mathematical instrument without involving momentum.

If I have a function f(x), and I move this function to the left by a distance d, the new function fd(x) will be equivalent to f(x+a).
f(x+a) can be expanded into a Taylor series and it can be proven to be equivalent to exp(a d/dx)f(x).
I haven't read any thing that would indicate that this is only true for certain functions. It appears it would work for any function. Is that right?
Now, d/dx is called the generator of translations.
If we can plug any finite number for "a" and let the operator do its work for all values of x of the original function to obtain the new function that would be great, and I would understand why d/dx could be called "the generator of translations"
Now, if we think of the operator as a matrix, and divide the motion in small steps, then for each step we would apply the operator with a/N instead of a.
Each time we apply the operator, the function will move a little bit, so the result of d/dx for the same value of x the next time won't be the same. This shows that we can't just calculate the derivative for each value of x for the initial position and use that in the diagonal of the operator.
If I lost you here, I think I am just getting to the point (by a different route) that this operator does not have the position kets as its eigenvectors (using the language of QM.) So the matrix can't be diagonal when in the x-basis.
Now, I understand that exp(a d/dx)f(x) can be expanded into a Taylor series to carry out the calculations, but that doesn't seem to make it any easier.
What I am getting at, is that if I wanted to use the beautifully simple expression exp( a d/dx) to write a computer program to move f(x) by a finite distance, I doubt it would be of much help.
If on the other hand I had been able to write a diogonal matrix where the diagonal elements are a*exp(d/dx) and apply that matrix once, that would have been great. I call it a matrix because that's the way I am visualizing it, although I realize it would be more of a "machine" than a matrix and that's why in this case is better to call it an operator.
I hope I didn't confuse you too much. maybe I was expecting too much from this "generator of translations". If you understand my confusing post, maybe you can point towards mistakes in my reasoning that may be causing my difficulties.

Last edited: Jul 4, 2004
19. Jul 3, 2004

vanesch

Staff Emeritus

This is a subtle remark !
I guess you think of a force F = f(r,theta,phi) 1_r ; so it is a central force (only a 1_r component) but it is not spherically symmetric. Well, I guess that the clue is that you cannot derive such a force from a potential, and hence it cannot be formulated in a Lagrangian fashion, so Noether's theorem (which only applies to Lagrangian systems) doesn't work...

cheers,
Patrick.

20. Jul 3, 2004

turin

alex,
Two brief comments (I will refer to Shankar for more detail later; I remember this translation operator being mentioned in there, but I will need to refresh.):
For some reason, I think that there should be a √(-1) in the exp, but I'm not sure.
(Why) do you think that the translation operator should be diagonal in the position basis?

21. Jul 3, 2004

alexepascual

Turin:
I think the i (imaginary unit) only appears when you use momentum as the generator. If you look at the momentum operator in the x-basis, you'll see that it has an i. so the i you are mentioning cancels that i.
I am pretty sure you don't need the i unless you use the momentum operator. For the moment I am trying to keep things as simple as possible and not use momentum..
I don't think the matrix should be diagonal in the x-basis. I only said that if it were diagonal, it would be nice and the spatial trasnslation operator in it's exponential form would be computationally more useful.
I understand that the formalism does not necessarily seek computational efficiency but it also seeks simplicity (in it's form) and elegance.
The only method that I can think of to carry out the operation of the translation operator is to convert it back to a Taylor series.
Of course it would be easier to just shift the x, but I guess in Hilbert space that's not the way to do things.

22. Jul 4, 2004

turin

Of course . I was thinking about it with the momentum in there, and I was thinking that d/dx was the momentum. Whoops. I should have also realized that the i was not needed just by considering the Taylor expansion.

By its very nature it cannot be diagonal in the position basis, as it takes a position to another position. If it were diagonal, then it would be useless, because it would represent a trivial translation of a position to itself.

I don't quite understand what you are saying here.

23. Jul 4, 2004

Eye_in_the_Sky

What I am about to discuss is something which I would have thought is discussed in the references you are using. But just in case ...

Let X and P be the position and momentum observables for a particle moving in one dimension. Suppose that we take the physical measuring device which measures "x" and translate it by an amount "a" in the positive x-direction. Then, the new observable for position is given by

X' = X - a .

This is so because any measurement which would have yielded the "result" (eigenvalue) x, relative to the original observable X, now yields the "result" (new eigenvalue) x' = x -a, relative to the new observable X'.

As it turns out, the linear operator U(a) defined by

U(a) = e-iPa/h_bar

is a unitary operator, and it has the property

U(a)XU(a)t = X - a = X' .

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

Instead of considering a translation of the measuring device, we can alternatively consider a translation of the apparatus which prepares the quantum system to be in a state |f>. A translation of that apparatus through a distance "a" in the positive x-direction results in a new state |f'> which is given by

|f'> = U(a)|f> .

In position representation, it turns out that

f'(x) = <x|f'> =<x| (U(a)|f>) = (<x|U(a)) |f> = <x -a|f> = f(x -a) .

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

Next, consider the translation of an entire experimental arrangement through a distance "a" in the positive x-direction. That is, we move all instruments - those pertaining to both preparation and measurement of the quantum system. We then find that for any observable A measured in the experiment, the "matrix elements" of A are invariant; i.e.

<g'|A'|f'> = <g|A|f> .

This is so because

|f'> = U(a)|f>, |g'> = U(a)|g>, and A' = U(a)AU(a)t ,

and it happens that U(a) is a unitary operator. What we have, in effect, done is performed a "change of basis".

In relation to this scenario, the following four statements are equivalent:

(1) all matrix elements are invariant;
(2) U(a) is unitary;
(3) a translation of the entire experimental arrangement results in a "change of basis";
(4) the space we are working in has "translational symmetry".

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

Finally, note that, in the above discussion, the active perspective has been taken. That is, we have imagined that various "instruments" of an experimental arrangement have been "actively" moved from one location in space to another. In a scenario like the last one, in which all of the instruments are moved together as a unit, we can, in virtue of "translational symmetry", take the passive perspective, whereby the "coordinate system" itself is translated by the amount "-a".

24. Jul 4, 2004

Eye_in_the_Sky

The translation operator uses "d/dx" ... and "d/dx" is(!) precisely the operator "iP/hbar" in x-representation; i.e.

(Pf)(x) = <x|P|f> = -ihbar(df/dx) .

Nevertheless, you can still try to understand it without any explicit mention of "momentum"; i.e. just by talking about "d/dx".

Yes, what you say is (essentially) correct. (However, since we are talking about functions which are equal to their Taylor series expansion, we need to make a restriction to arbitrary analytic functions.)

Exactly! ... And Turin offers a very simple argument for why this so:

Here is another way of saying the same thing. If an operator is diagonal in the "x-basis", then it must commute with the x-operator. But (for the infinitesimal case (and, therefore, for the finite case)) this is obviously not so:

[x, d/dx]f = x(d/dx)f - (d/dx)(xf) = -f ;

that is,

[x, d/dx] = -1 .

Now ... if we look at all those "crazy" Taylor expansions and try to see what makes a linear operator L a suitable "candidate" for being a "generator" of translations, represented by e-iLa for the finite case, then we find that the necessary and sufficient condition is just

[X, L] = i .

This condition, in turn, uniquely defines L up to the addition of an arbitrary function of X (i.e. L and L + f(X) are the "same" in this regard). We simply "throw away" all those possible "extra" functions of X (since they commute with X), leaving us with the unique choice:

L = P/hbar .

And so ... "P/hbar" is the generator of translations.

Next:

The expansion into a Taylor series is not intended to be a tool for calculational purposes! It is the manner in which we prove(!) that

[ea(d/dx)f](x) = f(x + a)

within any domain over which the function f is analytic.

Finally:

At this juncture, Alex, I must compliment you on your exceptional powers of visualization.

... No, you didn't confuse me. The "machine" you describe is one for which you have effectively taken, so to speak, the "operator" out of the "eigenprojections" and put(!) into the "eigenvalues". I doubt, however, that this sort of construction will offer any improvements over the "standard" Theory of Linear Operators.

As it turns out, the diagonal form of the operator you seek, according to the definition U(a) = e-iPa/h_bar, is just

U(a) = Integral { e-ipa/h_bar |p><p| dp } .

Last edited: Jul 4, 2004
25. Jul 17, 2004

alexepascual

Eye_in_the_Sky:
I think your explanation is quite detailed and now I understand this topic better.
I should point out that at the beginnning of this thread I was focusing more on momentum and looking at the "physics". Later on, I after reading a little more, I was under the impression that there was a more general mathematical mechanism worth exploring.
To be more specific, I thought that exp(a d/dx) should be able to displace any function (analytic as you point oput) along the x-axis (this would be a mathematical fact not necessarily connected with physics) . I wanted to see how this exponential could do this.
You mention in one of your posts, that a Taylor series is not intended as a calculation tool. Probably you mean that it is not very efficient.
In my exploration of this topic (before your post), I took a function y=x^2 and tried to move it using the exponential. I could not see any other way the exponential could do it's work but with a Taylor series. As the Taylor series for x^2 is very short, it was easy to use. This may seem trivial to you, but I enjoyed seing how the function had been moved a distance "a".
I understand your suggestion that if we make a change to the momentum basis, then the operation can be done with a single diagonal matrix.
I was thinking how this idea could be interpreted if we are moving a function that doesn't have anything to do with physics. I realize that some of these mathematical methods become useful only when applied to solve problems in quantum mechanics, but if the same technique can be applied outside of physics, even if it is not efficient for that purpose, this could provide (at least for me) a better understanding of the underlying math.
I haven't worked this thing out, but I was thinking that outside of physics there is no momentum (well, you could argue against this, but you know what I mean) so we can't use momentum. But what you do when you change from the position basis to the momentum basis is a Fourier transform right?.
So, If I take my x^2 function and do a Fourier transform on it, I should be able to use whatever I get to construct my diagonal matrix and move the function (in a basis of complex sinusoidal functions) the required distance right? Then I would have to convert back to the position basis. I'll try this unless you warn me that it won't work.
Looking at the effect of the exponential exp(a d/dx) in the position basis, I noticed that when it acts on each component of the vector it is applied to, the result of that partial operation is placed in the same component of the result vector as it was in the vector that the operator acted on. In other words, you take the function at a particular value of x, say x=3, calculate all the derivatives at that point, and the result you get from the Taylor series is the new value of the function at x=3. This would give the appearence of certain "diagonality". But then I realize that when we talk about diagonality, normally we are talking about matrices, and in that case (matrices) we are multiplying the component or components of a vector by a number/numbers. And I see that in the case of the differential operator in the Taylor series form, we are multiplying times 1, and then adding the other terms that contain the derivatives, which is very different from the action of a matrix.
On the other hand, I guess if we wanted to use a matrix in the position basis to move the function, one way to do it would be with a matrix with all zeroes except for a line parallel to the diagonal which would be full of Dirac deltas.
This scheme could be discretized (getting a finite delta) for calculation with the computer.
Is that right?
Thinking outside the box, (forgive me for this, I am just thinking aloud) we could build a diagonal matrix that does the moving, but it would be uggly, conceptually and computationally, as there would be an infinity in the matrix anywhere the value of the original function is zero, and the computation errors would be large when the value of the original function is very small. So this would prove that most of the time thinking outside the box doesn't pay. But I still like doing it.
Except for the experiment (exercise) I mention above that I still want to do, I am kind of satisfied with my present understanding (thanks to your careful explanations) . I think I feel comfortable enough to continue studying. Now I am starting chpter 2 of Sakurai.
By the way, towards the end of chapter 1, I found a reference to the dimensionality of the wave function, question you answered for me on another thread.

Eye_in_the_Sky and Turin,
I know I can be kind of irritating sometimes. Thanks for being patient with me.
Oh! by the way, I am curious to know (I don't remember if I asked you before) in what country/state you guys live. (I am in California, USA)

-Alex-

Last edited: Jul 20, 2004