# 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?