Momentum operator as the generator of translations.

[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--

 

[robphy]

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

 

[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.

 

[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--

 

[turin]

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

 

[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 until 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.

 

[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.

I will try to think about this a little tonight.

 

[sol2]

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.

I have been reading your responses and have been conntemplating a about orientation.


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 ( 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 . Turn on your speakers if you like Don't forget to scroll down right away to the post in question.

 

[slyboy]

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

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

 

[sol2]

What do you mean by cryptography requires a "medium"?

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?

 

[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.

 

[sol2]

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

On second part for sure.


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

Quantum physics is in an especially interesting phase. Until recently, the conceptual difficulties were considered of limited importance, since they had no practical effects. This led John Bell to his famous declaration: QM is fine FAPP (For All Practical Purposes)! But today technology and the discovery of the power of quantum information processing have dramatically changed the picture. Today, conceptual questions have potential applications, and technological breakthroughs open the way to new fundamental tests of the theory. My goal is to be an active player in this exciting dialog between fundamental and applied physics.

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

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.

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

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)?

 

[tavi_boada]

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 spatial representation.

 

[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

 

[tavi_boada]

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!

 

[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.

 

[turin]

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

 

[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 f_d(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.
I'll appreciate your input.

 

[vanesch]

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.

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.

 

[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?

 

[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.

 

[turin]

If you look at the momentum operator in the x-basis, you'll see that it has an i.

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.

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.

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.

... it would be easier to just shift the x, but I guess in Hilbert space that's not the way to do things.

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

 

[Eye_in_the_Sky]

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.

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".

 

[Eye_in_the_Sky]

It appears that the math allows you to treat translations using an operator without having to plug-in the momentum.

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

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

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

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 f_d(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?

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.)

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.

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

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.

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/h_bar .

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

Next:

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, 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.
... 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.

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

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

within any domain over which the function f is analytic.

Finally:

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.
I'll appreciate your input.

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 } .

 

[alexepascual]

Eye_in_the_Sky:
I have read your second post, as well as your private message.
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-

 

[turin]

... 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?

I hope I don't reveal too much of my ignorance, but, what is the FT of x²? Off the top of my head, I would assume that it is not defined. Certainly the DC component would be undefined, but I suspect that the whole thing is undefined. I actually went through the calculation explicitly and got infinity. I could not find the transform in the few tables to which I have access at the moment.

 

[Eye_in_the_Sky]

... what you do when you change from the position basis to the momentum basis is a Fourier transform right?.

Yes, that is 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.

The function x² isn't a good choice. The Fourier integral is going to "bust"(!) since

x² --> infinity, as x --> +/- infinity .

But instead of looking at a specific example, why not just look at the general situation? We have:

[1] f(x) = 1/sqrt{2 pi} Integral { e^ikx F(k) dk } ,

[2] F(k) = 1/sqrt{2 pi} Integral { e^-ikx f(x) dx } .

Now, apply d/dx to equation [1], and get the correspondence

[3] [(d/dx)f](x) <--> ik F(k) .

Next, consider the Fourier transform G(k) of f(x+a). It's just

G(k) = 1/sqrt{2 pi} Integral { e^-ikx f(x+a) dx }

= 1/sqrt{2 pi} Integral { e^-ik(x-a) f(x) dx }

= e^ika F(k)

from which we get the correspondence

[4] f(x+a) <--> e^ika F(k) .

And this is just what you wanted to see: multiplying the Fourier transform
F(k) by e^ika will "shift" the original function f(x) by the amount "a", as required, to give f(x+a).

Finally, if you want to go one more step, compare [4] to [3] and get confirmation that

f(x+a) = [e^a(d/dx)f](x) .

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".

There appears to be "linguistic" problem here. Look at the phrase:

"... is the new value of the function at x=3" .

This phrase should read:

"... is the value of a new function at x=3" .

What you get from the Taylor series is a new function g(x) = f(x+a), and
g(3) = f(3+a).

If I have understood you correctly, then what I have just said should cause you to want to retract your concluding statement:

This would give the appearence of certain "diagonality".

Next, regarding:

... 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.

Observe that this "operation" is in fact represented by a linear operator, say L, where

L = Sigma_[n = 0 to infinity] { (1/n!) [a(d/dx)]ⁿ } .

But, any linear operator will have a matrix representation relative to the basis of your choice.

So, if we choose, for example the discrete basis of "simple" polynomials

p_n(x) = xⁿ

(note that these are not in our Hilbert space (since they are not square integrable)), then relative to this basis we will have a matrix representation for L. (... Do you know how to do this? (If not, and you'd like to, please ask.))

On, the other hand, we could instead choose a continuous basis, indexed by a continuous parameter x', say

D_x'(x) = Dirac_delta(x-x') .

This is just the usual |x> basis (in its own x'-space); i.e. <x'|x>.

Or, we could choose a different continuous basis, indexed by a continuous parameter k, say

h_k(x) = 1/sqrt{2 pi} e^ikx .

Relative to this basis, our operator L turns out to be diagonal.

(... Again, if you are unsure as to how to get the matrix representation relative to a given basis for a linear operator, and would like to know more, feel free to ask.)

 

[reilly]

Eye in the Sky has it right; the exponential of a derivative operator is a Taylor Series. This idea has its major home in the theory of Lie Groups,and/or continuous groups, where one talks about generators of infitesimal displacements. In physics, it used to be, at least, that the initial introduction to the ideas of generalized translations came from the study of rigid body mechanics and the theory of contact transformations. These studies helped to build familiarity with some of the more sophisticated ideas that often were reintroduced in QM in the study of angular momenta. In particular, the theory of and representations of finite rotations are of signal importance in both nuclear and particle physics -- one example is the great practicality of the Jacob and Wick helicity representions -- spin is quantized along the particle's direction of motion.
It get's worse: the Poincare group has ten generators, 3 momenta, one energy, 3 angular momenta, and 3 Lorentz boosts -- there's a good discussion of this in Chapter II of Weinberg's The Quantum Theory of Fields.

I'm very curious about the background's of those having difficulties, or providing episodic essays on the generator and exponential form of spatial translations. I say this with all due respect, but given the "right background" -- i.e knowing something about continuous groups, knowing about rotations in advanced mechanics, particularly involving contact transformations, the idea of a derivative as the generator of translations becomes quite straightforward.

I'm an old guy, trained in the '50s. I see things as I learned them and taught them. Do students these days do Hamiltonian mechanics, rigid body mechanics, E&M radiation theory, elementary group theory. ... prior to a first course in QM?

Thanks and regards,
Reilly

 

[Haelfix]

Typically it goes like so. Elementary mechanics/EM.. then

Mechanics (this includes both hamiltonian, and rigid bodies)
EM (including circuits)
Optics (sometimes optional)
Thermodynamics
then quantum mechanics.
and special/General relativity

Two problems with the standard curriculum.

One, very often very important stuff is left out for lack of time. Rigid body mechanics, Maxwell wave mechanics, Poynting vectors etc were all taught to me in a rush, and I had to relearn them later in my career.

The other problem is the math.

Unfortunately elementary group theory is usually not a prereq or even taught outside of the math divisions. Nor is some of the other stuff, like linear algebra and more importantly, differential geometry and functional analysis (much less real analysis).

Its actually IMO a problem for undergrad classes, you really, really want to to have the math background first. I hated learning physics math, on the fly. It made little sense to me at the time, was adhoc, often wrong or grossly simplified. In the math divisions, though, you often spend too much time 'proving' things, which is more or less irrelevant for an undergrad.

Everyonce and awhile I would have a good teacher, who would focus on calculating examples, without skipping too much key background definitions
, and giving a good feel to the students. Differential geometry was such an example. It made GR a breeze for me. Ditto with statistics, Stat mech was then completely trivial.

Unfortunately, I did not have functional analysis before quantum mechanics, and frankly the whole experience was really painful the first time around. I got a good grade, but I didn't learn anything really.

 

[alexepascual]

Response to Turin and Eye_in_the_Sky

Turin and Eye,
I haven't done anything with Fourier transforms in many years. It was silly of me to think you can get a FT of x².
I just checked the book and of course, the function has to obey the Dirichlet conditions and have a finite integral, which just makes sense.
So with this Turin, you have not revealed any ignorance (but I have). And that's OK with me. I am not claiming certain level of knowledge or intelligence. I am in this forum to learn and to help if someone ask an easy question.

Eye,
With respect to one of your previous posts, I don't think of myself as having any out-of-the-ordinary skills in visualization. I do try to look at problems from a different angle, but that is just because of my defficiencies in memory and symbol-processing. I have these handicaps, and I have to look for ways to absorb and interpret the subject that will produce long-lasting neuronal connections.

Now with respect to post #27:
First I would just like to comment that I noticed that even you can't get a FT of x², you can still move it by using the Taylor series.
With respect to your derivation of the generator of translations, I did understand well how you got this:
$$f(x + a) < - - > G(k) = e^{ika} \,F(x)\,dx \\$$

What I don't understand is how you get [3] and how you compare [4] to [3]

With respect to my statement:

This would give the appearence of certain "diagonality"

I would drop this, not necessarily retract it as I think the reason for the apparent inconsistency is that I was not able to communicate clearly. But this was just something I mentioned in passing and it doesn't boder me.
On the other hand, if it is not too much trouble, I would like to hear more about the "simple polinomials". Are we talking about polinomials of the form
a₀x⁰+a₁x¹+a₂x²+...a_nxⁿ..
where each xⁿ is a basis for the space?
Would in that case the operator d/dx consist of a line of numbers n-1 parallel to the diagonal? Am I on the right track?

 

[alexepascual]

Response to Reilly and Haelfix

Reilly,
I would agree with most of what Haelfix has said.
With respect to your position, I get the impression you propose that very complex and abstract math should be learned first. I see two problems with that.
In the first place, different people may have different modes of learning. The sequence you might propose (that would probably be close to the sequence of classes and topics you took) might work for some people and not for others. The very abstract concepts may not be digested very well by people that prefer to see concrete examples and then abstract from them.
For these people, the math might be understood better within the concept of physics. Probably that's why they usually have a course entitled "mathematical methods"
The other problem I see with the approach of learning very well all the math first is that if you don't apply it you forget it, and there might be quite a gap in time from learning the mathematical concepts to applying them to physical problems.
On the other hand I think physics courses in most schools are not very well organized and they don't guarantee that a minimum of simple math is learned before it is needed. For example, when I took quantum mechanics, there was no pre-requisite to study linear algebra before taking this class. I had studied linear algebra but we had not covered the part that uses complex numbers. So I had to study this on my own. I was never tought group theory, but I think most QM courses at the undergraduate level don't require it.
Now I'll be taking graduate-level courses, so I'll make sure I do understand these concepts.
Reilly and Haelfix,
Your discussion has helped me see some of the areas in mathematics where I may need more knowledge. I thank you for that.

 

[Eye_in_the_Sky]

Response to Alex

I just checked the book and of course, the function has to obey ...

A sufficient, but not necessary, condition for the Fourier transform of a function f(x) to exist is

(i) Integral_{[all space]} { |f(x)| dx } converges ;

(ii) f(x) has a finite number of discontinuities .

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

... you can't get a FT of x², you can still move it by using the Taylor series.

Yes, absolutely.

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

What I don't understand is how you get [3] and how you compare [4] to [3].

Recall:

[1] f(x) = 1/sqrt{2 pi} Integral { e^ikx F(k) dk } ,

[2] F(k) = 1/sqrt{2 pi} Integral { e^-ikx f(x) dx } .

Now, apply d/dx to equation [1], and get the correspondence

[3] [(d/dx)f](x) <--> ik F(k) .

To get [3], apply d/dx to both sides of [1], and on the right-hand-side, push d/dx "through" and "under" the integral; the only function under the integral there which depends on "x" is e^ikx, and d/dx of that is just the same thing multiplied by "ik"; thus, [1] becomes

df(x)/dx = 1/sqrt{2 pi} Integral { e^ikx ik F(k) dk } .

From this, [3] follows.

Now, [3] tells us that the "image" in k-space of d/dx is just multiplication by ik. It then follows that for any (analytic) function h(x), the "image" in k-space of h(d/dx) is just h(ik). We therefore have the correspondence

[e^a(d/dx)f](x) <--> e^ika F(k) .

Now, take [4] and compare it to this. We had

[4] f(x+a) <--> e^ika F(k) .

The left-hand-sides of the last two correspondences must be equal; i.e.

f(x+a) = [e^a(d/dx)f](x) .

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

Finally, regarding:

... I would like to hear more about the "simple polinomials". Are we talking about polinomials of the form
a₀x⁰+a₁x¹+a₂x²+...a_nxⁿ..
where each xⁿ is a basis for the space?
Would in that case the operator d/dx consist of a line of numbers n-1 parallel to the diagonal? Am I on the right track?

"Yes" (where the full set {xⁿ|n=0,1,2,...} is a basis), "yes", and "yes". At the next opportunity, I will attempt to post more on this matter.

 

[alexepascual]

I think I see the source of my difficulty.
You derive [3] by differentiating both sides of [1] (which is an equation, not a correspondence). Don't you get an equation when you take the derivative with respect to the same variable of both sides of an equation?. Shouldn't [3] be an equality instead of a correspondence?
I hope I understood correctly the meaning of correspondence in this context. I would have interpreted it as "is a Fourier transform of" which would make:
f(x) <-->F(k)
f(x+a)<-->G(k)=e^ikaF(k)
On the other hand your argument on your last post ends up in a compelling way when you compare [3] to [4], so at this point I am a little confused.

 

[Eye_in_the_Sky]

Don't you get an equation when you take the derivative with respect to the same variable of both sides of an equation?

Yes. In our case, the equation was just:

df(x)/dx = 1/sqrt{2 pi} Integral { e^ikx ik F(k) dk }

Call this equation [5].

------

Shouldn't [3] be an equality instead of a correspondence?

No. And you have given the reason why this is so:

I hope I understood correctly the meaning of correspondence in this context. I would have interpreted it as "is a Fourier transform of" which would make:
f(x) <-->F(k)
f(x+a)<-->G(k)=e^ikaF(k)

You understood correctly:

"-->" means "is the Fourier transform of" ;

"<--" means "is the inverse Fourier transform of" .

Looking at equation [5] above with this understanding of "<-->" gives us the correspondence:

[3] [(d/dx)f](x) <--> ik F(k)

Does this help to clear up the confusion?

 

[alexepascual]

I guess I'll have to think about it.
Thanks a lot Eye. I won't be home today, probably tomorrow I'll be able to look into this.