- #1

ShayanJ

Gold Member

- 2,809

- 604

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter ShayanJ
- Start date

- #1

ShayanJ

Gold Member

- 2,809

- 604

- #2

- 202

- 178

- #3

ShayanJ

Gold Member

- 2,809

- 604

But even if we take that into account, then we should note that angular variables don't represent a constant spread in length for all distances from origin. For example for a constant [itex] \Delta \varphi [/itex], if [itex] R_1 > R_2 [/itex] then [itex] R_1 \Delta \varphi>R_2 \Delta \varphi [/itex] so [itex] \Delta \varphi [/itex] alone can't represent the uncertainty in position! But by using the angle operator alone in the commutation relation, the corresponding uncertainty relation will be in terms of spread in angle only.

Last edited:

- #4

- 13,194

- 764

I don't find anything in that article related to uncertainty relations of any kind.

- #5

strangerep

Science Advisor

- 3,319

- 1,387

+1 on that.I don't find anything in that article related to uncertainty relations of any kind.

@Shyan: the paper is talking about generalized commutation relations. Given these, one derives uncertainty (aka indeterminacy) relations. Cf. Ballentine section 8.4.

- #6

ShayanJ

Gold Member

- 2,809

- 604

I don't find anything in that article related to uncertainty relations of any kind.

That's exactly what I mean. If we say e.g. [itex] [\varphi,m r^2 \dot \varphi]=i\hbar [/itex], then we'll have an uncertainty relation [itex] \Delta \varphi \Delta (mr^2 \dot\varphi) \geq \frac \hbar 2 [/itex]. What I'm saying is a particular value of [itex] \Delta\varphi [/itex] alone can't give us enough information about uncertainty in position, because such an angle uncertainty gives a larger length uncertainty the further we are from the origin. So if we have points 1 and 2 such that [itex] r_1>r_2 [/itex] and we somehow have a constant amount of [itex] \Delta\varphi [/itex], we know that the particle will have a larger amount of uncertainty in its position if its in position 1 than it had if it was in position 2 because [itex] r_1 \Delta \varphi> r_2 \Delta \varphi [/itex], so I think this should somehow be taken into account in the commutation relations so that the resulting uncertainty relations take this into account too.@Shyan: the paper is talking about generalized commutation relations. Given these, one derives uncertainty (aka indeterminacy) relations. Cf. Ballentine section 8.4.

Anyway, despite my argument, the response of you science advosors makes me think I'm missing something here. So I'm ready to hear!

- #7

DrDu

Science Advisor

- 6,235

- 886

- #8

ShayanJ

Gold Member

- 2,809

- 604

I came to this paper by accident and because it seemed to me that [itex] \vec p=\frac \hbar i \vec \nabla [/itex] is well established and also I saw the mentioned problem with the paper, I came here to ask others' ideas about this paper. Actually I didn't read the paper, I just glanced at the equations.

The main point is, the guy is either right or wrong. If he's wrong, then there should be some reason and I want to find it out because it seems to me it doesn't need a lot of time and effort so its OK. But if we can find no reason to say he's wrong, then maybe he has a point!

Can you mention some papers or books that actually prove [itex] \vec p=\frac \hbar i \vec \nabla [/itex] is OK in general? I always thought its somehow trivial but now I think its good to see at least a hand wavy proof.

Last edited:

- #9

DrDu

Science Advisor

- 6,235

- 886

- #10

- 12,574

- 4,937

And yet, you can measure the angle by measuring the position, which, in turn, is an operator. Or is it? How would you interpret the idea that in quantum mechanics you can measure something which is not described by an operator?an angle operator does not exist

- #11

ShayanJ

Gold Member

- 2,809

- 604

The important point here is that, in spherical coordinates, such a translation only involves the r coordinate and in cylindrical coordinates it only involves [itex] \rho [/itex] and z coordinates. So this way it seems that we get no canonical momentums conjugate to angle coordinates and so they aren't on an equal footing with the ones conjugate to distance coordinates.

- #12

DrDu

Science Advisor

- 6,235

- 886

Ok, so let us introduce an angle operator as ##\phi=\mathrm{atan2}(x,y) ## this operator will jump from pi to -pi at when x becomes negative for negative y. We already know the correct angular momentum operator to be ##-i (x\partial/\partial y -y \partial/\partial x)##, so this will introduce a delta function into the commutation relations, which are therefore no longer of the canonical form we were assuming.And yet, you can measure the angle by measuring the position, which, in turn, is an operator. Or is it? How would you interpret the idea that in quantum mechanics you can measure something which is not described by an operator?

Last edited:

- #13

ShayanJ

Gold Member

- 2,809

- 604

He then calculates [itex] [\hat L_z,\hat \Upsilon]=\hbar \hat \Upsilon [/itex] and [itex] \hat \Upsilon[/itex]'s matrix elements which makes clear that it is actually a raising operator, i.e. [itex]\hat \Upsilon=C_+^{-1} \hat L_+ [/itex] with [itex] C_+=\sqrt{(l-m_l)(l+m_l+1)} [/itex].Levin said:The fact that ## \hat\Upsilon ## is unitary rather than Hermitian is not a problem, since the angle is not an observable.

- #14

DrDu

Science Advisor

- 6,235

- 886

https://www.physicsforums.com/threads/follow-on-to-is-the-time-derivative-hermitian.792270/

If we take the angle operator to be a generator of shifts of the angular momentum, we have to take into account that angular momentum is quantized. However, this good, as ##\exp (im\phi)## will only transform a continuous function into a continuous function if m is integer.

So I think we can write some Weyl commutation relations ##UV=VU\exp(i\alpha m)## where ##U=\exp(i\alpha L_z)##, ##V=\exp(i m \phi)## with alpha real and m integer. For non-integer values of m, the Weyl commutation relations won't hold.

- #15

DrDu

Science Advisor

- 6,235

- 886

So one could define ##\Upsilon## in terms of ##L^2##, ##L_z##, and ##L_+##, although the representation is somewhat nasty.He then calculates [itex] [\hat L_z,\hat \Upsilon]=\hbar \hat \Upsilon [/itex] and [itex] \hat \Upsilon[/itex]'s matrix elements which makes clear that it is actually a raising operator, i.e. [itex]\hat \Upsilon=C_+^{-1} \hat L_+ [/itex] with [itex] C_+=\sqrt{(l-m_l)(l+m_l+1)} [/itex].

- #16

- 12,574

- 4,937

So thereOk, so let us introduce an angle operator as ##\phi=\mathrm{atan2}(x,y) ## this operator will jump from pi to -pi at when x becomes negative for negative y. We already know the correct angular momentum operator to be ##-i (x\partial/\partial y -y \partial/\partial x)##, so this will introduce a delta function into the commutation relations, which are therefore no longer of the canonical form we were assuming.

- #17

ShayanJ

Gold Member

- 2,809

- 604

- #18

DrDu

Science Advisor

- 6,235

- 886

Why should it be proportional to the identity? However, it should commute with L^2 and L_z.

- #19

ShayanJ

Gold Member

- 2,809

- 604

- #20

DrDu

Science Advisor

- 6,235

- 886

Ok, but this ##C_+## is an operator, too, which can be expressed in terms of L^2 and L_z.

- #21

- 20,106

- 10,846

$$\hat{\vec{p}}=-\mathrm{i} \vec{\nabla},$$

you have an coordinate-independent description. You can express the nabla operator in any curvilinear coordinates you like. It's a vector operator and thus independent of the choice of coordinates!

- #22

DrDu

Science Advisor

- 6,235

- 886

Yes, you were right about that.So thereisangle operator, but it does not satisfy the canonical commutation relation. Is that right?

- #23

ShayanJ

Gold Member

- 2,809

- 604

I don't think that's true! If ##C_+## was an operator, then Levin couldn't say ## \Upsilon ## is a raising operator because ## C_+ ## could do things which make ## \Upsilon ## very different from ## L_+ ##!Ok, but this ##C_+## is an operator, too, which can be expressed in terms of L^2 and L_z.

Also ## C_+=\sqrt{(l-m_l)(l+m_l+1)} ## and ## l \ and \ m_l ## are numbers!!!

I don't see how ## C_+ ## can be an operator!

Last edited:

- #24

DrDu

Science Advisor

- 6,235

- 886

##1=\Upsilon^\dagger \Upsilon=L_-A^\dagger A L_+=AL_+L_-A^\dagger =\Upsilon \Upsilon^\dagger=1##.

Now ##L_+L_-=L_x^2-i(L_xL_y -L_yL_x)+L_y^2=L^2-L_z^2+L_z## and ##L_-L_+=L^2-L_z^2-L_z##. Now multiply from the right with A and from the left with ##A^\dagger##: ##A^\dagger A=A^\dagger AL_+L_-A^\dagger A##, so we get ##A^\dagger A=(L_+L_-)^{-1}##.

More later.

- #25

DrDu

Science Advisor

- 6,235

- 886

- #26

ShayanJ

Gold Member

- 2,809

- 604

I'm confused because there are several things I don't understand:

1) Here you're assuming that ## \Upsilon## makes a transition from ## m_z ## to ##m_z+1##, like ## L_+ ##. But if we say that ## \Upsilon =A L_+ ## where A is an unknown operator we're trying to find, then we can't be sure of the first sentence unless we assume A is a number, not an operator!!!(Or an operator that commutes with ##L_+##?) So I should ask how else can you know that ## \Upsilon ## acts like ## L_+ ##? Are you assuming

2) I understand the calculation in your last post and it seems A is actually an operator but how can you reconcile it with the fact that ## A=\left[(l-m_l)(l+m_l+1)\right]^{-\frac 1 2}## ?

3) I somehow know the answer to the first question but the problem is its in contradiction with A being an operator. The point is, as Levin explains, the commutation relation ## [L_z,\Upsilon]=\hbar \Upsilon ## implies that the matrix elements ## \Upsilon_{m_l' m_l} ## are non-zero only when ## m_l'=m_l+1 ##. This way Levin concludes that ## \Upsilon ## is a raising operator

- #27

DrDu

Science Advisor

- 6,235

- 886

I'm confused because there are several things I don't understand:

1) Here you're assuming that ## \Upsilon## makes a transition from ## m_z ## to ##m_z+1##, like ## L_+ ##. But if we say that ## \Upsilon =A L_+ ## where A is an unknown operator we're trying to find, then we can't be sure of the first sentence unless we assume A is a number, not an operator!!!(Or an operator that commutes with ##L_+##?) So I should ask how else can you know that ## \Upsilon ## acts like ## L_+ ##? Are you assuming

2) I understand the calculation in your last post and it seems A is actually an operator but how can you reconcile it with the fact that ## A=\left[(l-m_l)(l+m_l+1)\right]^{-\frac 1 2}## ?

3) I somehow know the answer to the first question but the problem is its in contradiction with A being an operator. The point is, as Levin explains, the commutation relation ## [L_z,\Upsilon]=\hbar \Upsilon ## implies that the matrix elements ## \Upsilon_{m_l' m_l} ## are non-zero only when ## m_l'=m_l+1 ##. This way Levin concludes that ## \Upsilon ## is a raising operatorbut this can only mean that ## \Upsilon=D L_+ ## where D is a constant number!(Or any operator that commutes with ## L_+ ##, but Levin seems to accept that D is a constant number!)

1. is a consequence of the line of thought in post #14, i.e. the commutation relations which only hold for integer values of m.

2. This are exactly the eigenvalues of the operator A I indroduced in the basis ##|l,m_z\rangle##. In my last post, I showed that A is diagonal in that basis.

3. I can't follow your conclusion, and I think you yourself can't either. I suppose you can work out yourself how B must look like if ##AL_+=L_+B##.

- #28

ShayanJ

Gold Member

- 2,809

- 604

OK, let's get back to the beginning. You said that because ## \Upsilon ## is a raising operator, then it should be equal to ## A L_+ ## where A is diagonal.(Is it what you said?) But I'm saying this is not the only possibility because we may have ## \Upsilon=BL_+ ## where ## [B,L_+]=0 ##. This way surely ##\Upsilon## will be a raising operator too. How can you exclude this case?1. is a consequence of the line of thought in post #14, i.e. the commutation relations which only hold for integer values of m.

2. This are exactly the eigenvalues of the operator A I indroduced in the basis ##|l,m_z\rangle##. In my last post, I showed that A is diagonal in that basis.

3. I can't follow your conclusion, and I think you yourself can't either. I suppose you can work out yourself how B must look like if ##AL_+=L_+B##.

- #29

strangerep

Science Advisor

- 3,319

- 1,387

Since you're necessarily working in a particular Hilbert space here, I suspect it's only reasonable to expect the identity when sandwiched between the eigenstates. Cf. look at Levin's eq(10.99), i.e.,

This holds on this Hilbert space, but clearly ##\hat L_x \ne 0## if considered as an operator in isolation. I.e., one must beware of the distinction between "strong" and "weak" properties of operators -- "weak properties" are those which hold with respect to matrix elements on a specific Hilbert space. (This is actually a very important distinction in advanced QM and QFT.)Levin said:##\langle n,\ell,m_\ell| \hat L_x |n,\ell,m_\ell\rangle

~=~ \langle n,\ell,m_\ell| \hat L_y |n,\ell,m_\ell\rangle ~=~ 0 ~. ~~~~~~~~ (10.99)##

To amplify this point, note that Levin's relation ##\hat\Upsilon = C^{-1}_+ \hat L_+## was inferred by looking the matrix-element equation (10.115).

Dirac used a different equality sign to denote weak equality (though that was in the context of constrained classical dynamics). It would make some parts of quantum theory clearer if that same distinction were adopted to express strong and weak equality of operators.

To check all this, you could try evaluating your expression for ##L_+ L_+^\dagger## between such eigenstates. Of course, it would be messy and error-prone to work with all those spherical harmonic functions, etc.

BTW, this highlights the problems that can arise when one assumes that the CCRs can be represented on a finite-dimensional Hilbert space. Sometimes the

All this has its beginnings back in classical Hamiltonian dynamics: for a integrable system, it's often useful to express the system in terms of (so-called) generalized action-angle variables, which satisfy an affine-like Poisson bracket relation instead of the canonical Poisson bracket.

It's interesting that some things become easier if we attempt to quantize such classical systems by reference to their affine Poisson brackets, instead of the canonical brackets. In the present case, it allows us to bypass the multi-valuedness issues that an angle variable like ##\phi## introduces. Use of ##e^{i\phi}## locks such inconveniences inside the exponential, and we can get on with our business...

Last edited:

- #30

strangerep

Science Advisor

- 3,319

- 1,387

The authors apparently intend more general cases of canonical variable pairs. See, e.g., their brief remark about the EM minimal coupling case. Although I didn't get much out of this particular paper for myself, there have certainly been examples in the past where I thought a paper was total rubbish, but later realized this merely reflected my own embarrassing lack of understanding.

Share: