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ShayanJ

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ShayanJ

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ShayanJ

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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.

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

- #5

strangerep

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+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.

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ShayanJ

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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!

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DrDu

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ShayanJ

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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.

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DrDu

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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

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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

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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?

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ShayanJ

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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

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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.

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DrDu

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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].

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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.

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ShayanJ

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DrDu

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

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ShayanJ

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DrDu

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

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

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

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

- #23

ShayanJ

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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!

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DrDu

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##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.

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DrDu

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