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Classical -> Quantum Generator of Translation

  1. Jun 5, 2008 #1
    Hi all, I've posted a little bit here in the past but I don't think anyone's going to really remember me from those. I hope to come back more frequently now, especially since I've now learned how to use the TEX feature. There was a post with members asking somewhat similar questions a couple months back but not many answers were given, so I apologize for starting a new post rather than reviving the old one.

    Anyhow, on the most recent discussions in the following thread on sciforums.com, at http://www.sciforums.com/showthread.php?t=78957&page=10, we're trying to derive the momentum operator as the generator of spatial translations. In classical Hamiltonian mechanics we know that choosing p, the canonical momentum, as the generator of an infinitesimal canonical transformation leads to an infinitesimal translation of the system in space. The official result for how functions of x change due to this transformation (in 1 dimension) is given by [tex]f(x+dx)=f(x)+dx\{f,p\}=f(x)+dx\frac{\partial f}{\partial x}[/tex].
    Here, [tex]\{u,v\}[/tex] refers to the Poisson bracket, which is evaluated in general, for 1-dimensional systems, as [tex]\{u,v\}=\frac{\partial u}{\partial x}\frac{\partial v}{\partial p}-\frac{\partial u}{\partial p}\frac{\partial v}{\partial x}[/tex]. The correspondence between classical Poisson brackets and quantum commutators is [tex][u,v]=i\hbar\{u,v\}[/tex].

    The Hamiltonian can be used in an infinitesimal canonical transformation as the classical generator of inifinitesimal time evolution, and this method carries straight over from Poisson brackets into QM. Similarly, can anyone show me a way to carry the classical generator of translations over to QM? I.e. can anyone here derive from the correspondence principle that the the infinitesimal translation operator, [tex]\mathcal{T}(dx)[/tex] is to be represented in terms of the canonical momentum as [tex]1-\frac{i}{\hbar}\hat{p}dx[/tex]?
    Last edited: Jun 5, 2008
  2. jcsd
  3. Jun 5, 2008 #2
    It seems you have already done that. Take [itex] f(x+dx)=f(x)+dx\{f,p\}=f(x)+dx\frac{\partial f}{\partial x}[/itex] and perform the substitution [itex] [u,v]=i\hbar\{u,v\} [/itex]

    The thing is though, that quantum mechanics isn't "derived" from classical mechanics. The correspondence principle is just a sanity check, but it has exceptions. (When the [tex]\hbar \to 0[/tex] limit can't safely be taken.) Quantum mechanics is more "true" than classical mechanics, so just accept its axioms and derive classical mechanics from that.

    The point of the Poisson brackets is just to massage derivatives out of fundamental objects that are easily manipulated. You can say the same thing in quantum mechanics. As far as generators go, the real object is the Lie bracket and you're moving around on a manifold using the Lie bracket.

    I have some notes on how the Poisson bracket arises from quantum mechanics: http://www.physics.thetangentbundle.net/wiki/Quantum_mechanics/Poisson_bracket
  4. Jun 5, 2008 #3
    Thanks for your help. I went to the link you gave me and read through a couple of articles. What it seems you're showing, if I'm correct, is that there's a correspondence between the Lie algebra of quantum operator commutators and the Lie algebra of Poisson bracket derivatives, with the correspondence becoming exact as [tex]\hbar\mapsto 0[/tex] so that higher order powers of [tex]\hbar[/tex] vanish. Would that be correct?

    If my understanding is correct though, I don't see how this proves that momentum is the quantum generator of translations. Say I fortuitously choose to represent [tex]\hat{p}\mapsto -i\hbar\frac{\partial}{\partial x}[/tex] in the [tex]\left|x\right\rangle[/tex] basis, then I can both get the correct commutation relations and derive that momentum is the quantum generator of translations. I can then also check and see that my quantum commutator [tex][x,p][/tex] corresponds to the classical Poisson bracket [tex]\{x,p\}[/tex] if I multiply it by [tex]i\hbar[/tex]. However, my biggest problem is I don't see how this shows that there's a unique choice of representation for [tex]\hat{p}[/tex] in the [tex]\left|x\right\rangle[/tex] basis to get the right commutation relations, and if there is more than one choice, then I don't see why a priori it would be required to generate translations.

    I'm trying to piece together the historical development of QM and the info doesn't seem to be all that widespread, so my own understanding is rather vague. If I understand right, however, Dirac noticed the connection between the time evolution of operators in QM and the use of the classical Hamiltonian as the generator of infinitesimal time translations. So I'm guessing he must have also noticed or postulated a similar analogy between classical and quantum generators of various other isometries.

    As you agree, classically we can take [tex]p[/tex] to be the generator of an infinitesimal canonical translation, and we get [tex]f(x+dx)=f(x)+dx\{f,p\}[/tex]. However, if I simply substitute [tex]\{f,p\}\mapsto\frac{[f,\hat{p}]}{i\hbar}[/tex], I can't make any sense of this expression. We have [tex]\hat{p}[/tex] as an operator and [tex]f[/tex] as a function, so I don't know how to define such a commutator. If I try to replace [tex]f(x)[/tex] with the ket [tex]\left|x\right\rangle[/tex], then I get really close to what I'm looking for, but I end up with [tex]\big[\left|x\right\rangle,\hat{p}\big][/tex], and I don't know how to make any sense of such a commutator either. Finally, if I choose [tex]f[/tex] to be an operator [tex]\hat{f}[/tex], then the commutator makes sense but I can't make sense of the expression [tex]\hat{f(x)}[/tex], since I thought all quantum operators depended exclusively on time. Does this sound reasonable to you, am I making a mistake somewhere in my reasoning, or is this approach a total dead end?
  5. Jun 5, 2008 #4
    The fact that [tex][x,p]=i \hbar[/tex] has a unique representation (up to unitary transformations) is guaranteed by the Stone–von Neumann theorem, which I won't go into (wikipedia).

    The theorem breaks down for systems with infinitely many degrees of freedom, such as quantum field theory. There people get all hickuppy but I think renormalization solves the problem also.
  6. Jun 5, 2008 #5
    Thanks again, now I think this makes much more sense to me. So am I right then to say that there is pretty much only one way, up to unitarity, to represent [tex]\hat{p}[/tex] in the [tex]\left|x\right\rangle[/tex] basis, in order to get the correct commutation relations? And if this is indeed the case, are you then saying that there's no connection between momentum as the generator of infinitesimal canonical translations in classical mechanics and momentum as the generator of infinitesimal translations in QM? As in, the only direct connection between classical and QM generators is for time evolution, and for obtaining the commutation relations?

    Put another way, is there no QM analogue to [tex]f(x+dx)=f(x)+dx\{f,p\}[/tex]? If there is, please show me or at least refer me to somewhere I can look it up, because I couldn't get a sensible expression when I tried to switch from Poisson brackets to QM commutators. What I'd like to be able to derive by such an analogy, ideally, would be the relation [tex]\left|x+dx\right\rangle=\left(1-\frac{idx}{\hbar}\hat{p}\right)\left|x\right\rangle[/tex].

    I don't want to make you run in circles, I think most of my confusion is now cleared up about what comes first. In all the QM classes I've taken so far, including graduate Quantum Theory (1 semester), they just give us this magic translation operator and tell us how to express it in terms of momentum. The only justification they give is that the resulting operator, [tex]\hat{p}[/tex], has eigenvalues with the same dimensions as momentum, therefore we're supposed to just take it for granted. Where can I learn the detailed foundations of QM? Do you cover it when you learn QFT, or is there some sort of advanced class for this my university might not be offering, or is it just independent study?
    Last edited: Jun 5, 2008
  7. Jun 9, 2008 #6
    Ok, I think I've figured something out, though I'm sure it's not the complete picture and I have tons of reading left to do.

    Classically, we know that Poisson brackets can be used with momentum to generate infinitesimal spatial tranlations via the relation [tex]f(x+dx)=f(x)+dx\{f,p\}=f(x)+dx\frac{\partial f}{\partial x}[/tex]. Now for the quantum analogue, we're not working with functions, but with operators. Just as Heisenberg defined state vectors as being stationary and the operators themselves as being actively transformed, this is the approach I'm going to take in making the transition from classical to quantum mechanics.

    So let's postulate an operator [tex]\hat{u(dx)}[/tex] such that [tex]\left\langle x'\right|\hat{u(dx)}\left|x\right\rangle[/tex]=[tex](x+dx)\left\langle x'\right|x\left\rangle=(x+dx)\delta(x'-x)=(x'+dx)\left\langle x'\right|x\left\rangle=(x'+dx)\delta(x'-x)[/tex], with [tex]\hat{u(0)}=\hat{x}[/tex].

    By classically analogizing this operator to the functions in Poisson brackets, with the association [tex]\{a,b\}\mapsto\frac{[a,b]}{i\hbar}[/tex], we obtain the equation [tex]\hat{u(dx)}=\hat{u(0)}+dx\frac{[\hat{u(0)},\hat{p}]}{i\hbar}=\hat{x}+dx\frac{[\hat{x},\hat{p}]}{i\hbar}[/tex]

    Now naively, we can look at the classical Poisson bracket [tex]\{x,p\}=1[/tex] to infer that the quantum commutator should be [tex][\hat{x},\hat{p}]=i\hbar[/tex], leaving us with [tex]\hat{u(dx)}=\hat{x}+dx[/tex], which is trivial because we know that [tex]\hat{u(dx)}[/tex] adds a quantity [tex]dx[/tex] to every eigenvalue of the operator [tex]\hat{x}[/tex] operating on the [tex]\left|x\right\rangle[/tex] basis. However, we may instead be interested in how the active transformation acting on the operator [tex]\hat{x}[/tex] can be considered instead as a passive transformation on the set of basis states [tex]\left|x\right\rangle[/tex].

    So rather than calculating the commutator [tex][\hat{x},\hat{p}][/tex] in advance and being done with it, let's expand it out. We get:

    [tex]\hat{u(dx)}=\hat{x}+dx\frac{\hat{x}\hat{p}-\hat{p}\hat{x}}{i\hbar}[/tex], which we can then rearrange, neglecting terms of [tex]\mathcal{O}(dx^2)[/tex], to get [tex]\hat{u(dx)}=(1+\frac{i}{\hbar}dx\hat{p})\hat{x}(1-\frac{i}{\hbar}dx\hat{p})[/tex]. Now this looks much more like the generators we're used to in quantum physics! If we make use of the known commutator [tex][\hat{x},\hat{p}]=i\hbar[/tex], we find that [tex]\hat{x}(1-\frac{i}{\hbar}dx\hat{p})\left|x\right\rangle=(x+dx)(1-\frac{i}{\hbar}dx\hat{p})\left|x\right\rangle[/tex], so we know that [tex](1-\frac{i}{\hbar}dx\hat{p})\left|x\right\rangle=e^{i\gamma}\left|x+dx\right\rangle[/tex]. Taking the adjoint of this equation yields the expected action on the space of [tex]\left\langle x\right|[/tex] vectors. Thus (for 1 dimension) we may define our infinitesimal translation operator acting on kets as [tex]\hat{\mathcal{T}(dx)}=1-\frac{i}{\hbar}dx\hat{p}[/tex], and the adjoint of this operator acts on bras. From there, if we choose it as our starting point, we can then work backwards and derive that [tex][\hat{x},\hat{p}]=i\hbar[/tex], and we can also derive that, as the generator of translations, [tex]\hat{p}[/tex] must have the representation [tex]-i\hbar\frac{\partial}{\partial x}[/tex].

    So what do you think? Does this argument sound reasonable and logical? If so, you still might not find it a useful or practical method for introducing the operator [tex]\hat{p}[/tex] into the bra and ket formalism, but it would at least show one of the many connections between quantum physics and classical Hamiltonian dynamics. I suppose another approach is to go with the Stone-Von Neumann approach as suggested by lbrits, and I'm definitely interested in reading up on it.

    Any comments would be appreciated, I feel like I might finally have this monkey off my back, in which case I can go back to sciforums.com and pass on the knowledge.
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