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Position operator in momentum space generator

How to get the position operator in the momentum representation from knowing the momentum operator in the position representation? derived the position operator in momentum space using commutators.

I want to extend the analogy of equivalence between position and momemtum space. In position space, we say momentum is the generator of translations. This can potentially suggest the form of the momentum operators. Is there a generator of some kind of infinitesimal symmetric transformation in momentum space that, when repeated several times, gives you position?

In other words, consider the statement "momentum is the generator of translation in position space." I want to know if there is an equivalent statement saying "X is the generator of translation in momentum space" i.e. X is a generator of momentum change. I tentatively say this has something to do with force because in Newtonian physics, force changes momentum. But I also see two reasons to think that this new generator should actually be something to do with position:

1 the position operator in momemtum psace looks very similar to the momentum operator in position space.

2 momentum and position are supposed to be connected deeply because wavenumber k space is the Fourier transform of position. It is supposed to be equally informative to work in both position and momentum space.

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Yes, there is. Let $\psi$ be a position-space wavefunction. We can always write $$\psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int \mathrm dp \ e^{ipx/\hbar} \tilde \psi(p)$$

where $\tilde \psi$ is the corresponding momentum-space wavefunction. A momentum-space translation $p\mapsto p+q$ looks like this: $$\psi(x) \mapsto \frac{1}{\sqrt{2\pi\hbar}}\int \mathrm dp \ e^{ipx/\hbar}\tilde \psi(p+q) = \frac{1}{\sqrt{2\pi\hbar}} \int \mathrm dp \ e^{i(p-q)x/\hbar} \tilde \psi(p) = e^{-iqx/\hbar} \psi(x)$$ which is straightforwardly implemented by the unitary momentum-translation operator $U_q :=e^{-iq\hat X/\hbar}$; the generator of such translations is by definition $$ \frac{\hbar}{i} \lim_{q\rightarrow 0} \frac{U_q - \mathbb I}{q} = -\hat X$$

i.e. the generator of spatial translations is $\hat P$, and the generator of momentum translations is $-\hat X$, validating your suspicions.


I tentatively say this has something to do with force because in Newtonian physics, force changes momentum.

This is an easy mistake to make. However, when we say "momentum is the generator of spatial translations," we're not talking about dynamics; that is, we're not saying "if you run time forward, momentum is what causes position to change."

Rather, is a deep statement specifically about Hamiltonian mechanics. As a light-speed overview, in Hamiltonian mechanics any differentiable function $F$ of the phase space variables gives rise to a vector field $\mathbf X_F$. Such a vector field has integral curves, which you might also call field lines in electrostatics. From there, we can define a map $\Phi_F^\lambda$ (called the flow generated by $F$) which takes a point $(x,p)$ in phase space and pushes it along its field line by a distance $\lambda$. So to recap, smooth functions$\rightarrow$vector fields$\rightarrow$flows.

The flow generated by the Hamiltonian function itself represents time evolution. However, the flow generated by the momentum function simply shifts the positions of all of the points in phase space, and it is in this sense that we say momentum generates spatial translations. In precisely the same sense, you would find that position generates momentum shifts.

This algebraic structure between observable quantities (which are now self-adjoint operators rather than smooth functions of the phase space variables) carries over to quantum mechanics; the association between unitary operators such as the translation operator and the momentum shift operator (which are analogous to the flows) and the momentum and position operators (analogous to the generators) is given by Stone's theorem.

For more on this structure in Hamiltonian mechanics, you might be interested in my related answers here and here.

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  • $\begingroup$ The second section is a lot to take in (but it seems important and I'm grateful) but the first part is exactly what I was hoping for, and it's much simpler than I thought it would be. Thank you so much! Do your two linked answers (at the end of this article) give a full non-light-speed discussion of the topic from the paragraph starting "rather, it is a deep statement..."? Or is there a section of a textbook you can recommend for the slow explanation of this material? $\endgroup$
    – user310291
    Aug 1 at 3:12
  • $\begingroup$ @user310291 Yes, that is largely the content of the two answers I linked to. The second includes some images and a gif that you might find helpful. Unfortunately I cannot provide a resource which covers this formulation of Hamiltonian mechanics but doesn’t heavily utilize differential geometry; the second linked answer attempts to provide an operational understanding which is not so abstract. And of course you are welcome to ask further clarifying questions on this site. $\endgroup$
    – J. Murray
    Aug 1 at 3:16
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"To be equally informative to work in both position and momentum space", this can only be achieved in the systems of free particle and simple harmonic oscillation. Other than these two systems, he potential form makes it very difficult to work in momentum space.

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  • $\begingroup$ I think this is wrong. There are many examples of strong applications of momentum space computations: for instance, studies of brillouin zones for several solid state systems, and free electron gasses are done in k space. $\endgroup$
    – user310291
    Aug 1 at 2:57
  • $\begingroup$ That is a wrong interpretation. The k of Bloch wave vector denotes the periodic boundary condition for the r-space wave function, $\psi(x+a) = e^{ika} \psi(x)$. The computation near an atom is in r-space for the screened coulomb potential. $\endgroup$
    – ytlu
    Aug 1 at 3:05

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