I'm wanting to understand the dynamics of a momentum eigenstate $| p \rangle$ governed by a harmonic oscillator Hamiltonian. Consider $\hat{H} = \hat{p}^2 + \hat{q}^2$. Then inserting a completeness relation in an attempt to simplify things (setting all constants to 1), \begin{align} \hat{U}(t) &= e^{-it(\hat{p}^2 + \hat{q}^2)} = \int\mathrm{d} p \: e^{-it(\hat{p}^2 + \hat{q}^2)} | p \rangle\langle p| \end{align} Is there a meaningful way to simplify this expression i.e. act that exponentiated operators on $| p \rangle$ to obtain a $c$-number (i.e. phase factor)? That is given $\hat{p} | p \rangle = p | p \rangle$ and $\hat{q} | p \rangle = \frac{\mathrm{d}^2}{\mathrm{d}p^2}\delta(p) | p \rangle$ (the latter result using the representation of $| p \rangle$ as a Fourier transform of $| x \rangle$) yielding: \begin{align} \hat{U}(t) &= \int\mathrm{d} p \: e^{-it ( p^2 + \mathrm{d}^2/\mathrm{d}p^2\delta(p))/\hbar} | p \rangle\langle p| \end{align} or does one need to use some Baker-Campbell-Hausdorff relation to split up the single exponential into two? I am afraid the latter will not work nicely since $[\hat{p},\hat{q}] \neq c$ where $c$ is a number, and so the BCH formula will feature an infinite series of commutators in non-closed form. Is it even meaningful to ask what the dynamics of $| p \rangle$ are if it is not an eigenstate of the Hamiltonian?


What I really want is to consider an interaction term between some motional and internal DoFs of a composite particle in the trap. So (neglecting the free Hamiltonian of the internal DoFs) $H = H_\mathrm{QHO} + H_\mathrm{int.} = (\hat{p}^2 + \hat{q}^2 ) + \hat{H}_0 \cdot \hat{p}^2$ where $\hat{H}_0 = E_g | g \rangle\langle g | + E_e | e \rangle\langle e |$ is the internal Hamiltonian. Then I wish to consider the dynamics of some initial state, say $| \psi(0) \rangle = | g \rangle | 0 \rangle$ where $| 0 \rangle$ is the ground state of the QHO. What I envision being the easiest way to go about this is to write $\hat{U}(t)$ in the momentum basis so that one can turn the $\hat{p}$ and $\hat{q}$ operators into $c$-numbers, so: \begin{align} \hat{U}(t) | \psi \rangle &= \int\mathrm{d}p \: e^{-it(\hat{p}^2 + \hat{q}^2) + \hat{H}_0 \cdot \hat{p}^2} | p \rangle\langle p | 0 \rangle | g \rangle \end{align} I am basically interested in the phase acquired by the state $| g \rangle$ after tracing out the COM DoFs, but having the differential form of $\hat{q}^2 = \partial_p^2$ in the exponent seems ugly and counter-productive to my goal.

  • $\begingroup$ The oscillator is symmetric between p and q. Are you cool with |x>? $\endgroup$ Sep 2, 2023 at 14:59
  • $\begingroup$ $\hat{q}^2 | p \rangle = \int\mathrm{d} p \: e^{ipq} \hat{q}^2 | q \rangle = \int\mathrm{d} p \: e^{ipx} q^2 | q \rangle = \frac{\mathrm{d}^2}{\mathrm{d}p^2} \delta(p)$ is not correct? Am I forgetting a basic result and over-complicating things? $\endgroup$
    – j.foobles
    Sep 2, 2023 at 15:10
  • $\begingroup$ Absorb ℏ in your units and thus set it to 1. Goodbye ℏ. Then you know $\hat q= i\int dp ~~|p\rangle \partial_p\langle p|$. Repeat to $\hat q^2= -\int dp ~~|p\rangle \partial_p^2\langle p|$ $\endgroup$ Sep 2, 2023 at 15:18

1 Answer 1


I'm not sure what you are really after (smells like an attempt to compute the Mehler kernel unconventionally), so I'll assume you are seeking the evolution matrix elements in momentum rep, $\langle p'| \hat{U}(t) |p\rangle$. Set ℏ=1 by absorbing it into your units (nondimensionalization).

Recall, or prove$^\natural$, the SU(1,1) van Kortryk identity, $$ e^{-i t(\hat{p}^2 + \hat{q}^2)} = e^{-\frac{i}{2}\tan (t)~\hat{p}^2} ~~ e^{-\frac{i}{2}\sin(2 t)~~ \hat{q}^2} ~~ e^{-\frac{i}{2}\tan (t)~\hat{p}^2}, $$ so that, $$\langle p'| \hat{U}(t) |p\rangle=e^{-\frac{i}{2}\tan t~(p'^2+p^2)} \langle p'| e^{-\frac{i}{2}\sin 2t~~ \hat{q}^2} |p\rangle \\ = e^{-\frac{i}{2}\tan ( t)~(p'^2+p^2)} \langle p'| e^{\frac{i}{2}\sin (2t)~ \partial_p^2} |p\rangle .$$

This is a good rotation in phase-space to review. Condon's epic paper shows that the oscillator hamiltonian generates uniform rotations in phase space.

$^\natural$Note the doublet irrep $\hat p ^2 \mapsto \sigma^+$ and $\hat x ^2\mapsto \sigma^-$ is faithful, so all group identities hold if they hold for the doublet, $$ e^{-it(\sigma^++\sigma^-)}= e^{-\frac{i}{2}\tan t~\sigma^+} e^{-\frac{i}{2}\sin(2t) ~\sigma^-}e^{-\frac{i}{2}\tan t~\sigma^+} , $$ trivial to confirm explicitly.

  • $\begingroup$ Thanks for the note, I've added some further information on my original post. My basic qualm is what to do with the $\partial_p^2$ in the exponent for computing an actual phase? Usually this would act on some test function, so does one then express $e^{(i/2)\sin(2t) \partial_p^2}$ as a series and then act this on the $\langle p' | p \rangle$ element? $\endgroup$
    – j.foobles
    Sep 2, 2023 at 19:37
  • $\begingroup$ Yes, you could do that, especially for small t ; Normally, you'd Fourier transform $|p\rangle= \int dx ~~|x\rangle\langle x|p\rangle$ and turn it into some type of Weierstrass transform. You might work out a few simple examples. $\endgroup$ Sep 2, 2023 at 20:24
  • $\begingroup$ $\langle p|0\rangle$ is a momentum Gaussian, the bridge between momentum and Fock space. $\endgroup$ Sep 3, 2023 at 22:22
  • $\begingroup$ Perhaps something to add: if one does not want the derivatives in the exponent one can insert an identity resolved in terms of $|q\rangle$, which would then lead to factors like $\langle q|p\rangle=\exp(ipq)$ and its c.c. leading the Fourier integrals that can be evaluated. $\endgroup$ Sep 5, 2023 at 3:34

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