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In classical mechanics, the orbital angular momentum of a particle is defined as $\textbf{L}=\textbf{r}\times\textbf{p}$. This is zero in the rest frame of the particle where $\textbf{p}=0$.

Quantum mechanically, $\textbf{p}$ is an operator. So putting $\hat{\textbf{p}}=0$ in $\hat{\textbf{L}}=\hat{\textbf{r}} \times\hat{\textbf{p}}$ and claiming that the orbital angular momentum of a quantum particle is zero in its rest frame does not make sense. One must look at the value of $\hat{\textbf{L}}^2$ on the "wavefunction in the rest frame" of the particle.

How does one find the wavefunction of a particle in its rest frame?

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    $\begingroup$ Hint: Show that the commutator $[L_i,p_j]$ is proportional to $p_k$. Next put $p=0$. $\endgroup$
    – Qmechanic ♦
    Commented Aug 13, 2017 at 16:24
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    $\begingroup$ Who says that a quantum particle (in anything that's not a plane-wave momentum eigenstate) has a rest frame to begin with? $\endgroup$
    – Emilio Pisanty
    Commented Jan 26, 2018 at 20:09
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    $\begingroup$ Your approach is not the correct one. The idea is to start assuming the existence of the generator $J$ of global rotations around the origin thus satisfying the Lie algebra relations of $SO(3)$ and next define $S_k= J_k - (X \wedge P)_k$. It is easy to prove, from CCR of $X$ and $P$ that the $S_k$ still define a representation Lie algebra relations of $SO(3)$ and commute with $X$ and $P$. Possibly this representation is trivial (i.e. there is no spin) otherwise the $S_k$ define the intrinsic angular momentum. $\endgroup$
    – Valter Moretti
    Commented Jan 27, 2018 at 9:12
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    $\begingroup$ Also known (perhaps improperly) as the angular momentum in the rest frame of the system. Actually, at quantum level the description is more delicate, and the existence of nontrival $S_k$ correspond (using Stone von Neumann theorem) to a factorization of the Hilbert space $H_{orb}\otimes H_{intrinsic}$. The former factor describes the orbital state, where $X$ and $P$ are defined, the latter describes the properties of the system independent form the orbital state and includes the operators $S_k$ (but also the charge for insatnce). $\endgroup$
    – Valter Moretti
    Commented Jan 27, 2018 at 9:16
  • $\begingroup$ This is a better quantum interpretation of the rest frame of the system. $\endgroup$
    – Valter Moretti
    Commented Jan 27, 2018 at 9:19

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The rest-frame wavefunction $\psi(\boldsymbol x,t)$ is the one such that $$ \boldsymbol 0\equiv\langle \boldsymbol p\rangle=\int_{\mathbb R^3}\psi^*(\boldsymbol x,t)(-i\boldsymbol \nabla)\psi(\boldsymbol x,t)\ \mathrm d\boldsymbol x $$

If $\boldsymbol k\equiv\langle \boldsymbol p\rangle$ is non-zero, we just need to redefine the wave-function: $$ \psi(\boldsymbol x,t)\to\mathrm e^{-i\boldsymbol k\cdot\boldsymbol x}\psi(\boldsymbol x,t) $$ which satisfies $\langle\boldsymbol p\rangle\equiv \boldsymbol 0$ by construction. This is just a translation in momentum space, $$ \tilde\psi(\boldsymbol p,t)\to \tilde\psi(\boldsymbol p-\boldsymbol k,t) $$ which obviously has zero mean.

More generally, if you have a system of many particles, the rest-frame of the system is, by definition, the one where $\langle\boldsymbol p\rangle\equiv\boldsymbol 0$, where $\boldsymbol p$ denotes the total linear momentum: $$ \boldsymbol p=\sum_i \boldsymbol p_i $$

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  • $\begingroup$ I would expect the "rest frame" of a given particle to also obey $\langle p^2\rangle=0$ (and thus be impossible to find in general). Is this some standard bit of (mis)terminology that I'm missing out on? Putting yourself in a frame where the expected momentum is zero is nice, but that doesn't mean that the particle "isn't moving" in that frame, as implied by the "rest" in "rest frame". $\endgroup$
    – Emilio Pisanty
    Commented Jan 26, 2018 at 20:11
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    $\begingroup$ @EmilioPisanty Rest frame means $\langle \boldsymbol p\rangle=0$ only. It does not mean $\langle \boldsymbol p^2\rangle=0$. I'll try to find a good reference. For now, it's just terminology. "Rest frame" means zero mean momentum; it does not mean "not moving at all" (whatever that means). In more classical terms, the rest frame of a collection of particles is the frame where the total momentum vanishes; but there is a non-zero dispersion, $\Delta p\neq 0$. Similarly, in QM, the rest frame is the one where the total momentum vanishes; but there is a non-zero dispersion too, $\Delta p\neq0$. $\endgroup$
    – AccidentalFourierTransform
    Commented Jan 26, 2018 at 20:22
  • $\begingroup$ The best I could find for now is Cohen-Tannoudji, Quantum Mechanics, Volume 1, Chapter VII, Section B. I'll see if I can find something better. $\endgroup$
    – AccidentalFourierTransform
    Commented Jan 26, 2018 at 20:30
  • $\begingroup$ $\langle p\rangle=0$ as the rest frame of a collection of particles makes sense, but a quantum particle isn't an ensemble. That use of the term seems completely bonkers to me (but that says nothing about whether it's in use or not). Oh well. $\endgroup$
    – Emilio Pisanty
    Commented Jan 26, 2018 at 20:38
  • $\begingroup$ @AccidentalFourierTransform Does your second equation represent the transformation of the wavefunction under Galilean boost or translation? $\endgroup$
    – SRS
    Commented Jan 27, 2018 at 11:00

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