For a harmonic oscillator in one dimension, there is an uncertainty relation between the number of quanta $n$ and the phase of the oscillation $\phi$. There are all kinds of technical complications arising from the fact that $\phi$ can't be made into a single-valued and continuous operator (Carruthers 1968), but roughly speaking, you can write an uncertainty relation like this:

$\Delta n \Delta \phi \ge 1$

The fact that the right-hand side is 1 rather than $\hbar$ is not just from using natural units such that $\hbar=1$; we can see this because $n$ and $\phi$ are both unitless. This suggests that this uncertainty relation is a classical one, like the time-frequency uncertainty relation that is the reason for referring to data transmission speeds in terms of bandwidth.

However, the only physical interpretation I know of seems purely quantum-mechanical (Peierls 1979):

[...] any device capable of measuring the field, including its phase, must be capable of altering the number of quanta by an indeterminate amount

If this uncertainty relation is classical, what is its classical interpretation? If it's not classical, then why doesn't the restriction vanish in the limit $\hbar\rightarrow0$?

related: Do we always ignore zero energy solutions to the (one dimensional) Schrödinger equation?

Carruthers and Nieto, "Phase and Angle Variables in Quantum Mechanics," Rev Mod Phys 40 (1968) 411 -- can be found online by googling

Peierls, Surprises in theoretical physics, 1979

  • 1
    $\begingroup$ Isn't the apparent absence of $\hbar$ just related to the fact that the number of quanta n goes to infinity in the classical limit? I would suppose that the uncertainty in the number of quanta also diverges in a suitably defined limit, and correspondingly $\Delta \phi \geq 1/\Delta n \rightarrow 0$ in the classical limit? $\endgroup$ Jun 13, 2013 at 16:14
  • $\begingroup$ There is a well known argument that you cannot have $[n,\phi]=i$ as an operator identity (on the whole Hilbert space), because this will imply $[\sin (\alpha n),\phi] = i~\alpha~ cos(\alpha n)$, and there is a contradiction with $\alpha = \pi$, because the LHS is zero, but not the RHS (we can take a basis where $n$ is diagonal to see that). So it is said that we have to take a "dense set" of the states of the Hilbert space, to make sense. $\endgroup$
    – Trimok
    Jun 13, 2013 at 16:22
  • $\begingroup$ @Trimok: That was what I was referring to as "technical complications," and that's why I gave the reference to Carruthers. I don't think these issues materially affect the question. Carruthers gives more rigorously correct uncertainty relations in equations 5.47-48, in terms of $C=\cos\phi$ and $S=\sin\phi$, which are valid operators. The same issue arises, which is that there is no $\hbar$. $\endgroup$
    – user4552
    Jun 13, 2013 at 17:07
  • $\begingroup$ To my taste, even the classical field itself - a coherent state - is a macroscopic quantum mechanical creature since it is capable of interference $\endgroup$
    – Slaviks
    Jun 13, 2013 at 17:11
  • $\begingroup$ @JaschaUlrich: I would think that the classical limit would require $\Delta n/n\rightarrow 0$, but that could happen with $\Delta n\rightarrow 0$, $\Delta n\rightarrow \infty$, or just about anything else. $\endgroup$
    – user4552
    Jun 13, 2013 at 17:14

1 Answer 1


I think I have at least one possible answer to my own question. Let's write $E=n\hbar\omega$ for the energy of a classical wave, and then the uncertainty relation becomes $\Delta E\Delta \phi \ge \hbar\omega$, which has $\hbar$ in it and is manifestly quantum mechanical.

I think this is similar to the relationship between the classical uncertainty relation $\Delta f \Delta t \ge 1$ and the quantum-mechanical one $\Delta E \Delta t \ge h$.

  • $\begingroup$ I agree. I was going to give the same opinion as comment... $\endgroup$
    – Trimok
    Jun 13, 2013 at 17:23

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