For example, in quantum mechanics, the commutator of the position and momentum is $$[\hat{P_i} ;\hat{Q_j} ] =i\hbar\delta_{ij}\neq 0, i\neq j$$ I know that the position space representation of the wave function is the Fourier transform of the momentum space representation. From what I know, this also applies for other variables, such as energy and time, but I never read anything like that about other pairs of variables.
On the other hand, the commutator of other variables, such as x-angular momentum and y-angular momentum is different from 0. Does this mean that:

  1. $J_x$ and $J_y$ are a pair of conjugate variables?
  2. There exists a Fourier transform between the two?
  3. For any two variables in quantum mechanics that have a commutator different from 0, 1. and 2. are true?
  4. In general (not necessarily in quantum mechanics), to every pair of conjugate variables there is an associated Fourier transform?
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    $\begingroup$ What is your working definition for "conjugate variables"? $\endgroup$
    – DanielC
    May 15, 2020 at 19:25
  • $\begingroup$ I do not know of any definition where $J_x$ and $J_y$ are considered conjugate pair. If anything the commutator is not constant, unlike $P$ and $Q$. $\endgroup$ May 15, 2020 at 20:01
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    $\begingroup$ Related: physics.stackexchange.com/q/281009/2451 $\endgroup$
    – Qmechanic
    May 15, 2020 at 20:02
  • $\begingroup$ @DanielC I don't have a deep understanding of them. In my view, they are a pair of somehow co-dependent variables. By asking this (apparently faulty) question, I was thinking of making sense of them by relating them to the Fourier transform $\endgroup$
    – Anthill
    May 15, 2020 at 20:04
  • $\begingroup$ @ZeroTheHero Now that I think of it this way, it does make sense that it's a different situation because the commutator is not constant. I only thought of it from my perspective of "co-dependent variables" $\endgroup$
    – Anthill
    May 15, 2020 at 20:07

2 Answers 2


The meaning of conjugate variables is usually that they are related by the Fourier transform (other transforms are possible, but I am not sure they are interesting). In quantum mechanics, this is simply a change of basis in Hilbert space. The relation between energy and time is similar, but time is a parameter in quantum mechanics, not an observable (this is even true in relativistic quantum mechanics). The meaning of the energy-time uncertainty relationship is consequently quite different from that of position-momentum.

  • $\begingroup$ actually in class. mech they are defined by their Poisson bracket being $1$. $\endgroup$ May 15, 2020 at 21:10
  • $\begingroup$ @ZeroTheHero, I should have just said that conjugacy refers to change of basis in Hilbert space. Notwithstanding the historical importance of canonical quantisation, commutation relations are properly derived from Hilbert space, not imposed by analogy with Poisson brackets, in spite of the apparent similarity of form. Arguments by analogy, and the application of the language of one structure to a different structure, are not helpful to deep understanding. $\endgroup$ May 15, 2020 at 21:52

It depends very much what you mean by conjugate pairs, especially in quantum mechanics.

In classical mechanics the most common definition is based on the Poisson bracket: \begin{align} \{P_i,Q_j\}=\delta_{ij} \tag{1} \end{align} and thus in this sense canonical transformations (which can be non-linear transformations in the original coordinates) take you from a set of conjugate variables to another.

Alternatively, one can use Darboux’ theorem and the definition of the (closed) canonical symplectic form \begin{align} \omega = \sum_i dp_i\wedge dq_i \end{align} to define the canonical pair $\{p_i,q_i\}$.

Since (1) is at the root of Dirac quantization, where the quantum commutator is equal to $i\hbar \times$(classical Poisson bracket), it is sensible to think of two operators $\hat P,\hat Q$ as conjugate if their commutator is $i\hbar$.

Note that canonical transformations involving non-linear functions of operators is a mine field because of non-commutativity issues.

The plot immediately thickens with energy and time, since time is not a quantum mechanical operator. (Deriving energy-time uncertainly relation can be tricky and requires care.) Thus if two classical quantities are Fourier pairs, there is no guarantee they will be conjugate “quantumly”.

Moreover, there are examples - for instance $\phi$ and $L_z=-i\hbar\frac{d}{d\phi}$ - which “look” conjugate in that they satisfy the correct commutation relations but in fact have issues of operator domains. $\phi$ and $L_z$ are Fourier pairs in the sense that a rotation about $\hat z$ is of the form $e^{-i\phi L_z}$. (This is an Abelian group so some technical hurdles linked to harmonic analysis are avoided.)

The notion of conjugate varibles in QM is also muddled by the related notion of complementary variables. One may naively understand this as follows. Given the uncertainty relation between $\hat x$ and $\hat p$ one may (loosely speaking) state that, if everything is known about one of the observable, nothing is known about the other. An alternative statement would be that, if you are in an eigenstate of one operator, then the probability of outcome for the other is constant. In this sense $\sigma_x$ and $\sigma_y$ (and indeed $\sigma_z$) are complementary as the outcomes of measuring $\sigma_y$ in any eigenstate of $\sigma_x$ (for instance) are equiprobable: if you know everything about the measurements of $\sigma_x$ (they have no fluctuations on eigenstates of $\sigma_x$), then you know nothing of the outcomes of measurements of either $\sigma_y$ or $\sigma_z$ as all the outcomes are equally probable. This is the starting point for the notion of mutually unbiased bases.

  • $\begingroup$ The $L_z$ and $\phi$ are still conjugate variables, right? I mean the domain of $\phi$ is different from $x$ but correspondingly, the domain of $L_z$ is also different from $p$. And there exists a Fourier relation between them via Fourier series rather than a Fourier transform. $\endgroup$
    – user87745
    May 15, 2020 at 22:23
  • $\begingroup$ @DvijD.C. in the classical sense yes, but it is difficult to think of them as conjugate in the quantum sense. see physics.stackexchange.com/a/338057/36194 for a discussion in a different context. $\endgroup$ May 15, 2020 at 23:29
  • $\begingroup$ Yes, I'm aware of the issue you point out but I think the operator can be made self-adjoint by being careful about its domain: physics.stackexchange.com/a/233311/20427. You can't obtain the usual uncertainty relations, of course, but the commutation relations still hold over the appropriate domain. I've added the aforementioned link to the post you referenced as it's relevant there independently of this discussion I guess. $\endgroup$
    – user87745
    May 15, 2020 at 23:43
  • $\begingroup$ There is currently no satisfactory definition of a self-adjoint phase operator with the right properties (2$\pi$ cyclic etc). It’s still an open question. It’s not even clear if one could measure such an operator as what appears is $e^{i\phi}$ rather than $\phi$ itself. $\endgroup$ May 15, 2020 at 23:44
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    $\begingroup$ @DvijD.C. that solution excludes plane waves since $\psi$ at the end point is $0$ so eigenfunctions of $L_z$, which are $e^{-i m \phi}$ are also excluded. The solution is valid but not really applicable to phases. Biedenharn and Louck in their “Angular momentum in quantum physics” book (or a closely related text v.g. Racah algebra) discuss this at length. Also related: Bonneau, Guy, Jacques Faraut, and Galliano Valent. "Self-adjoint extensions of operators and the teaching of quantum mechanics." American Journal of physics 69, no. 3 (2001): 322-331. for a discussion on this general topic. $\endgroup$ May 16, 2020 at 0:08

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