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The well known Heisenberg commutator relation $$[p,q]=\cfrac{\hbar}{i} \cdot \mathbb{I}$$ introduces the imaginary unit $i$ into quantum mechanics. I ask for the deeper reason:

Why does the correspondence with real coordinates q and p introduces complex numbers for the commutator? Is the reason from physics or from mathematics?

Aside: I'm familiar with complex numbers and with the fact, that some results from the real domain find a satisfactory explanation not until generalization to the complex domain.

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    $\begingroup$ Take the Hermitean conjugate of the relation. $\endgroup$ Commented Apr 21, 2022 at 10:42
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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/11396/50583, physics.stackexchange.com/q/46015/50583, physics.stackexchange.com/q/12983/50583 and their linked questions (which are many) $\endgroup$
    – ACuriousMind
    Commented Apr 21, 2022 at 11:49
  • $\begingroup$ truthfully im out of depth here but @JG's answer mentioned that $i$ here is related to the $i$ in $e^{i \frac{pq}{h}}$ and it is the case that anytime you want to describe periodic motion you probably want to invoke complex exponentials. So a really long winded and half baked thought would be: There is an $i$ in the commutator relation because some other part of the theory involves waves/periodic motion. $\endgroup$ Commented Apr 22, 2022 at 17:40
  • $\begingroup$ @SidharthGhoshal Good catch. Unitary transformations have an obvious connection to such periodic waves. Indeed, such factors can be interpreted as the de Broglie relation. $\endgroup$
    – J.G.
    Commented Apr 22, 2022 at 18:16
  • $\begingroup$ @SidharthGhoshal Like you I got the impression that a deep reason for the imaginary unit ‚i‘ is the omnipresence of waves. The mathematical tool for waves is the complex exponential exp(it) and Fourier analysis. Indeed, the seminal papers of Heisenberg from 1925 clearly show the search for an explanation of the spectral law as one of his goals. $\endgroup$
    – Jo Wehler
    Commented Apr 22, 2022 at 18:20

3 Answers 3

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Because operators $p$ and $q$ represent physical observables (i.e. they have real eigen-values), they need to be Hermitean (i.e. $p^\dagger=p$ and $q^\dagger=q$).

From this it is easy to show that their commutator $[p,q]$ is anti-Hermitean. $$[p,q]^\dagger = (pq-qp)^\dagger = (pq)^\dagger-(qp)^\dagger = q^\dagger p^\dagger-p^\dagger q^\dagger = qp-pq = -[p,q]$$

You can get a Hermitean operator from this anti-Hermitean $[p,q]$ only by multiplying it with $i$. $$(i[p,q])^\dagger = i[p,q]$$

So you can write Heisenberg's commutator relation also as $$i[p,q]=\hbar\mathbb{I}$$ with Hermitean operators on both sides. The operator on the right side corresponds to the very trivial physical observable, which always gives the same measurement value $\hbar$.

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    $\begingroup$ But this begs the question of why does the commutator need to be Hermetian? $\endgroup$
    – Dave
    Commented Apr 22, 2022 at 16:59
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    $\begingroup$ @Dave I think it's not that you "need" the commutator to be hermitian (it's anti-hermitian anyway). Rather, because $i[p,q]$ is hermitian, $[p,q]$ is $i$ times a hermitian operator. $\endgroup$ Commented Jul 19, 2022 at 15:41
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Is the reason from physics or from mathematics?

You can argue it either way. To deem it mathematical, see @ThomasFritsch's answer. But here's a physical insight, even if it requires mathematics to explain it. I'll work in $1$ dimension for simplicity. The real classical observable $p$ corresponds to a Hermitian $\hat{p}$, and leads to $e^{ipq/\hbar}$ factors of eigenvalue $p$. Why that factor, though? Because we need to tie momentum to arbitrary space translations ($[p,\,q]=-i\hbar\mathbb{I}$ is related to the Poisson bracket $\{p,\,q\}=-1$), quantum mechanics needs to be able to e.g. square-root unitary transformations (if I can move something a metre I can move it a half-metre, then another half-metre). This is arguably the main reason QM involves complex numbers. Luckily, if $\hat{O}$ is a dimensionless Hermitian operator, $\exp(i\hat{O})$ is unitary.

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    $\begingroup$ Can't understand the argument here completely, but I am interested in the idea. Please consider elaborating on this to make it more clear. $\endgroup$ Commented Apr 27, 2022 at 22:47
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This, bizarrely, is something I lectured on yesterday. The i emerges from the Fourier transformations and we don't need to know anything about position, momentum, or the wavefunction, for it to emerge completely naturally from the mathematics where it comes from a differential of position. I include the two pages of my lecture notes below:

page 1 of lecture notes page 2 of lecture notes

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    $\begingroup$ would be best if you could replace the image and cut-and-paste the LaTeX source so your answer then becomes searchable. $\endgroup$ Commented Apr 23, 2022 at 22:34

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