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I gained a lot of physical intuition about commutators by reading this topic. What is the physical meaning of commutators in quantum mechanics?

I have similar questions about the anti-commutators. What does it mean physically when two operators anti-commute ?

khalid
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2 Answers2

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Sorry but the analysis of what commutators mean (in the given link) although very good, does not provide intuition and does not generalise to anti-commutators.

Commutators used for Bose particles make the Klein-Gordon equation have bounded energy (a necessary physical condition, which anti-commutators do not do).

On the other hand anti-commutators make the Dirac equation (for fermions) have bounded energy (unlike commutators), see spin-statistics connection theorem.

In this sense the anti-commutators is the exact analog of commutators for fermions (but what do actualy commutators mean?). nice and difficult question to answer intuitively.

In a sense commutators (between observables) measure the correlation of the observables. Thus is also a measure (away from) simultaneous diagonalisation of these observables.

Elaborating a little on this.

Lets say we have a state $\psi$ and two observables (operators) $A$, $B$. When these operators are simultaneously diagonalised in a given representation, they act on the state $\psi$ just by a mere multiplication with a real (c-number) number (either $a$, or $b$), an eigenvalue of each operator (i.e $A\psi=a\psi$, $B\psi=b\psi$).

We know that for real numbers $a,b$ this holds $ab-ba=0$ identicaly (or in operator form $(AB-BA)\psi=0$ or $\left[A,B\right]\psi=0$) so the expression $AB-BA=\left[A,B\right]$ (the commutator) becomes a measure away from simultaneous diagonalisation (when the observables commute the commutator is identicaly zero and not-zero in any other case).

Another way to see the commutator expression (which is related to previous paragraph), is as taking an (infinitesimal) path from point (state) $\psi$ to point $A \psi$ and then to point $BA \psi$ and then the path from $\psi$ to $B \psi$ to $AB \psi$. If the operators commute (are simultaneously diagonalisable) the two paths should land on the same final state (point). If not their difference is a measure of correlation (measure away from simultaneous diagonalisation).

When talking about fermions (pauli-exclusion principle, grassman variables $\theta_1 \theta_2 = - \theta_2 \theta_1$), the commutators have to be adjusted accordingly (change the minus sign), thus become anti-commutators (in order to measure the same quantity).

Continuing the previous line of thought, the expression used was based on the fact that for real numbers (and thus for boson operators) the expression $ab-ba$ is (identicaly) zero.

However fermion (grassman) variables have another algebra ($\theta_1 \theta_2 = - \theta_2 \theta_1 \implies \theta_1 \theta_2 + \theta_2 \theta_1=0$, identicaly).

So what was an identical zero relation for boson operators ($ab-ba$) needs to be adjusted for fermion operators to the identical zero relation $\theta_1 \theta_2 + \theta_2 \theta_1$, thus become an anti-commutator.

Un-correlated observables (either bosons or fermions) commute (or respectively anti-commute) thus are independent and can be measured (diagonalised) simultaneously with arbitrary precision. If not, the observables are correlated, thus the act of fixing one observable, alters the other observable making simultaneous (arbitrary) measurement/manipulation of both impossible.

PS. See how the previous analysis can be generalised to another arbitrary algebra (based on identicaly zero relations), in case in the future another type of particle having another algebra for its eigenvalues appears.

Nikos M.
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  • I did not understand well the last part of your analysis. Are you saying that Fermion operators which anticommute can be measured simultaneously? (I do ot think so) Since it is obviously false. More strongly, Fermi field operators are not observables since they violate causality. You cannot directly measure the Fermi field. What you can measure are Boson currents associated to Fermi fields, I mean something like $:\overline{\psi}(x)\gamma^\mu \psi(x):$. – Valter Moretti Oct 24 '14 at 21:50
  • @ValterMoretti, sure you are right. Let me rephrase a bit. For the lorentz invariant quantities of fermion fields (which are constructed from pairs of fermion fields) the analogy stated in the last part holds Fermionic field) – Nikos M. Oct 25 '14 at 00:14
  • *Commutators used for Bose particles make the Klein-Gordon equation have bounded energy (a necessary physical condition, which anti-commutators do not do).

    On the other hand anti-commutators make the Dirac equation (for fermions) have bounded energy (unlike commutators)* Do they both have bounded energy?

    – MatterGauge Dec 08 '21 at 14:17
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    @MatterGauge Presumably Nikos meant bounded from below. – J. Murray Dec 08 '21 at 14:33
  • @MatterGauge, bounded from below is what is meant, exactly as J.Murray pointed out – Nikos M. Dec 08 '21 at 17:24
  • So bosons have bounded energy from below, while fermions have bounded energy from below? – MatterGauge Dec 09 '21 at 10:00
  • @MatterGauge, quantizing klein-gordon equation for bosons using anti-commutators leads to an energy calculation that is not bounded from below, which is un-physical. Similarly when quantising dirac equation with commutators. So commutators and anti-commutatiors are used for specific equations (bosonic or fermionic respectively) in order to produce physical results. Energy must be bounded from below in every case and the choice of appropriate commutators makes it so. – Nikos M. Dec 09 '21 at 10:04
  • So both have no energy bound from below? The electron is prevented to enter the negative energy states because they are occupied already by the electron sea. – MatterGauge Dec 09 '21 at 10:19
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    @MatterGauge, energy not bounded from below can mean, among other things, that entities can enter into arbitrarily large negative energies thus becoming a free source of infinite energy, which is an un-physical deduction. So the equations must be quantised in such way (using appropriate commutators/anti-commutators) that prevent this un-physical behavior. Hope this is clear – Nikos M. Dec 09 '21 at 10:25
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    @MatterGauge yes indeed, that is why two types of commutators are used, different for each one – Nikos M. Dec 09 '21 at 11:01
  • Yes that's clear. But doesn't this hold for both bosons and fermions? I think fermions have no lower bound, while bosons do. Giving rise to particles and antiparticles for fermions. There are no boson antiboson particle pairs. – MatterGauge Dec 09 '21 at 11:06
  • Thanks for your time! – MatterGauge Dec 09 '21 at 11:07
  • @MatterGauge, anti-particles in the case of fermions, do not exclude by themselves the occupation of arbitrarily large negative energies if dirac equation is quantised using commutators instead of anti-commutators. One reason is that if use commutators for the dirac equation, Pauli exclusion principle is violated. Hope this is clear – Nikos M. Dec 09 '21 at 11:08
  • Clear but still a bìt confusing. Anti commutation (with associated anti symmetric states only) entails occupation of negative energy states because all negative states are occupied. How does this translate to bosons? Aren't negative energy states a possibility in that case? – MatterGauge Dec 09 '21 at 11:34
  • @MatterGauge Quantizing with commutators the Klein Gordon equation simply expresses a physical fact of nature. Same for Dirac equation and anti commutators. No one really knows the mechanism, whether it is anti particles or something else. – Nikos M. Dec 09 '21 at 13:11
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Commutators and anticommutators are ubiquitous in quantum mechanics, so one shoudl not really restrianing to the interpretation provdied in the OP. There's however one specific aspect of anti-commutators that may add a bit of clarity here: one often u-ses anti-commutators for correlation functions.

Indeed, the average value of a product of two quantum operators depends on the order of their multiplication. We can however always write: $$AB = \frac{1}{2}[A, B]+\frac{1}{2}\{A, B\},\\ BA = \frac{1}{2}[A, B]-\frac{1}{2}\{A, B\}.$$ In the classical limit the commutator vanishes, while the anticommutator simply become sidnependent on the order of the quantities in it. One therefore often defines quantum equivalents of correlation functions as: $$ K_{AB}=\left\langle \frac{1}{2}\{A, B\}\right\rangle.$$

As an example see the use of anti-commutator see [the quantum version of the fluctuation dissipation theorem][1], where $$ S_{x}(\omega)+S_{x}(-\omega)=\int dt e^{i\omega t}\left\langle \frac{1}{2}\{x(t), x(0)\}\right\rangle$$ (I am trying to adapt to the notation of the Wikipedia article, but there may be errors in the last equation.)

Roger V.
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