In the attached Quantum Field image we have the mode expansions (for a Dirac field), with annihilation operators (for both particle and antiparticle) attached to an exponential with a (-ikt) while the creation operators are attached to a positive (+ikt).

If we switch these so that the annihilation operators are attached to an exponential with a (+ikt) and the creation operators are attached to a negative (-ikt), we wind up with the following commutation relations:

$$\boxed{[a(k),a^{\dagger}(p)]_{}=[b(k),b^{\dagger}(p)]_{} = -\delta^{3} (k-p),} $$

which varies from the standard commutation relations only by the negative Dirac delta function: $${{-\delta^{3} (k-p).}}$$

Is this fundamentally objectionable? Are there any specific reasons (physical or theoretical) that this wouldn't work.

The same question applies to the use of a complex scalar quantum field instead of the Dirac quantum field.

Quantum Field Operators

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    $\begingroup$ "In the attached Quantum Field image" - I was quite disappointed when I realized that the image was not of a Quantum Field. $\endgroup$ – Alfred Centauri May 20 at 16:06
  • $\begingroup$ Does the exponent $(ikt)$ look right to you? $\endgroup$ – Alfred Centauri May 20 at 17:13
  • $\begingroup$ Also, the time dependence of the operators are determined by the Hamiltonian and the Heisenberg equation of motion: $\dot{\hat{O}} = -i[\hat{O},\hat{H}]$. Is it valid to just switch which operator is associated with which exponential? $\endgroup$ – Alfred Centauri May 20 at 17:16

Ah, these conventions.

Reminds me of an exam, where I put the wrong sign in the Schrödinger equation. After a long discussion, the professor agreed that you can always replace every $i$ with $-i$. The physics stays the same.

To be consistent, you should then also change the $i$s in $u$ and $v$, and $\bar{\psi}$, probably resulting in positive $\delta$ again.

Think of $i$ not simply as a number, rather as a generator of a phase-transformation $U(1)$. So if you change the sign, it has to be consistent.

Maybe there is yet another convention/choice for the plane wave solutions you use. Could be independent.

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  • $\begingroup$ DoctorNuu, when you say, "also change the "i" in the a and b," do you mean change the sing of the "k" and "p" such as in a(k) and b(p) in the commutation relation I give above? That is what I presumed. Thanks. $\endgroup$ – Wolf May 20 at 16:20
  • $\begingroup$ Sorry, got (of course) confused again. I meant the spinors $u$ and $v$. And obviously in the inverse transformations from $\psi$ to $a$ and $b$. Also there is a hidden $i$ in $\bar{\psi}$. Only mess with these things if you're really ready for it ;-) And, no, no need to touch the signs of the c-numbers $p$ and $q$ $\endgroup$ – user257090 May 20 at 17:10

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