# Quantum Operator Fields - Consequences of switching the sign of the phases attached to the creation/annihilation operators

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. • "In the attached Quantum Field image" - I was quite disappointed when I realized that the image was not of a Quantum Field. – Alfred Centauri May 20 at 16:06
• Does the exponent $(ikt)$ look right to you? – Alfred Centauri May 20 at 17:13
• 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? – Alfred Centauri May 20 at 17:16

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.
• 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 $q$ – user257090 May 20 at 17:10