One of the principal concepts in QFT is to consider the expasion of the field $$\phi(x)=\int{\frac{d^3 \vec{p}}{2(2\pi)^3\omega_\vec{p}}}(a(\vec{p})e^{-ipx}+b^{\dagger}(\vec{p})e^{+ipx}),$$ with amplitudes $a(\vec{p}), b^{\dagger}(\vec{p})$ and then "quantize" it by considering them as operators.

Congenitally, one thinks of them as annihilation and creation operators, but it it racks my brain whether

  • it is necessary, or, instead, only a matter of convention that the exponential $e^{ipx}$ (resp $e^{-ipx}$) must stand by the side of the creation (annihilation) operator $b(\vec{p})^{\dagger}$ (resp $a(\vec{p})$)?

Is there a way to prove it?


Yes, it's rather simple to show this. The point is that a creation operator for a free field always increases the energy, while an annihilation operator decreases the energy. From that point on, the reasoning is identical to the simple case of the harmonic oscillator. The Heisenberg equation is $$\frac{d}{dt} a^\dagger = \frac{i}{\hbar} [H, a^\dagger] = \frac{i \omega}{\hbar} a^\dagger, \quad a^\dagger(t) = e^{i \omega t / \hbar} a^\dagger(0).$$ Assuming the standard $(+---)$ signature, this dependence goes with $e^{ipx}$, so the coefficient of a creation operator must be $e^{ipx}$.

  • $\begingroup$ (+1), of course this is the most clear way to explain this. Is my expanation below correct or not in your opinion? $\endgroup$ – MRT Aug 6 '18 at 10:23
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    $\begingroup$ @MRT Yes, it looks totally fine to me! I think it’s fundamentally the same proof too: everything boils down to the commutation relations between creation and annihilation operators. $\endgroup$ – knzhou Aug 6 '18 at 11:04
  • $\begingroup$ @knzhou: One point stays unclear. Indeed, using the expresion for $a^\dagger$ in terms of $\phi$ and $\pi$ as MRT explained below one get the relation $$\frac{d}{dt} a^\dagger = \frac{i}{\hbar} [H, a^\dagger] = \frac{i \omega}{\hbar} a^\dagger$$ and so we get indeed $$ \quad a^\dagger(t,p) = e^{i \omega t / \hbar} a^\dagger(0,p)$$ But the argument that by considering the standard $(+---)$ (so $x^{\mu} =(ct, -\vec{x})$) signature, this dependence "goes with $e^{ipx}$" isnt clear to me. Why it cant be also $e^{-ipx}$. Must the exponential fulfil this compatibility with four vectors? $\endgroup$ – Tim Grosskreutz Aug 6 '18 at 13:36

I think i have understand where my error was, i edited.
I anticipate that this is not a convention, it has to be like that.
I use the notation $x^{\mu} = (t,\vec{x}) = x$.
Transferring the Poisson brackets properties to the commutators (i.e quantizing) one has that the equal-time commutation rules between the field operators and the conjugate momenta $\pi = \partial_{0} \phi^{\dagger}$ and $\pi^{\dagger} = \partial_{0} \phi$ are: $$[\phi(x) , \pi(y)]_{t} = [\phi^{\dagger}(x) , \pi^{\dagger}(y)]_{t} = i \delta^{3} (x - y) (1) $$ The other commutators are all zero and here,the sub-index $t$, means that the commutators are equal time computed.
The correct way to write the fields and the conjugate momenta is only: $$\phi(x) = \frac{1}{(2\pi)^{3}} \int \frac{d^{3}p}{\sqrt{2 \omega_{p}}}\left(a(p) e^{-ipx} + b^{\dagger} (p) e^{ipx} \right) (2) $$ $$\phi^{\dagger}(x) = \frac{1}{(2\pi)^{3}} \int \frac{d^{3}p}{\sqrt{2 \omega_{p}}}\left(a^{\dagger}(p) e^{+ipx} + b (p) e^{-ipx} \right) (3) $$ $$\pi^{\dagger}(x) = \partial_{0} \phi(x)=\frac{-i}{(2\pi)^{3}} \int \frac{d^{3}p}{\sqrt{2 \omega_{p}}} \omega_{p} \left( a(p) e^{-ipx} - b^{\dagger} (p) e^{ipx} \right) (4) $$ $$\pi(x) = \partial_{0} \phi^{\dagger}(x)=\frac{-i}{(2\pi)^{3}} \int \frac{d^{3}p}{\sqrt{2 \omega_{p}}} \omega_{p} \left( -a^{\dagger}(p) e^{ipx} + b(p) e^{-ipx} \right) (5)$$ From these expressions you can obtain the two types operators (note that up to now i'm not calling it "creation" and "annihilation" operators) in terms of the fields and the conjugate momenta: $$a(p)= \int \frac{d^{3} x}{\sqrt{2\omega_{p}}}\left(i \pi^{\dagger} (x) + \omega_{p} \phi(x)\right) e^{ipx} $$ $$a^{\dagger}(p)= \int \frac{d^{3} x}{\sqrt{2\omega_{p}}}\left(-i \pi(x) + \omega_{p} \phi^{\dagger}(x)\right) e^{-ipx}$$ $$b(p)= \int \frac{d^{3} x}{\sqrt{2\omega_{p}}}\left(i \pi(x) + \omega_{p} \phi^{\dagger}(x)\right) e^{ipx}$$ $$b^{\dagger}(p) = \int \frac{d^{3} x}{\sqrt{2\omega_{p}}}\left(-i \pi^{\dagger} (x) + \omega_{p} \phi(x)\right) e^{-ipx}$$ With the expressions above, one can compute the commutators between the operators $a$ and $b$ (and their dagger), finding that the only non-zero commutators in momentum space are: $$\boxed{[a(k),a^{\dagger}(p)]_{t}=[b(k),b^{\dagger}(p)]_{t} = \delta^{3} (k-p)} $$ The point is that only now, looking to the boxed relation above, you can identify $a$, ($a^{\dagger}$),$b$,($b^{\dagger}$) as the annihilation (creation) operators, referred to two different family of harmonic oscillators. Simply beacause the boxed relations are just those well-known of the harmonic oscillator.
In the end, the only way to obtain the boxed relations, and so, the only way to identify the creation and annihilation nature of the operators, is to construct your field theory with the definition of the fields and conjugate momenta (2),(3),(4),(5). If you invert the sign of the exponentials, you will not find the boxed relations, and so you can't talk about creation and annihilation operators.

  • $\begingroup$ downvoted without any explanation? What’s wrong with my answer? $\endgroup$ – MRT Aug 6 '18 at 6:17
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    $\begingroup$ I edited my old answer, in any case it would be nice if it was a general rule to write the motivation of the downvotes under the comments... $\endgroup$ – MRT Aug 6 '18 at 8:50
  • $\begingroup$ So essentially the only reason for THIS asigning of operators $a$, ($a^{\dagger}$),$b$,($b^{\dagger}$) to the exponentials $e^{\pm ipx}$ and not the inverted one is just to reach the commutator relations $[a(k),a^{\dagger}(p)] $ as occuring for harmonic oscillator,right? as I understood you correctly the inverted choice would just dont work mathematically or is there another reason? $\endgroup$ – Tim Grosskreutz Aug 6 '18 at 14:40
  • $\begingroup$ yes, if you want the commutation rules as in the case og the harmonic oscillator, you have to choose only in this way. $\endgroup$ – MRT Aug 6 '18 at 14:56

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