I have a general question about heuristic way of QFT to introduce creation and annihilation operators: The Klein-Gordon field is introduced as continuous interference of plane waves $\mathrm{e}^{i(\omega_kt-\vec{k}\cdot\vec{x})}$ with positive energy (resp $\mathrm{e}^{-i(\omega_kt-\vec{k}\cdot\vec{x})}$ with negative energy).

The annihilation operator is $\hat{b}_{\mathbf{p}}^{\dagger}$ and creation operator is $\hat{a}_{\mathbf{p}}$.

The KG field expanded in terms of creation/annihilation operators is

$$\varphi(\vec{x},t) = \int\frac{d^Dk}{\sqrt{(2\pi)^D2\omega_k}}\left[\hat{a}(\vec{k})\mathrm{e}^{-i(\omega_kt-\vec{k}\cdot\vec{x})} + \hat{b}^\dagger(\vec{k})\mathrm{e}^{i(\omega_kt-\vec{k}\cdot\vec{x})}\right].$$

My question is why do the anhilation operator correspond to the plane wave with positive energy and the creation operator to the plane wave with negative energy? How can it be heuristically explained?

Or equivalently why it is told that the anhilation operates on plane waves with positive energy?

  • $\begingroup$ Closely related: physics.stackexchange.com/q/321327/92058 $\endgroup$ – tparker Oct 5 '18 at 20:52
  • $\begingroup$ @KarlPeter I think you interchanged the positive and negative energy plane waves .$\hat{E}=i\frac{\partial}{\partial t}$ gives the opposite sign of energy than the one you assumed $\endgroup$ – aitfel Sep 11 '19 at 11:22

It has to do with Feynman-Stuckelberg interpretation of negative energy solutions as outgoing antiparticles.

A plane wave mode with negative energy is given by $$a(\vec p)e^{-i(Et-\vec p\cdot\vec x)},\quad E<0,$$ which can be written as $$a(\vec p)e^{i(Et+\vec p\cdot\vec x)},\quad E>0.$$ Since the plane wave expansion is a sum over all 3-momenta, we can write it as $$\int d^3pa(\vec p)e^{i(Et+\vec p\cdot\vec x)}=\int d^3pa(-\vec p)e^{i(Et-\vec p\cdot\vec x)}=\int d^3pa(\vec p)d^3pe^{ipx},\quad E>0.$$

However, all particles in the theory must have positive energy so we need to given an interpretation to negative energy/frequency modes. Changing sign of time is equivalent to changing sign of charge and reflecting 3-momentum of the particle. Hence, if we second quantize the theory and the coefficients $a,a^\dagger$ are associated to particles, then the coefficients corresponding to negative frequencies are associated to antiparticles, $b,b^\dagger$. Moreover, these modes correspond to a reflected 3-momentum (negative time) and we interpret this as outgoing antiparticles as long as we interpret the positive frequency modes as incoming particles. By outgoing we mean a particle that is created in the system and then leaves it, therefore must be associated to a creation operator. An incoming particle enters the system and is absorbed by it, thus being associated to a annihilation operator.

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  • $\begingroup$ I can't understand what you did in the last equation where you're doing integration in 6 dimension and the same dummy variable is used twice. Is that a typo where you meant to use the factor $e^{E\times(-t)}$ i.e. you inverted the sign of t. $\endgroup$ – aitfel Sep 11 '19 at 11:23

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