# How to explain the field result of Klein Gordon field in QFT?

I derived the equations of Klein Gordon field, and I find a statement like this:

In quantum field theory, the wave functions that could have had both positive and negative probabilities are used as prefactors in formulae for quantum fields and the positive-energy (and positive-probability) and negative-energy (and negative-probability) solutions for the wave function are treated asymetrically. The latter must be multiplied by the creation operator and the former by the annihilation operators.

In effect, it means that in quantum field theory, we may "create" an arbitrary number of particles in wave functions that are allowed by the first-quantized (one-particle) quantum mechanical theory but we are only allowed to use the positive-energy (positive-probability) wave function to excite the vacuum. The negative-energy ones are multiplied by annihilation operators which annihilate the vacuum so we get no state. (How does QFT interpret the Negative probability problem of the real scalar fields' Klein-Gordon equation?)

The similar statement can also be found in QFT by Peskin P. 26.

My questions are as follows:

$$\phi \left(\overset{\rightharpoonup }{x},t\right)=\int \frac{d^3\overset{\rightharpoonup }{p}}{(2 \pi )^3}\cdot \left[e^{i p\cdot x} a^{\dagger }{}_{\overset{\rightharpoonup }{p}}+a_{\overset{\rightharpoonup }{p}} e^{-i p\cdot x}\right]\cdot \frac{1}{\sqrt{2 e_{\overset{\rightharpoonup }{p}}}}$$

1. I understand that creation operator is the coefficient of negative frequency single wave function, but there is still negative frequency, how this $$e^{i p\cdot x} a^{\dagger }{}_{\overset{\rightharpoonup }{p}}$$ quantity works?
2. I have tried other combinations, and find out that the above solution with creation times negative and annihilation times positive is the only possible to insure the commutator. but why this is the only possible way? Any deep reasons?
3. I still don't see how QFT solved the negative energies except for it doesn't mention this issue.
• Actually, if you read the passage in Peskin carefully, you'll probably understand why it is like this. Oct 5, 2016 at 4:38
• could u explain this a bit more. i have read it many times, but i can't capture its meaning. Oct 5, 2016 at 6:18
• I think Xavier has done that. Oct 5, 2016 at 8:37

Although the wave function $\phi_0$ in the old formalism and the field operator $\phi$ in QFT both satisfy the K-G equation, their consequences are very different. As a wave function, the expansion of $\phi_0$ in energy eigenstates has the form $\phi_0(x)=\Sigma c_n(x) e^{-iE_n t}$, So a term like $a^*(p)e^{ip\cdot x}$ means the existence of negative energy. But in QFT, the operator (where $a^\dagger_c$ corresponds to the antiparticle) $$\phi(x)\propto\int \frac{d^3 p}{\sqrt{2E}}\left[a(p) e^{-ip\cdot x}+a_c^\dagger(p) e^{ip\cdot x})\right]$$ has no such interpretation. In free theory, if we act $\phi^\dagger$ on the vacuum state: $$\phi^\dagger(x)|0\rangle\propto\int \frac{d^3 p}{\sqrt{2E}} e^{ip\cdot x}|\vec{p}\rangle=\int \frac{d^3 p}{\sqrt{2E}} e^{-i\vec{p}\cdot \vec{x}}\left(e^{iEt}|\vec{p}\rangle\right)$$ we get the superposition of momentum eigenstates in the non-relativistic limit, with weighting factor $e^{-i\vec{p}\cdot\vec{x}}$. So $\phi^\dagger(x)|0\rangle$ is just the position eigenstate of a particle in the Heisenberg picture. Similarly, $\phi(x)|0\rangle$ is the position eigenstate of an antiparticle. Therefore the wave function for a particle should be $$\langle x|\psi\rangle_{particle}=\langle0|\phi(x)|\psi\rangle_{particle}$$ which satisfies the K-G equaiton but only $a(p)$ and positive energy are involved here. And the wave function for an antiparticle is $$\langle x|\psi\rangle_{antiparticle}=\langle0|\phi^\dagger(x)|\psi\rangle_{antiparticle}$$ where only $a_c(p)$ and positive energy are involved.