Quantification of Klein-Gordon field: problem of sign in the creators?

Question

I would like to understand where I am wrong in my proof of second quantization for Klein-Gordon field.

This is what I have done:

I start with the K.G equation:

$$ (\Box +m^2) \phi=0.$$

I write its Fourier decomposition: $$\phi(t,\vec{x})=\int \frac{d^3k}{(2 \pi)^3} \phi_k(t) e^{i \vec{k} \vec{x}}.$$

Using K.G equation I end up with:

$$ (\partial_t^2 +\vec{k^2}+m^2) \phi_k(t)=0.$$

I write $$\omega_k=\sqrt{\vec{k^2}+m^2}.$$

From here I understand that $\phi_k(t)$ satisfies the equation of an harmonic oscillator at each $k$.

Thus I can use my Q.M background to know that I can write:

$$\phi_k(t) \rightarrow \widehat{\phi_k}(t) = \frac{1}{\sqrt{2 \omega}}(a_k(t)+a_k(t)^\dagger).$$ (the t dependance is because I decide to work in Heisenberg picture with my operators). From now I will not put the "hat" on operators, the $a_k$ are operators.

I can prove that $a_k(t)=e^{-i \omega_k t}a_k(0)=e^{-i \omega_k t}a_k$ (it just comes from the commutation with the Hamiltonian).

I plug it in the equation I had at the beginning:

$$\phi(t,\vec{x})=\int \frac{d^3k}{(2 \pi)^3} \phi_k(t) e^{i \vec{k} \vec{x}} \rightarrow \int \frac{d^3k}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega}}(e^{-i \omega_k t}a_k+e^{+i \omega_k t}a_k^\dagger) e^{i \vec{k} \vec{x}} $$

$$\phi(t,\vec{x})=\int \frac{d^3k}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega}}(e^{-ik.x}a_k+a_k^\dagger e^{+i \omega_k t + i \vec{k} \vec{x}})$$

And here I have a problem: in the second term I don't have a scalar product of 4-vectors. I could do a change of variables but then I would avec $a^{\dagger}_{-k}e^{+ik.x}$ in the second part. And for me it is a problem, I can't just redefine $a^{\dagger}_{-k}$ by $a^{\dagger}_{k}$ because the commutation relations would be wrong in such case.

Indeed, everything commutes for different k so for me there would be a problem. This question has already been asked here Quantization of a free field: Klein-Gordon case but for those reasons I don't get the answer.

Second question: Imagine that I no longer have my problem with $a_{-k}^{\dagger}$, I would end up with a field like this:

$$\phi(t,\vec{x})=\int \frac{d^3k}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega}}(e^{-ik.x}a_k+a_k^\dagger e^{+i k.x})$$

which is a real field ($\phi(t,\vec{x})^{\dagger}=\phi(t,\vec{x}))$. But why should it be real? I don't see where we would have made such an assumption? I know we can write a lagrangian for complex field but for me there is no reason for having an hermitic operator at the end?

Extra question: When we deal with interacting fields, we write them as:

$$\phi_{int}(t,\vec{x})=\int \frac{d^3k}{(2 \pi)^3} \frac{1}{\sqrt{2 \omega}}(e^{-ik.x}a_k(t)+a_k(t)^\dagger e^{+i k.x}).$$

When I first read it I thought we did an assumption to deal with such an expression. But in fact, isn't it just a general decomposition in Fourier basis of any scalar field? It is probably a basic question but I would like to be sure of it.

Answer 1

Let's go step by step. The easiest way to get the right answer is: forget at first about quantizing. Solve the classical problem $(\Box +m^2)\phi = 0$.

In this case, Fourier transforming in position only you get (with $\hat{\phi}$ meaning the Fourier transform)

$$(\partial_t^2+\omega^2)\hat{\phi}=0.$$

where $\omega^2 = k^2+m^2$. Now with $\mathbf{k}$ fixed, solve this equation. It is a rather simple equation with solution

$$\hat{\phi}(\mathbf{k},t)=a(\mathbf{k})e^{-i\omega t}+b(\mathbf{k})e^{i\omega t}$$

Now here you can impose in the classical case the condition for $\phi$ being real. Let us do it: the reality condition for the Fourier transform is $\hat{\phi}(-\mathbf{k},t)=\hat{\phi}(\mathbf{k},t)^\ast$. In our case this becomes:

$$a(-\mathbf{k})e^{-i\omega t}+b(-\mathbf{k})e^{i\omega t}=a^\ast(\mathbf{k})e^{i\omega t}+b^\ast(\mathbf{k})e^{-i\omega t}$$

By linear independence of the exponentials you must have $a(-\mathbf{k})=b^\ast(\mathbf{k})$ and $b(-\mathbf{k})=a^\ast(\mathbf{k})$. Thus your solution is:

$$\hat{\phi}(\mathbf{k,t})=a(\mathbf{k})e^{-i\omega t}+a^\ast(-\mathbf{k})e^{i\omega t}.$$

Apply Fourier inversion to recover the original field. You get

$$\phi(\mathbf{x},t)=\int \dfrac{d^3\mathbf{k}}{(2\pi)^3}\dfrac{1}{\sqrt{2\omega_\mathbf{k}}}(a(\mathbf{k})e^{-i\omega t}+a^\ast(-\mathbf{k})e^{i\omega t})e^{i\mathbf{k}\cdot \mathbf{x}}$$

In other words we get

$$\phi(\mathbf{x},t)=\int \dfrac{d^3\mathbf{k}}{(2\pi)^3}\dfrac{1}{\sqrt{2\omega_\mathbf{k}}}(a(\mathbf{k})e^{-(i\omega t-\mathbf{k}\cdot \mathbf{x})}+a^\ast(-\mathbf{k})e^{i(\omega t+\mathbf{k}\cdot\mathbf{x})})$$

Now you might think the second term is wrong but it isn't. Split the two integrals for a moment. On the second integral perform the change of variables $\mathbf{k}\to -\mathbf{k}$. By our definition $\omega_\mathbf{k} = |\mathbf{k}|^2+m^2$ so it is unchanged. The measure is unchanged. But $\mathbf{k}\cdot \mathbf{x} \to -\mathbf{k}\cdot \mathbf{x}$. Hence

$$\phi(\mathbf{x},t)=\int \dfrac{d^3\mathbf{k}}{(2\pi)^3}\dfrac{1}{\sqrt{2\omega_\mathbf{k}}}(a(\mathbf{k})e^{-ikx}+a^\ast(\mathbf{k})e^{ikx})$$

By defining $k = (\omega_\mathbf{k},\mathbf{k})$. Quantizing means now you turn $a,a^\dagger$ the creation and annihilation operators in a Fock space.

For the complex case, you have two fields actually, $\phi,\phi^\ast$. Remove the constraint I imposed. Start with the two fields $\phi,\phi^\ast$ as independent, solve the equations, and connect then by requiring that $(\phi(x,t))^\ast = \phi^\ast(x,t)$. You will find the correct expansion and quantization is the same.