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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 thoose 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.

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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.

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  • $\begingroup$ Thank you for your answer. I just have two questions. If I remember well, once we have the exact solution for the classical field we do the process of quantization by saying that the impulsion and position fields now turnd in operators must have the same commutator as the poisson bracket for the classical case. It allows us to find that $[a,a^\dagger]=I$. Is it enough to ensure that it is creation & annihilation operator ? Just because they follow the nice commutation relation ? $\endgroup$ – StarBucK Oct 6 '17 at 23:13
  • $\begingroup$ And second question : When we turn the $\phi$ in operator, we only say that the $a(\vec{k})$ (and its conjugate) are operators. Why for example the $e^{-ikx}$ would'nt also become an operator ? Ok the question may be strange but I think it would help me to precisely understand what we must and what we must not turn in operator when we quantize in a general case. $\endgroup$ – StarBucK Oct 6 '17 at 23:14
  • $\begingroup$ I know this is an old question, but this is driving me insane, @user1620696: it might be a stupid question but why is the measure unchanged in the change of variable $\mathbf{k}\rightarrow\mathbf{-k}$? it seems to me that it should change sign $\endgroup$ – user2723984 Oct 11 '18 at 8:14

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