# Quantizing Klein Gordon Field: Sign Problem

I'm trying to re-derive the Quantization of the Klein Gordon Field but I'm running into sign problems.

My starting point is: $$\phi(x,t) = \frac{1}{(\sqrt{2 \pi})^3} \int \tilde{\phi}(k,t) e^{i kx} dk \tag1$$ Where $$x$$ and $$k$$ are 3D vectors.

The idea then is to see $$\tilde{\phi}(k,t)$$ as the position operator of a quantum harmonic oscillator of frequency $$\omega_k = \sqrt{m^2 + |k|^2}$$ in the Heisenberg picture.

We know from Quantum mechanics that in the Schrodinger picture we have: $$\tilde{\phi}(k) = \frac{1}{\sqrt{2 \omega_k}}(a^{+}_{k} + a_{k})\tag2$$

Adding the Time dependency we get: $$\tilde{\phi}(k,t) = \frac{1}{\sqrt{2 \omega_k}}(a^{+}_{k} e^{i\omega_k t} + a_{k}e^{-i\omega_k t})\tag3$$

By plugging this expression into the first integral, we get:

$$\phi(x,t) = \frac{1}{(\sqrt{2 \pi})^3} \int \frac{dk}{\sqrt{2 \omega_k}}(a^{+}_{k} e^{i\omega_k t + i kx} + a_{k}e^{-i\omega_k t + i kx}) \tag4$$

Then we make the change of variables $$k \to -k$$ in the first part of the integral and get:

$$\phi(x,t) = \frac{1}{(\sqrt{2 \pi})^3} \int \frac{dk}{\sqrt{2 \omega_k}}(a^{+}_{-k} e^{i\omega_k t -i kx} + a_{k}e^{-i\omega_k t + i kx}) \tag5$$

Questions:

1. What is wrong with this derivation ? In all textbooks and course notes, it seems that people are getting $$a^{+}_{-k}$$ and not $$a^{+}_{k}$$ in that last expression. What did I miss ?

2. Another ingredient that people use is that since they want $$\phi(x,t)$$ to be Hermitian then this implies that $$\tilde{\phi}(k,t)^{+} = \tilde{\phi}(-k,t)$$. I believe this implies that $$a^{+}_{-k} = a^{+}_{k}$$ which would solve my first question. But then I wonder why don't people mention that more explicitly?

This would also imply that $$\tilde{\phi}(k,t) = \tilde{\phi}(-k,t)$$? which means that creating a particle of momentum $$k$$ is the same as creating a particle of momentum $$-k$$, this seems very odd to me, any interpretations? Does that mean that when a particle of momentum $$k$$ is created another one of momentum $$-k$$ is automatically created? Would this make it impossible to create just one particle of momentum $$k$$.

Please don't show me other derivations, I just want to understand why this one doesn't seem to work.

• – AccidentalFourierTransform Jul 14 at 21:00
• It's not really related. – curiosity Jul 15 at 1:15
• Related $\neq$ duplicate – Kyle Kanos Jul 16 at 12:25

While your computation is valid, the reason for this difference is in your ansatz \begin{align} \phi(x,t) = \frac{1}{(\sqrt{2 \pi})^3} \int \tilde{\phi}(k,t) e^{i kx} dk, \end{align} which together with the Hermiticity requirement $$\phi(x, t) = \phi^*(x, t)$$ implies that: \begin{align} \tilde{\phi}(k, t) = \tilde{\phi}^*(-k, t). \end{align}

Using \begin{align} \tilde{\phi}(k) = \frac{1}{\sqrt{2 \omega_k}}(a^{+}_{k} + a_{k}) \end{align} it is easy to see that Hermiticity condition is satisfied if: \begin{align} (a_{-k}^{+} + a_{-k})^\dagger = a_k^{+} + a_k. \end{align}

Assuming that $$(a_{q}^{+})^\dagger \equiv a_k$$ for some $$q$$ and $$k$$ (i.e., that $$a$$ and $$a^{+}$$ a creation-annihilation operators) we get: \begin{align} \left( a_{-k}^{+} \right)^\dagger & = a_k, \\ \left( a_{-k} \right)^\dagger &= a_k^{+}. \end{align}

It makes sense then to rename: \begin{align} b_k & \equiv a_k, \\ b_k^{+} & \equiv a_{-k}^{+} \end{align} so that $$(b_k)^\dagger = b_k^\dagger$$, and your last equation becomes: \begin{align} \phi(x,t) = \frac{1}{(\sqrt{2 \pi})^3} \int \frac{dk}{\sqrt{2 \omega_k}}\left( b^{+}_k e^{i\omega_k t -i kx} + b_{k} e^{-i\omega_k t + i kx} \right) \end{align}

• I don't get why this should be mandatory, I should be able to impose some conditions on $\tilde{\phi}$ at the end. Also it is not clean to me that I can continue my derivation from that. $f$ is not the solution of a harmonic oscillator but $f(k,t)-f^{*}(-k,t)$ is which is very different. – curiosity Jul 14 at 20:36
• Sorry, you are absolutely right. I have rewrote my answer to address this. – Darkseid Jul 14 at 20:48
• Additionally, I must ask where your argument about harmonic oscillator comes from? Almost all QFT textbooks make reference to harmonic oscillators (because of creation-annihilation operators), but I never seen the form of defining equation for $\tilde{\phi}$ from your question before. – Darkseid Jul 14 at 20:51
• I don't understand your argument after the "Assuming" part, typos ? – curiosity Jul 14 at 22:25
• The argument is that in order for your solution to give hermitian $\phi(x, t)$, $(a_k)^\dagger$ has to be equal to $a_{-k}^\dagger$. Perhaps I should replace $\dagger$ with $+$ for clarity. – Darkseid Jul 14 at 22:36

Let me start by giving a reason why $$a_k$$ and not $$a_{-k}$$ is multiplied by $$e^{-ikx}$$. Let's think of the transnational operator $$U(b)$$ that translate the system by some distance $$b$$. translating the field $$\phi(x,t)$$, $$U^\dagger(b) \phi(x,t) U(b) = \phi(x - b,t).$$ It's also not hard to show that, $$U^\dagger(b) a^\dagger_k U(b) = e^{ikb} a^\dagger_k,$$ $$U^\dagger(b) a_k U(b) = e^{-ikb} a_k,$$ which can be seen using that $$a^\dagger_k |0> = |k>$$ and that $$U^\dagger(b)|k> = e^{ikb} |k>$$. Putting these two facts together we can find that, $$U^\dagger(b) a^\dagger_{-k} U(b) e^{-ikx} = a^\dagger_{-k} e^{-ik(x+b)},$$ which is not how we would expect the field $$\phi(x)$$ to transform. This alone can be used to determine the expansion of the field $$\phi(x)$$ in terms of $$a_k$$ and $$a^\dagger_{k}$$.

Now let's look at the Hamiltonian for the KG equation, $$H = (1/2)\int dx \ \pi^2(x,t) + (\nabla \phi)^2 + m^2\phi^2(x,t),$$ where $$\pi(x,t)$$ is the momentum conjugate to $$\phi(x,t)$$. FT this expression we would have, $$H = (1/2) \int dk \ \tilde \pi(k,t) \tilde \pi(-k,t) + w_k^2 \tilde \phi(k,t) \tilde \phi(-k,t).$$ Now we can use the regular trick of defining rasing and lowering operators, and you also see the problem with your equation 2. If you want the Hamiltonian to have the usual form of $$w_k a^\dagger _k a_{k}$$, you need to have, $$\tilde \phi(k,t) = \frac{1}{\sqrt{w_k}} (a_k + a^\dagger_{-k}).$$ Try it!