Reality condition for creation and annihilation operators in QFT

Question

Consider a real scalar field, $$\phi(x)=\int\frac{d^3k}{(2\pi)^{3/2}}\left(a_{\vec{k}}v_{\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}+a_{\vec{k}}^\dagger v^*_{\vec{k}}(t)e^{-i\vec{k}\cdot\vec{x}}\right),$$ here the function $v_{\vec{k}}(t)$ satisfies the normalization condition, $$v_{\vec{k}}\dot{v}^*_{\vec{k}}-v^*_{\vec{k}}\dot{v}_{\vec{k}}=i.$$ Now, changing the variable $\vec{k}\rightarrow-\vec{k}$ in the fourier expansion we have,$$\phi(x)=\int\frac{d^3k}{(2\pi)^{3/2}}\left(a_{-\vec{k}}v_{-\vec{k}}(t)e^{-i\vec{k}\cdot\vec{x}}+a_{-\vec{k}}^\dagger v^*_{-\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}\right).$$ Comparing the coeeficeints of $e^{i\vec{k}\cdot\vec{x}}$ and its complex conjugate we have, $$a_{\vec{k}}v_{\vec{k}}=a^\dagger_{-\vec{k}}v^*_{-\vec{k}}.$$ From this can it be said that $v_{\vec{k}}=v^*_{-\vec{k}}$ and $a_{\vec{k}}=a^\dagger_{-\vec{k}}$ because both the operators and functions should be identified separately? If yes then why and if not then why?

Edit: As a comment mentioned that it is not clear what two expressions, I am comparing, I shall write it explicitly, $$\begin{align} \phi(x)&=\int\frac{d^3k}{(2\pi)^{3/2}}\left(a_{\vec{k}}v_{\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}+a_{\vec{k}}^\dagger v^*_{\vec{k}}(t)e^{-i\vec{k}\cdot\vec{x}}\right)\\&=\int\frac{d^3k}{(2\pi)^{3/2}}\left(a_{-\vec{k}}v_{-\vec{k}}(t)e^{-i\vec{k}\cdot\vec{x}}+a_{-\vec{k}}^\dagger v^*_{-\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}\right)\end{align}$$. Thus we can write, $$a_{\vec{k}}v_{\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}+a_{\vec{k}}^\dagger v^*_{\vec{k}}(t)e^{-i\vec{k}\cdot\vec{x}}=a_{-\vec{k}}v_{-\vec{k}}(t)e^{-i\vec{k}\cdot\vec{x}}+a_{-\vec{k}}^\dagger v^*_{-\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}.$$ Notice that this equality holds for each $k$ mode separately and there is no integration anymore. Now we compare the coefficients of the modes from LHS and RHS of the equation.

Answer 1

Just to encourage physical intuition over mathematical manipulation, I make the following note. Let's assume, for the sake of argument, the claim made by OP is correct (which is not true). Then we have a creation operator identified to an annihilation operator. This is problematic. To see this first define a vacuum state as the state annihilated by the annihilation operator, as usual, $$a_k|0\rangle=0,\quad\forall k\in\mathbb{R}^3.$$ Now try to create a one particle state by acting on the vacuum with a creation operator, i.e., $a_k^\dagger|0\rangle$. But wait, we have an identification between creation and annihilation operator, thus, $$a_k^\dagger|0\rangle=a_{-k}|0\rangle=0.$$ Therefore in OP's theory the Fock space consists of only one state and that is the vacuum. There is no other state in the theory. Therefore, on physical grounds we must reject the identification made by OP if we want a viable QFT with particles. Now what mathematical mistake did led OP to this erroneous conclusion, I believe @yu-v has done a good job explaining that.

Answer 2

I believe (also thanks to @SubhaneilLahiri comments in the discussion) that your mistake likes in the double expansion of $\phi$ both in Fourier base and in the creation and annihilation operators. When writing

$$ \phi(x) = \int \frac{d^3k}{(2\pi)^{3/2}} \left[a_{\vec{k}}v_{\vec{k}}(t)e^{-i\vec{k}\vec{x}}+a^{\dagger}_{\vec{k}}v^*_{\vec{k}}(t)e^{i\vec{k}\vec{x}}\right]$$

we explicitly expand the coefficients for the annihilation and creation operators each in Fourier basis. Since we integrate over all $k$, as you noted, it is just a dummy variable. Equally, we could have written

$$ \phi(x) = \int \frac{d^3k}{(2\pi)^{3/2}} \left[a_{\vec{k}}v_{\vec{k}}(t)+a^{\dagger}_{-\vec{k}}v^*_{-\vec{k}}(t)\right]e^{-i\vec{k}\vec{x}} \;\;\;(1)$$

this is exactly the same integral. Equating the expansion coefficients of $a_k$ and $a^{\dagger}_k$ separately miss the fact that both get contribution from $k$ and $-k$. Now, in order for the field to be real-valued, we just need that $$\left[a_{\vec{k}}v_{\vec{k}}(t)+a^{\dagger}_{-\vec{k}}v^*_{-\vec{k}}(t)\right]^{\dagger}= \left[a_{-\vec{k}}v_{-\vec{k}}(t)+a^{\dagger}_{\vec{k}}v^*_{\vec{k}}(t)\right]$$ which naturally holds (it is not surprising - we constructed it in such a way).

Edit following the clarification in the question: your assertion that "we can write, $$a_{\vec{k}}v_{\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}+a_{\vec{k}}^\dagger v^*_{\vec{k}}(t)e^{-i\vec{k}\cdot\vec{x}}=a_{-\vec{k}}v_{-\vec{k}}(t)e^{-i\vec{k}\cdot\vec{x}}+a_{-\vec{k}}^\dagger v^*_{-\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}.$$ Notice that this equality holds for each $k$ mode separately" is incorrect. You are ignoring the fact that you also have, in the integral in question, the terms $$a_{-\vec{k}}v_{-\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}+a_{-\vec{k}}^\dagger v^*_{-\vec{k}}(t)e^{i\vec{k}\cdot\vec{x}}$$ on the left-hand-side as well. You don't get to choose which expansion coefficient you compare, you must consider all the terms which mutiply $e^{ikx}$. In that sense, the writing of the integral as a single expansion in $k$ as I wrote in eq.(1) should make it clearer.

Answer 3

The argument presented in the question would work, if we were dealing with a Fourier expansion of a scalar function, where we have a uniqueness theorem for the expansion coefficients. In this case we are in addition in the operator space: \begin{equation} \phi(x) = \psi(x) + \psi^\dagger (x) \end{equation} If you think of $\psi(x)$ and $\psi^\dagger(x)$ (or $a$ and $a^\dagger$) as 2-by-2 matrices, one with the only non-zero element in its lower left corner and the other with the non-zero element in the upper right corner, it is clear that they cannot be equated: \begin{equation} \hat{a} = \begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix}, \hat{a}^\dagger = \begin{pmatrix} 0 & 0 \\ 1 & 0 \\ \end{pmatrix} \end{equation}