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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.

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  • $\begingroup$ I'm not sure I understand - it seems that $\phi$ is an operator, composed of the annihilation and creation operators $a$ and $a^{\dagger}$. You need to compare it with its Hermitian conjugate, not with its complex conjugate, which also transforms $a$ to $a^{\dagger}$, no? $\endgroup$ – user245141 Mar 10 '20 at 8:08
  • $\begingroup$ @yu-v Here $\vec{k}$ is a dummy variable therefore both the fourier expansion given for $\phi$ are equal. Thus I proceeded to compare the coefficients of each mode. $\endgroup$ – fogof mylife Mar 10 '20 at 8:36
  • $\begingroup$ the point is not $\vec{k}$ but how you treat $a_{\vec{k}}$ and more fundamentally $\phi$. It seems you think of them as numbers, and not operators $\endgroup$ – user245141 Mar 10 '20 at 8:43
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    $\begingroup$ The coefficient of $e^{i\vec{k}\cdot\vec{x}}$ is $(a_{\vec{k}} v_{\vec{k}} + a^\dagger_{-\vec{k}} v^*_{-\vec{k}})$. The coefficient comparison argument on its own doesn't imply that. $\endgroup$ – Subhaneil Lahiri Mar 10 '20 at 8:44
  • $\begingroup$ @yu-v - OP didn't take any sort of conjugate in the question, complex or Hermitean. $\endgroup$ – Subhaneil Lahiri Mar 10 '20 at 8:47
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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.

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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.

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  • $\begingroup$ Why should the fact that $a_k$ and $a^\dagger_k$ receiving contribution both from $k$ and $-k$ prevent us from comparing the coefficients of orthogonal modes? Can you state the condition mathematically? The relation $a_k=a^\dagger_{-k}$ is consistent with all the expression of expansion you've and I've written. So what mathematical condition prevents us, can you please clarify? $\endgroup$ – fogof mylife Mar 10 '20 at 13:01
  • $\begingroup$ Also see that, assuming $v_k(t)=e^{i\omega_kt}$, one can define $$\begin{aligned} a(k) &=\int d^{3} x e^{i k \cdot x} i \stackrel{\leftrightarrow}{\partial_{0}} \phi(x) \\ a^{\dagger}\left(k^{\prime}\right) &=\int d^{3} x \phi(x) i \stackrel{\leftrightarrow}{\partial_{0}} e^{-i k^{\prime} \cdot x} \end{aligned}$$. Thus the condition $a_{-k}=a_k^\dagger$ is consistent with these definitions. $\endgroup$ – fogof mylife Mar 10 '20 at 13:06
  • $\begingroup$ it doesn't prevent comparing coefficient for orthogonal modes. it just says that you must take into account all the contribution to each mode. what exactly are you comparing? can you write it explicitly? $\endgroup$ – user245141 Mar 10 '20 at 13:12
  • $\begingroup$ Please see edit in the question. $\endgroup$ – fogof mylife Mar 10 '20 at 13:21
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    $\begingroup$ I've edited my response. I think that you miss the fact that the integral introduces $e^{ikx}$ twice for each value of $k$. Therefore both this contributions must be taken into account. Just re-writing the integral as I wrote in my Eq.(1), which I hope you can see is exactly the same integral, makes it clearer $\endgroup$ – user245141 Mar 10 '20 at 13:49
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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}

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  • $\begingroup$ Dagger means taking hermitian conjugate: transposing the matrix (not its elements) and taking the complex conjugate. $\endgroup$ – Vadim Mar 10 '20 at 11:30
  • $\begingroup$ When talking about a two-level system $a$ and $a^\dagger$ correspond to raising and lowering operators. This is not QFT - it is basic quantum mechanics. $\endgroup$ – Vadim Mar 10 '20 at 11:44
  • $\begingroup$ A scalar field is a collection of harmonic oscillator of all possible momenta, $\vec{k}$ ranging from $-\infty$ to $\infty$. Now a creating a particle with momentum $\vec{k}$ is equivalent in field theory to annihilating an anti-particle with momentum $-\vec{k}$. And due to the fact that for a real scalar field both particles and anti-particles are the same, I am led to believe that, physically one can equate the creation operator with the annihilation operator with opposite momentum. Thus I am not sure whether the matrix representation you are suggesting works for a real scalar field. $\endgroup$ – fogof mylife Mar 10 '20 at 12:00
  • $\begingroup$ I am not suggesting anything - just reminding you the basic QM. I recommend using the corresponding volume of the Feynmann's lectures for getting the solid background. Otherwise you will get completely lost in the QFT. $\endgroup$ – Vadim Mar 10 '20 at 12:07

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