# Showing commutation relations

• JD_PM
PS: I've been studying a really similar problem (M&S chapter 1) and they indeed make the student to first derive and expression for ##a##. I am trying to solve this problem in a completely analogous way and any help would be highly appreciatedf

#### JD_PM

Homework Statement
Given the commutation relations

\begin{equation*}
[\chi(\eta, \vec x), \chi(\eta, \vec y)] = [\dot \chi(\eta, \vec x), \dot \chi(\eta, \vec y)] = 0,
\end{equation*}

\begin{equation*}
[\chi(\eta, \vec x), \dot \chi(\eta, \vec y)] = i \delta(\vec x - \vec y)
\end{equation*}

Where ##\eta## stands for conformal time and ##\chi## is given by

\chi = \int \frac{d^3 k}{(2 \pi)^{3/2}} \left( a_{\vec k} \chi_{\vec k} e^{i \vec k \cdot \vec x} + a^{\dagger}_{\vec k}\chi^*_{\vec k} e^{-i \vec k \cdot \vec x} \right) \tag{1}

Show that

\begin{equation*}
[a_{\vec k}, a_{\vec k'}]=0, \quad [a_{\vec k}, a_{\vec k'}^{\dagger}]= \delta^{(3)}(\vec k -\vec k')
\end{equation*}
Relevant Equations
N/A
I would say we first need to take the inverse Fourier transform of ##\chi## and associated quantities i.e.

\begin{equation*}
\chi_{\vec k} = \int d^3 \vec x \left( a_{\vec k} \chi e^{-i \vec k \cdot \vec x} + a^{\dagger}_{\vec k}\chi^* e^{i \vec k \cdot \vec x} \right) \tag{2}
\end{equation*}

\dot \chi_{\vec k} = \int d^3 \vec x \left( a_{\vec k} \dot \chi e^{-i \vec k \cdot \vec x} + a^{\dagger}_{\vec k}\dot \chi^* e^{i \vec k \cdot \vec x} \right) \tag{3}

\chi_{\vec k}^* = \int d^3 \vec x \left( a_{\vec k}^{\dagger} \chi^* e^{i \vec k \cdot \vec x} + a_{\vec k} \chi e^{-i \vec k \cdot \vec x} \right) \tag{4}

\dot \chi_{\vec k}^* = \int d^3 \vec x \left( a_{\vec k}^{\dagger} \dot \chi^* e^{i \vec k \cdot \vec x} + a_{\vec k} \dot \chi e^{-i \vec k \cdot \vec x} \right) \tag{5}

At this point I should perform a linear combination of ##(2), (3), (4), (5)##, obtaining ##a_{\vec k}## in terms of ##\chi## and ##\dot \chi## (as well as ##a_{\vec k}^{\dagger}## in terms of ##\chi## and ##\dot \chi##).

However, I do not see such linear combination. Might you please shed some light? I should be able to continue once I get such expressions.

Thank you!

I haven't worked out the problem, but what do you get if you plug (1) and its derivative into the commutation relations? It's something you could try.

Hi @vela, my apologies for the late reply.

That is an approach we could indeed try but I've been hinted to first find ##a_{\vec k}## in terms of ##\chi## and ##\dot \chi##. The latest I've been trying is assuming the ansatz

$$a = \int d^3 x (b(k, \eta) \chi_{\vec k}(x) + c(k, \eta) \dot{\chi}_{\vec k}(x)) e^{-ik \cdot x} \tag{*}$$

Then plugging ##(2)## and ##(3)## into ##(*)##. Doing the same with the conjugate of ##(*)## should be enough to find the functions ##b(k, \eta)## and ##c(k, \eta)## but I am not succeeding.

My question is, does this method look correct? If yes I'll post all steps so that we find my mistake.

PS: I've been studying a really similar problem (M&S chapter 1) and they indeed make the student to first derive and expression for ##a##. I am trying to solve this problem in a completely analogous way and any help would be highly appreciated

Solution

In your original post, are the exponentials supposed to be ##e^{i \vec k \cdot \vec x}## where both ##\vec k## and ##\vec x## are three-vectors, or ##e^{i k \cdot x}## where ##k \cdot x = k_{\mu}x^{\mu}##? If it's not the latter, where does the dependence on ##\eta## come in?

(It's been over 20 years since I've done calculations of this sort, so I'm just kind of throwing stuff out there for you to think about.)

JD_PM
I think @vela is right, if your expression depends on ##e^{i\vec{k}\vec{x}}## then shouldn't ##\dot{\chi}=0##?
I would propose that, first of all, you compute the expression for ##\dot{\chi}(x, t)## in terms of ##a## and ##\chi_k##.

JD_PM
In your original post, are the exponentials supposed to be ##e^{i \vec k \cdot \vec x}## where both ##\vec k## and ##\vec x## are three-vectors, or ##e^{i k \cdot x}## where ##k \cdot x = k_{\mu}x^{\mu}##? If it's not the latter, where does the dependence on ##\eta## come in?
It is indeed not the latter case, that is why I am struggling to follow the same approach M&S used to solve the analogous problem.

The ##\eta## dependence comes from how ##\chi## is defined i.e. ##\chi := a \delta \phi##, where ##a## is simply a function of time ##\eta## and ##\delta \phi## is a function of time ##\eta## and space ##x##.

Here's some background information

(It's been over 20 years since I've done calculations of this sort, so I'm just kind of throwing stuff out there for you to think about.)

No problem, your help is appreciated

Hi @Gaussian97 is nice to come across you again

I think @vela is right, if your expression depends on ##e^{i\vec{k}\vec{x}}## then shouldn't ##\dot{\chi}=0##?

I do not think so. It is true that the time derivative does not involve the exponential but it does hit ##\chi_{\vec k}## (which is defined ##\chi := a \delta \phi##) so we get

\begin{equation*}
\dot \chi = \int \frac{d^3 k}{(2 \pi)^{3/2}} \left( a_{\vec k} \dot \chi_{\vec k} e^{i \vec k \cdot \vec x} + a^{\dagger}_{\vec k}\dot \chi^*_{\vec k} e^{-i \vec k \cdot \vec x} \right) \neq 0
\end{equation*}

I would propose that, first of all, you compute the expression for ##\dot{\chi}(x, t)## in terms of ##a## and ##\chi_k##.

I think you are hinting at the same approach I wrote at the bottom of my OP: to obtain ##a_{\vec k}## in terms of ##\chi## and ##\dot \chi##. The issue is that I do not see the particular linear combination one should take to get such an expression.

Mmm... Okay, then I have no idea how to do it.
One idea could be to compute ##\ddot{\chi}## and use equation 43, I don't think this will give any useful formula, but you can try it if you want.

Another idea is to prove the opposite, so starting with the commutator for the ##a## prove the commutator for the fields, maybe this can give you some intuition on the intermediate steps you need to prove first.

JD_PM
Another idea is to prove the opposite, so starting with the commutator for the ##a## prove the commutator for the fields, maybe this can give you some intuition on the intermediate steps you need to prove first.

I found the reverse statement much easier to prove, as it is more plugging-and-chugging/algorithmic.

Let us show ##[\chi(\vec x), \dot \chi (\vec y)] = i \delta(\vec x - \vec y)##. Plugging the given mode expansion for ##\chi## (and ##\dot \chi##)

\begin{align*}
[\chi(x), \dot\chi(y)] &= \int \frac{d^3 \vec k d^3 \vec k'}{(2\pi)^3}\Big( a_{\vec k}a_{\vec k'}\chi_{\vec k} \dot \chi_{\vec k'} e^{i(\vec k \cdot \vec x + \vec k' \cdot \vec y)} + a_{\vec k}^{\dagger} a_{\vec k'}^{\dagger} \chi_{\vec k}^* \dot \chi_{\vec k'}^* e^{-i(\vec k \cdot \vec x + \vec k' \cdot \vec y)} + a_{\vec k}a_{\vec k'}^{\dagger}\chi_{\vec k} \dot \chi_{\vec k'}^* e^{i(\vec k \cdot \vec x - \vec k' \cdot \vec y)} + a_{\vec k}^{\dagger} a_{\vec k'}\chi_{\vec k}^* \dot\chi_{\vec k'} e^{i(-\vec k \cdot \vec x + \vec k' \cdot \vec y)}\\
&-a_{\vec k'}a_{\vec k} \dot \chi_{\vec k'} \chi_{\vec k} e^{i(\vec k' \cdot \vec y + \vec k \cdot \vec x)} - a_{\vec k'}^{\dagger} a_{\vec k}^{\dagger} \dot \chi_{\vec k'}^* \chi_{\vec k}^* e^{-i(\vec k' \cdot \vec y + \vec k \cdot \vec x)} - a_{\vec k'}^{\dagger}a_{\vec k} \dot \chi_{\vec k'}^* \chi_{\vec k} e^{i(\vec k \cdot \vec x - \vec k' \cdot \vec y)} - a_{\vec k'}a_{\vec k}^{\dagger} \dot\chi_{\vec k'} \chi_{\vec k}^* e^{i(-\vec k \cdot \vec x + \vec k' \cdot \vec y)}\Big) \\
&= \int \frac{d^3 \vec k d^3 \vec k'}{(2\pi)^3}\Big( [a_k, a_{k'}]\chi_{\vec k} \dot \chi_{\vec k'} e^{i(\vec k \cdot \vec x + \vec k' \cdot \vec y)} + [a_k^{\dagger}, a_{k'}^{\dagger}]\chi_{\vec k}^* \dot \chi_{\vec k'}^* e^{-i(\vec k \cdot \vec x + \vec k' \cdot \vec y)} + [a_k, a_{k'}^{\dagger}]\chi_{\vec k} \dot \chi_{\vec k'}^* e^{i(\vec k \cdot \vec x - \vec k' \cdot \vec y)} + [a_k^{\dagger}, a_{k'}]\chi_{\vec k}^* \dot\chi_{\vec k'} e^{i(-\vec k \cdot \vec x + \vec k' \cdot \vec y)} \Big)\\
&= \int \frac{d^3 \vec k d^3 \vec k'}{(2\pi)^3} \left( \delta^{(3)}(k-k')\chi_{\vec k} \dot \chi_{\vec k'}^* e^{i(\vec k \cdot \vec x - \vec k' \cdot \vec y)} - \delta^{(3)}(k-k')\chi_{\vec k}^* \dot\chi_{\vec k'} e^{i(-\vec k \cdot \vec x + \vec k' \cdot \vec y)} \right) \\
&= \int \frac{d^3 \vec k}{(2\pi)^3} \left( \chi_{\vec k} \dot \chi_{\vec k}^* e^{i \vec k \cdot( \vec x - \vec y)} - \underbrace{\chi_{\vec k}^* \dot\chi_{\vec k} e^{-i \vec k \cdot (\vec x - \vec y)}}_{\text{Next we use} \ \vec k \to -\vec k} \right)\\
&= \int \frac{d^3 \vec k}{(2\pi)^3} \left( \chi_{\vec k} \dot \chi_{\vec k}^* - \chi_{-\vec k}^* \dot\chi_{-\vec k} \right) e^{i \vec k \cdot( \vec x - \vec y)} \\
&= i \int \frac{d^3 \vec k}{(2\pi)^3} e^{i \vec k \cdot( \vec x - \vec y)}\\
&= i \delta^{(3)}(\vec x - \vec y)
\end{align*}

Where I used the given identity in the lecture notes ## \chi_{\vec k} \dot \chi_{\vec k}^* - \chi_{-\vec k}^* \dot \chi_{-\vec k} = i##

Mmm... Okay, then I have no idea how to do it.
If you have the time and motivation for this particular problem I would kindly suggest to have a look at #3, where I posted a really similar problem and its solution (M&S 3.1; instead of ##\chi## field we deal with the KG field).

If not I completely understand and hope to come across you in another thread soon.

To prove the original statement of this thread I am convinced it has to be possible to follow essentially the same steps one needs to solve M&S 3.1. These are

1) Find the explicit expression of ##a## in terms of ##\chi## and ##\dot \chi## (for us it is a bit harder because we are not given the explicit expression of ##a## in terms of ##\chi## and ##\dot \chi##).

2) Evaluate the ##a## commutators with the expression obtained at 1).

So the issue is 1). Studying how it is done for 3.1. one sees that, when taking the time derivative of the KG field, as the Fourier-expansion exponentials of the KG field depend on time, we can essentially take the linear combination ##\phi + \dot \phi## and we get the desired expression. However, in our particular problem this "trick" does not work because, as you both noticed, the Fourier-expansion exponentials of the ##\chi## do not depend on time... So we need another "trick".

I found the reverse statement much easier to prove, as it is more plugging-and-chugging/algorithmic.

Let us show ##[\chi(\vec x), \dot \chi (\vec y)] = i \delta(\vec x - \vec y)##. Plugging the given mode expansion for ##\chi## (and ##\dot \chi##)

\begin{align*}
[\chi(x), \dot\chi(y)] &= \int \frac{d^3 \vec k d^3 \vec k'}{(2\pi)^3}\Big( a_{\vec k}a_{\vec k'}\chi_{\vec k} \dot \chi_{\vec k'} e^{i(\vec k \cdot \vec x + \vec k' \cdot \vec y)} + a_{\vec k}^{\dagger} a_{\vec k'}^{\dagger} \chi_{\vec k}^* \dot \chi_{\vec k'}^* e^{-i(\vec k \cdot \vec x + \vec k' \cdot \vec y)} + a_{\vec k}a_{\vec k'}^{\dagger}\chi_{\vec k} \dot \chi_{\vec k'}^* e^{i(\vec k \cdot \vec x - \vec k' \cdot \vec y)} + a_{\vec k}^{\dagger} a_{\vec k'}\chi_{\vec k}^* \dot\chi_{\vec k'} e^{i(-\vec k \cdot \vec x + \vec k' \cdot \vec y)}\\
&-a_{\vec k'}a_{\vec k} \dot \chi_{\vec k'} \chi_{\vec k} e^{i(\vec k' \cdot \vec y + \vec k \cdot \vec x)} - a_{\vec k'}^{\dagger} a_{\vec k}^{\dagger} \dot \chi_{\vec k'}^* \chi_{\vec k}^* e^{-i(\vec k' \cdot \vec y + \vec k \cdot \vec x)} - a_{\vec k'}^{\dagger}a_{\vec k} \dot \chi_{\vec k'}^* \chi_{\vec k} e^{i(\vec k \cdot \vec x - \vec k' \cdot \vec y)} - a_{\vec k'}a_{\vec k}^{\dagger} \dot\chi_{\vec k'} \chi_{\vec k}^* e^{i(-\vec k \cdot \vec x + \vec k' \cdot \vec y)}\Big) \\
&= \int \frac{d^3 \vec k d^3 \vec k'}{(2\pi)^3}\Big( [a_k, a_{k'}]\chi_{\vec k} \dot \chi_{\vec k'} e^{i(\vec k \cdot \vec x + \vec k' \cdot \vec y)} + [a_k^{\dagger}, a_{k'}^{\dagger}]\chi_{\vec k}^* \dot \chi_{\vec k'}^* e^{-i(\vec k \cdot \vec x + \vec k' \cdot \vec y)} + [a_k, a_{k'}^{\dagger}]\chi_{\vec k} \dot \chi_{\vec k'}^* e^{i(\vec k \cdot \vec x - \vec k' \cdot \vec y)} + [a_k^{\dagger}, a_{k'}]\chi_{\vec k}^* \dot\chi_{\vec k'} e^{i(-\vec k \cdot \vec x + \vec k' \cdot \vec y)} \Big)\\
&= \int \frac{d^3 \vec k d^3 \vec k'}{(2\pi)^3} \left( \delta^{(3)}(k-k')\chi_{\vec k} \dot \chi_{\vec k'}^* e^{i(\vec k \cdot \vec x - \vec k' \cdot \vec y)} - \delta^{(3)}(k-k')\chi_{\vec k}^* \dot\chi_{\vec k'} e^{i(-\vec k \cdot \vec x + \vec k' \cdot \vec y)} \right) \\
&= \int \frac{d^3 \vec k}{(2\pi)^3} \left( \chi_{\vec k} \dot \chi_{\vec k}^* e^{i \vec k \cdot( \vec x - \vec y)} - \underbrace{\chi_{\vec k}^* \dot\chi_{\vec k} e^{-i \vec k \cdot (\vec x - \vec y)}}_{\text{Next we use} \ \vec k \to -\vec k} \right)\\
&= \int \frac{d^3 \vec k}{(2\pi)^3} \left( \chi_{\vec k} \dot \chi_{\vec k}^* - \chi_{-\vec k}^* \dot\chi_{-\vec k} \right) e^{i \vec k \cdot( \vec x - \vec y)} \\
&= i \int \frac{d^3 \vec k}{(2\pi)^3} e^{i \vec k \cdot( \vec x - \vec y)}\\
&= i \delta^{(3)}(\vec x - \vec y)
\end{align*}

Where I used the given identity in the lecture notes ## \chi_{\vec k} \dot \chi_{\vec k}^* - \chi_{-\vec k}^* \dot \chi_{-\vec k} = i##
Mmm... Interesting, at least now we know the relation ## \chi_{\vec k} \dot \chi_{\vec k}^* - \chi_{-\vec k}^* \dot \chi_{-\vec k} = i##, maybe that can help, although I don't see how.
Just in case, have you done the other 3 commutators? Maybe there appear some other interesting relations that can help us.

1) Find the explicit expression of ##a## in terms of ##\chi## and ##\dot \chi## (for us it is a bit harder because we are not given the explicit expression of ##a## in terms of ##\chi## and ##\dot \chi##).

I asked for a hint and the answer should look as follows

\begin{equation*}
a_{\vec k} = -i \int d^3 \vec x \left[ \dot \chi_{\vec k}^*(\eta) \chi(\eta, \vec x) -\chi_{\vec k}^*(\eta) \dot \chi(\eta, \vec x) \right] e^{-i \vec k \cdot \vec x}
\end{equation*}

\begin{equation*}
a_{\vec k}^{\dagger} = i \int d^3 \vec x \left[ \dot \chi_{\vec k}(\eta) \chi(\eta, \vec x) -\chi_{\vec k}(\eta) \dot \chi(\eta, \vec x) \right] e^{i \vec k \cdot \vec x}
\end{equation*}

How to get these expressions still remains a mystery to me [...]

Ok, so now you should be able to prove the commutation relations, right?

I asked for a hint and the answer should look as follows

\begin{equation*}
a_{\vec k} = -i \int d^3 \vec x \left[ \dot \chi_{\vec k}^*(\eta) \chi(\eta, \vec x) -\chi_{\vec k}^*(\eta) \dot \chi(\eta, \vec x) \right] e^{-i \vec k \cdot \vec x}
\end{equation*}

\begin{equation*}
a_{\vec k}^{\dagger} = i \int d^3 \vec x \left[ \dot \chi_{\vec k}(\eta) \chi(\eta, \vec x) -\chi_{\vec k}(\eta) \dot \chi(\eta, \vec x) \right] e^{i \vec k \cdot \vec x}
\end{equation*}

How to get these expressions still remains a mystery to me [...]
All you need to prove this is
$$\int_{\mathbb{R}} \mathrm{d}^3 \vec{x} \exp(\pm \mathrm{i} \vec{x} \cdot \vec{k})=(2 \pi)^3 \delta^{(3)}(\vec{k})$$
and the dispersion relation for the momentum eigenmodes,
$$\omega(\vec{k})=\sqrt{\vec{k}^2+m^2}.$$

JD_PM
Ok, so now you should be able to prove the commutation relations, right?
Indeed but I should first learn how to derive the expressions at #12 :)

Indeed but I should first learn how to derive the expressions at #12 :)
Well, I don't know if there's a direct way. For sure, now that you have the expressions, you should indeed compute the two integrals to prove that indeed the two equations are correct.

A way to "prove" those relations without knowing the correct expression, I think it could be to start by computing the integrals
$$\int \chi e^{ikx} d^3 x$$
and similar, and maybe then it is easier to see why the exact combination is that one.

Again, I haven't done the integrals so maybe not, but I'm just telling you the ideas I would do if I need to solve the problem myself.

JD_PM
Hi Hendrik, thanks for the reply.

All you need to prove this is
$$\int_{\mathbb{R}} \mathrm{d}^3 \vec{x} \exp(\pm \mathrm{i} \vec{x} \cdot \vec{k})=(2 \pi)^3 \delta^{(3)}(\vec{k})$$
and the dispersion relation for the momentum eigenmodes,
$$\omega(\vec{k})=\sqrt{\vec{k}^2+m^2}.$$

Oh, so we will need to use such relations.

The issue I have is that I do not see an algorithmic method to get the expressions at #12.

What I have been trying so far is to linearly combine ##(2), (3), (4), (5)## so that we end up solving ##a## in terms of ##\chi## and ##\dot \chi## but I did not succeed.

Do you think such method should work?

What do you mean by "algorithmic"? You can just check these formula for the inversion of the "relativistic Fourier transform", if you wish to call it like this.

What do you mean by "algorithmic"? You can just check these formula for the inversion of the "relativistic Fourier transform", if you wish to call it like this.

Let me express myself better. I was not necessarily looking for a purely algorithmic method but one that "did the job" (i.e. derive equations at #12).

Thanks to an extended discussion with a very talented and patient colleague I understand how to derive them .

We start off by taking the Fourier transform of the operators ##\chi## and ##\dot \chi##

\begin{equation*}
\tilde \chi = \int \frac{d^3 \vec x}{(2\pi)^{3/2}} \chi e^{-i \vec k \cdot \vec x}
\end{equation*}

\begin{equation*}
\tilde{\dot \chi} = \int \frac{d^3 \vec x}{(2\pi)^{3/2}} \dot \chi e^{-i \vec k \cdot \vec x}
\end{equation*}

Plugging them into the Fourier expansion (1) in the OP we get

\begin{align*}
\tilde \chi &= \int \frac{d^3 \vec x}{(2\pi)^{3/2}} \int \frac{d^3 \vec k'}{(2 \pi)^{3/2}} \left( a_{\vec k'} \chi_{\vec k'} e^{i \vec k' \cdot \vec x} + a^{\dagger}_{\vec k'}\chi^*_{\vec k'} e^{-i \vec k' \cdot \vec x} \right) e^{-i \vec k \cdot \vec x} \nonumber \\
&= \int \frac{d^3 \vec x d^3 \vec k'}{(2\pi)^3} \left(a_{\vec k'} \chi_{\vec k'} e^{i (\vec k' - \vec k ) \cdot \vec x} + a_{\vec k'}^{\dagger} \chi_{\vec k'}^* e^{-i (\vec k' + \vec k ) \cdot \vec x} \right) \nonumber \\
&= \int \frac{d^3 \vec k'}{(2\pi)^3}\left(a_{\vec k'} \chi_{\vec k'} \delta^{(3)}(\vec k' -\vec k) + a_{\vec k'}^{\dagger} \chi_{\vec k'}^* \delta^{(3)}(\vec k' +\vec k) \right) \nonumber \\
&= a_{\vec k} \chi_{\vec k} + a_{-\vec k}^{\dagger} \chi_{-\vec k}^*
\end{align*}

Similarly we get

\begin{equation*}
\tilde{\dot \chi} = a_{\vec k} \dot \chi_{\vec k} + a_{-\vec k}^{\dagger} \dot \chi_{-\vec k}^*
\end{equation*}

At this point we notice we have a ##2 \times 2## system of equations and all we have to do is solve it for ##a_{\vec k}## and ##a_{\vec k}^{\dagger}##. First we notice that ##\chi_{\vec k} = \chi_{-\vec k}## (this is because the equation of motion for ##\chi_{\vec k}## (equation (43) in #7) depends on the square of momentum ##k##). So our system of equations can be equivalently written to be

\begin{equation*}
\tilde \chi = a_{\vec k} \chi_{\vec k} + a_{-\vec k}^{\dagger} \chi_{\vec k}^*
\end{equation*}

\begin{equation*}
\tilde{\dot \chi} = a_{\vec k} \dot \chi_{\vec k} + a_{-\vec k}^{\dagger} \dot \chi_{\vec k}^*
\end{equation*}

By taking a particular linear combination and making use of the identity I used at #9 we get

\begin{equation*}
a_{\vec k} = -i \dot \chi _{\vec k}^* \tilde \chi + i \chi_{\vec k}^* \tilde{\dot \chi}
\end{equation*}

It follows that

\begin{equation*}
a_{-\vec k}^{\dagger} = i \dot \chi_{\vec k} \tilde \chi - i \chi_{\vec k} \tilde{\dot \chi}
\end{equation*}

The Fourier transforms of ##a_{\vec k}## and ##a_{\vec k}^{\dagger}## are hence given by

\begin{equation*}
\boxed{a_{\vec k} = -i\int \frac{d^3 \vec x}{(2\pi)^{3/2}}\left[\dot \chi_{\vec k}^* \tilde \chi - \chi_{\vec k}^* \tilde{\dot \chi} \right] e^{-i \vec k \cdot \vec x}}
\end{equation*}

\begin{equation*}
\boxed{a_{\vec k}^{\dagger} = i\int \frac{d^3 \vec x}{(2\pi)^{3/2}}\left[\dot \chi_{\vec k} \tilde \chi - \chi_{\vec k} \tilde{\dot \chi} \right] e^{i \vec k \cdot \vec x}}
\end{equation*}

Where we took ##-k \to k## on the last equation, as it is customary to work with +ive ##k##.