How to Fourier transform to find the operator coefficients in the solution to the real Klein-Gordon scalar field in the Heisenberg picture?

Question

From this site for the university of Nottingham QFT notes in the notes for Lecture 3 there is an exercise on page 4 to show that $$\hat a(\boldsymbol{k})=\int d^3x\left[E(\boldsymbol{k})\hat\phi(\boldsymbol{x})+i\hat \pi(\boldsymbol{x})\right]e^{i\boldsymbol{k}\cdot \boldsymbol{x}}\tag{1}$$ $$\hat a^\dagger(\boldsymbol{k})=\int d^3x\left[E(\boldsymbol{k})\hat\phi(\boldsymbol{x})-i\hat \pi(\boldsymbol{x})\right]e^{-i\boldsymbol{k}\cdot \boldsymbol{x}}\tag{2}$$ by taking the Fourier transform and solving for the operator coefficients $\hat a(\boldsymbol{k}),\,\hat a^\dagger(\boldsymbol{k})$ in $$\hat \phi (\boldsymbol{x})=\int\frac{d^3\boldsymbol{k}}{(2\pi)^32E(\boldsymbol{k})}\left(\hat a(\boldsymbol{k})e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+\hat a^\dagger(\boldsymbol{k})e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right)\tag{3}$$

$$\hat \pi (\boldsymbol{x})=\frac{1}{2i}\int\frac{d^3\boldsymbol{k}}{(2\pi)^3}\left(\hat a(\boldsymbol{k})e^{i\boldsymbol{k}\cdot\boldsymbol{x}}-\hat a^\dagger(\boldsymbol{k})e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}\right)\tag{4}$$

So to try to prove eqns. $(1)$ and $(2)$ I start by taking the Fourier transform of eqn. $(3)$ by multiplying both sides by $e^{-i\boldsymbol{k}\cdot\boldsymbol{x^\prime}}$ and integrating over $\boldsymbol{x^\prime}$, (I have chosen $\boldsymbol{x^\prime}$ as an integration variable to distinguish it from $\boldsymbol{x}$):

$$\mathcal{F}\left[\phi (\boldsymbol{x})\right] =\int d^3 \boldsymbol{x^\prime} e^{-i\boldsymbol{k}\cdot\boldsymbol{x^\prime}}\hat \phi (\boldsymbol{x})$$

$$=\int d^3 \boldsymbol{x^\prime} e^{-i\boldsymbol{k}\cdot\boldsymbol{x^\prime}}\int\frac{d^3\boldsymbol{k}}{(2\pi)^32E(\boldsymbol{k})}\hat a(\boldsymbol{k})e^{i\boldsymbol{k}\cdot\boldsymbol{x}}+\int d^3 \boldsymbol{x^\prime} e^{-i\boldsymbol{k}\cdot\boldsymbol{x^\prime}}\int\frac{d^3\boldsymbol{k}}{(2\pi)^32E(\boldsymbol{k})}\hat a^\dagger(\boldsymbol{k})e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}$$

$$=\int d^3 \boldsymbol{x^\prime} \int\frac{d^3\boldsymbol{k}}{(2\pi)^3}e^{i\boldsymbol{k}\cdot\left(\boldsymbol{x}-\boldsymbol{x^\prime}\right)}\color{red}{\frac{\hat a(\boldsymbol{k})}{2E(\boldsymbol{k})}}+\int d^3 \boldsymbol{x^\prime} \int\frac{d^3\boldsymbol{k}}{(2\pi)^3}e^{-i\boldsymbol{k}\cdot\left(\boldsymbol{x}+\boldsymbol{x^\prime}\right)}\color{red}{\frac{\hat a^\dagger(\boldsymbol{k})}{2E(\boldsymbol{k})}}\tag{A}$$

$$=\int d^3 \boldsymbol{x^\prime} \frac{\hat a(\boldsymbol{k})}{2E(\boldsymbol{k})}\delta^{(3)}\left(\boldsymbol{x}-\boldsymbol{x^\prime}\right)+\int d^3 \boldsymbol{x^\prime} \frac{\hat a^\dagger(\boldsymbol{k})}{2E(\boldsymbol{k})}\delta^{(3)}\left(\boldsymbol{x}+\boldsymbol{x^\prime}\right)\tag{B}$$

$$=\frac{\hat a(\boldsymbol{k})+\hat a^\dagger(\boldsymbol{k})}{2E(\boldsymbol{k})}\tag{C}$$

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There are two steps where I'm doubtful of the logic I used in arriving at $$\mathcal{F}\left[\phi (\boldsymbol{x})\right]=\frac{\hat a(\boldsymbol{k})+\hat a^\dagger(\boldsymbol{k})}{2E(\boldsymbol{k})}\tag{5}$$

1. For $(\mathrm{A})\to (\mathrm{B})$, I used the exponential form of the $\delta$ function $$\delta(x-\alpha)=\frac{1}{2\pi} \int_{-\infty}^\infty e^{ip(x-\alpha)}\,dp$$ which for my case, in 3d, is $$\delta^{(3)}(\boldsymbol{x}-\boldsymbol{x^\prime})=\frac{1}{(2\pi)^3} \int e^{i\boldsymbol{k}\cdot(\boldsymbol{x}-\boldsymbol{x^\prime})}\,d^3\boldsymbol{k}$$ $$=\frac{1}{(2\pi)^3}\int e^{-i\boldsymbol{k}\cdot(\boldsymbol{x}-\boldsymbol{x^\prime})}\,d^3\boldsymbol{k}=\delta^{(3)}(\boldsymbol{x^\prime}-\boldsymbol{x})\tag{6}$$ where I have assumed that since the Dirac-delta is an even 'function' I can replace $i$ with $-i$ in the complex exponential and the result doesn't change. But the problem is that the integrand in both terms of $(\mathrm{A})$ have factors (marked red) that depend on the integration variable, $\boldsymbol{k}$, so I'm not sure I can apply eqn. $(6)$ to $(\mathrm{A})$.

2. For $(\mathrm{B})\to (\mathrm{C})$ I used the fact that $$\int_{-\infty}^\infty d^3 \boldsymbol{x^\prime}\delta^{(3)}(\boldsymbol{x}-\boldsymbol{x^\prime})=\int_{-\infty}^\infty d^3 \boldsymbol{x^\prime}\delta^{(3)}(\boldsymbol{x}+\boldsymbol{x^\prime})=1$$ But I'm not sure if such a manipulation is defined in the 3 dimensional Dirac-delta function.

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To summarize, I am trying to derive eqns. $(1)$ and $(2)$ by taking the Fourier transform of $(3)$ and $(4)$, respectively, but I fail to see how this approach is getting me closer to proving eqns. $(1)$ and $(2)$. Even if the logic used is valid, all I have shown so far is that $$\mathcal{F}\left[\phi (\boldsymbol{x})\right]=\frac{\hat a(\boldsymbol{k})+\hat a^\dagger(\boldsymbol{k})}{2E(\boldsymbol{k})}$$ I could arrive at a similar expression by taking the Fourier transform of $(4)$, but I don't see the point as eqn. $(5)$ does not seem to be of any use to show eqns. $(1)$ and $(2)$.

Could someone please explain where I am going wrong or address the questions I asked regarding steps $(\mathrm{A})\to (\mathrm{B})$ and $(\mathrm{B})\to (\mathrm{C})$?

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Update:

I have been given two answers, in the comment below the answer by @Mark Viola, I asked if there is a way of arriving at the results, $(1)$ and $(2)$ without interchanging the order of integration, as in that answer it is mentioned explicitly mentioned that this is NOT legitimate.

I haven't received any response since, so I will have to assume that there is no way of arriving at $(1)$ and $(2)$ with the order of integration maintained. So as a final question, what is the justification for swapping the order of integration?

Answer 1

Starting from eqs. (3) and (4) of your original question, you find $$\phi(\mathbf{x})=\int \!\frac{d^3k}{(2\pi)^3 2 E(\mathbf{k})}\left[a(\mathbf{k})e^{i \mathbf{k}\cdot \mathbf{x}}+a^\dagger(\mathbf{k})e^{-i \mathbf{k}\cdot \mathbf{x}}\right]=\int \!\frac{d^3k}{(2\pi)^3}\frac{a(\mathbf{k})+a^\dagger(\mathbf{-k})}{2E(\mathbf{k})}\, e^{i\mathbf{k}\cdot \mathbf{x}} $$ and $$ \pi(\mathbf{x})=\frac{1}{2i} \int \! \frac{d^3k}{(2 \pi)^3}\left[a(\mathbf{k})e^{i \mathbf{k} \cdot \mathbf{x}}-a^\dagger(\mathbf{k})e^{-i \mathbf{k}\cdot\mathbf{x}} \right] = \frac{1}{2i}\int \! \frac{d^3 k}{(2\pi)^3}\left[a(\mathbf{k})-a^\dagger(-\mathbf{k})\right]e^{i \mathbf{k}\cdot \mathbf{x}}, $$ where $E(\mathbf{k}):=\sqrt{\mathbf{k}^2+m^2}=E(-\mathbf{k})$. In the second terms of the original integrands, the transformation of variables $\mathbf{k} \to \mathbf{-k}$ was performed in order to arrive at the final integral representation of the field $\phi(\mathbf{x})$ and its canonically conjugated field $\pi(\mathbf{x})$. Written in this form, you can immediately read off the inverse Fourier transforms $$\frac{a(\mathbf{k})+a^\dagger(-\mathbf{k})}{2E(\mathbf{k})}= \int \! d^3x \, \phi(\mathbf{x})\, e^{-i \mathbf{k}\cdot \mathbf{x}}, \qquad \frac{a(\mathbf{k})-a^\dagger(-\mathbf{k})}{2i}= \int \!d^3 x \, \pi(\mathbf{x}) \, e^{-i \mathbf{k}\cdot{x}}$$ without the necessity of any lengthy calculation. Combining these two equations, you arrive at $$ a(\mathbf{k})= \int \! d^3 x\, \left[E(\mathbf{k}) \phi(\mathbf{x})+i \pi(\mathbf{x})\right]\, e^{-i \mathbf{k}\cdot\mathbf{x}}, \quad a^\dagger(\mathbf{k})=\int \! d^3 x \, \left[E(\mathbf{k}) \phi(\mathbf{x})-i \pi(\mathbf{x}) \right] e^{i \mathbf{k}\cdot \mathbf{x}},$$ showing that there were simply two typos ($e^{i \mathbf{k} \cdot \mathbf{x}} \leftrightarrow e^{-i \mathbf{k}\cdot \mathbf{x}}$) in eqs. (1) and (2) of your question. These typos are already present in the lecture notes.

Answer 2

Well, the way you attempted is not the way to rigorously prove the Fourier Inverse Transform Theorem. If we wish to wave rigor, then here is an illegitimate way to arrive at the coveted result.

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IMPORTANT NOTE:

$$E(\boldsymbol{k})=E(-\boldsymbol{k})$$

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Let $\hat \phi$ be defined as

$$\hat \phi (\boldsymbol{x})=\frac1{(2\pi)^3}\int_{\mathbb{R^3}}\frac{\hat a(\boldsymbol{k'})}{2E(\boldsymbol{k'})}e^{i\boldsymbol{k'}\cdot\boldsymbol{x}}\,d^3\boldsymbol{k'}+\frac1{(2\pi)^3}\int_{\mathbb{R^3}}\frac{\hat a^\dagger(\boldsymbol{k'})}{2E(\boldsymbol{k'})}e^{-i\boldsymbol{k'}\cdot\boldsymbol{x}}\,d^3\boldsymbol{k'}\tag1$$

Then, we multiply $(1)$ by $e^{ i\boldsymbol{k}\cdot\boldsymbol{x}}$ and integrate over $\boldsymbol{x}$. Proceeding, we have

$$\begin{align} \int_{\mathbb{R^3}}\hat \phi (\boldsymbol{x})e^{ i\boldsymbol{k}\cdot\boldsymbol{x}}\,d^3\boldsymbol{x}&=\frac1{(2\pi)^3}\int_{\mathbb{R^3}}\int_{\mathbb{R^3}}\frac{\hat a(\boldsymbol{k'})}{2E(\boldsymbol{k'})}e^{i\boldsymbol{k'}\cdot\boldsymbol{x}}\,d^3\boldsymbol{k'}e^{ i\boldsymbol{k}\cdot\boldsymbol{x}}\,d^3\boldsymbol{x}\\\\ &+\frac1{(2\pi)^3}\int_{\mathbb{R^3}}\int_{\mathbb{R^3}}\frac{\hat a^\dagger(\boldsymbol{k'})}{2E(\boldsymbol{k'})}e^{-i\boldsymbol{k'}\cdot\boldsymbol{x}}\,d^3\boldsymbol{k'}e^{ i\boldsymbol{k}\cdot\boldsymbol{x}}\,d^3\boldsymbol{x}\tag2 \end{align}$$

Next, we interchange the order of integration in $(2)$ (this is NOT legitimate) to find that

$$\begin{align} \int_{\mathbb{R^3}}\hat \phi (\boldsymbol{x})e^{ i\boldsymbol{k}\cdot\boldsymbol{x}}\,d^3\boldsymbol{x}&=\int_{\mathbb{R^3}}\frac{\hat a(\boldsymbol{k'})}{2E(\boldsymbol{k'})}\frac1{(2\pi)^3}\int_{\mathbb{R^3}}e^{i\boldsymbol{(k'+ k)}\cdot\boldsymbol{x}}\,d^3\boldsymbol{x}\,d^3\boldsymbol{k^\prime}\\\\ &+\int_{\mathbb{R^3}}\frac{\hat a^\dagger(\boldsymbol{k'})}{2E(\boldsymbol{k'})}\frac1{(2\pi)^3}\int_{\mathbb{R^3}}e^{-i\boldsymbol{(k'- k)}\cdot\boldsymbol{x}}\,d^3\boldsymbol{x}\,d^3\boldsymbol{k'}\\\\ &=\frac{\hat a(- \boldsymbol{k})}{2E( \boldsymbol{k})}+\frac{\hat a^\dagger( \boldsymbol{k})}{2E( \boldsymbol{k})}\tag 3 \end{align}$$

Applying the same approach to the Fourier Transform of $\hat \pi (\boldsymbol{x})$, we find that

$$\int_{\mathbb{R^3}}\hat \pi (\boldsymbol{x})e^{ i\boldsymbol{k}\cdot\boldsymbol{x}}\,d^3\boldsymbol{x}=\frac{\hat a(- \boldsymbol{k})}{2i}-\frac{\hat a^\dagger( \boldsymbol{k})}{2i}\tag 4$$

Now, multiply $(3)$ by $E( \boldsymbol{k})$, multiply $(4)$ by $i$, subtract the results and find that

$$\hat a^\dagger( \boldsymbol{k})=\int_{\mathbb{R^3}}\left[E( \boldsymbol{k}) \hat \phi (\boldsymbol{x})-i \hat \pi (\boldsymbol{x})\right]e^{ i\boldsymbol{k}\cdot\boldsymbol{x}}\,d^3\boldsymbol{x}$$

Can you find the expression for $\hat a(\boldsymbol{k})$?

Answer 3

I feel it's better if I make an answer to my own question to illustrate the confusion I have.

Following the comments given below my post and Marks' answer. I now multiply each side of eqn. $(3)$ in my original question by $e^{-i\boldsymbol{k^{\prime}}\cdot\boldsymbol{x}}$ and integrate over $\boldsymbol{x}$ (the Fourier transform convention I'm using has a negative exponent):

$$\mathcal{F}\left[\phi (\boldsymbol{x})\right] =\int d^3 \boldsymbol{x} e^{-i\boldsymbol{k^{\prime}}\cdot\boldsymbol{x}}\hat \phi (\boldsymbol{x})$$

$$=\int d^3 \boldsymbol{x} e^{-i\boldsymbol{k^\prime}\cdot\boldsymbol{x}} \int\frac{d^3\boldsymbol{k}}{(2\pi)^3}e^{i\boldsymbol{k}\cdot \boldsymbol{x}}\frac{\hat a(\boldsymbol{k})}{2E(\boldsymbol{k})}+\int d^3 \boldsymbol{x} e^{-i\boldsymbol{k^\prime}\cdot\boldsymbol{x}} \int\frac{d^3\boldsymbol{k}}{(2\pi)^3}e^{-i\boldsymbol{k}\cdot \boldsymbol{x}}\frac{\hat a^\dagger(\boldsymbol{k})}{2E(\boldsymbol{k})}$$

$$=\int d^3 \boldsymbol{x} \int\frac{d^3\boldsymbol{k}}{(2\pi)^3}e^{i\left(\boldsymbol{k}-\boldsymbol{k^\prime}\right)\cdot \boldsymbol{x}}\frac{\hat a(\boldsymbol{k})}{2E(\boldsymbol{k})}+\int d^3 \boldsymbol{x} \int\frac{d^3\boldsymbol{k}}{(2\pi)^3}e^{-i\left(\boldsymbol{k}-\boldsymbol{k^\prime}\right)\cdot \boldsymbol{x}}\frac{\hat a^\dagger(\boldsymbol{k})}{2E(\boldsymbol{k})}$$

$$=\int d^3 \boldsymbol{k} \frac{\hat a(\boldsymbol{k})}{2E(\boldsymbol{k})}\left(\frac{1}{(2\pi)^3}\int d^3\boldsymbol{x}e^{i\left(\boldsymbol{k}-\boldsymbol{k^\prime}\right)\cdot \boldsymbol{x}}\right)+\int d^3 \boldsymbol{k} \frac{\hat a^\dagger(\boldsymbol{k})}{2E(\boldsymbol{k})}\left(\frac{1}{(2\pi)^3}\int d^3\boldsymbol{x}e^{-i\left(\boldsymbol{k}+\boldsymbol{k^\prime}\right)\cdot \boldsymbol{x}}\right)$$

$$=\int d^3 \boldsymbol{k} \frac{\hat a(\boldsymbol{k})}{2E(\boldsymbol{k})}\delta^{(3)}\left(\boldsymbol{k}-\boldsymbol{k^\prime}\right)+\int d^3 \boldsymbol{k} \frac{\hat a^\dagger(\boldsymbol{k})}{2E(\boldsymbol{k})}\delta^{(3)}\left(\boldsymbol{k}+\boldsymbol{k^\prime}\right)$$ $$=\frac{\hat a(\boldsymbol{k})}{2E( \boldsymbol{k})}+\frac{\hat a^\dagger(- \boldsymbol{k})}{2E( \boldsymbol{k})}$$

Rewriting this gives $$\bbox[yellow]{-i\int d^3 \boldsymbol{x} E(\boldsymbol{k})\phi (\boldsymbol{x})e^{-i\boldsymbol{k}\cdot\boldsymbol{x}}=\frac{\hat a(\boldsymbol{k})+\hat a^\dagger(-\boldsymbol{k})}{2i}}\tag{a}$$

Now applying a similar logic to eqn. $(4)$ in my original post:

$$\mathcal{F}\left[\hat\pi (\boldsymbol{x})\right] =\int d^3 \boldsymbol{x} e^{-i\boldsymbol{k^{\prime}}\cdot\boldsymbol{x}}\hat \pi (\boldsymbol{x})$$

$$=\frac{1}{2i}\int d^3 \boldsymbol{x} e^{-i\boldsymbol{k^{\prime}}\cdot\boldsymbol{x}}\int\frac{d^3\boldsymbol{k}}{(2\pi)^3}\hat a(\boldsymbol{k})e^{i\boldsymbol{k}\cdot \boldsymbol{x}}-\frac{1}{2i}\int d^3 \boldsymbol{x} e^{-i\boldsymbol{k^{\prime}}\cdot\boldsymbol{x}}\int\frac{d^3\boldsymbol{k}}{(2\pi)^3}\hat a(\boldsymbol{k})e^{-i\boldsymbol{k}\cdot \boldsymbol{x}}$$

$$=\frac{1}{2i}\int d^3\boldsymbol{k} \hat a(\boldsymbol{k}) \left(\frac{1}{(2\pi)^3}\int d^3\boldsymbol{x}e^{i\left(\boldsymbol{k}-\boldsymbol{k^\prime}\right)\cdot \boldsymbol{x}}\right)-\frac{1}{2i}\int d^3\boldsymbol{k} \hat a^\dagger(\boldsymbol{k}) \left(\frac{1}{(2\pi)^3}\int d^3\boldsymbol{x}e^{-i\left(\boldsymbol{k}+\boldsymbol{k^\prime}\right)\cdot \boldsymbol{x}}\right)$$

$$=\frac{1}{2i}\int d^3 \boldsymbol{k} \hat a(\boldsymbol{k})\delta^{(3)}\left(\boldsymbol{k}-\boldsymbol{k^\prime}\right)-\frac{1}{2i}\int d^3 \boldsymbol{k} \hat a^\dagger(\boldsymbol{k})\delta^{(3)}\left(\boldsymbol{k}+\boldsymbol{k^\prime}\right)$$

$$=\frac{\hat a(\boldsymbol{k})}{2i}-\frac{\hat a^\dagger(-\boldsymbol{k})}{2i}$$

Rewriting this gives

$$\bbox[yellow]{\int d^3 \boldsymbol{x}e^{-i\boldsymbol{k}\cdot \boldsymbol{x}}\hat \pi(\boldsymbol{x})=\frac{\hat a(\boldsymbol{k})-\hat a^\dagger(-\boldsymbol{k})}{2i}}\tag{b}$$

Adding $(\mathrm{a})$ to $(\mathrm{b})$ gives

$$\hat a(\boldsymbol{k})=\int d^3 \boldsymbol{x}\left(E(\boldsymbol{k})\hat \phi(\boldsymbol{x})+i\hat \pi (\boldsymbol{x})\right)e^{\color{red}{-}i \boldsymbol{k}\cdot \boldsymbol{x}}\ne \text{eqn. (1)}\tag{c}$$

While subtracting $(\mathrm{b})$ from $(\mathrm{a})$ gives

$$\hat a^\dagger(\boldsymbol{\color{red}{-}k})=\int d^3 \boldsymbol{x}\left(E(\boldsymbol{k})\hat \phi(\boldsymbol{x})-i\hat \pi (\boldsymbol{x})\right)e^{-i \boldsymbol{k}\cdot \boldsymbol{x}}\ne \text{eqn. (2)}\tag{d}$$

Eqns. $(\mathrm{c})$ and $(\mathrm{d})$ are very similar to $(1)$ and $(2)$, respectively. What's causing the problem is the incorrect sign of the exponent in $(\mathrm{c})$ and the negative argument for the creation operator, $\hat a^\dagger(-\boldsymbol{k})$ in $(\mathrm{d})$.

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So not only do I fail to derive the correct relations (eqns. $(1)$ and $(2)$), but I have interchanged the order of integration, this is unacceptable and wrong.

What I would love to see is a rigorous way of deriving those relations, without this unjustified nonsense of pulling out factors depending on the integration variable of the integral and treating them as though they were constants.