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I recently graduate with a bachelor's in physics, and I've been trying to take the next steps toward learning QFT. To this end, I have been working through Peskin and Schroeder's textbook step-by-step. Currently, I am confused on a detail from chapter two, where the authors solve the Klein-Gordon equation using a Fourier transformation of the field.

In particular, I have looked through online notes and other Stack Exchange answers to understand how to work with the following integral, but I haven't found an answer that goes through the intermediate steps to evaluate it properly.

$$(\Box + m^2)\int e^{i\vec{p}\cdot\vec{x}}\phi(\vec{p},t)d^3p =0$$

I can following the process by which we exchange differentiation and integration and then use the linearity of the operator inside of the integral. As such, I can arrive at the integrand:

$$\Box\left[e^{i\vec{p}\cdot\vec{x}}\phi(\vec{p},t)\right]$$

The operator here is the sum of second-order derivatives, and so I think we can use a product rule for higher derivatives:

$$\Box\left[e^{i\vec{p}\cdot\vec{x}}\phi(\vec{p},t)\right]= \Box\left[e^{i\vec{p}\cdot\vec{x}}\right]\phi(\vec{p},t)+2\partial\cdot\left[e^{i\vec{p}\cdot\vec{x}}\right]\partial\cdot\left[\phi(\vec{p},t)\right] + e^{i\vec{p}\cdot\vec{x}}\Box\left[\phi(\vec{p},t)\right]$$

Tackling the first term, the exponential component doesn't depend on $t$ so this term should evaluate to $|\vec{p}|^2e^{-\vec p\cdot\vec x}\phi(\vec p, t)$ after taking the spacial derivatives and summing right?

Then the second term is similar, but we only need first order derivatives so I evaluated it as $2i(p_1+p_2+p_3)e^{i\vec p\cdot \vec x}\partial\cdot\phi(\vec t, t)$.

Then the third term would stay the same so factoring out the exponential we would in total we should get

$$\int e^{i\vec p\cdot\vec x}(\partial_t^2 - \nabla^2+|\vec p|^2 +m^2)\phi(\vec p, t)d^3p + 2i\int e^{i\vec p\cdot\vec x}(p_1+p_2+p_3)\partial\cdot\phi(\vec p, t)d^3p$$

I know that I should have gotten

$$\int e^{i\vec p\cdot\vec x}(\partial_t^2 +|\vec p|^2 +m^2)\phi(\vec p, t)d^3p$$

But I cannot figure out the steps in the middle. Am I over-complicating the derivation and heading in the wrong direction or are there terms that I can cancel somewhere?

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    $\begingroup$ $\phi(p,t)$ does not depend upon $x$ $\endgroup$ – jacob1729 Oct 7 '19 at 14:14
  • $\begingroup$ Note a typo: $\phi(\vec t, t)$, after "I evaluated it as". $\endgroup$ – Emilio Pisanty Oct 7 '19 at 14:58
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As mentioned in the comments, $\phi(p,t)$ does not depend on $x$. That means that in your expression $$ \int e^{i\vec p\cdot\vec x}(\partial_t^2 - \nabla^2+|\vec p|^2 +m^2)\phi(\vec p, t)d^3p + 2i\int e^{i\vec p\cdot\vec x}(p_1+p_2+p_3)\partial\cdot\phi(\vec p, t)d^3p $$ the $\nabla^2$ vanishes.

The $\partial\cdot \phi$ term there also vanishes, but you're already wrong by then $-$ the issues come from upstream, where you set $$ 2\partial\cdot\left[e^{i\vec{p}\cdot\vec{x}}\right]\partial\cdot\left[\phi(\vec{p},t)\right] = 2i(p_1+p_2+p_3)e^{i\vec p\cdot \vec x}\partial\cdot\phi(\vec t, t). $$ Here your notation is nonstandard and you don't define it in full, so it's impossible to say that the LHS is "right" or "wrong" (but the RHS is definitely wrong). The LHS can be made to work, so long as you (i) interpret $\nabla \cdot$ as a four-dimensional (i.e. 4-vector) gradient, and then (ii) take suitable care when taking the inner product between those 4-vector gradients of $e^{i\vec p\cdot \vec x}$ and $\phi(\vec p, t)$ as a spacetime inner product, i.e. with opposite signs on the space and time indices, according to whatever signature you're using.

However, you don't really need to do any of that: this cross product must vanish, because $e^{i\vec p\cdot \vec x}$ doesn't depend on time and $\phi(\vec p, t)$ doesn't depend on space, so the cross products of their derivatives vanish.

That's enough to pull you out to where you want to go.

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  • $\begingroup$ Then, the upstream problems stem from the third term containing only first derivatives. Taking this for all components, we would get the sum $2(\partial^\mu f)\nabla_{\mu\nu}(\partial^\nu g)$, correct? That is what I have tried to convey in that term. Taking $f=\exp{i\vec p\vec x}$ and $g=\phi(\vec p, t)$, the first term cancels due to the time independence of $f$ and the rest of the three terms vanish due to the space-independence of $g$. I think I understand now! Is that the point that you are making? If so, I think it's clear to me now. $\endgroup$ – Noah M Oct 7 '19 at 16:22
  • $\begingroup$ Then, the upstream problems stem from the third term containing only first derivatives. Taking this for all components, we would get the sum $2(\partial^\mu f)\eta_{\mu\nu}(\partial^\nu g)$, correct? That is what I have tried to convey in that term. Taking $f=\exp{(i\vec p\cdot\vec x)}$ and $g=\phi(\vec p, t)$, the first term cancels due to the time independence of $f$ and the rest of the three terms vanish due to the space-independence of $g$. I think I understand now! Is that the point that you are making? If so, I think it's clear to me now. $\endgroup$ – Noah M Oct 7 '19 at 16:30
  • $\begingroup$ Sorry, in the first comment I accidentally wrote $\nabla_{\mu\nu}$ when I meant the metric $\eta_{\mu\nu}$. $\endgroup$ – Noah M Oct 7 '19 at 17:17

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