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I have two basic questions on the solution of the Klein Gordon equation.

The Lagrangian of the Klein Gordon field is $$\mathcal{L}=\frac12\partial_\mu\phi\partial^{\mu}\phi-\frac12m^2\phi^2 $$

Peskin & Schroeder in their Introduction to quantum field theory argue, in analogy with a simple harmonic oscillator $$ \phi(x)=\int\frac{d^3\mathbf{p}}{(2\pi)^3}\frac{1}{\sqrt{2\omega_\mathbf{p}}}(a_\mathbf{p}e^{i\mathbf{p}\cdot\mathbf{x}}+a_{\mathbf{p}}^\dagger e^{-i\mathbf{p}\cdot\mathbf{x}})\\ \pi(x)=-i\int\frac{d^3\mathbf{p}}{(2\pi)^3}\sqrt{\frac{\omega_\mathbf{p}}{2}}(a_\mathbf{p}e^{i\mathbf{p}\cdot\mathbf{x}}-a_{\mathbf{p}}^\dagger e^{-i\mathbf{p}\cdot\mathbf{x}})$$

where $\omega_{\mathbf{p}}=\sqrt{\vec{p}^2+m^2}$, i.e. $\mathbf{p}=(\omega_\mathbf{p}, \vec{p})$. $\pi(x)$ is supposed to be the conjugate momentum of the field, defined as $$\pi(x)=\frac{\partial\mathcal{L}}{\partial(\partial_0\phi)}=\partial_0\phi= i\int\frac{d^3\mathbf{p}}{(2\pi)^3}\sqrt{\frac{\omega_\mathbf{p}}{2}}(a_\mathbf{p}e^{i\mathbf{p}\cdot\mathbf{x}}-a_{\mathbf{p}}^\dagger e^{-i\mathbf{p}\cdot\mathbf{x}})$$

What of the minus sign introduced by the analogy with the harmonic oscillator?

Second question: right after, they say that they'll often use these solutions in the form

$$ \phi(x)=\int\frac{d^3\mathbf{p}}{(2\pi)^3}\frac{1}{\sqrt{2\omega_\mathbf{p}}}(a_\mathbf{p}+a_{\mathbf{-p}}^\dagger )e^{i\mathbf{p}\cdot\mathbf{x}} $$

I fail to see why is this equivalent to the form above, I understand that it is obtained by a change of variable $\mathbf{p}\rightarrow\mathbf{-p}$ in the second part of the integral, but I think this should introduce a minus sign in the measure, because if $p_i\rightarrow -p_i$ for $i=1,2,3$ then $dp_1dp_2dp_3\rightarrow-dp_1dp_2dp_3$ yielding the form

$$ \phi(x)=\int\frac{d^3\mathbf{p}}{(2\pi)^3}\frac{1}{\sqrt{2\omega_\mathbf{p}}}(a_\mathbf{p}-a_{\mathbf{-p}}^\dagger )e^{i\mathbf{p}\cdot\mathbf{x}}. $$

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The sign change in the measure is compensated by the change in the limits of integration: the integral changes from $\int_{-\infty}^\infty$ to $\int_\infty^{-\infty}$; to go back to the original bounds, just change the sign.

When dealing with multiple variables, it is usually simpler to think of integrals as being over a certain region instead of being from one number to another. In this setting, if we are integrating over a region $D \subseteq \mathbb{R}^3$ and do a change of variables $q = T(p)$, the transformation is

$$\int_D d^3p\ f(\mathbf{p}) = \int_{T(D)} d^3q\ |\det(DT)|\ f(T^{-1}(\mathbf{q})),$$

with the absolute value of the Jacobian determinant.

Edit: I missed the first part of the question! Technically your expressions are wrong if the field is time-dependent: $a_p e^{i\mathbf{p}\cdot\mathbf{x}}$ should be changed to $a_p e^{-ipx}$, with $px = -\omega_p t + \mathbf{p}\cdot\mathbf{x}$. This accounts for the minus sign when taking the time derivative.

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  • $\begingroup$ woah, forgot how to math for a minute there, you're right, thank you. Any insight on the first part? $\endgroup$ – user2723984 Oct 11 '18 at 18:00
  • $\begingroup$ On your edit: there's a confusion here because I assumed that Peskin & Schroeder meant $p_\mu x^{\mu}$ when they wrote $\mathbf{x}\cdot\mathbf{p}$, I just stuck with their notation. Am I interpreting this right that the problem is in the metric signature choice? I would have said $p_\mu x^{\mu}=\omega_p t- \vec{p}\cdot \vec{x}$, if yes, how does the definition of conjugate momentum depend on the signature? $\endgroup$ – user2723984 Oct 11 '18 at 18:10
  • $\begingroup$ And what about The answer to this question that I frankly hadn't found before, but I'm glad because it makes me even more confused? $\endgroup$ – user2723984 Oct 11 '18 at 18:13
  • $\begingroup$ @user2723984 Typically we use bold face to represent 3D vectors, and plain letters to represent four-vectors. The problem is not with the signature, it's that the formulas you wrote are the fields at $t=0$. It can be seen that for nonzero $t$, $a_\mathbf{p}$ gets a factor of $e^{-i\omega t}$, hence the exponent $e^{-ipx}$. $\endgroup$ – Javier Oct 11 '18 at 18:18
  • $\begingroup$ Sorry, I think I'm getting it but not quite yet. You're saying that I misinterpreted the notation and really in the book all those expressions are time independent, and to make them time dependent I should do as in the answer I linked, right? But the Hamiltonian is not $\omega$, if it was it would actually make sense $\endgroup$ – user2723984 Oct 11 '18 at 18:24

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