# Sign confusions in solution to Klein Gordon's equation

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

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

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.

• woah, forgot how to math for a minute there, you're right, thank you. Any insight on the first part? – user2723984 Oct 11 '18 at 18:00
• 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? – user2723984 Oct 11 '18 at 18:10
• 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? – user2723984 Oct 11 '18 at 18:13
• @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 $e^{-ipx}$. – Javier Oct 11 '18 at 18:18
• 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 – user2723984 Oct 11 '18 at 18:24