In Section 2.3, Peskin & Schroeder discusses the quantization of real scalar field in Schrodinger picture. He writes Eq. (2.25) as follows.

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

After that he rearranges it and write Eq. (2.27) as follows. $$\phi(\textbf{x}) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_\textbf{p}}} \left( a_\textbf{p} + a_\textbf{-p}^\dagger\right) e^{i\textbf{p} \cdot \textbf{x}}$$

I am not sure how to do this rearrangement. I manipulate the second term in Eq. (2.25) in the following way.

Let $-\textbf{p} = \textbf{q}$. Then the second term in Eq. (2.25) becomes \begin{eqnarray} && \int \frac{-d^3q}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{-\textbf{q}}}} a_{-\textbf{q}}^\dagger \, e^{i\textbf{q} \cdot \textbf{x}} \\ &=& \color{red}{-}\int \frac{d^3q}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{\textbf{q}}}} a_{-\textbf{q}}^\dagger \, e^{i\textbf{q} \cdot \textbf{x}} \\ &=& \color{red}{-} \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_{\textbf{p}}}} a_{-\textbf{p}}^\dagger \, e^{i\textbf{p} \cdot \textbf{x}}, \end{eqnarray} where in the first step I have used the fact that $d^3p = -d^3q$ , in the second step I have used the fact that $\omega_{-\textbf{q}} = \sqrt{|-\textbf{q}|^2 + m^2} = \sqrt{|\textbf{q}|^2 + m^2} = \omega_{\textbf{q}}$ and in the third step I performed a change of variable: $\textbf{q} \rightarrow \textbf{p}$.

My Question

But as we can see, in Eq. (2.27), there is no minus sign before the second term. What am I missing here?

  • 7
    $\begingroup$ Check overall sign after substituting -p with q. The lower and upper limits of the integral also gets flipped. $\endgroup$
    – KP99
    Sep 20, 2021 at 6:45
  • 1
    $\begingroup$ Thanks! My bad! I should have noticed. $\endgroup$
    – rainman
    Sep 20, 2021 at 7:06
  • 2
    $\begingroup$ Withdraw the question, then? $\endgroup$ Oct 22, 2021 at 19:54
  • 2
    $\begingroup$ But this.. well maybe someone also had a brain shortcircuit and they can search and found out the simple answer here? $\endgroup$
    – Rescy_
    Apr 26, 2023 at 10:00

1 Answer 1


$$\int_{k=-\infty}^{k=\infty}d k=\int_{p=\infty}^{p=-\infty}d (-p)=\int_{-\infty}^{\infty}dp $$


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