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. $\endgroup$
    – paul230_x
    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


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.