Covariant commutation relations in Mandl and Shaw

In page 47 of Mandl and Shaw, the $\Delta$-function can be written as $$\Delta(x) = \frac{-1}{(2 \pi)^3} \int \frac{d^3k}{\omega_k} \sin(kx) \tag{3.43}$$ and as equation $$\Delta(x) = \frac{-i}{(2 \pi)^3} \int d^4k\delta(k^2-\mu^2) \epsilon(k_0) e^{-ikx} \tag{3.45}$$ where $\epsilon(x)$ is the sign function. The book states that both definitions are equivalent if we perform a $k_0$-integration on (3.45), using relation $$\delta(k^2-\mu^2) = \frac{1}{2 \omega_k} \left[\delta(k_0 + \omega_k) + \delta(k_0 - \omega_k)\right] \tag{3.47}$$

I want to prove this, but don't know how to perform the integration in order to get equation (3.43). I have to compute $$\Delta(x) = \frac{-i}{(2 \pi)^3} \int d^4k \frac{1}{2 \omega_k} \left[\delta(k_0 + \omega_k) + \delta(k_0 - \omega_k)\right] \epsilon(k_0) e^{-ikx}$$ but this becomes $$\Delta(x) = \frac{-i}{(2 \pi)^3} \int d^3k \frac{1}{2 \omega_k} \left[e^{-i\omega_kt} - e^{ i \omega_k t}\right] e^{i \vec{k} \cdot \vec{x}} = \frac{-1}{(2 \pi)^3} \int d^3k \frac{1}{\omega_k} \sin(\omega_kt) e^{i \vec{k} \cdot \vec{x}}$$

So there is no way I can relate this result to $\sin(kx)$ of equation (3.43) with the spatial part $e^{i \vec{k} \cdot \vec{x}}$. Where is the mistake on my calculations?

• Please include the relevant equations into the post instead of referring to them by numbers. Commented Apr 30, 2015 at 20:29
• Also note that "check my work"-type questions are off-topic and liable to be closed. Commented Apr 30, 2015 at 20:39

So there is no way I can relate this result to Sin(kx) of equation (3.43) with the spatial part $e^{i\vec k \cdot \vec x}$. Where is the mistake on my calculations?

Underneath the $d^3k$ integral you can rewrite $$e^{i\vec k\cdot \vec x}sin(\omega_k t)$$ as $$\sin(k\cdot x)$$ with the understanding that $k_0=\omega_k$. This is true because you can split the integral back into two pieces (one with a $e^{i\omega_k t}$ and one with a $e^{-i\omega_k t}$ and then in one of the pieces change integration variables from $\vec k$ to $-\vec k$. This doesn't effect the measure $d^3k$.

• If you change $\vec{k}$ to $-\vec{k}$, measure $d^3k$ changes to $-d^3k$ Commented Apr 30, 2015 at 22:54
• No, it does not. Think about what you are saying.... Or more formally... what is the Jacobian of the transformation, and how is that related to the volume element...
– hft
Commented May 1, 2015 at 5:29
• I didn't realize any volume element changes by the absolute value of the Jacobian matrix. Thank you very much. Commented May 1, 2015 at 13:19
• I'm not sure I agree with this - the negative sign of the Jacobian is important, I think Commented May 1, 2015 at 16:23
• @innisfree, I'm sure you are wrong. Check out any basic calculus text that covers 3 dimensional integration and change of variables.
– hft
Commented May 1, 2015 at 17:06

You are attempting the integral: \begin{align} I &= \int_{-\infty}^\infty d^3k \sin(\omega_kt) e^{i \vec{k} \cdot \vec{x}}\\ &= 2i\left[\int_{-\infty}^\infty d^3k e^{i \vec{k} \cdot \vec{x} + i\omega_kt} - \int_{-\infty}^\infty d^3k e^{i \vec{k} \cdot \vec{x} - i\omega_kt}\right] \end{align} On the first integral only, make a change of variables $\vec k \to - \vec k$ (this also flips the integration limits): \begin{align} I &= 2i\left[-\int_{\infty}^{-\infty} d^3k e^{-i \vec{k} \cdot \vec{x} + i\omega_kt} - \int_{-\infty}^\infty d^3k e^{i \vec{k} \cdot \vec{x} - i\omega_kt}\right]\\ &= 2i\left[\int_{-\infty}^{\infty} d^3k e^{-i \vec{k} \cdot \vec{x} + i\omega_kt} - \int_{-\infty}^\infty d^3k e^{i \vec{k} \cdot \vec{x} - i\omega_kt}\right] \end{align} Note that the negative sign of the Jacobian is important: it meant that we could flip the integration limits back again. Finalizing, we have that \begin{align} I &= 2i\left[\int_{-\infty}^{\infty} d^3k e^{i kx} - \int_{-\infty}^\infty d^3k e^{-i kx}\right]\\ &= \int_{-\infty}^{\infty} d^3k \sin(kx) \end{align} Finishing your proof of (3.43).

• As mentioned in the lengthy comment thread of my answer. The sign of the Jacobian is irrelevant. You are not integrating "\$d^3k" from "-infinity" to "infinity", you are are integrating over a volume. You do not "flip" any limits of integration because of the sign of the jacobin. Consider the two dimensional or the four dimensional case. The Jacobian determinant is positive in those cases, and it is negative in the three dimensional case. Clearly it is the absolute value of the Jacobian determinant that is important. This is discussed in basic calculus textbooks.
– hft
Commented May 3, 2015 at 22:14
• ...So, basically, you copied my answer and inserted some incorrect language about the sign of the Jacobian... really great...
– hft
Commented May 3, 2015 at 22:14
• I'm still not sure, surely the sign of the Jacobian indicates whether the orientation of the said volume ought to be reversed? Of course, one can also take the absolute value of the Jacobian and never reverse the orientation... Commented May 4, 2015 at 7:40
• in any case, thanks for your comments - i've read them and don't have much more to say on the matter Commented May 4, 2015 at 8:02