Covariant commutation relations in Mandl and Shaw

Question

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?

Answer 1

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 $d^3k$.

Answer 2

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).