- #1
JD_PM
- 1,131
- 158
- Homework Statement
- Given that [tex]\Delta (x) = \frac{-i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{-ik \cdot x}.[/tex]
Show that [itex]\Delta^{*}(x) = \Delta (x)[/itex] and [itex]\Delta(-x) = - \Delta (x)[/itex].
HINT: You may need to use the following Lorentz identities
(i) [itex]\ (\Lambda k) \cdot (\Lambda x) = k \cdot x[/itex]
(ii) [itex]\ p \cdot (\Lambda x) = (\Lambda^{-1}p) \cdot x[/itex]
(iii) [itex]\ \left(\Lambda^{-1}k \right)^{2} = k^{2}[/itex]
- Relevant Equations
- Please see below
The following exercise was proposed by samalkhaiat here.
The given Lorentz identities were proven here.
We first note that ##d^4 k = d^3 \vec k dk_0##, the ##k_0## integration is over ##-\infty < k_0 < \infty## and ##\epsilon (k_0)## is the sign function, which is defined as
$$\epsilon (k_0)=\frac{k_0}{|k_0|}=
\begin{cases}
1, & \text{if} \ \ \ \ k_0>0 \\
-1, & \text{if} \ \ \ \ k_0<0
\end{cases} $$
OK let's now start with [itex]\Delta^{*}(x) = \Delta (x)[/itex]
Based on the hint, one may think the way to approach the problem is to show that ##\Delta (x)## is Lorentz invariant; we go term by term
1) $$e^{-i (\Lambda k) \cdot (\Lambda x)}=e^{-ik \cdot x}$$
2) I've been reading about why the given delta function is invariant, but I do not see what properties of the above were applied to get it.
3) I also have difficulties when trying to show that ##\epsilon (k_{0}) ## is invariant. Mandl & Shaw say that the invariance of the sign function is obvious 'since proper LTs do not interchange past and future', but I do not really understand what they meant here.
Is this the way to proceed (i.e. first trying to show invariance of ##\Delta (x)## and then try to show [itex]\Delta^{*}(x) = \Delta (x)[/itex]?
Once I understand how to show [itex]\Delta^{*}(x) = \Delta (x)[/itex], showing [itex]\Delta(-x) = - \Delta (x)[/itex]. should be easier.
Any hint would be appreciated.
Thank you.
The given Lorentz identities were proven here.
We first note that ##d^4 k = d^3 \vec k dk_0##, the ##k_0## integration is over ##-\infty < k_0 < \infty## and ##\epsilon (k_0)## is the sign function, which is defined as
$$\epsilon (k_0)=\frac{k_0}{|k_0|}=
\begin{cases}
1, & \text{if} \ \ \ \ k_0>0 \\
-1, & \text{if} \ \ \ \ k_0<0
\end{cases} $$
OK let's now start with [itex]\Delta^{*}(x) = \Delta (x)[/itex]
Based on the hint, one may think the way to approach the problem is to show that ##\Delta (x)## is Lorentz invariant; we go term by term
1) $$e^{-i (\Lambda k) \cdot (\Lambda x)}=e^{-ik \cdot x}$$
2) I've been reading about why the given delta function is invariant, but I do not see what properties of the above were applied to get it.
3) I also have difficulties when trying to show that ##\epsilon (k_{0}) ## is invariant. Mandl & Shaw say that the invariance of the sign function is obvious 'since proper LTs do not interchange past and future', but I do not really understand what they meant here.
Is this the way to proceed (i.e. first trying to show invariance of ##\Delta (x)## and then try to show [itex]\Delta^{*}(x) = \Delta (x)[/itex]?
Once I understand how to show [itex]\Delta^{*}(x) = \Delta (x)[/itex], showing [itex]\Delta(-x) = - \Delta (x)[/itex]. should be easier.
Any hint would be appreciated.
Thank you.