Showing properties of a propagator given certain Lorentz identities

[JD_PM]

Homework Statement

Given that $$\Delta (x) = \frac{-i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{-ik \cdot x}.$$



Show that $\Delta^{*}(x) = \Delta (x)$ and $\Delta(-x) = - \Delta (x)$.



HINT: You may need to use the following Lorentz identities



(i) $\ (\Lambda k) \cdot (\Lambda x) = k \cdot x$



(ii) $\ p \cdot (\Lambda x) = (\Lambda^{-1}p) \cdot x$



(iii) $\ \left(\Lambda^{-1}k \right)^{2} = k^{2}$

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 \Delta^{*}(x) = \Delta (x)

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 \Delta^{*}(x) = \Delta (x)?

Once I understand how to show \Delta^{*}(x) = \Delta (x), showing \Delta(-x) = - \Delta (x). should be easier.

Any hint would be appreciated.

Thank you.

 

[nrqed]

Homework Statement:: Given that \Delta (x) = \frac{-i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{-ik \cdot x}.
Show that \Delta^{*}(x) = \Delta (x) and \Delta(-x) = - \Delta (x).
HINT: You may need to use the following Lorentz identities
(i) \ (\Lambda k) \cdot (\Lambda x) = k \cdot x
(ii) \ p \cdot (\Lambda x) = (\Lambda^{-1}p) \cdot x
(iii) \ \left(\Lambda^{-1}k \right)^{2} = k^{2}
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 \Delta^{*}(x) = \Delta (x)

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.

How does $k^2$ transform? How does $m^2$ transform?

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.

Since $k_0/|k_0|$ is just the sign of $k_0$, you have to address the question: under a proper Lorentz transformation, how does the zeroth component of a four-vector transform?

 

[JD_PM]

Thank you, I think I got it.

How does $k^2$ transform? How does $m^2$ transform?

They transform as scalars, so based on (iii) we see why the given delta function is invariant

$$\delta((\Lambda^{-1} k)^2 - (\Lambda^{-1} m)^2)=\delta(k^2 - m^2)$$

Since $k_0/|k_0|$ is just the sign of $k_0$, you have to address the question: under a proper Lorentz transformation, how does the zeroth component of a four-vector transform?

The zeroth component of a four vector is a scalar, so under continuous LT we expect to get

$$\epsilon (\Lambda^{-1} k_0)=\epsilon (k_0)$$

Besides, we know that the following integral is Lorentz Invariant (see Tong's notes, page 32)

$$\int d^4 k \delta (k^2 - m^2)$$

 

[JD_PM]

Above we've discussed why $\Delta(x)$ is Lorentz invariant. However, I did not discussed how to show
$\Delta^{*}(x) = \Delta (x)$ and $\Delta(-x) = - \Delta (x)$.

What I've been thinking is that we could use non-continuous proper Lorentz Transformations; i.e. time reversal $(t, \vec x) \rightarrow (-t, \vec x)$ to show those properties.

Let's first try to show that $\Delta^{*}(x) = \Delta (x)$

We have

$$\Delta^{*} (x) = \frac{i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{ik \cdot x}.$$

So the idea was applying time reversal to $\Delta^{*} (x)$; we get

$$f(k) = \frac{-i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{-ik \cdot x}.$$

Note that here $k=(-k_0, \vec k)$ due to time reversal. Thus we get $k \cdot x = -k_0 x_0 -\vec k \cdot \vec x$, which justifies the sign of the exponent. Besides $\epsilon (k_0) \rightarrow -\epsilon (k_0)$, which justifies the sign of the integral.

As $\Delta^{*} (x)$ is Lorentz invariant, the expression we get (which I labelled $f(k)$) after applying time reversal to $\Delta^{*} (x)$ must be equal to $\Delta^{*} (x)$. We also note that $f(k)=\Delta (x)$. Thus

$$\Delta (x)=\Delta^{*} (x)$$

QED.

Do you agree?

 

[JD_PM]

Regarding $\Delta(-x) = - \Delta (x)$ I used the same idea; we start with

$$\Delta (-x) = \frac{-i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{ik \cdot x}.$$

Applying time reversal (i.e. $\epsilon (k_0) \rightarrow -\epsilon (k_0)$ and $k=(-k_0, \vec k)$) we get

$$g(k) = \frac{i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{-ik \cdot x}.$$

Clearly $g(k) = -\Delta(x)$, so

$$\Delta(-x) = - \Delta (x)$$

QED.

 

[nrqed]

Thank you, I think I got it.
They transform as scalars, so based on (iii) we see why the given delta function is invariant

$$\delta((\Lambda^{-1} k)^2 - (\Lambda^{-1} m)^2)=\delta(k^2 - m^2)$$
The zeroth component of a four vector is a scalar, so under continuous LT we expect to get

$$\epsilon (\Lambda^{-1} k_0)=\epsilon (k_0)$$

Besides, we know that the following integral is Lorentz Invariant (see Tong's notes, page 32)

$$\int d^4 k \delta (k^2 - m^2)$$

Watch out, the zeroth component of a four vector is *not* a scalar. A scalar does not change under LT. $k_0$ changes value under a LT. What you must show that is the the *sign* of $k_0$ does not change under proper, ortho synchronous Lorentz transformations.

 

[JD_PM]

What you must show that is the the *sign* of $k_0$ does not change under proper, ortho synchronous Lorentz transformations.

Oh I see.

What I understand so far from the discussion in the relativity thread is that the sign of $k_0$ does not change under proper, ortho synchronous Lorentz transformations because when dealing with time-like vectors with $k_0 >0$ these remain $k_0 >0$ after the transformation (which means that they stay in the upper part of the cone) and when dealing with time-like vectors with $k_0 <0$ these remain $k_0 <0$ (which means that they stay in the lower part of the cone).

The above is a conceptual explanation; Is there a way to prove this mathematically?

 

[JD_PM]

I tried to show that $\Delta^{*}(x) = \Delta (x)$ and $\Delta(-x) = - \Delta (x)$ at #3 and #4 and failed.

Could you please give me a hint on what approach should I take to show them?

Thanks

 

[TSny]

Let's first try to show that $\Delta^{*}(x) = \Delta (x)$

We have

$$\Delta^{*} (x) = \frac{i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{ik \cdot x}.$$

Ok

So the idea was applying time reversal to $\Delta^{*} (x)$; we get

$$f(k) = \frac{-i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{-ik \cdot x}.$$

Note that here $k=(-k_0, \vec k)$ due to time reversal. Thus we get $k \cdot x = -k_0 x_0 -\vec k \cdot \vec x$, which justifies the sign of the exponent.

Wouldn't time reversal also change the sign of $x_0$ as well as change the sign of $\vec k$? ($\vec k$ is like a momentum. )So under time reversal wouldn't $k \cdot x = k_0 x_0 - \vec k \cdot \vec x$ become $+k_0 x_0 + \vec k \cdot \vec x$ ?

Instead of trying to introduce a symmetry transformation such as time reversal or a Lorentz transformation, go back to $$\Delta^{*} (x) = \frac{i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{ik \cdot x}.$$

Change the integration variables in $\Delta^* (x)$ from $k_\mu$ to $k'_\mu$ where $k'_\mu = - k_\mu$. This should lead to $\Delta^* (x) =\Delta (x)$ (unless I'm missing something).

 

[samalkhaiat]

The above is a conceptual explanation; Is there a way to prove this mathematically?

You don’t need that if you know the meaning of the proper Lorentz transformations, but yes you can prove it as follows: Lorentz transformation consists of rotations and boosts. Rotations leaves the time-component of vectors invariant. So you only need to consider boosts. Now, for a boost along the z-axis, you will have (k^{0})^{\prime} = \gamma (k^{0} - \beta k^{3}). But, you know that \gamma &gt; 0 and \beta &lt; 1. So, for k^{0} &gt; 0, you will have k^{0} &gt; k^{3} &gt; \beta k^{3} \ \Rightarrow \ (k^{0})^{\prime} &gt; 0. Similarly, k^{0} &lt; 0 \ \Rightarrow \ (k^{0})^{\prime} &lt; 0.

To prove that \Delta (x) is Lorentz invariant, calculate \Delta (\Lambda x) and use k \cdot \Lambda x = \Lambda^{-1}k \cdot x. Now define a new integration variable by p = \Lambda^{-1}k, \ \Rightarrow \ p^{2} = (\Lambda^{-1}k)^{2} = k^{2}. Since the proper Lorentz transformation leaves both the sign function and the 4-volume invariant, you can write \epsilon (k_{0}) = \epsilon (p_{0}), \ d^{4}k = d^{4}p. So, you end up with \Delta (\Lambda x) = \frac{-i}{(2 \pi)^{3}} \int d^{4}p \ \epsilon (p_{0}) \delta (p^{2} - m^{2}) \ e^{- i p \cdot x} = \Delta (x).

To prove \Delta (-x) = - \Delta (x), as well as \Delta^{*}(x) = \Delta (x) change integration variable as k = - p, then use k^{2} = p^{2}, \ \epsilon (k_{0}) = - \epsilon (p_{0}) and d^{4}k = d^{4}p.

 

[JD_PM]

Thank you both for your replies

Wouldn't time reversal also change the sign of $x_0$ as well as change the sign of $\vec k$? ($\vec k$ is like a momentum. )So under time reversal wouldn't $k \cdot x = k_0 x_0 - \vec k \cdot \vec x$ become $+k_0 x_0 + \vec k \cdot \vec x$ ?

You are right, my bad.

You don’t need that if you know the meaning of the proper Lorentz transformations, but yes you can prove it as follows: Lorentz transformation consists of rotations and boosts. Rotations leaves the time-component of vectors invariant. So you only need to consider boosts. Now, for a boost along the z-axis, you will have (k^{0})^{\prime} = \gamma (k^{0} - \beta k^{3}). But, you know that \gamma &gt; 0 and \beta &lt; 1. So, for k^{0} &gt; 0, you will have k^{0} &gt; k^{3} &gt; \beta k^{3} \ \Rightarrow \ (k^{0})^{\prime} &gt; 0. Similarly, k^{0} &lt; 0 \ \Rightarrow \ (k^{0})^{\prime} &lt; 0.

To prove that \Delta (x) is Lorentz invariant, calculate \Delta (\Lambda x) and use k \cdot \Lambda x = \Lambda^{-1}k \cdot x. Now define a new integration variable by p = \Lambda^{-1}k, \ \Rightarrow \ p^{2} = (\Lambda^{-1}k)^{2} = k^{2}. Since the proper Lorentz transformation leaves both the sign function and the 4-volume invariant, you can write \epsilon (k_{0}) = \epsilon (p_{0}), \ d^{4}k = d^{4}p. So, you end up with \Delta (\Lambda x) = \frac{-i}{(2 \pi)^{3}} \int d^{4}p \ \epsilon (p_{0}) \delta (p^{2} - m^{2}) \ e^{- i p \cdot x} = \Delta (x).

Oh I see it now.

Instead of trying to introduce a symmetry transformation such as time reversal or a Lorentz transformation, go back to $$\Delta^{*} (x) = \frac{i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{ik \cdot x}.$$

Change the integration variables in $\Delta^* (x)$ from $k_\mu$ to $k'_\mu$ where $k'_\mu = - k_\mu$. This should lead to $\Delta^* (x) =\Delta (x)$ (unless I'm missing something).

To prove \Delta (-x) = - \Delta (x), as well as \Delta^{*}(x) = \Delta (x) change integration variable as k = - p, then use k^{2} = p^{2}, \ \epsilon (k_{0}) = - \epsilon (p_{0}) and d^{4}k = d^{4}p.

Ahhh so the trick was to introduce the change of variables $k := - p$ (I feel I bit silly now to be honest )

- Let's show $\Delta^{*}(x) = \Delta (x)$

We start with

$$\Delta^{*} (x) = \frac{i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{ik \cdot x}.$$

Applying $k = - p$, k^{2} = p^{2}, \ \epsilon (k_{0}) = - \epsilon (p_{0}) and d^{4}k = d^{4}p to $\Delta^{*} (x)$ we get

$$\frac{-i}{(2\pi)^{3}} \int d^{4}p \ \epsilon (p_{0}) \delta (p^{2} - m^{2}) \ e^{-ip \cdot x}=\Delta (x)$$

QED.

- Let's show \Delta (-x) = - \Delta (x)

We start with

$$\Delta (-x) = \frac{-i}{(2\pi)^{3}} \int d^{4}k \ \epsilon (k_{0}) \delta (k^{2} - m^{2}) \ e^{ik \cdot x}.$$

Applying $k = - p$, k^{2} = p^{2}, \ \epsilon (k_{0}) = - \epsilon (p_{0}) and d^{4}k = d^{4}p to $\Delta (-x)$ we get

$$\frac{i}{(2\pi)^{3}} \int d^{4}p \ \epsilon (p_{0}) \delta (p^{2} - m^{2}) \ e^{-ip \cdot x} = - \Delta (x)$$

QED.

 

[JD_PM]

Nice extra insight on Lorentz Transformation properties