Question related to the Lorentz Invariance

In summary, the conversation discusses the use of two methods to show the result of problem 2.9c in the book "Quantum Field Theory" by Mark Srednicki. The first method involves expanding lambda to infinitesimal form using Taylor expansion, while the second method uses the original exponential form. When values of S^12 are inserted, the first method gives a value of 0, while the second method gives a value of 1. The correct answer is 0, but there is a discrepancy between the two methods. The conversation then provides a detailed explanation using matrix calculations, and concludes that the exponential form of e^{-i \theta S} is equal to \begin{bmatrix} 1 & 0
  • #1
lhcQFT
5
0
I have a question related to the Lorentz invariance.

(on the book of Mark Srednicki Quantum Field Theory, page 35 prob. 2.9 c)

There are representations of [itex]\Lambda[/itex] and S.

In order to show that result of problem, I use number of two ways.

1. I expanded [itex]\Lambda[/itex] to infinitesimal form using Taylor expansion.
2. Just original shape of exponential form.

When I inserted values of S^12, the first method gives me that the value is 0.(by specific element, that is rho = 0, tau =1) But the second method gives me that the value is 1.

The correct answer is 0. because most of components are 0 except for middle row and middle column(2x2). I can't understand this discrepancy. Please explain this error.

(Sorry about that I can't use Latex. because I'm a novice of this web site..)
 
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  • #2
you don't have to use infinitesimal expansions... Just use the series form for the exponential:

[itex] e^{A} = \sum_{n} \frac{A^{n}}{n!}[/itex]

I use [itex] \hbar =1 [/itex] although it doesn't matter to state since you can also rename [itex]S^{12} \equiv \frac{1}{\hbar} S^{12}[/itex] and using the appropriate form for [itex]S^{12}[/itex] repr, as I do below - no hbar, all to keep exponential argument dimensionless. Also some expressions might look terrible, but if you are accustomed to these calculations, you can jump them - i just did everything almost explicitly because I don't know you.

[itex] e^{-i \theta S^{12}} = \sum_{n} \frac{(-i \theta S^{12})^{n}}{n!}[/itex]

[itex]e^{-i \theta S^{12}} = 1_{4 \times 4} + \sum_{p=1} (-i)^{2p} (S^{12})^{2p} (\frac{\theta}{2p!})^{2p}+\sum_{p=0} (-i)^{2p+1} (S^{12})^{2p+1} (\frac{\theta}{2p+1!})^{2p+1} [/itex]

Now [itex][(S^{12})^{2}]^{\rho}_{\omega} = \sum_{\sigma} (S^{12})^{\rho}_{\sigma} (S^{12})^{\sigma}_{\omega} = \sum_{\sigma} (-i)^{2} (g^{1 \rho} \delta^{2}_{\sigma}- g^{2 \rho} \delta^{1}_{\sigma}) (g^{1 \sigma} \delta^{2}_{\omega}- g^{2 \sigma} \delta^{1}_{\omega})= - \sum_{\sigma}[ g^{1 \rho} \delta^{2}_{\sigma} g^{1 \sigma} \delta^{2}_{\omega} - g^{1 \rho} \delta^{2}_{\sigma}g^{2 \sigma} \delta^{1}_{\omega}- g^{2 \rho} \delta^{1}_{\sigma}g^{1 \sigma} \delta^{2}_{\omega}+ g^{2 \rho} \delta^{1}_{\sigma}g^{2 \sigma} \delta^{1}_{\omega}][/itex]

Now from this -due to summing over the delta kroenecker and metric is diagonal- the 1st and last term vanish, while the other two survive... So you have [doing the sum of sigma]:

[itex][(S^{12})^{2}]^{\rho}_{\omega} = g^{1 \rho} g^{22} \delta^{1}_{\omega}+ g^{2 \rho} g^{11} \delta^{2}_{\omega}[/itex]

From here you see that the squared [itex] S^{12} [/itex] is diagonal (try putting [itex] \rho \ne \omega[/itex] ) and has entries 1 at [itex] \rho = \omega = 1,2 [/itex]. I'll drop 12 of S since we are only caring about them, and with [itex]S[/itex] I'll mean the matrix of [itex]S^{12}[/itex]

[itex] S^{2} = \begin{bmatrix}
0 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0\\
\end{bmatrix} [/itex]

So every power [itex]S^{2k} = S^{2},~ (k>0) [/itex]. This can also be seen without doing any calculations if you could have noticed the similarity of [itex]S[/itex] to the Pauli matrice in a block form. What about the [itex]S^{2k+1}[/itex] ? Well you just have to multiply [itex]S^{2}[/itex] with the [itex]S[/itex] matrix...
[itex]
S \cdot S^{2} = -i \begin{bmatrix}
0 & 0 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & 0 & 0\\
\end{bmatrix}

\begin{bmatrix}
0 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 0\\
\end{bmatrix} =

\begin{bmatrix}
0 & 0 & 0 & 0\\
0 & 0 & -i & 0\\
0 & i & 0 & 0\\
0 & 0 & 0 & 0\\
\end{bmatrix} = S

[/itex]

And finally you can find the exponential:

[itex]e^{-i \theta S} = 1_{4 \times 4} + S^{2} \sum_{p=1} (-i)^{2p} (\frac{\theta}{2p!})^{2p}+ S \sum_{p=0} (-i)^{2p+1} (\frac{\theta}{2p+1!})^{2p+1}= 1_{4 \times 4} + S^{2} \sum_{p=1} (-1)^{p} (\frac{\theta}{2p!})^{2p}- i S \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} [/itex]

which in matrix form is:

[itex]e^{-i \theta S} =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & 1+\sum_{p=1} (-1)^{p} (\frac{\theta}{2p!})^{2p} & - \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} & 0\\
0 & \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} & 1+\sum_{p=1} (-1)^{p} (\frac{\theta}{2p!})^{2p} & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}
[/itex]

That's the [itex]\Lambda[/itex] you have... just see the series expansions of cos and sin in your expression...

[itex]e^{-i \theta S} =
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & \sum_{p=0} (-1)^{p} (\frac{\theta}{2p!})^{2p} & - \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} & 0\\
0 & \sum_{p=0} (-1)^{p} (\frac{\theta}{2p+1!})^{2p+1} & \sum_{p=0} (-1)^{p} (\frac{\theta}{2p!})^{2p} & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}

=
\begin{bmatrix}
1 & 0 & 0 & 0\\
0 & \cos \theta & - \sin \theta & 0\\
0 & \sin \theta & \cos \theta & 0\\
0 & 0 & 0 & 1\\
\end{bmatrix}

[/itex]
 
Last edited:
  • #3
Ah.. Thank you for your guide. I make sense how to calculate. :-)
 

1. What is Lorentz Invariance?

Lorentz Invariance is a fundamental principle in physics that states that the laws of physics should remain the same for all observers in any inertial reference frame, regardless of their relative motion.

2. Why is Lorentz Invariance important?

Lorentz Invariance is important because it is a key principle in Einstein's theory of Special Relativity, which revolutionized our understanding of space and time. It also plays a crucial role in modern physics, including quantum mechanics and particle physics.

3. How is Lorentz Invariance related to the speed of light?

The speed of light, c, is the same for all observers in any inertial reference frame, according to the principle of Lorentz Invariance. This means that no matter how fast an observer is moving, they will always measure the speed of light to be c. This led to the famous equation, E=mc^2, and the realization that the speed of light is the ultimate speed limit in the universe.

4. Can Lorentz Invariance be violated?

Currently, there is no evidence to suggest that Lorentz Invariance can be violated. It has been extensively tested and confirmed by numerous experiments. However, some theories, such as string theory, propose that Lorentz Invariance may break down at very small scales.

5. How does Lorentz Invariance impact our daily lives?

Although we may not directly notice the effects of Lorentz Invariance in our daily lives, it has greatly influenced our understanding of the universe and has led to many technological advancements. GPS systems, for example, use the principles of Lorentz Invariance to accurately determine location and time. Without it, our modern world would look very different.

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