How to Derive the Equality Involving u^s(p) and Spin 4-Vector?

In summary, the author is having a problem with deriving the following equality: \begin{equation}u^{s}(p) \bar{u}^{s}(p) = 1/2 ((\slashed{p} + m)(1+\gamma^5 \slashed{s})) \end{equation}
  • #1
Suzie
3
0
Hello everyone, I have a problem with deriving following equality:
\begin{equation}
u^{s}(p) \bar{u}^{s}(p) = 1/2 ((\slashed{p} + m)(1+\gamma^5 \slashed{s}))
\end{equation}

where s is spin 4-vector. I know how to calculate this tensor product when there is spin sum in front of it, but without the sum, I am clueless. Can someone help me, please? Thank you.
 
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  • #2
Suzie said:
Hello everyone, I have a problem with deriving following equality:
\begin{equation}
u^{s}(p) \bar{u}^{s}(p) = 1/2 ((\slashed{p} + m)(1+\gamma^5 \slashed{s}))
\end{equation}

where s is spin 4-vector. I know how to calculate this tensor product when there is spin sum in front of it, but without the sum, I am clueless. Can someone help me, please? Thank you.
This identity cannot be true if there is no sum over the "s".
 
  • #3
No, it is true only without the sum over s. Now I see there might have appeared small misconception: the "s" on the left side is not the same "s" as on the right side of the equality. On the right side, the "s" is four vector.
For example, in particle's rest frame,

\begin{equation}
s = (0, \vec{s})
\end{equation}

where the 3-vector is a unit vector in the direction of the spin of a particle.
With the sum, we would have the completeness relation

\begin{equation}
\sum_{s} u^{s}(p) \bar{s}^s (p) = \slashed p + m
\end{equation}
 
  • #4
Where did you find it? How can it be true if you don't sum over spin degrees of freedom? Calling it "s" or "r", it does not matter. As well as the completeness relation
[tex]\sum_{s} u^{ s }( p ) \bar{ u }^{s} ( p ) = p_{ \mu } \gamma^{ \mu } + m ,[/tex]
you also have the followings
[tex]\frac{1}{2} ( 1 + \gamma_{5} \gamma^{ \mu } s_{ \mu } ) u^{ s } ( p ) = u^{ s } ( p ) ,[/tex]
[tex]\frac{1}{2} \bar{ u }^{ s } ( p ) ( 1 - s_{ \mu } \gamma^{ \mu } \gamma_{ 5 } ) = \bar{ u }^{ s } ( p ) .[/tex]
 
  • #5
I think you can find the answer in Bjorken&Drell's famous book "Relativistic Quantum Mechanics", I remenber it's in the first few chapters.
 
  • #6
Thank you, there is not exactly what I needed, but it helped me to understand at least part of it.
I still don't see how to get from those projection operators from Bjorken&Drell to my expression.
 

FAQ: How to Derive the Equality Involving u^s(p) and Spin 4-Vector?

1. What is the spin 4-vector and how is it used in deriving equality?

The spin 4-vector is a mathematical concept used in quantum mechanics to describe the spin of a particle. In deriving equality, the spin 4-vector is used to represent the spin states of two particles and their interactions.

2. What is the significance of deriving equality with spin 4-vector?

Deriving equality with spin 4-vector allows us to understand the fundamental properties and interactions of particles at a quantum level. It also helps us to develop more accurate models and predictions in the field of quantum mechanics.

3. Can you explain the mathematical process of deriving equality with spin 4-vector?

The mathematical process involves using the Dirac equation, which describes the behavior of spin 1/2 particles, to derive the equality between two particles with different spin states. This involves manipulating the spin 4-vector and its corresponding matrices to achieve the desired equality.

4. What are some real-world applications of deriving equality with spin 4-vector?

The understanding of spin 4-vector and its role in deriving equality has led to advancements in various fields such as materials science, telecommunications, and medical imaging. It has also helped in the development of new technologies such as quantum computers and spin-based electronic devices.

5. Are there any limitations or challenges in using spin 4-vector to derive equality?

One limitation is that the spin 4-vector only applies to spin 1/2 particles, so it cannot be used for particles with different spin values. Additionally, the mathematical calculations involved in deriving equality can be complex and may require advanced knowledge of quantum mechanics and mathematics.

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