Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,082 questions , 2,232 unanswered
5,353 answers , 22,789 comments
1,470 users with positive rep
820 active unimported users
More ...

  Evaluation of the spin-sum for a massive spin-1 particle?

+ 0 like - 0 dislike
2883 views

Assuming that a massive spin-1 particle has momentum only in the z-direction, the polarization vectors are given by

$$\varepsilon_{\mu}(J_z = +1) = (0,-\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0 )$$

$$\varepsilon_{\mu}(J_z = 0) = (\frac{p}{m},0,0, \frac{E}{m})$$

$$\varepsilon_{\mu}(J_z = -1) = (0,\frac{1}{\sqrt{2}},-\frac{i}{\sqrt{2}},0 )$$

The so-called spin-sum is the claimed to be

$$
\sum\limits_{J_z = -1,0,+1} \varepsilon_{\mu}\varepsilon_{\nu}^* = g_{\mu\nu} + \frac{p_{\mu}p_{\nu}}{m^2}
$$

I absolutely dont understand how this spin-sum is evaluated.
What does $\varepsilon_{\mu}\varepsilon_{\nu}^*$ even exactly mean? Is it a scalar product between two of the three above polarization vectors, or is it a "tensor-product" between the components of a single polarization vector which results in a 4x4 matrix and one has finally to sum all such matrices for the three possible values of $J_z$?
I would really appreciate it if somebody can explain to me what this spin-sum exactly means and how it is evaluated step-by-step.869

asked Jul 29, 2015 in Theoretical Physics by anonymous [ no revision ]

It is a tensor-product, as you could have seen from the right hand side. Just evaluate the three matrices explicitly and sum them to get the result. It is pure linear algebra.

To add to Arnold, here might be a even straighter way for OP to see this is a elementary linear algebra calculation: Think of $\varepsilon_\mu(J_z)$ as a 4 by 3 matrix $\varepsilon$, rows labeled by $\mu$ and columns labeled by $J_z$, then the spin sum is nothing but $\varepsilon \varepsilon^\dagger$, which is a 4 by 4 matrix.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysicsOve$\varnothing$flow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...