The Gamma Matrix Trace Paradox: A Conundrum in Spinor Calculations

malawi_glenn
Science Advisor
Messages
6,735
Reaction score
2,431
gamma matrix trace Paradox??

Hello, i tried to evaluate this particular little guy:

\text{Tr} (\gamma ^0 p_\mu \gamma ^\mu \gamma ^0 q_\nu \gamma ^\nu )

using these identities:

\gamma^0 \gamma^0 = I

\text{Tr} (\gamma^\mu \gamma^\nu \gamma^\rho \gamma^\sigma) = 4 (g^{\rho \sigma} g^{\mu \nu} - g^{\nu \sigma} g^{\mu \rho} + g^{\mu \sigma}g^{\nu \rho} )

\text{Tr} (\gamma^\mu\gamma^\nu) = 4\eta^{\mu\nu}

g^{00} = 1, \quad g^{ii} = -1using that second relation, I get:

p_\mu q_\nu \text{Tr} (\gamma ^0 \gamma ^\mu \gamma ^0 \gamma ^\nu ) = p_\mu q_\nu 4 (g^{0\mu} g^{0 \nu} - g^{0 0} g^{\mu \nu } + g^{\mu 0}g^{\nu 0} ) =

p_\mu q_\nu (8\delta ^{0\mu}\delta ^{0\nu} - 4g^{\mu \nu } ) = 4p^0q^0 + 4\vec{q}\cdot \vec{p}

Using the first and third, and the fact the traces are invariant under cyclic permutations of matrices.

p_\mu q_\nu\text{Tr} (\gamma^0 \gamma ^0 \gamma ^\mu \gamma ^\nu ) = p_\mu q_\nu 4g^{\mu \nu } = 4p^0q^0 - 4\vec{q}\cdot \vec{p}

What happened?
 
Physics news on Phys.org


I found the solution to the paradox, I did not do a CYCLIC permutation ;-)
 


I think it is much easier to substitute p_\mu\gamma^\mu = p_0\gamma^0 + p_j\gamma^j. Then separate into two sums. You get two sums. One has (\gamma^0)^3 which you can reduce easily. The other has a product \gamma^0\gamma^j\gamma^0. To reduce this, note that \gamma^0 and \gamma^j anticommute. Now do the same thing with the other gamma product.

I never could figure out why students are taught the hard way to do these problems. One should always rely on the facts of the Clifford algebra, that is, anticommutation, and squaring to +1 or -1 for the gammas.
 


I don't know, we are taught to use anticommutation relations with the metric and using cyclic invariant of trace.

I see the strength in your advice although. Thank you
 


I think the reason they teach it to you that way is because it is more general. That is, they are teaching you things according to the principles of symmetry and this will work for other spin cases than spin 1/2. The trick I showed you only works for spin 1/2.

By the way, there is a whole nother method that uses density matrix principles instead of spinors and I like those methods too.
 
Toponium is a hadron which is the bound state of a valance top quark and a valance antitop quark. Oversimplified presentations often state that top quarks don't form hadrons, because they decay to bottom quarks extremely rapidly after they are created, leaving no time to form a hadron. And, the vast majority of the time, this is true. But, the lifetime of a top quark is only an average lifetime. Sometimes it decays faster and sometimes it decays slower. In the highly improbable case that...
I'm following this paper by Kitaev on SL(2,R) representations and I'm having a problem in the normalization of the continuous eigenfunctions (eqs. (67)-(70)), which satisfy \langle f_s | f_{s'} \rangle = \int_{0}^{1} \frac{2}{(1-u)^2} f_s(u)^* f_{s'}(u) \, du. \tag{67} The singular contribution of the integral arises at the endpoint u=1 of the integral, and in the limit u \to 1, the function f_s(u) takes on the form f_s(u) \approx a_s (1-u)^{1/2 + i s} + a_s^* (1-u)^{1/2 - i s}. \tag{70}...

Similar threads

Replies
2
Views
3K
Replies
13
Views
5K
Replies
2
Views
1K
Replies
4
Views
3K
Replies
0
Views
1K
Replies
1
Views
2K
Back
Top