Wednesday, 6 February 2019

homework and exercises - Pauli matrices identities


There are some definitions and properties for Pauli matrices and their combinations:


εαβ=ε˙α˙β=(0110)αβ,ε˙α˙β=εαβ=εαβ,

(σμ)α˙α=(ˆE,ˆσ)μα˙α,(˜σμ)˙ββ=εαβε˙α˙β(σμ)α˙α=(ˆE,ˆσ)μ,˙ββ,
(σμν)αβ=14((σμ˜σν)αβ(σν˜σμ)αβ),(˜σμν)˙α˙β=14((˜σμσν)˙α˙β(˜σνσμ)˙α˙β),
(˜σμ)˙αα(σμ)β˙β=2δ˙α˙βδαβ.
How to show, that (σαβ)ab(˜σμν)˙c˙dgαμ=0?
(It helps to show, that spinor irreducible representation of the generators of Lorentz group expands on two spinor subgroups).


My attempt.


I tried to show this, but only got following. (σαβ)ab(˜σμν)˙c˙dgαμ=116((σα)a˙n(˜σβ)˙nb(˜σμ)m˙c(σν)m˙d(σα)a˙n(˜σβ)˙nb(˜σν)m˙c(σμ)m˙d)

+116((σβ)a˙n(˜σα)˙nb(˜σμ)m˙c(σν)m˙d+(σβ)a˙n(˜σα)˙nb(˜σν)m˙c(σμ)m˙d)gαμ
After that I transformed each summand like (σα)a˙n(˜σβ)˙nb(˜σμ)m˙c(σν)m˙dgαμ=(σα)a˙n(˜σα)m˙c(˜σβ)˙nb(σν)m˙d=
=ε˙c˙γ(σα)a˙n(˜σα)˙γm(˜σβ)˙nb(σν)m˙d=2ε˙c˙γδ˙γ˙mδna(˜σβ)˙nb(σν)m˙d=
2ε˙c˙m(˜σβ)˙mb(σν)a˙d=2(σβ)b˙c(σν)a˙d.
Finally, I got (σαβ)ab(˜σμν)˙c˙dgαμ=18((σβ)b˙c(σν)a˙d+(σβ)b˙d(σν)a˙c+(σβ)a˙c(σν)b˙d+(σβ)a˙d(σν)b˙c).
What to do next?



Answer



Note, that, on the third line, the β indice of σμν and the ˙β indice of ˜σμν must be raised for indice coherence. Same error for the following lines.


The formula you want to demonstrate is certainly false. Take β=ν=0, and noting that σα0=12σα,˜σμ0=12σμ, if the formula was exact, it would imply (with g11=g22=g33), and in short tensorial notation :


14σσ=0



which is obviously false.


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