Verify that q ·ε=0 for each λ, and show that, for a spacelike photon (q^2<0), Σ(-1)^λ+1 ε^* * ε^

Step 1: **Understanding the notation and context** - $q^mu$ is the four-momentum of the photon.

Question

Verify that $q \cdot \varepsilon=0$ for each $\lambda$, and show that, for a spacelike photon $\left(q^2<0\right)$, $$ \Sigma(-1)^{\lambda+1} \varepsilon^{* *} \varepsilon^r=-g^{\mu r}+\frac{q^\mu q}{q^2}, $$ where the sum runs over the three polarization states of (8.49) and (8.50). For $q^2>0$ the factor $(-1)^{\lambda+1}$ is omitted and the sum (8.51) with $q^2$ replaced by $M^2$ is identical to that over the spin states of a massive vector particle, see Section 6.12.

   Verify that $q \cdot \varepsilon=0$ for each $\lambda$, and show that, for a spacelike photon $\left(q^2<0\right)$,
$$
\Sigma(-1)^{\lambda+1} \varepsilon^{* *} \varepsilon^r=-g^{\mu r}+\frac{q^\mu q}{q^2},
$$
where the sum runs over the three polarization states of (8.49) and (8.50). For $q^2>0$ the factor $(-1)^{\lambda+1}$ is omitted and the sum (8.51) with $q^2$ replaced by $M^2$ is identical to that over the spin states of a massive vector particle, see Section 6.12.

Quarks and leptons: introductory course in modern particle physics

Francis Halzen, Alan… 1st Edition

Chapter 8, Problem 14

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Step 1

- $\varepsilon^\mu$ represents the polarization vector of the photon. - $\lambda$ indexes the polarization states. - $g^{\mu\nu}$ is the metric tensor of spacetime. - $q^2 = q_\mu q^\mu$ is the square of the four-momentum, which determines the nature of the Show more…

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Verify that $q \cdot \varepsilon=0$ for each $\lambda$, and show that, for a spacelike photon $\left(q^2<0\right)$, $$ \Sigma(-1)^{\lambda+1} \varepsilon^{* *} \varepsilon^r=-g^{\mu r}+\frac{q^\mu q}{q^2}, $$ where the sum runs over the three polarization states of (8.49) and (8.50). For $q^2>0$ the factor $(-1)^{\lambda+1}$ is omitted and the sum (8.51) with $q^2$ replaced by $M^2$ is identical to that over the spin states of a massive vector particle, see Section 6.12.

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