 Solution

ID: past-exam-of-the-advanced-quantum-field-theory-for-particle-physics-course-of-the-university-of-oxford/2018/3/a/i/solution

Because

(γ_{5})^{2} = 1

, its even and odd powers are

γ_{5}^{2 r} = 1

and

γ_{5}^{2 r + 1} = γ_{5}

. Separate the matrix exponential into these two power series:

e^{i α γ_{5}} = 1 r = 0 \sum \infty \frac{( i α ) ^{2 r}}{( 2 r )!} + γ_{5} r = 0 \sum \infty \frac{( i α ) ^{2 r + 1}}{( 2 r + 1 )!} = 1 cos α + i γ_{5} sin α .

No representation of the gamma matrices is needed. Hermiticity of

γ_{5}

also makes this exponential unitary for real

α

, while the projectors

(1 \pm γ_{5}) /2

acquire opposite phases.

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