 Solution

ID: past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2021/section-b/b4/g/solution

P_{\pm}^{2} = P_{\pm}

P_{+} P_{-} = 0

and

P_{+} + P_{-} = I

. Define the uniform-medium generator

K = γ^{+} P_{+} + γ^{-} P_{-} = (γ - i δ i δ γ) .

Expanding its matrix exponential by its power series and using these projection identities gives

e^{- Kz} = e^{- γ^{+} z} P_{+} + e^{- γ^{-} z} P_{-} = J (z) .

Equivalently, with

Q = (0 - i i 0)

, one has

Q^{2} = I

K = γ I + δ Q

and

e^{- Kz} = e^{- γ z} [I cosh (δz) - Q sinh (δz)]

. This is exactly the corrected matrix of part (d). The entries of

K

have units of inverse length, and the map satisfies

d J / d z = - K J

J (0) = I

and the homogeneous propagation law

J (z_{1} + z_{2}) = J (z_{1}) J (z_{2})

The extra

sech^{2}

factor in the printed version violates this last law for nonzero

δ

, so it cannot be generated by one constant

K

with the stipulated eigenvalues. The constant-

K

description assumes uniform coefficients and fixed circular eigenmodes.

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