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 Incoming links: Transfer matrix of a delta potential

Assume

a > 0

and put

μ = m g / ℏ^{2}

q = k /ℏ

, and

γ = μ / q

x_{0}

, integration of the Schrödinger equation gives the matching conditions

ψ (x_{0}^{+}) = ψ (x_{0}^{-}), ψ^{'} (x_{0}^{+}) - ψ^{'} (x_{0}^{-}) = 2 μ ψ (x_{0}) .

(A_{L}, B_{L})

and

(A_{R}, B_{R})

gives

(A_{R} - A_{L}) e^{i q x_{0}} + (B_{R} - B_{L}) e^{- i q x_{0}} = 0,

i q [(A_{R} - A_{L}) e^{i q x_{0}} - (B_{R} - B_{L}) e^{- i q x_{0}}] = 2 μ (A_{L} e^{i q x_{0}} + B_{L} e^{- i q x_{0}}) .

(A_{R} B_{R}) = D (x_{0}) (A_{L} B_{L}), D (x_{0}) = (1 - iγ iγ e^{2 i q x_{0}} - iγ e^{- 2 i q x_{0}} 1 + iγ), det D (x_{0}) = 1.

x < - a

- a < x < a

, and

x > a

. Successive matching gives

M = D (a) D (- a)

between the two exterior pairs. For real

q

M = (d^{*} i F - i F d), d = (1 + iγ)^{2} + γ^{2} e^{4 i q a}, F = 2 γ [cos (2 q a) + γ sin (2 q a)] .

The determinant identity

det M = 1

reads

∣ d ∣^{2} - F^{2} = 1

. Left incidence means

(t, 0)^{T} = M (1, r)^{T}

, whence

t = \frac{1}{d} = \frac{q ^{2}}{( q + i μ ) ^{2} + μ ^{2} e ^{4 i q a}}, r = - \frac{i F}{d} .

Here the first relation also follows from

t = M_{11} - M_{12} M_{21} / M_{22} = det M / M_{22}

k

, are

T (k) = \frac{1}{1 + F ( k ) ^{2}}, R (k) = \frac{F ( k ) ^{2}}{1 + F ( k ) ^{2}}, F (k) = \frac{2 μ ℏ}{k} [cos \frac{2 ka}{ℏ} + \frac{μ ℏ}{k} sin \frac{2 ka}{ℏ}] .

They obey

R + T = 1

. As checks,

g = 0

gives

T = 1

, and the coincident limit

a \to 0

gives

T = q^{2} / (q^{2} + 4 μ^{2})

, the answer for a single delta potential of strength

2 g