 Solution

ID: past-exam-of-the-theoretical-physics-ii-course-of-the-university-of-cambridge/2012/paper-1/5/d/solution

After integrating

x_{1}, \dots, x_{N - 1}

, the remaining exponent contains

(x_{T} - x_{0})^{2} / N

(\frac{m}{2 πi ℏ ϵ})^{N /2} k = 1 \prod N - 1 (\frac{2 πi ℏ ϵ}{m} \frac{k}{k + 1})^{1/2} = \frac{m}{2 πi ℏ ϵ} \frac{1}{N} = \frac{m}{2 πi ℏ T} .

Since

N ϵ = T

K (x_{T}, x_{0}; T) = \frac{m}{2 πi ℏ T} exp [\frac{im ( x _{T} - x _{0} ) ^{2}}{2ℏ T}] .

For

T > 0

, the square root has phase

e^{- iπ /4}

. Its short-time limit is the position delta function in the distributional sense, consistent with unitary evolution.

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