 Solution

ID: past-exam-of-the-quantum-theory-course-of-the-university-of-oxford/2021/1/a/iii/solution

Since

P ψ_{n} = n ℏ ψ_{n}

, applying the Hamiltonian operator gives

H ψ_{n} = \frac{P ^{2}}{2 m} ψ_{n} = \frac{n ^{2} ℏ ^{2}}{2 m} ψ_{n} .

Thus the energy eigenvalues and corresponding full stationary states are

E_{n} = \frac{n ^{2} ℏ ^{2}}{2 m}, Ψ_{n} (x, t) = e^{- i n^{2} ℏ t / (2 m)} ψ_{n} (x) .

ψ_{0}

with energy zero and quantum degeneracy one. For each

n = 1, 2, \dots

, the positive energy level

n^{2} ℏ^{2} / (2 m)

span {ψ_{n}, ψ_{- n}}

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