ϕ_{n} (x) = \frac{2}{a} sin \frac{nπ x}{a}, E_{n} = \frac{n ^{2} π ^{2} ℏ ^{2}}{2 m a ^{2}}, 0 < x < a .

The sine identity expands the initial wavefunction as

ψ_{0} = \frac{3 ϕ _{1} - ϕ _{3}}{10} .

The normalization factor

4/ 5 a

9/10

for

E_{1}

and

1/10

for

E_{3} = 9 E_{1}

. Thus the energy expectation is

⟨ E ⟩ = \frac{9}{10} E_{1} + \frac{1}{10} E_{3} = \frac{9 π ^{2} ℏ ^{2}}{10 m a ^{2}} .

ii

Solution

Each energy eigenstate acquires the phase

e^{- i E_{n} t /ℏ}

under unitary time evolution. Consequently at the requested time $t = 1$ ,

ψ (x, 1) = \frac{1}{10} (3 e^{- i E_{1} /ℏ} ϕ_{1} (x) - e^{- i 9 E_{1} /ℏ} ϕ_{3} (x)), 0 < x < a,

and the wavefunction is zero outside the well. Equivalently, in elementary functions,

ψ (x, 1) = \frac{1}{5 a} (3 e^{- i π^{2} ℏ/ (2 m a^{2})} sin \frac{π x}{a} - e^{- i 9 π^{2} ℏ/ (2 m a^{2})} sin \frac{3 π x}{a}) .

The distinct phases retain their relative sign from the initial expansion; an overall global phase would not change measurement probabilities.

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