Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2018 / Paper 2 / Section A(ii) / 14 / ii

Only $n=1,2$ have energies below $6ϵ$. Squaring the corresponding expansion coefficients gives $0.2^2+0.3^2=0.13$.

The printed rounded coefficients have normThus the expected rounded answer is $P(E<6ϵ)\simeq 0.13$. If the decimal coefficients are treated as literal exact inputs and the state normalized, the exact probability for that input is $0.13/S\simeq 0.129916$. This distinction avoids treating rounding error as a physical departure from wavefunction normalization.

Orthonormal energy eigenstates contribute incoherently to the expectation of energy:Therefore $⟨E⟩\simeq 13.2018ϵ$ for the normalized literal decimals. Ignoring the small rounding-induced norm error gives $13.2104ϵ$, conventionally reported at this precision as about $13.2ϵ$. The mean need not equal an individual energy eigenvalue.

Each energy component acquires its own phase:The $1/\sqrt{S}$ factor normalizes the printed rounded input; it can be omitted if the coefficients are regarded as an already normalized rounded representation.

The energy probabilities and $⟨E⟩$ remain constant under this time-independent hamiltonian because phase factors have unit modulus. position probabilities can nevertheless vary: interference between different energy components generally changes with their relative phases. A stationary energy distribution is not the same thing as a stationary spatial probability distribution.

The result $16ϵ$ projects the state onto the corresponding nondegenerate energy eigenstate. Immediately after an ideal measurement,for an unobservable overall phase $\chi$. Subsequent isolated evolution adds $e^{-16iϵt/\hslash }$ but does not mix in other energy eigenstates. A repeat ideal energy measurement gives $16ϵ$ with probability one, provided the hamiltonian is unchanged and no intervening disturbance occurs.