Quantum Mechanics | Part IB, 2001

A quantum system has a complete set of orthonormalised energy eigenfunctions ψn(x)\psi_{n}(x) with corresponding energy eigenvalues En,n=1,2,3,E_{n}, n=1,2,3, \ldots

(a) If the time-dependent wavefunction ψ(x,t)\psi(x, t) is, at t=0t=0,

ψ(x,0)=n=1anψn(x)\psi(x, 0)=\sum_{n=1}^{\infty} a_{n} \psi_{n}(x)

determine ψ(x,t)\psi(x, t) for all t>0t>0.

(b) A linear operator S\mathcal{S} acts on the energy eigenfunctions as follows:

Sψ1=7ψ1+24ψ2Sψ2=24ψ17ψ2Sψn=0,n3\begin{aligned} &\mathcal{S} \psi_{1}=7 \psi_{1}+24 \psi_{2} \\ &\mathcal{S} \psi_{2}=24 \psi_{1}-7 \psi_{2} \\ &\mathcal{S} \psi_{n}=0, \quad n \geqslant 3 \end{aligned}

Find the eigenvalues of S\mathcal{S}. At time t=0,St=0, \mathcal{S} is measured and its lowest eigenvalue is found. At time t>0,St>0, \mathcal{S} is measured again. Show that the probability for obtaining the lowest eigenvalue again is

1625(337+288cos(ωt))\frac{1}{625}(337+288 \cos (\omega t))

where ω=(E1E2)/\omega=\left(E_{1}-E_{2}\right) / \hbar.

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