Quantum Mechanics | Part IB, 2002

A quantum mechanical system has two states χ0\chi_{0} and χ1\chi_{1}, which are normalised energy eigenstates of a Hamiltonian H3\mathrm{H}_{3}, with

H3χ0=χ0,H3χ1=+χ1.H_{3} \chi_{0}=-\chi_{0}, \quad H_{3} \chi_{1}=+\chi_{1} .

A general time-dependent state may be written

Ψ(t)=a0(t)χ0+a1(t)χ1,\Psi(t)=a_{0}(t) \chi_{0}+a_{1}(t) \chi_{1},

where a0(t)a_{0}(t) and a1(t)a_{1}(t) are complex numbers obeying a0(t)2+a1(t)2=1\left|a_{0}(t)\right|^{2}+\left|a_{1}(t)\right|^{2}=1.

(a) Write down the time-dependent Schrödinger equation for Ψ(t)\Psi(t), and show that if the Hamiltonian is H3\mathrm{H}_{3}, then

ida0dt=a0,ida1dt=+a1.i \hbar \frac{d a_{0}}{d t}=-a_{0}, \quad i \hbar \frac{d a_{1}}{d t}=+a_{1} .

For the general state given in equation (1) above, write down the probability to observe the system, at time tt, in a state αχ0+βχ1\alpha \chi_{0}+\beta \chi_{1}, properly normalised so that α2+β2=1|\alpha|^{2}+|\beta|^{2}=1.

(b) Now consider starting the system in the state χ0\chi_{0} at time t=0t=0, and evolving it with a different Hamiltonian H1H_{1}, which acts on the states χ0\chi_{0} and χ1\chi_{1} as follows:

H1χ0=χ1,H1χ1=χ0H_{1} \chi_{0}=\chi_{1}, \quad H_{1} \chi_{1}=\chi_{0}

By solving the time-dependent Schrödinger equation for the Hamiltonian H1H_{1}, find a0(t)a_{0}(t) and a1(t)a_{1}(t) in this case. Hence determine the shortest time T>0T>0 such that Ψ(T)\Psi(T) is an eigenstate of H3H_{3} with eigenvalue +1+1.

(c) Now consider taking the state Ψ(T)\Psi(T) from part (b), and evolving it for further length of time TT, with Hamiltonian H2H_{2}, which acts on the states χ0\chi_{0} and χ1\chi_{1} as follows:

H2χ0=iχ1,H2χ1=iχ0H_{2} \chi_{0}=-i \chi_{1}, \quad H_{2} \chi_{1}=i \chi_{0}

What is the final state of the system? Is this state observationally distinguishable from the original state χ0\chi_{0} ?

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