A quantum system has Hamiltonian $H$ with normalised eigenstates $\psi_n$ and corresponding energies $E_n$ ($n = 1,2,3...$). A linear operator $Q$ is defined by its action on these states: $$ Q\psi_1 = \psi_2 $$ $$ Q\psi_2 = \psi_1 $$ $$ Q\psi_n = 0, n>2 $$ Show that $Q$ has eigenvalues 1 and -1 and find the corresponding normalised eigenstates $\zeta_1$ and $\zeta_2$, in terms of energy eigenstates. Calculate $\langle H \rangle$ in each of the states $\zeta_1$ and $\zeta_2$.

A measurement of $Q$ is made at time=0, and the result 1 is obtained. The system is then left undisturbed for a time $t$, at which instant another measurement of $Q$ is made. What is the probability that the result will again be 1? Show that the probability is 0 if the measurement is made after a time $T = \pi \hbar/(E_2 - E_1)$, assuming $E_2 - E_1> 0$.

I found $$ \zeta_1 = (\psi_1 + \psi_2)/\sqrt{2} $$ $$ \zeta_2 = (\psi_1 - \psi_2)/\sqrt{2} $$ and $\langle H \rangle = (E_1 + E_2)/2$ for both.

I have trouble doing the second part.

Doesnt the system collapse into $ \zeta_1$ given we know this is the state at time =0? so probability will be 1?


1 Answer 1


Doesnt the system collapse into ζ1 given we know this is the state at time t=0?

True, but after the measurement at $t_0$ the system follows the unitary dynamics given by $H$. As you can easily check, $\zeta_1$ is not an eigenstate of the Hamiltonian, iff $E_1\neq E_2$. Hence the system will not remain in the state $\zeta_1$ during time-evolution.

If you were to perform another measurement directly after the first one, the outcome would be the same. Here and generally this is not true for arbitrary times between measurements.

As a hint, I remind you that the time-evolution of an eigenstate of the Hamiltonian looks like $$ U(t,t_0)\psi_n = \exp\left[-i(t-t_0)E_n / \hbar \right] \psi_n $$

  • $\begingroup$ thanks. so ψ(x,0) = (ψ1 + ψ2)/√2, and U(t,0) = exp(-iHt/ħ), then ψ(x,t) = U(t,0)ψ(x,0) = exp(-iHt/ħ)(ψ1 + ψ2)/√2 taking inner product : ∫ [exp(iHt/ħ)((ψ1)* + (ψ2)*)/√2][(ψ1 + ψ2)/√2] dx = exp(iHt/ħ) is this correct? but i dont know what H is here, is it (E1+E2)/2? $\endgroup$ Commented Sep 19, 2014 at 8:19
  • $\begingroup$ @MartinCheung Well, you know how $H$ acts on the $\psi_n$... Btw.: there is no variable $x$ in your problem. You're not dealing with position here. You can label your states by a quantum number $n$ as eigenstates of the Hamiltonian. So don't write $\psi(x,t)$ and better not represent the inner product as an integral. $\endgroup$
    – Nephente
    Commented Sep 19, 2014 at 8:27
  • $\begingroup$ so ψ(0) = (ψ1 + ψ2)/√2, ψ(t) = U(t,0)ψ(0) = exp(-iHt/ħ)(ψ1 + ψ2)/√2 = [exp(-iE1t/ħ)ψ1 + exp(-iE2t/ħ)ψ]/√2. then find the probability amplitude of (ψ1 + ψ2)/√2 in this time-varying state by taking the inner product ((ψ1 + ψ2)/√2 , ψ(t)), squared it to find the probability = [1 + cos((E2-E1)t/h)]/2, then sub in T = pi*h/(E2-E1) indeed got probability = 0. but why after T it'll stay 0? doesnt it cycle again by the consine term? $\endgroup$ Commented Sep 20, 2014 at 2:48
  • $\begingroup$ @MartinCheung Your analysis seems correct and indeed, the probability is periodic. It does not stay zero. As far as i understand, the problem doesn't claim this to be the case. It asks for a very specific time, not all times larger than some time. $\endgroup$
    – Nephente
    Commented Sep 21, 2014 at 8:52
  • $\begingroup$ @MartinCheung Did you solve the problem or do you need further clarification? If not, consider upvoting/accepting my answer. $\endgroup$
    – Nephente
    Commented Sep 23, 2014 at 8:08

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