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A non-degenerate two-level system is described by a Hamiltonian $\hat H$ with $\hat H|n\rangle = \epsilon_n|n\rangle$, where $n = 1, 2$. An observable $\hat B$ has eigenvalues $\pm 1$, with the corresponding orthonormal eigenfunctions $|\pm\rangle$ related to $|n\rangle$ by

$|+\rangle = (|1\rangle + 2|2\rangle)/\sqrt{5}$

$|−\rangle = (2|1\rangle − |2\rangle )/\sqrt{5}$

What are the possible outcomes, together with their probabilities, of a measurement of $\hat B$ in a state of the system with the energy $\epsilon_2$? Would you get the same result if you made a second measurement of $\hat B$ some time after the first?

I think i can do the first part:

P(+) = |<2|+>|^2 = 4/5

P(-) = |<2|->|^2 = 1/5

But after making a measurement the wavefunction collapses into either one or the other so should the the probability be 1 and 0, how does this depend on time?

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  • $\begingroup$ Does $\varepsilon$ depend on $n$? $\endgroup$ – probably_someone May 9 '18 at 13:50
  • $\begingroup$ Yes, should be ε subscript n, i.e ε1 doesnt equal ε2 $\endgroup$ – jjp1996 May 9 '18 at 13:53
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If you measurement yields $+$, your state will be in the eigenstate of $\hat B$ with eigenvalue $+$. Thus, for instance, if you obtain $+$ at $t=0$ your system will collapse to $\vert +\rangle$. This becomes your new initial state for the evolution. The time-evolution is obtained by expanding $$ \vert\Psi(0)=\vert +\rangle = \vert 1\rangle \langle 1\vert +\rangle +\vert 2\rangle \langle 2\vert +\rangle $$ from which $$ \vert\Psi(t)\rangle= e^{-i\epsilon_1 t}\vert 1\rangle \langle 1\vert +\rangle +e^{-i\epsilon_2 t}\vert 2\rangle \langle 2\vert +\rangle $$ Since the time-evolution does not produce an eigenstate of $\hat B$, a second measurement of $\hat B$ would produce $+$ with probability $$ P(+,t)=\vert \langle +\vert\Psi(t)\rangle\vert^2 =\vert e^{-i\epsilon_1 t}\langle +\vert 1\rangle \langle 1\vert +\rangle +e^{-i\epsilon_2 t}\langle +\vert 2\rangle \langle 2\vert +\rangle\vert^2 \ne 1 $$ and likewise for $P(-,t)$.

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  • $\begingroup$ so this will give: P(+,t) = |exp(-iE1t/h)/5 + 4exp(-iE2t/h)/5|^2, i take it then that the exp are not cacelled by the mod squared as this would just give a probability of 1. $\endgroup$ – jjp1996 May 9 '18 at 15:34
  • $\begingroup$ correct. $\vert A+B\vert^2 \ne \vert A\vert^2+ \vert B\vert^2$ so the exponentials will "interfere" and will produce a time-dependent probabilities. $\endgroup$ – ZeroTheHero May 9 '18 at 15:51

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