Given are three energy states: $$|\psi_1\rangle = \begin{pmatrix} \frac{1}{\sqrt2}\\ \frac{-1}{\sqrt2}\\ 0 \end{pmatrix} \,,\quad |\psi_2\rangle = \begin{pmatrix} \frac{1}{\sqrt6}\\ \frac{1}{\sqrt6}\\ \frac{-2}{\sqrt6} \end{pmatrix} \,,\quad |\psi_3\rangle = \begin{pmatrix} \frac{1}{\sqrt3}\\ \frac{1}{\sqrt3}\\ \frac{1}{\sqrt3} \end{pmatrix} \,. $$ The above three matrices are the eigenstates of a three state system with energy $E$, $E$, $E-3g$ respectively The system is initially in the state $$ |\mathrm i \rangle = \begin{pmatrix} 1\\ 0\\0 \end{pmatrix} \,. $$ The probability that it will be in a state $$|\mathrm f \rangle = \begin{pmatrix} 0\\ 0\\1\end{pmatrix} $$ is given by what?


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  • $\begingroup$ I see what you mean, let me write an answer ... $\endgroup$ – Martin Ueding Oct 1 '16 at 11:52

You have three eigenstates of the Hamiltonian: $$|\psi_1\rangle = \begin{pmatrix} \frac{1}{\sqrt2}\\ \frac{-1}{\sqrt2}\\ 0 \end{pmatrix} \,,\quad |\psi_2\rangle = \begin{pmatrix} \frac{1}{\sqrt6}\\ \frac{1}{\sqrt6}\\ \frac{-2}{\sqrt6} \end{pmatrix} \,,\quad |\psi_3\rangle = \begin{pmatrix} \frac{1}{\sqrt3}\\ \frac{1}{\sqrt3}\\ \frac{1}{\sqrt3} \end{pmatrix} \,. $$ You know their energies, $E_1 = E$, $E_2 = E$, and $E_3 = E-3g$ where $g$ is some constant. This means that we can write the stationary Schrödinger equation $\hat H |\psi_i\rangle = E_i$.

The states $|\mathrm i\rangle = (1, 0, 0)$ and $|\mathrm f\rangle = (0, 0, 1)$ that you give are not eigenstates of the Hamiltonian $\hat H$. Therefore they are not stationary. They evolve with the time evolution operator in spectral representation like this: $$ |\mathrm i(t)\rangle = \sum_j \mathrm e^{- \mathrm i E_j t / \hbar} | \psi_j \rangle\langle \psi_j | \mathrm i \rangle \,. $$

The only interesting question would be to ask for the probability to find that a system prepared in the initial state end up in the final state after waiting a certain time $t$. What we want to compute is the scalar product $\langle \mathrm f | U(t) | \mathrm i \rangle$. Plugging everything in gives \begin{align} \langle \mathrm f | U(t) | \mathrm i \rangle &= \langle \mathrm f |\mathrm i(t)\rangle \\ &= \sum_j \langle \mathrm f | \mathrm e^{- \mathrm i E_j t / \hbar} | \psi_j \rangle \langle \psi_j | \mathrm i \rangle \\ &= \sum_j \mathrm e^{- \mathrm i E_j t / \hbar} \langle \mathrm f | \psi_j \rangle \langle \psi_j | \mathrm i \rangle \\ &= \sum_j \mathrm e^{- \mathrm i E_j t / \hbar} \langle \psi_j | \mathrm f \rangle^* \langle \psi_j | \mathrm i \rangle \,. \end{align}

The expressions $\langle \psi_j | \mathrm i \rangle$ are just the coefficients given for the states above. They are all real, so the complex conjugation does not make any difference. Then you can plug in the energy values $E_j$ and compute the probability for a given time $t$.

Note that for $t = 0$ the expression falls down to $\langle \mathrm f | \mathrm i \rangle$ which is just zero. The two states are orthogonal to begin with. Only if you let one of them evolve in time, it develops an overlap with the final state.

  • $\begingroup$ Exactly. one state doesnt evolve into a state which is orthogonal to it correct? $\endgroup$ – Prasad Mani Oct 1 '16 at 12:21
  • $\begingroup$ If you look at $\langle \mathrm f(t) | \mathrm i(t) \rangle$ you see that it is just $\langle \mathrm f(0) | \mathrm i(0) \rangle$ because the time evolution would just cancel like $U^\dagger U = \mathbf 1$. The state $|\mathrm i\rangle$ is not an eigenstate of $\hat H$, so it does change with time. And then it has a chance to become overlapping with some particular final state (which does not evolve in time because one holds it fixed). $\endgroup$ – Martin Ueding Oct 1 '16 at 12:24
  • $\begingroup$ To an energy eigenstate will not evolve into something else, it will always be orthogonal to the other energy eigenstates. But a mixed state can change over time. $\endgroup$ – Martin Ueding Oct 1 '16 at 12:26
  • $\begingroup$ okay so it has to evolve since it is not a stationary(eigen) state,vut in this case it doesnt evolve into the final state mentioned since they are orthogonal? is that the reasoning?...is that always true? "a state does not evolve into some other state which is orthogonal to it?" $\endgroup$ – Prasad Mani Oct 1 '16 at 12:28
  • $\begingroup$ A state can evolve into a state which is orthogonal to it, if the state in question is not an energy eigenstates. As you see in the big equation with the explicit time evolution, the relative components of the state $|\mathrm i(t)\rangle$ do change over time! A system in state $|\psi_1\rangle$ will always stay in that state and always be orthogonal to $|\psi_2\rangle$. $|\mathrm i\rangle$ however can evolve to be non-orthogonal to $|\mathrm f\rangle$ (if that state is held fixed). The overlap is calculated in the multi-line equation. Since the $E_j$ are not all the same, it changes in time. $\endgroup$ – Martin Ueding Oct 1 '16 at 12:32

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