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We have the following Hamiltonian

$\hat{H}=a|u_{1}\rangle\langle u_{2}|+a|u_{2}\rangle\langle u_{1}|$

with a $\in \mathbb{R}$ and $|u_{1}\rangle,|u_{2}\rangle$ an orthonormal system

The matrix representation of $\hat{H}$ in that system is

\begin{pmatrix} 0 & a\\ a & 0 \end{pmatrix}

the eigenvalues are a and -a, and the corresponding orthonormal eigenvectors are $|\phi_{1}\rangle = \frac{1}{\sqrt2}(1,1)^\top$ and $|\phi_{2}\rangle = \frac{1}{\sqrt2}(1, -1)^\top$.

Let $|\Psi(t=0)\rangle = |u_{1}\rangle$ be the state at the moment $t=0$.

1. What are the energy values that we can measure at the moment $t=0$ and with what probabilities?

2. What is the state at the moment t? Which energy values can we measure at that moment and with what probabilities? I've found the state at time t to be $|\Psi(t)\rangle=\frac{1}{\sqrt2}\exp\left(-\frac{iE_{1}}{\hbar}t\right)|\phi_{1}\rangle + \frac{1}{\sqrt2}\exp\left(-\frac{iE_{2}}{\hbar}t\right)|\phi_{2}\rangle$

So, my question is, are there different values of the energy depending on the moment? For what I know, the energy values that we can measure are the eigenvalues of the Hamiltonian. So in both questions 1 and 2 the energy values that we can measure are a and -a, right? And both with probability $\frac{1}{2}$.

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Yes, you are right. First of all about the fact that the only allowed measurement outcomes for an energy measurement are the eigenvalues of $\hat{H}$, $a$ and $-a$, regardless of the time instant in which the measurement is performed. Secondly about the fact that the probabilities are both equal to $\frac{1}{2}$ and constant over time, since only the phases of the coefficients of the time-dependent wavefunction oscillate over time. These phases square out in the absolute values squares that yield the probabilities.

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  • $\begingroup$ Okay, thank you very much. So the only values we can measure for energy are the eigenvalues of the Hamiltonian. However, the system can be found in a state that does not necessarily have to be an eigenvector of the Hamiltonian, right? $\endgroup$
    – schrodingal
    Commented Oct 18, 2020 at 15:30
  • $\begingroup$ No. Well it depends. As long as you measure the energy, the only allowed post-measurement states are energy eigenstates. However, if you measure some other observable, the post-measurement state is one of the eigenstates of the given observable $\endgroup$
    – Milarepa
    Commented Oct 18, 2020 at 15:36
  • $\begingroup$ That bothers me because in an exercise I did, there was this question: what is the probability of finding the system at the moment t in the state $|u_{2}\rangle$? But the state $|u_{2}\rangle$ is not an eigenstate. $\endgroup$
    – schrodingal
    Commented Oct 18, 2020 at 16:25
  • $\begingroup$ This does not contradict any of our statements above: given any two orthonormal states defining a basis which spans the Hilbert space you can perform a projective measurement in this basis. These two states are related to some observable represented by an hermitian operator, whose eigenvalues are the allowed measurement results for the projective measurement. $\endgroup$
    – Milarepa
    Commented Oct 18, 2020 at 16:36
  • $\begingroup$ So $|u_{2}\rangle$ is an allowed post-measurement state because we can express it as a linear combination of $|\phi_{1}\rangle$ and $|\phi_{2}\rangle$, which are the eigenstates of the Hamiltonian. Did I understand right? $\endgroup$
    – schrodingal
    Commented Oct 18, 2020 at 16:43

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