0
$\begingroup$

Let $H$ be the Hamiltonian of a system with a set of degenerate energies, so that $H |1⟩ = E_1 |1⟩$, $H |2⟩ = E_1 |2⟩$ and $H |3⟩ = E_2 |3⟩$. Given a linear combination of these three states, I want to know what is the state after you measure the energy of the system and get $E_1$.

For example, if you have a normalized state $|\psi⟩ = \frac{1}{\sqrt{14}}(2|1⟩ + |2⟩ - 3|3⟩)$, and $E_1$ is measured, is the state immediately after $$|\psi⟩ = \frac{1}{\sqrt{2}}(|1⟩ + |2⟩),$$

where the probabilities of $|1⟩$ and $|2⟩$ are the same?

Or does the state preserve the relative probabilities between $|1⟩$ and $|2⟩$ and it's just the original renormalized state without the $|3⟩$ component? i. e. $$|\psi⟩ = \frac{1}{\sqrt{5}}(2|1⟩ + |2⟩)?$$

I am interested in the general case where the probabilities of the degenerate eigenstates are different in the initial state.

$\endgroup$
3
$\begingroup$

The appropriate projection operator for this type of measurement is $\Pi_{E_1}=\vert 1\rangle \langle 1\vert + \vert 2\rangle \langle 2\vert$, i.e. simple unweighted sum over all the projectors corresponding to energy $E_1$. Of course you will need to renormalize your ket after the projection.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.