I have another point in QM that I would like clarified. Suppose $$\{|n\rangle\}$$ is a set of eigenstates of both the Hamiltonian $H$ and another operator $\hat O$ corresponding to an observable also. We have a state $$|\psi\rangle=\sum_{n=1}^3 a_n |n\rangle$$ If it is such that $$\hat O|1\rangle=1,\,\,\, \hat O|2\rangle=\hat O|3\rangle=-1$$ and a measurement of $O$ at time $t=0$ gives $-1$. How then does the system evolve In other words, what is $$|\psi(t)\rangle$$?

I know that $$|\psi(t)\rangle=\exp(-i\hat H t/\hbar)|\psi(0)\rangle$$

But then does it mean it is $$|\psi(t)\rangle=N(\exp(-iE_2 t/\hbar)a_2|2\rangle+\exp(-iE_3 t/\hbar)a_3|3\rangle)$$ with some normalization constant in front?

So we can never get $|1\rangle$ back?


1 Answer 1


Yes, this looks correct, except that the energy of the state $|2\rangle$ in $|\psi(t)\rangle$ should be $E_2$. I also don't think you want hats on the energies $E_2$ and $E_3$. Such hats are usually used in basic quantum mechanics to indicate operators, bur $E_i$ are the eigenvalues of the operator $\hat{H}$, and hence just ordinary numbers.

Since $\hat{H}$ and $\hat{O}$ have the same eigenvectors these operators commute, which means that the eigenvalues of $\hat{O}$ are preserved under time evaluation. At time $t=0$ measure $\hat{O}$ and obtain the value $-1$. This value is preserved in time, so you will never find an overlap with the state $|1\rangle$ at any later time since this would mean a later measurement of $\hat{O}$ could give the value $1$ instead of $-1$.


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