Trouble understanding how quantum observables evolve in time

Question

I'm at that stage of a beginners QM course where after the Schrodinger equation 100 times you go back and learn the formalism, but I'm starting to discover that I misunderstood how some things really work.

For instance, in stationary states (wavefunctions of the form $\phi(x)T(t)$), $\phi(x)$ is an eigenfunction of the Hamiltonian. We can show that for stationary states, they evolve in time by $T=e^{-iEt/\bar h}$. It is then a postulate that any wavefunction can be written as a linear combination of stationary states $\Phi(x,t)=\sum a_n\phi_n(x)e^{-iE_nt/\bar h}$. So far so good.

This next bit is what I'm struggling with so I'll try and explain it and hopefully you can tell me where its wrong. I am told now that we can do this for any Hermitian observable. For instance, if we have some observable $Q$ with eigenfunctions $q_n$ then we can write $\Phi(x,t)=\sum b_n q_nT_n(t)$
Question: Can it? it was said in lectures that you can for a time independent state($\psi(x)=\sum b_nq_n)$, but I'm guessing that we can multiply each $q_n$ by some time dependence function $T_n$ like we could for the eigenstates of the Hamiltonian

I have a follow on question too. In the case of the Hamiltonian, it turns out that $|T_n|=1$. Is this true for any operator? If it isn't, could I not increase $T_1$ and decrease $b_1$ to make the probabilities whatever I like(which must not be true)? Maybe to ensure normality I would have to increase $b_2$ a bit but whatever, I hope you see the point.

Please note that I don't know what bras or kets are.

Answer 1

If Q and H commute then the eigenfunctions of Q are the same as the eigenfunctions of H. If they do not commute and the eigenfunctions are different, then you can write $q_{n} = \sum a_{i}\phi_{i}$ which evolves as $\sum a_{i}\phi_{i} e^{-2\pi{}iE_{n}t/h}$.

The reason this works for eigenfunctions of the Hamiltonian is because the Schrodinger Equation is

$\frac{d}{dt}\Psi = \frac{2\pi}{ih}\hat{H}\Psi$.

For an eigenfunction of $\hat{H}$,

$\frac{d}{dt}\phi_{n} = \frac{2\pi}{ih}\hat{H}\phi_{n} = \frac{2\pi}{ih}E_{n}\phi_{n}$.

So that

$\phi_{n}(x,t) = \phi_{n}(x,0)e^{-2\pi{}iE_{n}t/h}$.

In general, this doesn't work for an eigenfunction $q_{n}$ for operator Q which does not commute with the Hamiltonian.

$T_{n}(x,t) = \frac{q_{n}(x,t)}{q_{n}(x,0)} = \frac{\sum a_{i}\phi_{i} (x)e^{-2\pi{}iE_{i}t/h}}{\sum a_{i}\phi_{i}(x)}$

You can see that, now that we add multiple $\phi_{i}$ together, constructive and destructive interference make $T_{n}$ a mess that depends on $x$.

Answer 2

Given a Hamiltonian H of a system, we can write the time-dependent Schrodinger equation for a state $\psi(t)$ as, $$H \psi(t) = i \hbar\frac{\partial}{\partial t}\psi(t)$$ Given the initial time to be $t_0$, and if H is independent of time,

$$\psi(t) = e^{-iH(t-t_0)/\hbar}\psi(0)$$

Here, $U(t,t_0) = e^{-iH(t-t_0)/\hbar}$ is the time-evolution operator. You can check that this is a Unitary Operator if the Hamiltonian is Hermitian (which it should be for Hermitian Quantum mechanics), that is, time evolution will keep the wavefunction normalized.

Given any state $\Phi(x)$ (Say an eigenfunction of the Operator Q), we can expand it in the eigenbasis {$\phi_n(x)$} of the Hamiltonian as,

$$\Phi(x) =\sum _{n} a_n \phi_n(x)$$ where, $H \phi_n(x) = E_n \phi_n(x)$

Now, we can compute how the state $\Phi(x)$ evolves in time by keeping track of how the basis states evolve. For a basis state $\phi_n$ (which is an eigenfunction of the Hamiltonian), using the time evolution operator,

$$\phi_n(x,t) = U(t,t_0)\phi_n(x,t_0) = e^{-iH(t-t_0)/\hbar} \phi_n(x,t_0) = e^{-iE_n(t-t_0)/\hbar} \phi_n(x,t_0) $$

Putting all of it together and taking $t_0 = 0$, $$\Phi_n(x,t) = \sum_n a_n e^{-iE_nt/\hbar} \phi_n(x,0) $$

Note: This kind of a time evolution is only when Hamiltonian is independent of time, when H is time-dependent, another complication arises i.e Do the Hamiltonian operators at different times commute? .