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Say I have a qubit that can be in two charge eigenstates, $|0\rangle$ and $|1\rangle$. The qubit also has two distinct energy levels with eigenstates $|E_0\rangle$ and $|E_1\rangle$, which each have a probability of 50% of being measured, i.e: $$ |0\rangle = \frac{1}{\sqrt{2}}(|E_0\rangle +i|E_1\rangle) $$ $$ |1\rangle = \frac{1}{\sqrt{2}}(|E_0\rangle -i|E_1\rangle) $$ Say I measure the energy of the qubit when it is in the charge state $|1\rangle$ at $t = 0$ and I find out it is in the energy state $E_1$, would subsequent measurements of the energy after this initial measurement yield the same value? I understand that immediately after this measurement (i.e still $t = 0$) that the probability of measuring the same energy state is 100% since the wave-function has collapsed, but what about when $t \neq 0$? Would the probability remain 100%? What about the charge states $|0\rangle$ and $|1\rangle$, would it remain in the $|1\rangle$ state since I initially measured the energy in that state, or would it still have it's respective probabilities?

I think I am a little bit confused. I have tried to be very careful with my wording, any help will be greatly appreciated.

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This is a problem of time evolution of a state. If your initial state is $|\psi\rangle(t=0)$, then the state at later time $t$ is given by, $$|\psi\rangle(t)=e^{\frac{i\hat{H}t}{\hbar}}|\psi\rangle(0)$$ where $\hat{H}$ is the hamiltonian operator. Moving to your problem, first you measure the sate for energy and end up with the energy eigenstate $|E_1\rangle$. This means that at $t=0$, $|\psi\rangle(0)=c_1|E_1\rangle$ (which is an eigenstate of the Hamiltonian), where $c_1$ is some amplitude. So at some later time $t$, your state should evolve like, $$|\psi\rangle(t)=e^{\frac{i\hat{H}t}{\hbar}}c_1|E_1\rangle$$ $$|\psi\rangle(t)=e^{\frac{iE_1t}{\hbar}}c_1|E_1\rangle$$ You will be able to write last equation beacuse $|E_1\rangle$ is the eigenstate of the Hamiltonian (If you want ot derive this, Taylor expand the $e^{\frac{i\hat{H}t}{\hbar}}$ about $t=0$). But now you can clearly see that even after the time evolution of your initial state, $|\psi\rangle(t)$ remains eigenstate of the Hamiltonian as $e^{\frac{iE_1t}{\hbar}}$ is just a phase. Therefore, the probability of measurement of in state $|E_1\rangle$ remains 100%.

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The generic case is that the state $\vert E_1\rangle$ resulting from the initial measurement will evolve in time as per $U(t)\vert E_1\rangle $ with $$ U(t)=e^{i\hat H t/\hbar}\, . $$ Since in your specific example $\vert E_1\rangle$ is an eigenstate of $\hat H$, we have $$ U(t)\vert E_1\rangle = e^{-i E_1t/\hbar}\vert E_1\rangle $$ so that the probability of finding the system in the state $\vert \psi\rangle$ after time $t$ is $$ \vert \langle \psi\vert U(t)\vert E_1\rangle\vert^2= \vert \langle \psi\vert E_1\rangle e^{-i E_1t/\hbar}\vert^2 = \vert\langle \psi\vert E_1\rangle\vert^2 \tag{1} $$
So if $\vert\psi\rangle$ is the energy state $\vert E_1\rangle$ then simply sub in this in (1).

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Energy eigenstates have trivial time evolution. Just a complex phase. Thus inner products are preserved. What this means is that once you’re in an energy eigenstate of the Hamiltonian, unless there’s a perturbation, you’ll remain in the energy eigenstate as time progresses.

Say initially the state was $|1\rangle$ and we measured the energy and got the reading $E_0$. So now our state is $|E_0\rangle$. Simple rearrangement of the two given relations will give you: $$|E_0\rangle= \frac{1}{\sqrt{2}}\left(|0\rangle+|1\rangle\right)$$

This then tells you once you measure energy, you’re equally likely to be in either of the charge states.

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