# What is the probability of obtaining the same measurement a finite time after causing a wave-function collapse with my initial measurement?

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.

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%.
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).
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: $$|E_0\rangle= \frac{1}{\sqrt{2}}\left(|0\rangle+|1\rangle\right)$$