Does it make sense to ask: what is the probability of a particle being found in a certain state at time $t>0$?

I am dealing with a problem which involves a quantum system of orthonormal two states, $$\left|\nu_1\right>$$ and $$\left|\nu_2\right>$$, which are eigenstates of a time-independent Hamiltonian, having the energies $$E_1$$ and $$E_2$$.

The question being asked is: what are the probabilities the particle being in the following states at time $$t>0$$, given that at $$t=0$$ it is in the state $$\left|\nu_1\right>$$: $$\left|\nu_1\right>,$$ $$\left|\nu_2\right>,$$ $$\left|\nu_e\right>=\cos\theta\left|\nu_1\right>+\sin\theta\left|\nu_2\right>\text{, and}$$ $$\left|\nu_\mu\right>=-\sin\theta\left|\nu_1\right>+\cos\theta\left|\nu_2\right>\text{?}$$

Of course, one can easily find the time-dependent state to be:

$$\left|\Psi,t\right>=e^{\frac{-iE_1t}{\hbar}}\left|\nu_1\right>.$$

This means that the particle always will be in the $$\left|\nu_1\right>$$ state, for all $$t$$, right? This seems trivial, but if instead the particle were initially in a superposition of energy eigenstates, like $$\left|\nu_e\right>$$ or $$\left|\nu_\mu\right>$$ (as they are defined above), does it even make sense that there is a probability of the particle being in a superposition of states after time $$t$$?

I've only ever dealt with finding the probabilities of a particle being in basis states, not superpositions of states, so am mightily confused by what is meant by this question, as in finding probabilities, we square the coefficients of basis states, and these probabilities are the probabilities of particles being found in those states, not being in those states.

It seems that we could never know the probability of a particle being in a superposition of states, but this just nullifies the second half of the question being asked, which doesn't seem to make sense...

Any help I would appreciate.

• I suspect this isn't your confusion but the phrasing "probability it is in the state $|\psi\rangle$" is a very poor (if common) choice of words. After all, you've written down the state at time $t$ and it isn't (generically) any of those. – jacob1729 Nov 2 '18 at 21:11
• @jacob1729: I understand that much, but where that argument breaks down is in the second part of the question, where the initial state is a superposition of states. My question is basically this: what is the probability of a particle being in a superposition of states at time $t>0$? – T. Zaborniak Nov 2 '18 at 21:23
• Are you sure that the problem doesn't state that the initial state is $\vert \nu _e \rangle$? That would make much more sense from the physical point of view, and also make a more interesting mathematical problem! – pppqqq Nov 3 '18 at 0:03
• @pppqqq: one of the questions has $\left|\nu_1\right>$ as being the initial state, and the other as being $\left|\nu_e\right>$ as being the initial state... however, my question still stands, as it does not seem possible to determine whether something is in a superposition of states at time $t>0$ given its initial state. – T. Zaborniak Nov 3 '18 at 1:26
• Ok sorry, I misread your question. Anyway, keep in mind that the concept of "a state being a superposition of states" is of course basis dependent and has no intrinsic meaning. At the purely formal level, you can always ask the question "What is the probability that, after evolving for a time $t$, the system will find itself in a state $\vert \alpha\rangle?", as discussed more precisely in Jacob's answer. – pppqqq Nov 3 '18 at 1:57 1 Answer Unfortunately, we need to modify the question until it makes sense. As it stands the state at time $$t$$ is (as you say): $$|\psi(t)\rangle = \exp(-iE_1t/\hbar)|\nu_1\rangle$$ And so since that isn't $$|\nu_2\rangle$$ the chance it is in state $$|\nu_2\rangle$$ at time $$t$$ is precisely zero, whilst the chance it is in state $$|\nu_1\rangle$$ is: $$P(\text{In state }|\nu_1\rangle) = \begin{cases} 1, & E_1t/\hbar = 2\pi k , k\in \mathbb{N} \\ 0, & \text{otherwise} \end{cases}$$ However this is a fairly common misuse of language and what is really meant is "if at time $$t$$ someone performed a measurement, thus collapsing the wavefunction, what is the probability it would end up in the following states". The measurement is crucial since without it the state just evolves deterministically. In your case, you have a two-state system with two states. Call them $$|1\rangle$$ and $$|2\rangle$$. You are being asked what the probability to measure the system to be in some superposition state $$|a\rangle=c_1|1\rangle+c_2|2\rangle$$ is. Thus you need to imagine actually doing the experiment where you measure some operator one of whose eigenvectors is $$|a\rangle$$ (since otherwise you will never get $$|a\rangle$$ anyway). In which case the obvious and correct thing to do is to resolve you initial state into the basis of eigenvectors for the operator that you are measuring (say $$|a\rangle$$ and $$|b\rangle$$). To finally answer your question then: the two superposition states you give are orthogonal: $$\langle \nu_e | \nu_\mu \rangle = 0$$ and so there is some Hermitian operator with them as eigenvectors for example: $$H = |\nu_e\rangle\langle\nu_e| - |\nu_\mu\rangle\langle\nu_\mu|$$ if we measure this operator and get the value 1 we will claim the system is in the state $$|\nu_e\rangle$$ and likewise for $$|\nu_\mu\rangle$$ if we measure -1. In the basis of eigenvectors of $$H$$ we can rewrite the initial state as: $$|\nu_1\rangle = \langle\nu_e|\nu_1\rangle|\nu_e\rangle + \langle\nu_\mu|\nu_1\rangle|\nu_\mu\rangle$$ And so the chance to measure the eigenvalue corresponding to either $$|\nu_e\rangle$$ or $$|\nu_\mu\rangle$$ is just the corresponding inner product squared. • Thank-you kindly for the in-depth response. I understand for the most part your argumentation, but something still seems amiss. Does your response suggest that if we were to measure the observable for which$\left|\nu_1\right>$and$\left|\nu_2\right>$form an eigenbasis, then the probability of finding it in$\left|\nu_1\right>$would be 1, as I found? And that if we were to measure the observable for which$\left|\nu_e\right>$and$\left|\nu_\mu\right>$form an eigenbasis, then the probability of finding it in$\left|\nu_1\right>$would be zero? – T. Zaborniak Nov 3 '18 at 2:11 • And if, as you acknowledge, the probability of finding the particle in$\left|\nu_1\right>$at$t>0$is precisely 1, how could it ever be in any other state, regardless of eigenbasis? – T. Zaborniak Nov 3 '18 at 2:25 • Never mind! I just started drawing parallels between this problem and that of the Stern-Gerlach experiment, and it all seems to make sense now. The$\left|S_z^+\right>$of that experiment can be thought of as being analogous to$\left|\nu_1\right>$, and$\left|S_z^-\right>$can be thought of as being analogous to$\left|\nu_2\right>\$. Based on that, of course it is necessary to consider the probabilities of the states being found in their respective eigenbases... silly me. – T. Zaborniak Nov 3 '18 at 3:13
• @T.Zaborniak your last comment is right: you can only measure it as being in an eigenstate of whatever operator you measure. The the first two (measure 1 or 2) requires you to measure in the 1/2 basis whilst the latter two (e,mu) in the e/mu basis. It's entirely analogous to changing basis from z to x when dealing with spins. – jacob1729 Nov 3 '18 at 11:05