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If we study a QM system and an observable that we can measure on it (like Energy etc). I know that for the probability that the system is in an eigenstate (discrete and not degenerated) we have something of this sort:

$ P(a_n)=|\langle U_n|\psi \rangle|^2 $ where $a_n$ is the coefficient and $U_n$ is the basis vector when we can express a state as a linear combination of the basis kets.

My question is, how do we find the probability of the system being in a whatever state (which can be a linear combination of the eigenstates) after some time $T$.

Can anyone explain that to me.

Because if I follow the above logic, then I need to do the inner product between a time-dependent eigenstate and the state, for which I am calculating the probability of the system being in it. I don't know any formula for that case.

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Let's say that the state of the system is $\vert\psi\rangle$ at time $t=0$. The OP has used a bit of unfortunate language in regards to what it is the probability of that is given by the Born rule. So, let me correct it first.

The correct statement is (assuming all the good things such as discrete and non-degenerate eigenspectra) that the probability that the system will be found to be in $\vert o_n\rangle$ upon measuring an operator $\hat{O}$ is given by $\vert\langle o_n\vert\psi\rangle\vert^2$ where $\vert o_n\rangle$ is an eigenstate of the operator. The statement is not that the system whose state is $\vert \psi\rangle$ is in $\vert o_n\rangle$ with a probability $\vert\langle o_n\vert\psi\rangle\vert^2$. This should be self-evident -- the system whose state is $\vert\psi\rangle$ is obviously in the state $\vert\psi\rangle$ with a $100\%$ probability and it is not at all in any other state that is different from $\vert\psi\rangle$. However, I do not blame an early learner of quantum mechanics for the abuse of language by the popular media. :-(

Now, due to the unfortunate language in which the OP has formulated the question, there are two ways to interpret the question posed by the OP.

Taken literally, the question is as to what will be the probability of the system described by $\vert\psi\rangle$ at time $t=0$ being in various different states at time $t$. The answer is simply that it will be in the state $e^{-i\hat{H}t}\vert\psi\rangle$ with a $100\%$ probability and will be in a state that is different from $e^{-i\hat{H}t}\vert\psi\rangle$ with a $0\%$ probability. The time evolution of a state in quantum mechanics is perfectly deterministic, it is given by the Schr$\ddot{\mathrm{o}}$dinger equation.

However, one can also interpret the question to be as to what is the probability that the system whose state is given by $\vert\psi\rangle$ at time $t=0$ will be found to be in $\vert o_n\rangle$ upon measuring an operator $\hat{O}$ at time $t$. Well, in order to answer that question, we simply need to find out as to what the state of the system will be at time $t$, and then we just apply the Born rule. As mentioned in the earlier paragraph, the state of a system at time $t$ will be $e^{-i\hat{H}t}\vert\psi\rangle$ if its initial state was $\vert\psi\rangle$. Now, we can just apply the Born rule and see that the probability that the measurement of the operator $\hat{O}$ will yield $\vert o_n\rangle$ is given by $\vert \langle o_n\vert e^{-i\hat{H}t}\vert \psi\rangle\vert^2$. Notice that these probabilities will remain the same regardless of the value of $t$ if $\vert o_n\rangle$ is an eigenstate of the Hamiltonian. But, generically, it is not true that they remain invariant with time.

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  • $\begingroup$ First of all thank you for your answer, After spending time and also reading your answer i was able to understand it, and also solve an exercise regarding this topic. One more question tho, i don't know how to formulate this one. I will simply say that in our lecture we were observing a sudden change of potential. Does this notation makes sense: Initial stationary state $|n^\circ (t) = exp(-iE_n^\circ/h)|n^\circ>$ . As you can notice we have to stationary states, one time dependent and on not. Is that possible ? $\endgroup$
    – imbAF
    Apr 13 '21 at 23:30
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OP has used a poor notation; so I'm kind of change the notation. Supposing the initial state $|\psi(0)\rangle $ to be $$|\psi(0)\rangle =\sum_n|n\rangle \langle n|\psi(0)\rangle $$

At any time $t$, the state can be written as $$|\psi(t)\rangle =U(t)|\psi(0)\rangle =\sum_nU(t)|n\rangle \langle n|\psi(0)\rangle$$ where $U(t)$ stands for time-evolution operator. Now the probability amplitude for state to be $|m\rangle$ given by $$\langle m|\psi(t)\rangle =\sum_n\langle m|U(t)|n\rangle \langle n|\psi(0)\rangle$$ Once probability amplitude is known you can find probability.

If hamiltonian is autonomous, then $$U(t)=e^{-iHt/\hbar}$$ $$\langle m|\psi(t)\rangle =\sum_n\langle m|e^{-iHt/\hbar}|n\rangle \langle n|\psi(0)\rangle$$

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  • $\begingroup$ This answer is wrong in multiple ways: the first is that the OP didn't only ask for the probabilities pertaining to energy eigenstates. The statement that the probabilities over different eigenvalues of an observable do not change over time is only true for energy eigenvalues. Secondly, the probabilities you are describing are not the probabilities pertaining to the state to be in some eigenstate but pertaining to finding the state in some eigenstate upon measurement of the relevant operator. As such the state will be in the state $U(t)\vert\psi(0)\rangle$ with $100\%$ probability. $\endgroup$
    – Dvij D.C.
    Apr 12 '21 at 21:16
  • $\begingroup$ I have corrected it. $\endgroup$ Apr 13 '21 at 4:09
  • $\begingroup$ I don't see how the new version resolves the issues I raised in my previous comment. $\endgroup$
    – Dvij D.C.
    Apr 13 '21 at 13:09

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