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Let a quantum mechanical system, at time $t=0$, be described by:

$$ |\psi(0)\rangle = c_1(0) |E_1\rangle + c_2(0) |E_2\rangle \;, $$

here $|E_1\rangle$, $|E_2\rangle$ are energy eigenstates.

Now, for later time,

$$ |\psi(t)\rangle = c_1(t) |E_1\rangle + c_2(t) |E_2\rangle \;, $$ where $c_j(t) = e^{-i\,E_j\,t/\hbar}$, $\, j=1,2$.

Now what I see is that probability of finding state to have collapsed into $\left|E_1\right.\rangle$ (on measurement) is the same at time $t=0$ and at later time $t$.

The same is true for the other energy state also. The phase factors in each expansion coefficient $c_j(t)$ cancel out while calculating the probabilities.

This I believe will also be true if we could expand the state vectors in terms of $n$ energy eigenkets. The $|\psi(t)\rangle$ state mentioned above seems not to be quite changing (physically speaking) compared to the original, as the probability of collapsing of the state into specific energy eigenstate (on measurement), is not changing. Also at the same time the state is not a state with definite energy.

So, I request an interpretation of what is going on. Can I say that since the probabilities are not changing, it is reasonable to consider the time-evolved state to be equivalent to the original state (physically)?

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    $\begingroup$ Can you please use MathJax and write in paragraphs to make all of this readable? $\endgroup$ Jan 22, 2023 at 19:31
  • $\begingroup$ I honestly referred to MathJax while writing this question(it took me good 40 minutes),but anyway I am very sorry for your inconvenience .Trying to rectify . $\endgroup$
    – The Theory
    Jan 22, 2023 at 19:35
  • $\begingroup$ Here is a Tutorial. You can have more than one symbol within the math environment. For example, you can write $|\psi(t)\rangle = c_1(t)\, |E_1\rangle$ (right-click on the math and then show tex commands will show you how I produced this). $\endgroup$ Jan 22, 2023 at 19:37
  • $\begingroup$ Thanks, but while I refer to the tutorial, I request you to look up the question(which I have modified slightly ). $\endgroup$
    – The Theory
    Jan 22, 2023 at 19:39
  • $\begingroup$ The rangle command is not working at all.I have been having issues with this for some time, I just can't write ket's properly. $\endgroup$
    – The Theory
    Jan 22, 2023 at 19:40

1 Answer 1

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  1. You are correct that the probability to measure the system to have energy $E_1$ is $|c_1|^2$ (assuming $|c_1|^2+|c_2|^2=1$), and that this probability does not depend on time.

  2. You are also correct that if you had started with a generic superposition of $N$ energy eigenstates \begin{equation} |\Psi(t)\rangle = \sum_{j=1}^N c_j(0) e^{i E_j t / \hbar} |E_j\rangle \end{equation} with $\sum_{j=1}^N |c_j|^2=1$, that the probability to measure the system to have energy $E_1$ is $|c_1|^2$, independent of time.

Facts 1 and 2 are a kind of quantum mechanical version of energy conservation. If you put the system into a (non-degenerate) energy eigenstate, then the system will remain in that eigenstate, and by linearity the same is true of superpositions of energy eigenstates.

  1. You cannot conclude that $|\Psi(t)\rangle$ and $|\Psi(0)\rangle$ are the same state. Let's take $N=2$ for simplicity, and then consider the average value of the position of a particle \begin{eqnarray} \bar{x}(t) &=& \langle \Psi(t) | x | \Psi(t) \rangle \\ &=& \Big(c_1^\star(t) \langle E_1| + c_2^\star(t) \langle E_2 | \Big) x \Big(c_1(t) | E_1\rangle + c_2(t) |E_2 \rangle \Big) \\ &=& |c_1|^2 \langle E_1 | x | E_2 \rangle + |c_2|^2 \langle E_2 | x | E_2 \rangle \\ && + \Big(c_1^\star(t) c_2(t) \langle E_1 | x | E_2 \rangle + c.c.\Big) \end{eqnarray} The last term in the third equation, $c_1^\star(t) c_2(t) \langle E_1 | x | E_2 \rangle + c.c.$, where $c.c.$ is the complex conjugate, will generally be time dependent. So for a general observable -- not the energy -- the probability distribution of that observable will change with time. Here, we have illustrated that general point through the change in expectation value of the position operator.
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  • $\begingroup$ Thank you for your answer. $\endgroup$
    – The Theory
    Jan 22, 2023 at 20:17

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