# Collapse of Wavefunction, and Subsequent Time Evolution

To keep it simple, suppose the system is the well-known particle in a 1D infinite potential well.

Suppose the wavefunction is $a|1\rangle + b|2 \rangle + c|3\rangle$, where the $|i\rangle$ are energy eigenfunctions.

Suppose I measure the energy, and I get $E_1$.

Does this mean that after the measurement, the wavefunction becomes $|1 \rangle$? This is my understanding of "collapsing the wavefunction."

What happens after the measurement?

The propagator (from Shankar) is the sum over $n$ of:

$$|n \rangle \langle n| \exp\left(\frac{-i E_n t} {\hbar} \right)$$

If I apply this to $|1\rangle$, only the inner product for $n = 1$ survives, so the answer is $|1\rangle \exp\left(\frac{-i E_1 t} {\hbar} \right)$.

So it stays in state $|1\rangle$ forever?

• Relevant. It is not explicitly addressed in this answer, but if (and only if) $| 1\rangle$ is an energy eigenstate, then - yes - it will stay in that state forever. Sep 5 '18 at 21:24

The expression of "collapse of the wavefunction" means that you are taking a measurement which means a change in the interactions and boundary conditions, which means a new wavefunction.

Let us take the hydrogen atom. How would you check that it is in the ground state? By sending a photon that would change it from the ground state to a higher state,(a different wave function) and getting the photon coming back down to the ground state. The wavefunction changes with each interaction, or decay, into another wavefunction fulfilling different boundary conditions. The original "collapses" , i.e. no longer describes the probabilities of the system.

• Thanks for your response. Part of Shankar's Postulate III is "The state of the system will change to . . . |1> as a result of the measurement." He has a footnote about requiring the experiment to be "ideal." I understand your response to suggest that there really aren't "ideal" experiments. So, I guess my question is, in an "ideal" experiment, would the particle stay in state |1> forever, after it's energy is measured to be E_1? Sep 5 '18 at 17:46
• well in a sense in my example, after detecting the relaxation photon back to the ground state, the atom will stay in the ground state, until it interacts again. One has gained the information. It is the "measured" that has to be defined in your ideal case. Sep 5 '18 at 18:01

In your specific example the state collapses to the energy eigenstate $\vert 1\rangle$. Since energy eigenstates are stationary, i.e. they evolve as $e^{-i E_n t/\hbar}\vert b\rangle$, and since there is only one state with energy $E_1$ in your system, the system will remain in $\vert 1\rangle$ once it has been measured to have energy $E_1$.

I interpret "ideal" in a slightly different way, as meaning that your outcome is $E_1$ without possibility of errors. A non-ideal detector would have some (presumably small) probability of "confusing" $E_2$ with $E_1$, in which case the system would have a (presumably small) probability of being "measured" as having energy $E_1$ but being projected to the $\vert 2\rangle$ eigenstate. This could happen for instance if the energy resolution of your detector was larger than the energy difference $\vert E_2-E_1\vert$. Obviously if the energy levels are separated by a few $eV$s and your detector has resolution of $0.01$ eV, your measurements will be close to ideal.

• Thanks. This makes a lot of sense. It seems the key here, is that I chose the energy eigenstates. In this example, energy is conserved, so energy "propagates" by staying constant. I went back to Shankar, and saw him use the propagator on the free particle problem, with the initial wave function being a Gaussian packet centered at x = 0, in the x basis. Indeed, applying the propagator changes the wavefunction, so that the packet center moves at the expected velocity. Sep 6 '18 at 0:59