# Collapse of the wavefunction after measurement

I learned during my study that when we make a measurement of a wavefunction then the wavefunction will collapse to one of its eigenfunctions with an eigenvalue.

It is also mentioned that when we instantly do the same measurement again on the same wavefunction, then we will get the same eigenvalue as the wavefunction has not evolved in time. I want to know what is the time the wavefunction takes to evolve such that it can switch to another eigenvalue on measurement (i.e. the probability of another eigenvalue increases more than the previous one on evolved wavefunction).

I am interested in the time scales into which the wavefunction typically evolves. Is it dependent on the type of system or/and type of measurement?

• You cannot do this. "It is also mentioned that when we instantly again do the same measurement on the same wavefunction". A measurement means an interaction, and after an interaction a new wavefunctions arises, not the same. Maybe this answer will help on the concept of measurement which necessarily causes "collapse" physics.stackexchange.com/questions/400162/… Aug 24 '20 at 3:42

Since the question explicitly asks for time scale involved in the evolution of states, I decided to expand my comment.

I am interested in the time scales into which the wavefunction typically evolves. Is it dependent on the type of system or/and type of measurement?

Let us consider the simplest case of a two (energy) state system. Let the energies be $$E_1$$ and $$E_2$$ with $$E_1. Let us measure some observable and let two of its eigenstates be given by $$|c_1(t=0)\rangle=\frac{1}{\sqrt 2}\Big(|E_1\rangle + |E_2\rangle\Big)$$ $$|c_2(t=0)\rangle=\frac{1}{\sqrt 2}\Big(|E_1\rangle - |E_2\rangle\Big)$$

Now the time evolution of each of the states will be governed by the time dependent Schrödinger equation. So we’ll have: $$|c_1(t)\rangle=\frac{1}{\sqrt 2}\Big(e^{-i\omega_1 t}|E_1\rangle + e^{-i\omega_2 t}|E_2\rangle\Big)$$ $$|c_2(t)\rangle=\frac{1}{\sqrt 2}\Big(e^{-i\omega_1 t}|E_1\rangle - e^{-i\omega_2 t}|E_2\rangle\Big)$$ where $$\omega_i=E_i/\hbar$$.

Now if we need to find the time $$t$$ such that one eigenstate goes to another, $$|c_1(t)\rangle\to|c_2(t)\rangle$$ we have the condition $$\frac{e^{-i\omega_1 t}}{e^{-i\omega_2 t}} = -1$$ The smallest time at which the above condition holds is at: $$t=\frac{\pi}{\omega_2-\omega_1}$$

It is also mentioned that when we instantly do the same measurement again on the same wavefunction, then we will get the same eigenvalue as the wavefunction has not evolved in time.

And for this part, you might be interested in the Quantum Zeno effect.

As stated in the comment made by @anna v, you can't perform a second measurement instantly. That would be the same as performing one measurement.
In the case of position measurement, the wavefunction will have evolved already when you make a second measurement. When making the second measurement a very short time after the first, the wavefunction has already evolved into a continuous set of position eigenstates.
Even when $$\Delta t$$ approaches $$dt$$ this will be the case. The set of eigenstates, in this case, will have risen too, but with an amount approaching zero.

What about other systems? The same picture. All quantum mechanical states evolve in the same way in time.
Take for example the case of the superposition of two states. After you have done the first measurement (resulting in one out of two eigenstates), the wavefunction will instantly start to develop into a new superposition of two eigenstates. And again it depends on how long after the first measurement you perform the second one.

• To add to this, the typical timescales of a system are governed by the energy eigenstates. Aug 24 '20 at 7:06
• I am unsure about your statement "After you have done the first measurement (resulting in one out of two eigenstates), the wavefunction will instantly start to develop into a new superposition of two eigenstates." Of course, in a sense you are probably correct in some realistic limit, but the OP's question (to me) seems to imply a very idealised situation with a closed system. In the ideal scenario, I don't believe that your quoted sentence holds for all measurements. After performing an energy measurement the system will stay in that energy eigenstate forever. Aug 24 '20 at 9:06
• Then how does a superposition of two eigenstates comes into being? Aug 24 '20 at 9:46
• @Mashy this is assuming the eigenstate isn’t simultaneously an energy eigenstate in which case the measured state is in a superposition of energy eigenstates which changes as a function of time. Otherwise of course it’ll remain in the same state (ideal measurement) indefinitely. Aug 24 '20 at 9:55

$$|\Psi (0)\rangle$$ is not necessarily an eigenstate of H , however, because H is time-independent, the evolution of $$|\psi \rangle$$ is

$$\displaystyle |{\Psi (t)}\rangle = e^{-iHt/\hbar } |{\Psi (0)}\rangle = \Big(1 + \frac{-iHt}{\hbar } + \frac{1}{2}\left(\frac{-iHt}{\hbar }\right)^2 + \cdots \Big) |{\Psi (0)}\rangle \, .$$
Thus,

$$\displaystyle \begin{split} \langle {\Psi (0)} | {\Psi (t)} \rangle & = 1 + \frac{-it}{\hbar } \langle {\Psi (0)}H|{\Psi (0)}\rangle + \frac{-t^2}{2\hbar ^2}\langle {\Psi (0)}H^2|{\Psi (0)}\rangle + O(t^3) \\ & = 1 + \frac{-it}{\hbar } \langle H\rangle + \frac{-t^2}{2\hbar ^2}\langle H^2\rangle + O(t^3) \\ & = 1 + \frac{-it}{\hbar } \langle H\rangle + \frac{-t^2}{2\hbar ^2}\Big((\Delta H)^2 + \langle H \rangle ^2\Big) + O(t^3) \end{split}$$
where expectation values $$\langle H \rangle$$ and $$\langle H^2 \rangle$$ are taken with respect to $$| \psi (0) \rangle$$, and in the last step I have used that $$(\Delta H)^2=\langle H^2 \rangle−\langle H \rangle^2$$.

Thus, to order $$t^2$$,

$$\displaystyle \begin{split} \big | \langle {\Psi (0)}|{\Psi (t)}\rangle \big |^2 & = \Big[1 + \frac{-it}{\hbar } \langle H\rangle + \frac{-t^2}{2\hbar ^2}\Big((\Delta H)^2 + \langle H \rangle ^2\Big)\Big] \\ & \qquad \qquad \Big[1 + \frac{it}{\hbar } \langle H\rangle + \frac{-t^2}{2\hbar ^2}\Big((\Delta H)^2 + \langle H \rangle ^2\Big)\Big] + O(t^3) \\ & = 1 + \frac{-t^2}{\hbar ^2}\Big((\Delta H)^2 + \langle H \rangle ^2\Big) + \frac{t^2}{\hbar ^2}\langle H \rangle ^2 + O(t^3) \\ & = 1 - \frac{t^2}{\hbar ^2}(\Delta H)^2 + O(t^3) \, . \end{split}$$

As an interesting side point, if t is very small then $$\big | \langle {\Psi (0)}|{\Psi (t)}\rangle \big |^2 \approx 1$$ (to order t ). In other words, if we measure whether $$\langle \psi (t) \rangle$$ is in the state $$\langle \psi (0) \rangle$$ after only letting the system evolve for a very short time, we expect with high probability to indeed find an outcome of $$| \psi (0) \rangle$$, i.e. the system effectively hasn't evolved at all! If, after letting the system evolve for another very short time, we measure again, we again likely expect to find the state to be $$| \psi (0) \rangle$$. This is the source of the so-called quantum Zeno effect; if you continually measure a quantum state, it does not change, or colloquially, a watched quantum pot never boils