# If a wave function collapses into one state, does it ever go back to a superposition of states?

It is my understanding that after an observation, the wavefunction collapses to one state. Thus, if you do an observation right after an observation (that collapsed the wavefunction), you get the same thing. This implies that taking the second observation after a long period of time might not get you the same conclusion.

Do wave functions go back into a superposition of states? If so, how does it happen?

• There's no big distinction between 'superposition' and 'normal' states. A spin $1/2$ particle with spin up is actually a superposition of spin left and spin right, or a superposition of spin forward and spin backward. A particle with a definite position is a superposition of all momenta, etc. – knzhou Mar 29 '17 at 0:01
• So the question doesn't make sense: every state can be regarded as a "collapsed" state, or a "superposition" state, depending on your basis choice. – knzhou Mar 29 '17 at 0:01
• If the question is really about decoherence, then the answer is that no, it cannot go back to superpositions after observation. – SDsolar Mar 29 '17 at 5:28
• I have 7 pets. 7 is 4 plus 3. If I get another two pets, so that I have 9 altogether, will the number of my pets ever go back to being a sum? – WillO Mar 29 '17 at 7:21
• Perhaps it would be helpful to address how the particle gets to be in a superposition (of basis states of the measured observable) in the first place. It is clearly possible to put a particle into a superposition as the initial conditions of a lab experiment - will that happen spontaneously? – Paul B Mar 29 '17 at 18:45

Unless the wavefunction collapses to an eigenstate of the Hamiltonian, the subsequent time-evolution will produce a superposition.

The postulates clearly state that, if you measure the observable $\Lambda$ and obtain the outcome $\lambda$ (assumed non-degenerate for simplicity), then the state collapses to the eigenstate $\vert\psi_{\lambda}\rangle$ of $\hat \Lambda$, and the subsequent evolution is given by $$\sum_{k}e^{-iE_k t/\hbar}\vert \Psi_{E_k}\rangle\langle \Psi_{E_k}\vert\psi_{\lambda}\rangle$$ where $\vert \Psi_{E_k}\rangle$ is an eigenstate of $H$ with eigenvalue $E_k$. Thus, unless $\langle \Psi_{E_k}\vert\psi_{\lambda}\rangle=\delta_{E_{k}\lambda}$, the system will revert to a superposition.

Edit: after the measurement the state $\vert \psi_{\lambda}\rangle$ functions as an initial state and its time development is obtained in the usual manner by expanding over a complete set of eigenstates of $H$ using $$\hat 1=\sum_k\vert\Psi_{E_k}\rangle\langle \Psi_{E_k}\vert$$ so that $$\vert\Psi(0)\rangle=\vert\psi_\lambda\rangle= \sum_k\vert\Psi_{E_k}\rangle\langle \Psi_{E_k}\vert\psi_\lambda\rangle$$ and evolving the $H$-eigenstates $$\vert\Psi(t)\rangle= \sum_k\,e^{-iE_{k}t/\hbar }\vert\Psi_{E_k}\rangle\langle \Psi_{E_k}\vert\psi_\lambda\rangle\, .$$

• @JohnForkosh Nothing changes. The argument still works. As you correctly point out the sum of $k$ is independent of the $\vert\psi_\lambda\rangle$. – ZeroTheHero Mar 29 '17 at 0:58

The way I like to understand this is the following: suppose you have one observable $A$ with spectrum $\sigma(A) = \{ a_n : n \in \mathbb{N}\}$ which we will assume discrete and non-degenerate for simplicity. In constructing the theory you would like to have states where the value of $A$ is indeed certain. Those states, the postulates of QM tell you to be the eigenstates of $A$.

So in the eigenstate $|a_i\rangle$ you are certain to measure $A$ with eigenvalue $a_i$. That is all fine.

Now imagine your system is prepared in the state $|\psi\rangle$ and evolves to $|\psi(t)\rangle$ after some interval of time $t$. In particular this means that the probability of measuring $a_i$ at time $t$ is $|\langle a_i |\psi(t)\rangle|^2$.

Thus in the state $|\psi(t)\rangle$ you are not certain of what value $A$ takes. The system could have any one of the allowed values of $A$ and this uncertainty is built into the state $|\psi(t)\rangle$.

At some time $t_1$ then you then measure $A$ and you find out that $A$ has value $a_i$. Now there's a problem: if your system continued to be in the state $|\psi(t_1)\rangle$ immediately after the measurement, the theory would not be consistent.

As you measure $A$ and find $a_i$ you are certain of the value of $A$ while in the state $|\psi(t)\rangle$ you have nonzero probabilities for other values of $A$ other than $a_i$. How could your system be in such a state, when you know that the probability for $a_i$ should be one and zero for all other values?

Measuring $A$ gives you new information about your system: you know the value of that physical quantity at that time. So the state must change to contain that information. There's however one specific state that achieves this, and that is $|a_i\rangle$. Thus you have now that imediately after the measurement the state should be $|a_i\rangle$, or:

$$\lim_{t\to t_1^+}|\psi(t)\rangle = |a_i\rangle$$

But the Hamiltonian contains the information about the influences on the system that makes it evolve in time, after all energy is the generator of time translations. Hence after the measurement the system will evolve because of the Hamiltonian. Thus your state at time $t > t_1$ will satisfy

$$i\hbar\dfrac{d|\psi(t)\rangle}{dt}=H|\psi(t)\rangle$$

with initial condition $|\psi(t_1)\rangle = |a_i\rangle$. Thus the evolution in time might make you depart from that state of "extra information" granted by measurement.

I say might because if $A$ commutes with the Hamiltonian the situation is another. In that case $A$ is a constant of motion, in the sense that it is a conserved quantity. So along the evolution, the value of $A$ doesn't change. Once you found it out at time $t_1$, it won't change anymore. Thus you will not change the state.

• Yes but the paragraph "Thus in the state $|\psi(t)\rangle$ you are not certain of what value $A$ takes. The system could have any one of the allowed values of $A$ and this uncertainty is built into the state $|\psi(t)\rangle$." should be changed to "Thus in the state $|\psi(t)\rangle$ you are not certain of what value $A$ will take when measured. The system could result in any one of the allowed values of $A$ and this uncertainty is built into the state $|\psi(t)\rangle$." – Bruce Greetham Mar 29 '17 at 12:00
• @BruceGreetham, this is a question of interpretation. You can interpret that: (1) the system simply hasn't a well defined $A$ quantity before the measurement and only take a stand when measured and (2) the system has a value, but we cannot know it until we measure, and the theory accounts just for the probabilities. Anyway, the only thing that matter is what you find out when you do measure, because the only way of knowing is actually measuring it, and in this point the theory has got you covered. As for the way of thinking, I tend to align myself with (2), that's why I wrote as I did. – user1620696 Mar 29 '17 at 13:46
• OK thanks - I could argue that it is a crucial aspect of QM that in any interpretation it is meaningless to talk about the value of a pure superposition before some decoherence process has effectively diagonalized the density matrix into a mixed state - but maybe this is going deeper than the OPs concern which I agree you have addressed. – Bruce Greetham Mar 29 '17 at 17:04

You are basically asking for Quantum decoherence. A wave function doesn't collapse on its own, but by interaction with something else (e.g. observation via a photon), which itself will afterwards also have a modified wave function. So unless you manage to exactly reverse the observation process (in time), part of the information required to "de-collapse" the wave function again is practically lost.

The particle is always in a superposition. It could be here, it could be there, it could be fast, it could be slow, etc.

The observation collapses the wave function so that it is less spread out, but there is always some uncertainty left.

Instead of an superposition between a particle being at one point and being far away, you get a superposition between being at some point and being at a very nearby point.

Likewise you get a superposition of having a certain speed and having almost the same speed. Same with direction, spin and anything else you might want to measure.

This uncertainty will gradually spread out until the particle is being all over the place again, but there is no sharp point in time when it goes from being in one state to being in a superposition.

• Excellent answer for non-physics students like me. Thank you. – joehinkle11 Jun 16 '18 at 6:02

The answers are no, and maybe.
The best example of no, is Schrodinger's cat experiment. If you observe that the cat is dead now, it will still be dead letter in time.
An example of maybe, a shorted battery. After a battery is shorted/drained, it might recover some of its capacity because the chemical process might reverse over time.

Consider a superposition of Live/Dead cats,

$$\vert\Psi(t)\rangle = \vert L\rangle\langle L\vert e^{-iHt}\vert D\rangle + \vert D\rangle\langle D \vert\ e^{-iHt}\vert D\rangle$$

Assuming an observation made at $t = 0$ resulted in a D-state, the superposition $\vert\Psi(t)\rangle$ would represent its subsequent time evolution as generated by the cat's atomic many body Hamiltonian under the assumption that the L/D states are not $H$ eigenvectors.

Because the square of $\langle L\vert e^{-iHt}\vert D\rangle$ is then finite, there would be a finite probability that an observation made at t would result in a revivified cat.