Wavefunction collapse to a state that seems impossible

Let $\Omega$ be an observable and let $\Psi = \sum a_i\phi_i$ be a decomposition of a state of a system (satisfying the Schrödinger equation) in eigenfunctions of $\Omega$ (assume for simplicity it is a finite linear combination and there is no degeneracy). Measuring the observable $\Omega$ will give the value $\lambda_i$, the eigenvalue of state $\phi_i$, with probability $|a_i|^2$, in which case the wavefunction will collapse to $\phi_i$.

However, there is no reason why $\phi_i$ should be a state of the system. How is that possible? A few ways out occur to me:

1. the state instantanously "evolves" to a new, valid state
2. such a measurement is somehow not possible

Concretely, consider a particle in a 1-D box, and let $\Omega$ be the momentum operator. Then the state can be written as an "linear combination" (in this case an integral combination) but no state of definite momentum is a solution.

If we imagine the walls of the box, the infinite potential barriers, to be finitely wide, then if 1) is the case, it looks like the measurement of the momentum could cause the particle to tunnel through an infinite barrier (it would collapse to a state with uniform position probability, and evolve to a state that is nonzero on both sides of the barrier).

What is actually going on? Or did I just totally misunderstand something?

EDIT: Strictly speaking such a $\phi_i$ always is a state of the system (only the time-evolution is governed by the Schrödinger equation). To be more precise would be to remark that it seems that measurement could make it go into a state of (extremely) much higher potential without any other effort than making that measurement.

• I don't think your example for measuring momentum of a particle in a box is a problem. Measurement is an interaction between system and equipment. If there is a measurement acting on particle in a box, the equipment has to be able to go from outside the box to inside the box. Therefore, the infinity barrier is invalidated anyway. If there is a technical method to break such box, then there is no problem for wavefunction becoming a plane wave. – user26143 Jun 13 '14 at 15:54
• ((continue from last comment), causing any possible tunneling for an infinity barrier) And in general, according to the standard version of quantum mechanics, the wavefunction after measurement becomes an eigenstate of a given Hermitian operator, it may not be related to Schrodinger equation (time-independent $H\psi=E\psi$, presumbly that what you mean) – user26143 Jun 13 '14 at 16:03
• – Bubble Jun 14 '14 at 1:32
• Don't forget to accept an answer if one is satisfactory. If none are satisfactory, you can always ask for further clarification. – DanielSank Jun 16 '14 at 9:14

However, there is no reason why $\phi_i$ should be a state of the system. How is that possible?

If you perform the measurement and find result $\lambda_i$, with zero uncertainty in the measurement then indeed the state is now $\phi_i$. You ask how this is possible. Any state is possible, not just energy eigenstates.

A few ways out occur to me:

1. the state instantanously "evolves" to a new, valid state

Almost. The system does indeed evolve to a new state, namely $\phi_i$, but that evolution is not instantaneous. It is very fast in some situations, but in others it can be quite slow.

Concretely, consider a particle in a 1-D box, and let Ω be the momentum operator. Then the state can be written as an "linear combination" (in this case an integral combination) but no state of definite momentum is a solution.

If we imagine the walls of the box, the infinite potential barriers, to be finitely wide, then if 1) is the case, it looks like the measurement of the momentum could cause the particle to tunnel through an infinite barrier (it would collapse to a state with uniform position probability, and evolve to a state that is nonzero on both sides of the barrier).

The thing is, you can't build a system which measures the momentum to arbitrarily high precision in a particle-in-a-box. The boundary conditions of the system prevent that.

The infinite wall potential is mathematically pathological (the wave functions have discontinuous derivatives at the edge of the box). Consider a less pathological system, such as a particle in a finite height box. In this system, it is possible in principle to measure the momentum to arbitrarily high accuracy. If you measure the momentum to be $p_0$ with an accuracy of $\sigma_p$, then the resulting wave function in the $p$ basis will be a reasonably sharp wave packet: $$\propto \exp \left[ -(p-p_0)^2 / 2 \sigma_p^2 \right]$$ In the position basis this is $$\propto \exp \left[ -x^2 / 2\sigma_x^2 \right]$$ where $\sigma_x = 1/\sigma_p$. As $\sigma_x$ is very large, the wave function in the $x$ basis is a very wide function.

The crucial thing here is that you never ever measure anything to infinite accuracy, so the wave functions resulting from your measurement are not exact eigen-states of what you think is the measurement operator. This is not just "experimental dirtiness". This is a fundamentally important aspect of QM which you should keep near your mental centre as you learn more.

• Thanks for an elaborate answer! Just a point I would like to verify for my own understanding. In my understanding, we never ever make a measurement that can produce an exact eigenstate of the momentum operator because there is no normalizable (and thus, physically realizable) eigenstate of the momentum operator. And that, in turn, is the reason why we can never ever measure the momentum precisely--no matter how arbitrarily small we can make the error-bars nonetheless. [...] – Dvij Mankad Sep 5 '18 at 23:41
• [...] Now, all of that is fine and fundamental and not a result of the experimental dirtiness--but in the cases where there are physically realizable eigenstates, any inability of a ''measurement'' to produce a true eigenstate must be credited to the experimental dirtiness, right? For example, for an operator with discrete non-generate spectrum, there seems to be no way for a measurement to produce anything but a specific true eigenstate. Correct me if I am wrong. – Dvij Mankad Sep 5 '18 at 23:41
• @DvijMankad why not pose that as a complete question and then leave a link here? I can take a look and try to provide an answer. – DanielSank Sep 6 '18 at 0:27
• Sure. I have finally put together my thoughts and posted a question here: physics.stackexchange.com/q/427195/20427 – Dvij Mankad Sep 6 '18 at 20:26

Daniel Sank gives the right answer from a physical perspective. From a mathematical perspective, the answer is that there is no self-adjoint momentum operator defined on the Hilbert space of a particle in a box that obeys the canonical commutation relations. The usual operator that acts as $-i \hbar\, \partial_x$ in the position basis is not self-adjoint on this Hilbert space, because when you integrate it by parts, you pick up surface terms that are in general nonzero. It's a very subtle point of functional analysis - see here for way more detail than you probably care about.

Basically you're asking how is it possible to measure some observable which does not have the same eigenvectors as the system's Hamiltonian. This is indeed possible. Any Hermitian operator is an observable and any observable can be measured at least in principle.

Take the state your system is in at the time of measurement, write that state in terms of eigenvectors of the observable you want to measure (they will be orthonormal since the corresponding operator is Hermitian), the modulus squared of these amplitudes will give the probabilities that you will find the system in this state and the result of the measurement will be the corresponding eigenvalue of the observable. It doesn't matter that the observable's eigenvector is not an eigenvector of the system's Hamiltonian because the system is no longer evolving unitarily under its' own Hamiltonian.

During measurement the environment of the system, which is macroscopic, and the system itself interact. The total evolution of both is still unitary according to the laws of quantum mechanics. However, the evolution of the much smaller system is not. In fact, the system is forced into an eigenstate of what you're measuring.
This is known as quantum decoherence, a phenomena which occurs when very large macroscopic systems with many degrees of freedom interact with microscopic systems, and explains how the classical world we see everyday arises from quantum physics.

After measurement, assuming that the state will still evolve under the same Hamiltonian, it will continue to do so. The state (which is one of the observable's eigenvectors) will evolve under unitary evolution generated by the Hamiltonian. That is, the observable's eigenvector written in the basis of the Hamiltonian's eigenvectors with energies as phases.

This is the quantum measurement paradox which has many interpretations (objective collapse, subjective collapse, no collapse, etc..).

According to quantum mechanics (and an interpretation the Copenhagen one) once a measurement has been made the system remains there with probability 1. This is the quantum collapse problem.

UPDATE: According to the popular/prevalent interpretation of QM (Copenhagen), a quantum description of a system is complete (sth Einstein famously did not subscribe to) and the only allowed measurements are the eigenvalues of an operator. In this sense the answer to your question is that the wavefunction never collapses to an eigenvalue that is not in the system description.

Now what physicists are doing when one quantum description fails to accomodate an experiment, is that they formulate another one, an ansatz (for example systems with spin) which accomodates the experiment and then assumed to be complete.

• I agree that there are many difficulties around the interpretation of wavefunction collapse, but in this question I just assume that. My question is what happens when it collapses into a state that is not a solution is the Schrödinger equation anymore – doetoe Jun 13 '14 at 21:37
• @doetoe This is exactly related to the measurement paradox, like this: If the wavefunction does not collapse (it answers your question in one way), if the wavefunction collapses (a-la copenhagen way), it also answers your question (in another way), since it remains there with probability 1. – Nikos M. Jun 13 '14 at 23:14
• @doetoe, remember that (according to Copenhagen Int.) the quantum description of a system is complete and only eignevalues of an operator are the only allowed measurements (and in extension, system states) – Nikos M. Jun 13 '14 at 23:17
• Only eigenvalues of Hermitian operators are allowed results of measurement. This has nothing to do with interpretation. – Bubble Jun 14 '14 at 0:00
• He's not asking about the decoherence process. He's asking eg. how you can measure a definite momentum, yielding a plane wave, if the system does not support plane waves, as in the infinite box potential. The answer is that you can't measure momentum to infinite precision in that system. See my answer. – DanielSank Jun 14 '14 at 0:34