# Does measurement change the evolution of wave function?

Basically any measurement is on wave function $|\psi\rangle$ is done by operator $X$ such that $X|\psi\rangle$ results observable $x$ with some probability.

But what happens to $|\psi\rangle$? Does it continue to evolve by the same hamiltonian $H$ as it was before? Or does the wave function suddenly change from the previous hamiltonian evolution and evolves differently from what it would have if not measured?

Here I am thinking $|\psi \rangle$ as an objective wave function, encoding probability distribution that would be right in infinite time, not our subjective knowledge of particles.

What is a wave function? It is the solution of a quantum mechanical equation ( with the appropriate potentials),on which boundary conditions are imposed to make it specific to a system . $|\psi\rangle$ by itself is not independent of the environment the way that the operators X are.

Thus the answer depends on the system under consideration.

I like using the single electron at a time double slit experiment because it shows how the boundary conditions that will define $|\psi\rangle$ have to be taken into account.

The wavefunction we need is the solution of the topology :plane wave single electron , field of two slits. The operator in this case is the (x,y) operator that acted on the screen to give the dots on the top image.

For each individual electron the $|\psi\rangle$ that describes its probability changes the minute the operator X operates (hit on the screen) . A completely different $|\psi\rangle$ will describe it from then on because the fields and boundary conditions are drastically different. If there were no screen and the electron could propagate to infinity the wavefunction describing its probabilistic manifestation mathematically is not changed.

The addition of many electrons describe the probability distribution in (x,y) for this setup, i.e. the complex conjugate square of the wavefunction, up to the z of the screen, where everything changes.

You are touching upon the measurement problem in quantum mechanics. The wave-function changes in two distinct manners:

1. Unitary evolution by the Hamiltonian in the Schrodinger equation.
2. Non-unitary measurement by Born's rule, i.e. what is often called "collapse of the wave-function."

To many people, the latter remains puzzling, though this is a contentious topic.

Prior to a measurement, a wave-function evolves according to the Schrodinger equation. Upon measurement, it "collapses" into a eigenstate of the measurement operator with probability according to Born's rule. After measurement, it again evolves according the the Schrodinger equation.

The act of measurement doesn't change the Hamiltonian - after measurement, the time-evolution is identical, though it is a different state that is evolving.

Basically any measurement is on wave function $|\psi\rangle$ is done by operator $X$ such that $X|\psi\rangle$ results observable $x$ with some probability.

is not entirely correct. In quantum mechanics a system is described by a state $|\psi\rangle\in\mathcal{H}$ and a set of self-adjoint operators $A_1,\ldots,A_n$ representing the observables that we want to measure. Moreover, we require the eigenstates of those operators to completely span the entire Hilbert space we started with (and this in a non-trivial requirement) such that $$|\psi\rangle = \sum_ac_a|e_a\rangle$$ with $|e_a\rangle$ being a set of eigenstates for the $k^{\textrm{th}}$ observable, $A_k|e_a\rangle=\lambda_a^{(k)}|e_a\rangle$. The evolution of the state in time is given, at any time, by the Schrödinger equation $$i\hbar\,\frac{\partial}{\partial t}|\psi\rangle = H\,|\psi\rangle$$ where $H$ is a given operator, plus initial and boundary conditions.

Given the state $|\psi\rangle$ and an observable $A$, before we perform the measurement each of the $|e_a\rangle$ is a possible outcome accompanied with probability $|c_a|^2$. After you have performed the measurement only one of the $|e_a\rangle$ is chosen and the state $|\psi\rangle$ becomes $|e_a\rangle$ with probability 1.

Once so, the state of the system has changed but nevertheless the Hamiltonian describing the system has not (and there is no reason why it should have). The example provided above by @anna is a very good explanation of the mechanism of the collapse of the state after measuring one of its quantities.

All evolution in Quantum Mechanics is done by the Hamiltonian.
However, measurements of an operator result in a single eigenvalue, and the wavefunction collapses (As to how, why, etc, it is unknown as of now, Copenhagen is the most common interpretation) to an eigenstate of the operator measured with the same eigenvalue.
Specifically, it collapses to the projection onto the space of eigenvectors with the same eigenstate for that operator. (The same eigenstate that was measured)

After the measurement, the new wave function evolves by the hamiltonian, and it's the same one. (Same hamiltonian)

• Eveni if it can it be the same hamiltonian, when the boundary conditions for the particle under consideration have changed, your famous "collapse", that particle/instance of the psi probability distribution has been changed and a new psi is needed for the specific particle. see the example in my answer – anna v Jul 2 '15 at 8:31
• It still does not change my answer. The wavefunction evolves purely by the hamiltonian. Collapse simply changes the wavefunction via projection. – Omry Jul 2 '15 at 8:44
• It is false to say that we don't know "how" the wave function collapse can be explained in terms of quantum mechanics. Decoherence is well understood. – CuriousOne Jul 2 '15 at 8:49
• What isn't understood is a mechanism to this collapse, i.e., something more physical than "We measure and something happens, it collapsed" – Omry Jul 2 '15 at 9:30
• @CuriousOne aren't you slightly exaggerating the insights from de-coherence? It can explain why a density matrix is diagonalized, but it still cannot tell us which one of the diagonals will be selected by a measurement and that's still a non-unitary process. It doesn't solve the measurement problem. – innisfree Jul 3 '15 at 10:55

When a quantum mechanical object is subjected to a measurement its wavefunction has to be coupled to a measurement device. That coupling will change both the wavefunction of the object and the wavefunction of the measurement device (as well as the wavefunction of the environment!). It is important to understand that a measured quantum object is not the same as a freely evolving quantum object and it will have a very different dynamic than the free object would have had without the measurement. The details of this process are usually summarized under "quantum decoherence". You may want to read "Decoherence and the transition from quantum to classical -- REVISITED" by Wojciech H. Zurek.

• Anybody wants to tell me why the downvotes, except for the angry iharob person who didn't like me? Any rational reasons? I would be interested on how to improve the answer! Thanks! – CuriousOne Jul 3 '15 at 0:45

## protected by Qmechanic♦Jul 2 '15 at 11:40

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).