0
$\begingroup$

For measurement of any observable associated with the particle, should the wavefunction after collapse be a solution of the time-dependent Schrodinger equation?

A general solution of the time dependent schrodinger equation is given by superposition of eigenfunctions of the hamiltonian operator. Here by eigenfunctions, I mean the complete wavefunction as a function of spatial and time co - ordinates.

But since the wavefunction collapses to an eigenfunction of the operator of an obsevable when a measurement of the observable associated with the particle is made, What I want to ask is the following:

Should the collapsed wavefunction which is an eigenfunction of that operator be general solution of the Time - dependent schrodinger equation as well?

|cite|improve this question
$\endgroup$
  • $\begingroup$ The wavefunction is always a solution of the Schrödinger equation, but not always an eigenfunction of the Schrödinger equiation. Perhaps you could clarify exactly what you are asking and explain why specifically you're considering the time dependent Schrödinger equation. $\endgroup$ – John Rennie Nov 18 '15 at 10:42
  • $\begingroup$ For example, when the momentum of the particle in some potential energy distribution is measured, the wavefunction collapses to a wavefunction of the oscillatory kind A e ^ (ikx - wt). However this collpsed wavefunction is not a solution of the time - dependent schrodinger equation for this potential energy distribution. So is this collapse wavefunction allowed?? $\endgroup$ – Siddharth Joshi Nov 18 '15 at 10:50
  • $\begingroup$ ...is not what? $\endgroup$ – Brionius Nov 18 '15 at 10:53
  • $\begingroup$ The collapsed state will be a superposition of the energy eigenfunctions and therefore a solution of the Schrodinger equation. $\endgroup$ – John Rennie Nov 18 '15 at 11:19
  • $\begingroup$ Can anyone precisely proof that any wavefunction (a function of both spatial and time co - ordinates) can be expressed as a superposition of energy eigenfunctions. If not possible, then are such collapsed wavefunctions allowed?? $\endgroup$ – Siddharth Joshi Nov 18 '15 at 11:30
1
$\begingroup$

The situation here is a bit complicated. The problem is that with a measurement process, you are not allowed to consider only the evolution of the measured system, but you have to take into account the evolution of the global system formed by the measured system and measurement apparatus.

For that system, the (global) wavefunction always evolves by means of the Schrödinger equation. Of course, the "restricted" wavefunction of the measured system alone would not obey for all times the "unperturbed" Schrödinger equation (the one of the system alone when not in contact with the measurement apparatus).

In the simple case of an instantaneous measurement (it never is), you would see a sharp discontinuity at the time of measurement $t^*$ for the partial wavefunction $\psi(t)$ of the measured system: for any $0\leq t<t^*$, the map is a (continuous in time) solution of the unperturbed Schrödinger equation with initial datum $\psi(0)$, and for any $t\geq t^*$ it is a (continuous in time) solution of the unperturbed Schrödinger equation with initial condition $\psi(t^*)$ (the wavefunction after measurement, that is in general different from $\lim_{t\uparrow t^* }\psi(t)$, since the measurement process perturbs instantaneously the system).

|cite|improve this answer
$\endgroup$
  • $\begingroup$ Since in this case the wavefunction is for the entire system and not only the particle, isn't it wrong to comment on the individual wavefunction of the particle. How can we say that the wavefunction of the particle collapses to an eigenfunction of the operator. $\endgroup$ – Siddharth Joshi Nov 18 '15 at 12:02
  • $\begingroup$ That is what you do assume your measurement process will do (since it is what you observe experimentally in concrete systems). It is (in my opinion) a rather remarkable mathematical result that for any observable it is always possible to devise (at least in principle) a measurement process that has the suitable properties of real concrete measurements (like the collapse). $\endgroup$ – yuggib Nov 18 '15 at 12:06
  • $\begingroup$ Am I right in my thinking that the collapse of the wavefunction is just a theoretical construct which is merely validated by our experimental results. $\endgroup$ – Siddharth Joshi Nov 18 '15 at 12:24
  • 1
    $\begingroup$ And what isn't? Is not the wavefunction a mental construct? That is the only way we know to do science: experimentation, and modelling; with models that are able to predict the outcome of experiments and can therefore be verified (or falsified if you like Popper's terminology more) $\endgroup$ – yuggib Nov 18 '15 at 12:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.