# Collapsing a wave function without hitting the detector

If a free particle is placed at the origin (in 1D) with a wave function that consists of a superposition of the particle moving in both the +/- X direction, and a single detector is placed on the positive X axis, what would happen?

I was thinking that the wave function would collapse at the time that the +X state would impinge on the detector, and there is a 50% chance of the detector being hit. If the particle does not hit the detector, the wave function collapses into the -X state. So, the wave function is collapsed even though the particle seems to have never interacted with the detector. Is this accurate? Would the -X state then be 'fixed', or can it evolve over time, become a superposition again, and eventually hit the detector at a later time?

So, the wave function is collapsed even though the particle seems to have never interacted with the detector. Is this accurate?

Totally. There is even a whole class of measurements called quantum nondemolition measurements, where you can detect whether or not a single photon could set off a trigger without actually setting off the trigger.

Would the -X state then be 'fixed', or can it evolve over time, become a superposition again, and eventually hit the detector at a later time?

That statement seems confused. If your wave separated and half ran away and half ran over to your detector then half the time your detector fires and now the running away branch is not an issue. Or else the other half of the time, the detector doesn't go off and the running away branch becomes the entire wave.

And as a wave it will then evolve according to the Schrödinger equation, like always.

One has to decide what the observable to measure is, and expand the state onto its eigenstates accordingly.

In your example one wants to measure the position of the particle, hence it is convenient to expand the state $|\psi\rangle$ as $$|\psi\rangle = \int \textrm{d}x'\,|x'\rangle\langle x'|\psi\rangle.$$ In order to measure the possible position of the particle we need to have a detector (meaning any possible device) interact with it. We place the detector at $|x_{\textrm{det}}\rangle$ (assuming this makes sense, for simplicity) and start performing measurements in any possible way (for example, the detector scatters photons all around in the universe waiting for them to be scattered back by the particle). The particle is now in its initial state, but because of this measurement being performed (i. e. the photons being scattered) its state gets modified and broken into $|x_{\textrm{pos}}\rangle$. Therefore we have $$|\psi_{\textrm{init}}\rangle = \int \textrm{d}x'\,|x'\rangle\langle x'|\psi\rangle \quad \xrightarrow{\textrm{photons}}\quad |x_{\textrm{pos}}\rangle$$ and we define $|x_{\textrm{pos}}\rangle$ as the outcome of the measurement.

In your example you want the measurement to be placed at the position of the detector, as we want the particle to pass through. Consequently: either the state collapses into $|x_{\textrm{pos}}\rangle = |x_{\textrm{det}}\rangle$ at some point in time (namely the particle hits the detector) or it does not; in the latter case the particle state is still a superposition of all possible other states, since no measurement has been performed that would make it collapse.

• 'since no measurement has been performed that would make it collapse' But a measurement has occurred, its just that all those other positions are in the same eigenspaces of that localized detector they are in the "unfired" state of the detector. – Timaeus Nov 8 '15 at 1:57
• Good to see you changed that horrible blue alien profile pic. – user36790 Nov 8 '15 at 5:01
• @Timaeus I don't understand how a measurement could occur in the latter case, since no interaction with the device has happened (by definition of measurement). – gented Nov 8 '15 at 15:18
• @GennaroTedesco At which point, each wave can act like it is the only wave, and thus one wave just went one way and the detector was silent but it becomes a world unto itself. And the other wave went to the detector and influenced a lot of things and became a world unto itself. And absolutely no one ever claimed that measurements happen because a particle was somewhere and interacted with something. Measurements happen when a wave gets separated into parts, each of which can thereafter act like it is the whole wave. – Timaeus Nov 8 '15 at 15:34
• @GennaroTedesco If you had a superposition of a packet going left L and a packet going right R then you can have an unfired position detector U on the right. We know the packet going left is unaffected. If the detector on the right is designed to change its state (to F) in response to a packet going right then we know as time passed that we get a state like (L+R)U to evolve into L'U or into R'F where the primes are evolved versions. And a measurement happens either way. Did you read my answer and the reference to nondemolition measurements? I'm not using strange definitions. – Timaeus Nov 8 '15 at 22:11