0
$\begingroup$

This seems to me it must be trivial, but I have not been able to grasp it.

As I understand it, the wavefunction crucially depends on its immediate surroundings, whether it be a nucleus, a box, etc. Energy levels are quantized accordingly.

And yet, there is no reduction of state needed to gain this information. The wavefunction does not collapse, there is no interaction. The potential seems an open book to be read anonymously.

Can we detect a particle in the same manner, without interaction, simply by reading its effect on another wavefunction?

I'm curious to know what I'm getting wrong

Edit after helpful comments: more specifically, what I am getting at is interactions that are not measurements, like the effect of slits on an evolving wavefunction. The wavefunction evolves into a form that considers the shape of the slits, and this shape can be inferred from measurements.

How can it be that the wavefunction and the slits can interact without a reduction of state? Are we really getting information about a system without having to do a direct measurement of the system itself? And does this also work for getting information from quantum systems?

Improve this question
$\endgroup$
3
  • $\begingroup$ I don't know enough to risk putting an answer out, but as a potential direction to look, I think you may be confounding the transient responses of the system with their steady state responses. Why is it that energy levels are quantized in the first place? $\endgroup$
    – Cort Ammon
    Commented Nov 15, 2016 at 0:19
  • $\begingroup$ I'm not really sure what you're asking. What "gain of information" are you talking about? The potential is part of the dynamics, it's not itself a quantum state whose information you could learn. $\endgroup$
    – ACuriousMind ♦
    Commented Nov 15, 2016 at 0:53
  • $\begingroup$ It does seem I was mixing up the dynamics with the quantum state itself. What confuses me is whether there are dynamics apart from interactions, whether a wavefunction can "hit walls" without turning into a particle, and whether the walls are affected at all by being "hit". $\endgroup$
    – Ketil Tunheim
    Commented Nov 15, 2016 at 16:08

2 Answers 2

Reset to default
1
$\begingroup$

No, the wavefunction does not have extra knowledge of its surroundings. If $\psi(x_0) = 0$ at some time, then the evolution of the wavefunction at that time does not depend on $V(x_0)$ at all.

You're getting confused because you've only looked at stationary states, in particular the 'standing waves' that can get set up in a potential. But this has nothing to do with how a particle outside a stationary state evolves. If you put a particle in a box, its wavefunction will gradually spread out, totally unaware of the walls until it hits them.

As a classical analogy, a string makes a note when plucked. But that doesn't mean that each atom in the string knows where all the other atoms are, the relevant wave equation is local.

Improve this answer
$\endgroup$
7
  • $\begingroup$ My confusion remains concerning the dynamics once an evolving wavefunction reaches an obstacle. It seems a wave function can dynamically respond to, say, slits, without having interacted with those slits. Is this not a form of stealth sensing of the slits (by us, once we see the pattern on the screen)? I am curious whether a wavefunction could dynamically sense other wavefunctions without interacting. A part of me has a hunch we're already doing this and it's trivial $\endgroup$
    – Ketil Tunheim
    Commented Nov 15, 2016 at 16:26
  • $\begingroup$ @KetilTunheim But it does interact with the slits. In particular, the quantum double slit experiment is the same as the double slit experiment for classical waves -- a water wave passes through both slits and does the same thing. It has to pass through both to know both are there. $\endgroup$
    – knzhou
    Commented Nov 15, 2016 at 16:47
  • $\begingroup$ So are we talking about an interaction that is not mediated by particles, one which does not require a reduction of state, and one in which the slits themselves are unaffected by the interaction? $\endgroup$
    – Ketil Tunheim
    Commented Nov 15, 2016 at 20:44
  • 1
    $\begingroup$ @KetilTunheim Yes. I know, it's pretty weird how exactly such an interaction doesn't count as a measurement! I was confused about the same point and asked a question about it here. It's for the slightly different case of a Stern-Gerlach apparatus but it's the same idea. The essential part is that the slit state is affected, but the new slit state is so much like the old one that it doesn't matter. $\endgroup$
    – knzhou
    Commented Nov 15, 2016 at 20:45
  • $\begingroup$ That is very interesting. So when a wavefunction evolves in accordance with an obstacle, the obstacle is also affected. I assume that if the slits were instead another wavefunction, this effect would be more significant. I also assume that wavefunctions have dynamics of interaction that do not require measurement, but that nevertheless change the state. If I got it right (possibly not), then what remains of my question is simply, what initiates decoherence. And that is a very different topic $\endgroup$
    – Ketil Tunheim
    Commented Nov 15, 2016 at 21:26
0
$\begingroup$

I would attempt an answer of this in the hope that I learn something from comments.

As I understand it, the wavefunction crucially depends on its immediate surroundings, whether it be a nucleus, a box, etc. Energy levels are quantized accordingly.

No offence intended, but you are only talking about bound states when you say this, I guess you are ignoring free electrons for example, unconfined by any local potential. Is that correct?

And yet, there is no reduction of state needed to gain this information. The wavefunction does not collapse, there is no interaction. The potential seems an open book to be read anonymously.

I don't follow your point here, sorry, we have not gained any information until we perform a measurement, otherwise we just have a series of equations with probably, many possible outcomes.

Can we detect a particle in the same manner, without interaction, simply by reading its effect on another wavefunction?

I would back go your assumption above that, (I hope I understand you correctly), we can essentially discover something real, by looking only at the math behind it. We can't do this, we must interact in some fashion.

There are plenty of people on this site who know far, far more than I do, so hopefully they will tell us something we can both learn from, but we can't detect a particle without interacting with it.

Entanglement may be what you mean by this last statement , but I am not, through ignorance of the subject, able to give you an answer on that.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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