6
$\begingroup$

I was wondering about something while studying quantum mechanics. If the wave function collapses when measuring a particle and assumes a single position, how do we know that it was a wave in the first place?

P.S.: sorry if it is absurdly simple, I was just confused and couldn't come up with any explanation.

$\endgroup$
  • $\begingroup$ The "collapse of the wave function" is nothing but a colorful expression for the Born rule. It doesn't have an objective counterpart (like the little implosion that people seem to be implying when they hear the term). That the wave function is a field is simply given by its mathematical definition. The field has some values before the measurement and some other values thereafter. That's all. $\endgroup$ – CuriousOne Dec 21 '14 at 16:05
  • $\begingroup$ Doesn't the collapse of a wave function mean it takes a definite position due to the act of measurement. So doesn't that mean that every time we measure the particle it takes a definite position so how do we know it was acting like a wave (according to the schrodingers equation )in the first place? - I'm confused $\endgroup$ – Matthew V Dec 21 '14 at 16:12
  • $\begingroup$ No, it doesn't mean that, at all. There is nothing that could take a definite position, to begin with. There are no little balls in there! The position that your detector shows you are three NUMBERS x, y, z. Absolutely nothing special is taking place at the coordinates x, y, z. The Born rule merely tells you that the next measurement that you make will behave AS IF the system had been given a fresh start at those coordinates. It would not only take an infinite amount of energy to actually localize the system at that point, there would also be an infinity of newly created particles coming out. $\endgroup$ – CuriousOne Dec 21 '14 at 16:16
  • 2
    $\begingroup$ @MatthewV There's a nice discussion on this provided by ACuriousMind on his blog, check it out here! $\endgroup$ – Phonon Dec 21 '14 at 16:48
  • $\begingroup$ @MatthewV: I disagree with the Curious Mind. I am a materialist, and I think that everything around us is MATTER. So, the w.f. is the description of the MATTER behavior. And this is not a so bad description, the QM is excellent description of the microscopic world. This is what gives us some confidence that the w.f. is quite a close description of what really we have in our apparatuses. That the w.f. doesn't answer to all our questions, well, we have to do more investigations. $\endgroup$ – Sofia Dec 21 '14 at 20:07
6
$\begingroup$

It was not a wave.

The wavefunction is not a wave. It fulfills the Schrödinger equation in the position representation, and although that looks similar to what one usually writes as "wave equation", and produces similar interference phenomena, it is not a wave in any physical sense. The wave function is not a physical object, it is merely a way of writing the coefficients for a quantum state in the position basis. It is not measureable, and there is, in general, no physical quantity oscillating that would be associated with it.

Any object in quantum mechanics is described by an abstract state in a Hilbert space, and the abstract Schrödinger equation tells you that there is a basis of "stationary states" that evolve in time just by being multiplied by a phase $\mathrm{e}^{\mathrm{i}Et}$, i.e. essentially doing nothing. If you now add several of these states with different energies/Hamiltonian eigenvalues $E$, the overall time evolution of the state is not a simple multiplication anymore, and the state is indeed changing. Essentially, this is all "interference" means - you have phases $\mathrm{e}^{\mathrm{i}Et}$ with different $E$ that can be added, and then some non-trivial kind of evolution appears. (Since usual solutions to wave equation also contain $\mathrm{e}^{\mathrm{i}\omega t}$ in this way, this explains the name)

But that doesn't mean quantum states are waves. It also doesn't mean they are particles. They are quantum objects, states in a Hilbert space. Not waves. Not particles. When we look at them in some ways, e.g. at their time evolutions, and their properties of superposition and interference, they look like our inutitive notion of waves. When we look at them in detectors, they often look like our intuitive notion of particles.1 They are neither.


1It should be noted that trying to describe that actual occurence of such measurements is still a topic of some debate. Nevertheless, "collapse" is not a necessary interpretation of the math - decoherence approaches to measurements/emergence of classical physics do not need that concept.

$\endgroup$
  • $\begingroup$ We don't know what is the wave-function (w.f.). But we have some confirmation that it describes quite well the reality that passes through our apparatuses. That's why the closest image that we can make to ourselves about that reality, is wave. The confirmation is that besides predicting well probabilities of measurements, all the intermediate steps before the detector (passing through fields, beam-splitters, etc.) are well-described with the concept of a complex wave possessing magnitude and phase (concept already known to us from e.m. waves). $\endgroup$ – Sofia Dec 21 '14 at 22:53
  • 2
    $\begingroup$ @Sofia: While it is possible to describe most of quantum mechanics in terms of wave functions, it is not the most advisable way: It doesn't generalize easily to QFT, it is difficult to explain what spin, angular momentum and the like are, and it lures people into thinking of quantum objects as classical waves. I can write down Hamiltonians where the solution to the Schrödinger equation is far from being any kind of wave. I can consider systems with finitely many degrees of freedom that will not be described by such a function on the reals at all. $\endgroup$ – ACuriousMind Dec 22 '14 at 14:50
  • 2
    $\begingroup$ This answer boils my experimentalist blood. In what sense are electromagnetic waves more of a "physical object" than a quantum wave function? They're both mathematical book-keeping tricks to reliably reproduce what we see in experiments. They're just useful constructs in our mental metaphor of Nature. $\endgroup$ – DanielSank Dec 22 '14 at 18:32
  • 1
    $\begingroup$ @DanielSank: EM fields/waves are physical in the sense that charges move in them (that's how we define an EM field, after all - by its action on test charges). There's no such interpretation for the "wavefunction" - it has no such test objects, instead it encodes a probability distribution that is equivalently encoded in many other things as well, while we'd be hard pressed to find such equivalent formulations for EM fields. $\endgroup$ – ACuriousMind Dec 23 '14 at 15:25
  • 1
    $\begingroup$ @DanielSank: The wavefunction is just the expansion coefficients in the position basis. Choose any other complete basis of the abstract Hilbert space (e.g. momentum), and their coefficients also encode the same information. Also, you could switch to the density matrix formalism. And, again, in systems with finitely many degrees of freedom - e.g. only spin - there's no such wavefunction (as in "function on the reals telling us what's gonna happen") at all. To pick one of these descriptions and say "That's what it really is" is giving ontological weight to it that is unnecessary. $\endgroup$ – ACuriousMind Dec 23 '14 at 15:37
2
$\begingroup$

This is a supplement to my original answer. For rigor, we have to make a distinction between the reality that travels in our apparatuses, and the mathematical description that we give it. However, the mathematical description proved to be so successful, that sometimes we place a sign of equality between them. About what happens with a quantum object when it interacts with a macroscopic apparatus, we don't know. At present, we don't have a better tool to handle this problem than the collapse (reduction postulate of von Neumann). And we simply use it because we have to go on, to work.

Now, the wave-form for the wave-function works well in some cases, and works badly in other cases. But in most of the cases in which interference in involves, it works well. For instance if we put on the way of the particle a beam-splitter, we believe that we get a splitting of the wave, into a reflected wave and a transmitted wave. I.e. although we speak of one particle, we believe that we get two waves. Then, if we redirect, with mirrors, the two waves to cross the path of one another, we get an interference pattern (see experiments with the Mach-Zender interferometer in Wikipedia) if in the crossing region we place a photographic plate.

However, the interference tableau doesn't appear for one single particle. We have to prepare many particles carefully, in an identical way, i.e. same type of particle, same velocity, etc.

So, interference pattern is produced by waves, while a single particle is detected on the photographic plate in a single place, as any particle.

Though, we incline to admit that before the detection on the plate, we had for each particle and particle, the two waves as I said above, and at the detection all the energy of the particle is delivered to one single molecule on the plate.

(The process of impressing the photographic plate is some more complicated but I restricted myself to a simple line. What is most important is that at the detection on the plate, the particle doesn't impart its energy to all the region covered by the interference pattern. No, the energy is delivered to a single point (e.g. a certain molecule is decomposed) ).

$\endgroup$
  • 2
    $\begingroup$ The wave function is a field. If it describes the system perfectly well, it's as good as any other description that gives the right answers. Are you happier with matrices? Mathematically matrix mechanics is equivalent and predicts the same outcome. Is that a better physical picture? $\endgroup$ – CuriousOne Dec 21 '14 at 16:22
  • $\begingroup$ I am back! But you have to tell me what you have in mind behind the statement that the w.f. is a field. Let me say frankly - to understand what is the w.f. is worthy a Nobel prize. But nobody knows. We can't say what was in the apparatus before we measure, because these tiny objects don't allow us to measure them without decohering them. Whatever we can say of the w.f. is that it is a good machine to predict probabilities. Thus, field, or wave, what does it matter? $\endgroup$ – Sofia Dec 21 '14 at 19:25
  • $\begingroup$ To understand what is the w.f. we need, maybe, apparatuses more delicate than these particles. But, probably, this is a NO-GO. $\endgroup$ – Sofia Dec 21 '14 at 19:28

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.