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Famously, the collapse of the wave function is considered one of the biggest puzzles of quantum mechanics and motivates people to take ideas like the many-worlds interpretation seriously.

Something I always found puzzling is that there seems to be a quite similar phenomenon in classical physics. In a purely classical system with uncertainty, we can use a description in phase space involving a probability density $\rho$. Using $\rho$, we can make probabilistic predictions. However, as soon as we measure the system, $\rho$ collapses instantaneously to a single point.

Is this also considered to be a puzzle? And if not, why?

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    $\begingroup$ Do you mean the measurement problem when you say "collapse of the wave function"? Not all interpretations agree that there even is collapse in the strict sense, so they could not consider it the biggest puzzle... $\endgroup$ – ACuriousMind Jun 3 at 15:38
  • $\begingroup$ Trying to follow the logic of the question... In the classical case, what do you mean by $\rho$ collapses instantly to a single point? If the original probability distribution is just a representation of our ignorance about how the system was prepared, then the "collapse" is likewise just a bookkeeping device to account for the new information we get from the measurement. If that's not what you mean by collapse (in the classical case), then what do you mean by probability density (in the classical case)? $\endgroup$ – Chiral Anomaly Jun 4 at 23:52
  • $\begingroup$ @ChiralAnomaly Yes, the original probability distribution is just a representation of our ignorance about the system's state. $\endgroup$ – jak Jun 5 at 5:35
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Without tackling the discussion of the measurement problem head-on, I'd like to describe how there is a distinction between the probability associated with classical phase space density $\rho$ and the probability associated with the quantum state $\psi$, especially by pointing out that the quantum-mechanical analog of the phase space density is the density matrix operator.

I'll be roughly working with the Copenhagen "interpretation" of quantum mechanics but I believe everything I say can be parsed without much of an issue to the subjective description of the world as described by an observer in Everettean QM. A committed Everettean might want to make Everettean flavor of the density matrix more explicit but I'd refrain from such an exercise.

  • The state of a quantum system is fully described by its state $\psi$ whereas the state of a classical system is fully described by a point in phase space $(q,p)$, not the phase space density $\rho$. Thus, the probabilities associated with the phase space density $\rho$ are purely a result of our ignorance, whereas the probabilities associated with the state $\psi$ are fundamental (in the sense that they are not arising out of our ignorance but are rather intrinsic to the nature of physics).

  • The quantum mechanical analog of the classical phase space density $\rho$ is the density matrix operator $\hat\rho$ which describes a quantum system about which we are ignorant. For example, imagine that you have a friend who prepares a spin half particle in either spin-up or spin-down state depending on the result of a coin toss. You get hold of this state but you don't know what was the result of the coin toss--so now, the system is truly in a specific quantum state, but you don't know which one so you describe it with a density matrix $\hat\rho$ which assigns different probabilities to the system being in different quantum states. Contrast this with the probabilities described by the quantum state $\psi$ itself which describes the probabilities of finding the system in some quantum states upon a measurement.

  • As one of the other answers mentions, just like the classical probabilities described by the phase space density, the quantum density matrix can immediately change without doing anything to the system if your friend, for example, just tells you the result of their coin toss. You then immediately know the actual quantum state of the system that they prepared the system in--and the density matrix immediately reduces to a pure-state density matrix. Contrast this to the scenario with a system described by the quantum state: the probabilities described by such a state only change in a non-unitary way when you actually make a measurement on the state (nobody can whisper into your ear what the outcome would be because there is no predetermined outcome, truly).
  • Finally, while the probabilities involved with the quantum state and the density matrix are different (in the sense that probabilities associated with a quantum state are not arising out of our ignorance), it must be pointed out that quantum and classical probabilities as such irretrievably mix up in a system described by a density matrix and you cannot tell apart one from the other in an invariant way.
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    $\begingroup$ I think this answer tries but fails to not invoke interpretation-specific beliefs: 1. You can use the phase space formulation of QM (cf. Wigner-Weyl transform) and then both classical and quantum states are described by functions $\rho(q,p)$ (but the quantum state is not necessarily a proper probability density). 2. The claim that QM isn't predetermined is interpretation-dependent - Bell's theorem excludes local hidden variables but it does not forbid all hidden variables and there certainly exist interpretations (e.g. Bohmian) that have hidden variables. $\endgroup$ – ACuriousMind Jun 3 at 15:46
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    $\begingroup$ 3. In "relational" interpretations of QM, the density matrix does represent observer knowledge, so this distinction also depends on you implicitly taking some variant of Copenhagen as "true" here. $\endgroup$ – ACuriousMind Jun 3 at 15:46
  • $\begingroup$ @ACuriousMind Yeah, I agree. My "interpretation free" pretense was supposed to be limited to not saying anything about whether I'm explicitly speaking in terms of Copenhagen or in terms of subjective experience of an observer in Evererttian QM. I'll make it more explicit. $\endgroup$ – Dvij D.C. Jun 3 at 15:51
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Take a situation where we have (classical) uncertainty about a situation. For example, I toss a coin and hide the outcome from you. When I reveal it, nothing about the coin changes. The coin either landed heads or it landed tails. All that changed was your state of knowledge about the outcome. Nothing remarkable there.

Now, consider the quantum analogue of this example. The heads and tails become states of a qbit, for example. Let’s call it a quantum coin. When we measure the state, is it still the case that all that changed was our state of knowledge about the state? That the state was always either heads or tails?

It can’t be so simple. Let’s consider a second property of the coin, it’s colour. So it has two measurable properties: the face, which may be heads or tails, and the colour, which let’s say may be gold or silver. Crucially, the quantum coin can’t have a definite colour and a definite face simultaneously, whereas the classical coin can of course have a definite face showing and a definite colour. This is the important complication relative to the classical case.

So, when we measure the face of the quantum coin and see heads, can we say that it was heads all along and we just learnt it? What if we’d chosen to measure the colour and seen gold? In that case, we’d be saying, ah I’ve seen gold so it must have been gold all along. In the classical case, that works. But in the quantum case, it can’t have a definite colour and face at the same time. So it appears that we are not simply learning properties of the quantum state.

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    $\begingroup$ It may be worth to mention that the physical realization of the "quantum coin" are typically spinful objects (e.g. silver atoms) in a Stern-Gerlach experiment, and "colour" or "face" are different directions of spin. $\endgroup$ – ACuriousMind Jun 3 at 15:48
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If we assume the existence of hidden variables (and why shouldn't we, when an outlandish concept as "many worlds" is proposed in the context of the collapse of a quantum mechanical wavefunction?), we can compare the motion of a Brownian particle with the motion of a quantum mechanical particle. In both cases probabilities are present.
The classical surroundings of the Brownian particle (early considered as continuous and later discovered to consist out of discrete particles) are in this case equivalent to the (continuous or not) hidden variables. Both classical surroundings as hidden variables "push" the particle around, in accordance with the probability distribution.
A quantum mechanical system is, when taking hidden variables into consideration, a deterministic system just as a classical system is. In both cases the probabilities are deterministic. When we measure the position of a Brownian particle, the position did already exist before the measurements, just as we determine the position of a quantum mechanical particle.
It's possible to construct an uncertainty relation for the Brownian particle, just like for a quantum mechanical particle. This can be read on pages 17-18 in this article about the entanglement of Brownian particles.
So the processes of measurement of the position (or momentum) of the Brownian particle and the measurement of the position (or momentum) of a quantum particle are the same kind of processes. Both involve a collapse.
The collapse of the classical probability distribution is not really a puzzle, just as it ain't in the hidden variables approach to quantum mechanics.
The article shows that there is uncertainty too in classical mechanics when measuring the position (or momentum), just like in quantum mechanics. The probability distribution doesn't collapse to a point in phase space.

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    $\begingroup$ "Why shouldn't we?" -- because they've been largely ruled out by Bell experiments and the other proposals violate basic principles of science such as local propagation of causes? $\endgroup$ – Dvij D.C. Jun 3 at 13:34
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    $\begingroup$ @DvijD.C. The usual response. Non-local hidden variables are not ruled out, and there is ongoing research to construct quantum mechanics based on hidden variables. So who knows what the future brings? $\endgroup$ – Deschele Schilder Jun 3 at 16:55
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    $\begingroup$ Alright! I get your point. But I think many worlds, for example, are even less appealing. Obviously, these will always stay an exercise in metaphysics, just to explain the wavefunction collapse. Non-local hidden variables (they are hidden, just like quarks will always stay hidden, but they do a damned good job in explaining, among other things, meson or baryon behaviour) can likewise do a good job in the underlying mechanism of QM, or in explaining (non-local) entanglement. $\endgroup$ – Deschele Schilder Jun 3 at 17:27
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    $\begingroup$ Maybe space, in which elementary particles are embedded, somehow constitutes the realm of hidden variables, like Brownian particles are embedded in a huge, almost continuous collection of atoms, pushing the particles around. $\endgroup$ – Deschele Schilder Jun 3 at 17:31
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    $\begingroup$ @DvijD.C. Bell's theorem (like any theorem) depends on specific assumptions and its interpretation is far from straight-forward. In particular, "non-local" is used in this context in a very specific sense. So it's not that easy to proclaim that "hidden variables have been ruled out". Have you looked at what Bell's theorem tells us about Brownian motion? Is Brownian motion "non-local" according to the defintion that is used in this context? $\endgroup$ – jak Jun 4 at 8:33
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This is not the same thing.

In case of a wave function collapse we are talking about changes in the physical system itself - a repeated measurement on the same system will find the system in its collapsed state. Some consider it a puzzle, because such behavior is difficult to interpret in everyday mechanistic terms, others simply accept that quantum laws are different from the classical ones, and then this is not a puzzle at all.

On the other hand, the probability theory refers to selecting one system out of an ensemble of systems. If the systems are classical, the measurement does not change their state - it only changes our knowledge about this state.

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  • $\begingroup$ But there are psi-epistemic interpretations of QM, aren’t there? This answer seems to be all about psi-ontic $\endgroup$ – innisfree Jun 3 at 10:54
  • $\begingroup$ @innisfree The main point is that wave function collapse is about a single object, whereas probability is about ensemble (at least in the frequentest view). $\endgroup$ – Vadim Jun 3 at 11:12
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    $\begingroup$ But the differences between classical and quantum don’t depend on us using a frequnetist interpretation of probability! $\endgroup$ – innisfree Jun 3 at 11:38
  • $\begingroup$ My interpretation of the question is that it is not about classical vs. quantum, but about wave function collapse vs. measurement in usual probabilistic sense. The latter takes place in both quantum and classical physics, but in classical physics it is not accompanies by a wave function collapse or any other features specific to a particular interpretation of QM. $\endgroup$ – Vadim Jun 3 at 11:49
  • $\begingroup$ Saying that "wave function collapse" = "changes in the physical system" seems like a big assumption (c.f.en.wikipedia.org/wiki/Mind_projection_fallacy). It's equally valid to say that the wave function simply expresses our knowledge about the system and then the scenario is equivalent to the "probability density collapse" scenario. This becomes even more obvious if we consider a phase space formulation of QM. $\endgroup$ – jak Jun 4 at 8:28

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