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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?

jak
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3 Answers3

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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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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.

innisfree
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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.

Deschele Schilder
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