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How does a wavefunction collapse into one state? More specifically, what conditions cause a wavefunction for a quantum particle to collapse?

Does this have to do with density matrices?

Please try to give a mathematical explanation that is simple and concise...

TanMath
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I agree in full with Marty Green except the explanations of chemistry in which I was unable to follow so well (that doesn't say that I disagree with them).

But, let me put the things in short. The collapse is a phenomenon that is supposed to occur when a quantum object comes in contact with a quantum system. For instance, a quantum particle falls on a beam-splitter and we try to decide if it was transmitted or reflected.

The wave function says that the particle behaves as a wave, s.t. part of the wave is transmitted, and part reflected. And that, for each particle and particle. This is what the wave-function says. But if we put detectors on both paths, the transmitted and the reflected, only one of the detectors gives a click.

Why so? If the wave-packet of each particle splits at the beam-splitter into a transmitted wave-packet and a reflected wave-packet, why only one of the two wave-packets produces a recording? And which one of them? And how is it decided which one?

The mathematician and physicist that dealt with this question was John von Neumann. But, to go straight to the conclusion, he wasn't able to give an answer. He introduced the phrase "wave-packet reduction", or in short, "collapse". It's just a name, because we don't know how the things work in fact.

Einstein believed that behind the fact that one of the wave-packets gives an answer, and the other doesn't stand some secret properties of the particle. He said the famous phrase "God does not play dice". The above properties are currently named in the literature, "hidden variables", hidden because we don't specifically which properties are these. Well, later experiments showed that Einstein was wrong. And this is a very deceiving fact, because we remain with the idea of collapse which contains no explanation.

But, even worse, a great physicist named John Bell, showed that if we insist that there are some hidden features that govern the behavior of the particle and which response it gives, these variables have to be non-local. Famous experiments conducted by Alain Aspect proved this. (These experiments were done on pairs of particles, but the details exceed the scope of my explanation.) For our case with the transmitted and reflected wave-packet, the things go as if there is an agreement between the two wave-packets about which one will respond. A famous experiment performed later by Grangier, Roger, and Aspect (known in the literature as "the GRA beam-splitter") showed that if one wave-packet gives a click, the other wave-packet never gives.

Well, this is the story, in very short form. But, before I end the answer, and in order that you get a complete picture, are you familiar with the von Neumann measurement? If not, I recommend that you ask what is that.

Good luck !

Sofia
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I am going to answer this in a hurry because the question is on the edge of being closed.

Quantum mechanics isn't just about "wavefunctions", it is also about "observables".

An observable is something like: energy, position, momentum... i.e. it includes all the properties of classical physics.

The wavefunction (or state vector or quantum state) is the thing which gives you the probabilities of observable properties.

There are wavefunctions which correspond to a particular observable having a definite value with 100% probability.

For example, a wavefunction that is a "Dirac delta function" peaked at a point x, might correspond to "particle has position x with probability 100%". (Since you like math details, I will remark that a delta function is not an ordinary function - the one I just mentioned will be "infinity" at x and zero everywhere else - but there are ways to make it a well-defined concept.)

Or, a wavefunction which is a "plane wave" of a specific uniform wavelength, might correspond to "particle has momentum p with probability 100%".

The way to apply wavefunctions boils down to this: You start out with some knowledge of observables - e.g. particles with particular positions or particular momenta. You use the wavefunctions which correspond to those observables having those values, and you evolve the combined wavefunction according to the Schrodinger equation for the physical system in question. You arrive at a new overall wavefunction at a later time, and then you can use that new wavefunction to calculate probabilities for whatever properties you want to know about at that later time.

To get those probabilities, you basically take the wavefunction, express it as a sum (or integral) over wavefunctions that are "100%-probability" wavefunctions for particular values of the property you're interested in, and then you can get the probabilities from the coefficients in the sum/integral.

For example, maybe you have arrived at a wavefunction that is smeared across space, and you want to know about a particle's position. In effect you are re-expressing that wavefunction as a weighted sum of delta functions - and recall that a delta function peaked at x, corresponds to "particle is definitely at point x".

The wavefunction psi(x), a complex-number function which varies from point to point, can be thought of as "integral over all values of x, of psi(x) Dirac-delta(x) dx".

So for example, the delta function at a point x0, contributes with weight psi(x0).

And then the probability rule is that the probability of the actual particle being at point x0, is |psi(x0)|^2.

Alternatively, if you cared about momentum, you would express wavefunction psi as a sum of plane waves of different wavelength. Again, each component has a coefficient, which we might write psi(p) - p for momentum - and the probability that the actual particle has momentum p0 is going to be |psi(p0)|^2.

All these rules have the odd result that you cannot have a wavefunction which corresponds to 100% probability for some particular position and 100% probability for some particular momentum. The definite-momentum wavefunction, a plane wave, is spread out across all positions, and the definite-position wavefunction, a delta function, is a sum over all plane waves if you Fourier-transform it.

That's the uncertainty principle.

What I have described are the bare bones of applied quantum mechanics. You're studying something, like electrons and protons interacting electromagnetically. Perhaps you have a "Hamiltonian" or "Lagrangian" which encodes how they interact in a formula. If this was classical physics, that formula would allow you to start with specific initial conditions - electron here, proton there - and deduce the behavior which follows deterministically from those initial conditions.

But in quantum mechanics, you use that formula differently - you use it to construct a Schrodinger equation for the proton and electron, which tells you how their wavefunctions develop. (Important technical note, this isn't a complete leap in the dark with respect to classical physics, there is a classical equation called the Hamilton-Jacobi equation which anticipates the Schrodinger equation.)

In your question you want to understand "wavefunction collapse". So the most important thing to understand first, is the perspective that maybe wavefunctions aren't real at all. It's position, momentum, and so on, which are the reality. The wavefunctions are "just" a mathematical recipe that happens to give accurate probabilities. The real question would then be, why does this recipe, quantum mechanics, work?

What "wavefunction collapse" refers to, is the part after you have used the Schrodinger equation to evolve a wavefunction through time, when you then calculate probabilities for physical properties. Maybe the result was, 75% probability the particle is at A, 25% probability the particle is at B. Meanwhile, you were also doing an experiment - which the wavefunction calculation was meant to describe - and the particle actually showed up at A.

If you were to then represent the situation "particle at A" by a wavefunction, you would use a Dirac delta function peaked at the point A - or more likely, since you only know A within some experimental error, you'd use a "gaussian" function that is sharply peaked around A.

So you've gone from "wavefunction with some waves at A and some waves at B" to "wavefunction concentrated around A". That is the so-called "collapse of the wavefunction", but the "collapse" here just means that you got some physical data, and started using the appropriate wavefunction. As Rod Vance says in a comment, this is like updating a probability distribution: Flip a coin, before you look you might say "50% heads, 50% tails", you look, it's tails, now you say "100% tails". The probability distribution "collapsed" onto tails.

Back to your question: Quantum mechanics says nothing about "how does a wavefunction collapse", because it says nothing about whether a wavefunction exists in the first place. The wavefunction is first of all part of a calculation, that gives you the odds of the particle arriving at A or B, given some initial condition.

The real question is, what is really going on? The question, what causes the wavefunction to collapse, already contains the assumption that wavefunctions are physical things and that they have collapsed by the time e.g. the particle is at a definite place. So it's a question that is appropriate for a particular attempt to get beyond applied quantum mechanics to some new physical theory or some understanding more fundamental than "do these calculations and it works"... namely, the path where you hypothesize that wavefunctions are real things.

Incidentally, I should mention decoherence. This is a thing that the Schrodinger equation can cause a wavefunction to do, but it is not the same as collapse. It doesn't take a particle whose wavefunction is at A and at B, and produce a wavefunction just at A (for example). What it does, is to take the combined wavefunction for e.g. a particle and a physical "pointer" with two values, the A-value and the B-value, and produces a wavefunction with a peak at "position A and pointer value A" and another peak at "position B and pointer value B". This means that decohering interactions are good for measurement, but the wavefunction evolution in itself still doesn't produce a single definite outcome, you still have to apply the probability rule to the decohered wavefunction, which in the case I just described is still a "superposition" over two possible outcomes.

I have written this informal tutorial about quantum mechanics because the ultimate fact is that no-one yet knows what is really going on. There are people who don't care about anything beyond applying the quantum formulas; there are people who somehow believe that reality isn't there before observation; there are people who try to make a classically objective theory just out of wavefunctions; and there are people who try to make a theory in some other way.

And meanwhile, people continue to develop new theories within the quantum framework, all the way up to string theory (which is this whole apparatus of wavefunctions and uncertainty principle and observables, applied to vibrating interacting "strings"). This progress within the quantum framework has gone very far and become very sophisticated. But, though lots of people have ideas, and lots of people will tell you that they already know the answer, the primordial question of what lies within or beyond quantum mechanics remains unanswered.

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    I'll add that using wavefunctions which evolve with time isn't even the only way to get the predictions of quantum mechanics. The methods of Heisenberg (operator algebra) and of Feynman (path integral) work differently or in a different order, compared to this Schrodinger picture, and for all we know might be more fundamental. – Mitchell Porter Dec 10 '14 at 04:08
  • To get at numbers you have to use solutions of the QM equations, the psi one operates on. No? – anna v Dec 10 '14 at 05:51
  • The Heisenberg picture is unitarily equivalent to that of Schroedinger, so it doesn't seem to make sense to say one is more fundamental. – Ruslan Dec 10 '14 at 09:24
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What is a wavefunction?

A wave function is a mathematical solution of one of the basic quantum mechanical equations: Schroedinger, Klein Gordon, Dirac.

By the postulates of quantum mechanics the square of this wavefunction gives the probability of finding the system under study when looking at $(x,y,z,t)$ or $(p_x, p_y, p_z,E)$ or similar four vector spaces.

How does it relate to measurements?

As the predictions are probabilistic, experiments accumulate information from many instances to display a probability distribution. Up to now the probability distributions predicted by the wavefunctions have been validated, which validates the theory of Quantum Mechanics.

Probability distributions have the same interpretation in classical and quantum mechanics, and even economics of populations studies.

Take a population plot by age. It has a distribution. A thirty year old individual is a instance at the "30 years axis". He/she is not spread out from zero age to a hundred.

In a similar way a given measurement gives an entry to the distribution function given by the square of the wavefunction, period.

How does a wavefunction collapse into one state?

The "collapse" is a fancy way to state that a measurement was made on a system of quantum mechanical dimensions, i.e. commensurate to the sizes given by h_bar. It is mathematically isometric, it is only a semantic differentiation.

More specifically, what conditions cause a wavefunction for a quantum particle to collapse?

A measurement, i.e. interaction with particles on mass shell in a Feynman diagram ( real in contrast to virtual which are a mathematical tool). An instance picked from the predicted probability distribution of the wavefunction squared.

Does this have to do with density matrices?

Density matrices are useful mathematical tools when the systems describe more than two or three particles.

With large numbers/dimensions/interactions the system coincides with the classical equation descriptions.

anna v
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The question you ask is essentially: How to solve the measurement problem? As you can see from that article (although I wouldn't say it's a very good one), there are several approaches to either get a theory where no collapse occurs (thereby rendering the question futile) or to explain the collapse. So far, nothing has been so satisfactory as to be a generally accepted solution.

Take a photon, put it in a superposition of left and right polarized light (this is no problem) and then send it through a right-polarizor. What you observe is: Either the photon goes through or it doesn't. You prepared it in a state that is neither left polarized nor right polarized but after the polarizer, it is polarized. That's the collapse. How to explain it? How does the photon choose whether it is right- or left polarized and subsequently passes the polarizer or doesn't? And how does it become one of the two options, if it is previously something different? As I said, we don't know.

Is it important? From an empirical perspective: probably not, because it seems that this knowledge would not let us predict anything knew (in particular, we cannot be able to predict in what way the photon will collapse, there are just no "hidden variables" that tell us this). From a metaphysical perspective: of course. But that's not physics.

Martin
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In theory, every example of wavefunction collapse should be explainable through a mechanism of normal time evolution of the wave function. The apparent collapse is only an illusion. People who agree with this idea like to talk about "decoherence", which is a fancy sounding word, but it doesn't really tell you anything. In fact, there is very little interest in working out actual mechanisms to explain what is going on in so-called "collapse" situations.

You cannot even find, on this forum or anywhere on the internet, a concise list of the Top Ten examples of "collapse of the wave function". I've asked for such a list in the past, and haven't gotten a single answer. I expect people would list things like the Stern Gerlach experiment, where a silver atom in a superposition of spin states suddenly chooses either the spin-up or the spin-down state. But people who like to talk about the collapse have a lot of trouble agreeing on where they thing the "collapse" occurs...in the magnet, or at the detector screen.

People also like to talk about an atom being in a superposition of energy states, but when you actually measure the energy, it jumps into one of the energy eigenstates. I've never seen these people explain how you actually measure the energy eigenstate of a single atom. Or how you produce a hydrogen atom in the pure 2p state.

For my money, the classic example of the collapse of the wave function is the light of a distant star, focused on a photographic plate, where the wave function (corresponding to the classical e-m wave) is obviously far too weak to stimulate the reduction of a silver atom in the silver bromide crystal. And yet an image of the star nevertheless develops, one silver atom at a time, even if it takes hours between detection events. The explanation? Collapse of the wave function into a "photon" which concentrates the energy previously spread over a huge volume of space into a single silver atom.

I recently posted a blog article which explains this phenomenon via the normal time evolution of the wave function. There are two critical points that need to be explained:

  1. The positive energy needed to make the chemical transition from silver bromide to metallic silver.

  2. The energy barrier needed to get the electron to the silver ion via the "conduction band" of the crystal.

The traditional explanation uses this energy diagram to justify the requirement for the photon's E = hf quantum of energy to drive the transition:

enter image description here

This diagram contains a serious error in the overall energy calucation. It is true that there is a positive Gibbs Free Energy associated with the reduction of silver. But this energy is calculated in the standard state at stoichiometric conditions. The actual photographic plate is very different. An exposed image may contain only parts per trillion of reduced silver. At these concentrations, the Gibbs Free Energy actually tilts in the negative direction, and the conversion becomes spontaneous. In other words, you don't need the external energy of the photon to drive the transition. The energy is already available in the chemistry of the silver bromide crystal.

That still doesn't explain how the electron gets from A to B, because there is still the energy barrier represented by the conduction band level. But I have discovered an ingenious explanation of a siphoning type of mechanism, where a small amount of energy driven into the conduction band can actually trigger a self-perpetuating system, whereby the energy released at the target silver atom site is pumped back into the body of the crytal, replenishing the conduction band in a continuous loop.

I explain this mechanism more fully in this blogpost, Wavefunction Collapse Explained by Quantum Siphoning.

Marty Green
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  • You can measure the energy eigenstate of a hydrogen atom by measuring the magnetic moment of the atom, which will depend on the angular momentum of the electronic state, the energy of the electron, and the alignment of the electron's spin. – Zo the Relativist Dec 09 '14 at 21:57
  • Can you show the quantum mechanics math behind it? Also, can you explain about density matrices (unless they are not related!) – TanMath Dec 09 '14 at 22:12
  • The density matrix is related but it's not exactly my territory. I would refer you to this interesting discussion where the people who understand density matrices show that you cannot distinguish between a system where 50% of the atoms are in an excited state versus a system where all the atoms are in a 50-50 superposition of states: http://physics.stackexchange.com/questions/8049/are-these-two-quantum-systems-distinguishable/13544#13544 – Marty Green Dec 09 '14 at 22:21
  • What for, density matrices? Why complicate the things? Would it explain the collapse? I remember that von Neumann had some proof about partitions of an ensemble, but the proof was criticized. Why get into all this? – Sofia Dec 10 '14 at 00:31
  • " concentrates the energy previously spread over a huge volume of space into a single silver atom" That one photon from a distant star raises an energy level is a proof of the particle nature of photons and at the same time a vindication of the PROBABILITY wave that describes particles quantum mechanically. The energy /mass etc are NOT described by a wave, but by a single quantum. "concentratin energy" over light years is agains special relativity and the velocity of light. – anna v Dec 10 '14 at 04:52
  • Anna, downvoting my answer doesn't mean that you're right and I'm wrong. – Marty Green Dec 10 '14 at 04:57
  • It is a signal to future readers to check the physics. If you could see my downvote record you would see that I very very seldom downvote, except at what I consider basic physics misinformation. – anna v Dec 10 '14 at 05:05
  • @jerrySchirmer no you can't measure the energy of a hydrogen atom by measuring its spin state. The 1s and 2s states have the same spin, so a hydrogen atom with a spin of 1/2 can be in any superposition of the 1s and 2s states. – Marty Green Dec 10 '14 at 15:03
  • @MartyGreen: they have the same spin but different orbital energy, which will translate to different orbital angular momenta. And also, once you factor in fine structure, they dont' have the same spin. – Zo the Relativist Dec 10 '14 at 21:07
  • Oh come on. They both have zero orbital energy. And dont' try to squirm out of it by invoking the fine structure constant. – Marty Green Dec 10 '14 at 21:28
  • @MartyGreen: I'm not invoking the fine structure constant. I'm invoking fine structure splitting of the energy levels, which gives the states different energies. Or is the spectroscopy of the 1930s too advanced to talk about? – Zo the Relativist Dec 15 '14 at 17:52
  • $<p^{2}>$ will be different for the two states, irrespectively, and this will couple to a magnetic field differently. – Zo the Relativist Dec 15 '14 at 17:53
  • Let's remember your starting position. Here is what you said: "You can measure the energy eigenstate of a hydrogen atom by measuring the magnetic moment of the atom, which will depend on the angular momentum of the electronic state, the energy of the electron, and the alignment of the electron's spin." – Marty Green Dec 15 '14 at 19:30