# Nature of collapse / does collapse exist?

• B
Gold Member
I learned that the moment a wavefunction collapse takes place is a matter of interpretation. So, I suppose the phenomenon 'wavefuntion collapse' is something that has to be witnessed by observation at some point to be able to establish it at all! So my question is: if collapse doesn't actually occur at whichever moment whatsoever, shouldn't we rather speak of a redistribution of probabilities, I mean that at the 'point' of measurement (the interaction of the measuring device with one single particle) the probability of the particle yielding a particular value mixes with the probability of the measuring device measuring a particular value, without any need for the concept of a 'collapse' at any point, for that is an arbitrary assignment? That is: is the probability the particle has the value and that a particular value is measured temporarily narrowed, for it (the wavefuntion) to spread out again? I subquestion would be: what actually is decoherence of the property of a single particle in the environment?

## Answers and Replies

atyy
Science Advisor
Without additional or different assumptions, there is no probability until observation. So you cannot speak of redistribution of probabilities.

bhobba
Mentor
So my question is: if collapse doesn't actually occur at whichever moment whatsoever

This idea of instantaneous collapse is an interpretive thing.

The modern idea is that observations where what is being observed is not destroyed (called filtering type observations) is simply a state preparation procedure. All states are is the equivalence classes of state preparation procedures (that equivalence class is determined by the Born rule ie two states are the same if it gives the same expectation for all observables) so obviously you get another state after. The issue of if an actual instantaneous collapse occurs is left up in the air.

Thanks
Bill

bhobba
Mentor
I subquestion would be: what actually is decoherence of the property of a single particle in the environment?

Properties do not decohere.

All decoherence is, is a mechanism for converting a superposition to a mixed state. Susskind explains it in his book I know you have studied.

Thanks
Bill

vanhees71
Science Advisor
Gold Member
Whether or not there is a collapse of states is a real think taking place in nature or not is, in my opinion, not part of science but of personal opinion. There are, however, so many scientific arguments against it that I have the strong opinion that it's not a real mechanism taking place in any way in nature. For physics is completely superfluous and it doesn't matter whether it's real or not, because it's simply not needed to apply the quantum-theoretical formalism to observations, and all there is in physics is whether a theory or model has observable consequences in nature and whether the predictions of the theory or model are in accordance with these observations or not. For QT the accordance is overwhelming. It's the best tested theory ever and it provides the best agreement between theory and experiment ever. You don't need a collapse, just Born's Rule, which is one of the fundamental postulates of quantum theory that can (most probably) not be derived from the other postulates (concerning kinemematics and dynamics of QT).

For a good review on interpretational problems, see

S. Weinberg, Lectures on Quantum Mechanics, Cambridge University Press

Note that Weinberg comes to a different, very salomonic conclusion: The interpretational issues for him are unresolved yet, and that's it. As far as I can see, in the remainder of the marvelous textbook he uses just the physics content of QT, i.e., just the minimal interpretation, taking Born's Rule seriously and not adding any other interpretational additions, which usually don't help much in understanding the physics behind the QT formalism.

Gold Member
Properties do not decohere.

All decoherence is, is a mechanism for converting a superposition to a mixed state. Susskind explains it in his book I know you have studied.
I reached chapter 8 after going for a second peruse of the book until that chapter. I still can't picture what is mixed state really is, and what it has to do with decoherence, an before going on the wrong track, I thought I might need some indications of which direction to go in what I picture in my mind.

bhobba
Mentor
I thought I might need some indications of which direction to go in what I picture in my mind.

Its entirely mathematical. As I know only too well it cant be explained in words.

Thanks
Bill

Gold Member
Its entirely mathematical. As I know only too well it cant be explained in words.
It's of course physical too, and I figure we have to have some idea of what we are doing when we do math or set up an experiment, right?

In other words: I think some level of abstraction is necessary?

bhobba
Mentor
It's of course physical too, and I figure we have to have some idea of what we are doing when we do math or set up an experiment, right? In other words: I think some level of abstraction is necessary?

Its purely mathematical. Sorry.

Thanks
Bill

Gold Member
Its purely mathematical. Sorry.
I respect your opinion, and I suspect you are right. But for you to reach such a conclusion, you must have some insight in what it is that makes conceptualisation fail in this matter. For instance, I wouldn't say that states, operators and entanglement, to name a few topics in TM, go without some abstract idea about them, at least to me! On the other hand, the Schroedinger equation is difficult if not impossible to picture to me at this stage.

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vanhees71
Science Advisor
Gold Member
I reached chapter 8 after going for a second peruse of the book until that chapter. I still can't picture what is mixed state really is, and what it has to do with decoherence, an before going on the wrong track, I thought I might need some indications of which direction to go in what I picture in my mind.
I don't want to go into the discussions about interpretion again. I consider them usually pretty useless, but I hope I can clarify what a "mixed state" describes from the minimal-interpertation point of view, which is, in my opinion, the only part of "interpretation" that is necessary to relate the QT formalism to observations of and doing experiments with real-world systems.

A pure state is represented by a vector ##|\Psi \rangle## in Hilbert space (more accurately, it's the corresponding ray in Hilbert space, but that's not important in this context). An observable ##A## is represented by a self-adjoint operator ##\hat{A}## in that Hilbert space. For simplicity let's assume that it's non-degenerate, i.e., for each eigenvalue or ##\hat{A}## the corresponding eigenspace is one-dimensional. Then the probability to measure the value ##a##, which necessarily is an einvalue of ##\hat{A}##, of the observable ##A## is
$$P(a)=|\langle a|\Psi \rangle|^2.$$
The properties of a system cannot be specified more accurately than by preparing it in a such described pure state. However, there is no other physical meaning of state kets than this probabilistic meaning, and thus the pure state in QT refers to ensembles of equally prepared systems to be in that state.

Now we assume that the system is not prepared in a pure state, but it's known that with probability ##p_j## it's described by the state ##|\Psi_j \rangle##. Tnis implies ##p_j \in [0,1]## and \sum_j p_j=1. This defines a "mixed state". Now the probability to find the value ##a## when measuring observable ##A##, provided it's prepared in state ##|\Psi_j \rangle## is given by the conditional probability
$$P(a|\Psi_j)=|\langle a|\Psi_j \rangle|^2,$$
and thus, due to Baye's theorem to find the value ##a## given the mixed state is
$$P(a)=\sum_j p_j P(a|\Psi_j)=\sum_j p_j |\langle a|\Psi_j \rangle|^2.$$
Now we rewrite this a bit to get a more crisp definition of how to describe such mixed states:
$$P(a)=\sum_j p_j \langle a|\Psi_j \rangle \langle \Psi_j|a \rangle= \langle a | \left (\sum_j p_j |\Psi_j \rangle \langle \Psi_j| \right)|a \rangle=:\langle a|\hat{\rho} a \rangle.$$
The operator
$$\hat{\rho}=\sum_j p_j |\Psi_j \rangle \langle \Psi_j|$$
is called the statistical operator of the system.

Of course, you can now also describe a pure state in terms of a statistical operator. It's just the "mixture", where the system is with probability 1 prepared in the pure state described by ##|\Psi \rangle##, and thus the pure state is represented by the statistical operator
$$\hat{\rho}_{\Psi} = |\Psi \rangle \langle \Psi|.$$

• Nugatory and bhobba
Dear Entropy1,

The mathematical formalism adopted by Schroedinger leaves clear that in the instant the location of the particle is made, all probabilities disappear. Strangely, since the formulation to this day, numerous discussions about the significance of this disappearance occur, maintaining that there is something mysterious in it. Nevertheless, when we have a dice in hand, before we throw it the possibility of each face falling upside is one to six. In the moment it falls upon the table and immobilize, it's clear one can no more speak in probabilities, as one of the faces was defined. It is no more probable, it's certain. Its obvious, there is nothing mysterious in it, as even Einstein and Niels Bohr concurred.

atyy
Science Advisor
Dear Entropy1,

The mathematical formalism adopted by Schroedinger leaves clear that in the instant the location of the particle is made, all probabilities disappear. Strangely, since the formulation to this day, numerous discussions about the significance of this disappearance occur, maintaining that there is something mysterious in it. Nevertheless, when we have a dice in hand, before we throw it the possibility of each face falling upside is one to six. In the moment it falls upon the table and immobilize, it's clear one can no more speak in probabilities, as one of the faces was defined. It is no more probable, it's certain. Its obvious, there is nothing mysterious in it, as even Einstein and Niels Bohr concurred.

This is wrong, because the state space of the dice is a simplex, but the quantum mechanical state space is not a simplex.

Gold Member
Thank you for taking the effort for such an extended reply, vanhees!

It looks as if in ##P(a)=\langle a|\hat{\rho}|a \rangle##, ##\hat{\rho}## takes the place of a standard observable: ##P(a)=\langle a|L|a \rangle##. Is the density matrix then in fact itself an observable?

It looks as if in ##P(a)=\langle a|\hat{\rho}|a \rangle##, ##\hat{\rho}## takes the place of a standard observable: ##P(a)=\langle a|L|a \rangle##. Is the density matrix then in fact itself an observable?

Mathematically, what is important for an observable is that it has a spectral decomposition (i.e. the ## |a\rangle ##'s in Vanhee's example form an orthogonal basis). Physically, this means that the outcomes corresponding to different eigenvalues of an observable are perfectly distinguishable.
What is important for a density matrix is that it is positive semidefinite (or positive in short) i.e. has only non-negative eigenvalues. Physically, this means that probabilities have to be non-negative.

In a complex vector space (which we are in in quantum mechanics), positivity of ## \rho ## implies that it is also Hermitian hence has a spectral decomposition. (This is not true in a real vector space.) But its physical meaning is still different from that of an observable.

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what it has to do with decoherence

My understanding is as follow.

Let's treat collapse as just a mathematical device. I can collapse a state ## |\psi \rangle ## in two ways: collapse it to the pure state ## |a\rangle ## that I observe, or to a mixed state composed of eigenvectors of ## \hat{A} ##. The second form is weaker and leads to the first form after postselecting the outcome that I observe. (This is almost your "redistribution" but, as atyy pointed out, is not quite correct because of the distinction between proper and improper mixed states.)

Since the observation a particular outcome is probabilistic, I don't have a deterministic unitary time evolution that brings ## |\psi \rangle ## to ## |a\rangle ##. Neither a unitary time evolution can bring the initial pure state to a mixed state.

The way out is that unitary time evolutions (the Schrödinger's equation) are only valid for a system uncorrelated with other system. So entanglement with other system can bring ## |\psi \rangle ## to a mixed state. Decoherence (as you already know from the other threat) is when this entanglement is macroscopic and for all intents and purposes, irreversible.

the Schroedinger equation is difficult if not impossible to picture to me at this stage.

For a qubit at least, a unitary time evolution is literally a rotation of the Bloch ball. So it can't bring pure states on the surface of the ball inside the ball where mixed states live.

Gold Member
but, as atyy pointed out, is not quite correct because of the distinction between proper and improper mixed states.)
What is a proper mixed state, and what is an improper mixed state?

What is a proper mixed state, and what is an improper mixed state?
It is an interpretational issue.

If we interpret a mixed state as saying that there is some set of (pure) states and a probability distribution over them due to our ignorance of the actual state the system is in, then a mixed state is the same as a classical probability distribution. And if we have such a mixed state, there would be no need for collapse, as you noted in the OP. This is called a proper mixed state.

An improper mixed state is a mixed state that we get by starting with a correlated joint state and then tracing out some subsystem. We can't say that the system described by the mixed density matrix is actually in some (pure) state that we don't know, because the system doesn't truly have a (pure) state of its own; it is correlated with other system. An improper mixed state is what decoherence gives us.

Quantum mechanics describes these two situations with the same mathematical object, the density matrix.

Gold Member
An improper mixed state is what decoherence gives us.
Why is that so?

Because tracing out the environment that is entangled with your system is what you do to get decoherence.

Gold Member
Because tracing out the environment that is entangled with your system is what you do to get decoherence. (emphasis added)
What system are you refering to?

A system of interest that is being measured / decoheres / (effectively) collapses.

Gold Member
So let me check if I got this straight (I fear not): if we 'measure' a (weighed mixture of) pure state(s), there is no decoherence (in the measuring device)?

Before we go on, I should also say that given any density matrix, it can be decomposed into a weighed mixture of pure states in infinitely many ways. For example, the maximally mixed state of a polarization can be written as 50% of vertically polarized light + 50 % of horizontally polarized light, or 50% of clockwise circularly polarized light + 50% of counterclockwise circularly polarized light etc.

So
if we 'measure' a (weighed mixture of) pure state(s), there is no decoherence (in the measuring device)?

Conceptually, this is possible, if we measure the correct observable e.g. if the mixture is of vertically polarized light and horizontally polarized light, then we measure whether it is vertically or horizontally polarized.

The problem is that quantum mechanics doesn't allow us to distinguish between a weighed mixture of pure states or a system entangled with another system (nor between infinitely many different weighed mixtures) if they are described by the same density matrix (unless we take into account other systems that could be entangled with our system).

Gold Member
I thought that any kind of measurement that yields a macroscopic result is the result of decoherence in the measuring device?