# 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?

atyy
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
Gold Member
2021 Award
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
Gold Member
2021 Award
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
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?

Gold Member
If we look at the combined system: "measured system - environment" as being a closed system, and we would measure a pure or proper mixed state, would coherence be conserved?

Sorry, I thought I said something really wrong so I deleted my latest reply. Now I don't remember the exact words I said but this new reply should be clearer. It might also answer your new question "would coherence be preserved?"

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

If the system is actually already in some pure state that is an eigenstate of the observable I measure, then I will just get that state. No decoherence is needed.

But to always interpret a mixed state to be such a correct proper mixed state is unfounded.

Gold Member
Thank you @Truecrimson Do you also know the answer to my question #26?

What do you mean by "would coherence be conserved"? The joint "(original) measured system + environment" would just become my new "measured system" and we go back to the same discussion.

Gold Member
What do you mean by "would coherence be conserved"? The joint "(original) measured system + environment" would just become my new "measured system" and we go back to the same discussion.
So a measurement is always prone to decoherence (heat bath)?

What I mean is this:
Wikipedia on decoherence said:
[..](although the combined system plus environment evolves in a unitary fashion).

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It's probably more correct to say that there is no measurement without decoherence. Decoherence is what provides the measurement record.

Now you might wonder what about the proper mixed states that I talked about. My point is just that, proper mixed states are introduced to be conceptually the same as classical probability distribution, and classically we don't need decoherence. But QM is different.

If we look at the combined system: "measured system - environment" as being a closed system, and we would measure a pure or proper mixed state, would coherence be conserved?

If you make a measurement on the combined system, it's not closed anymore.

Gold Member
So making a measurement can't be done in a closed system, and that is why decoherence comes in play, right?

Or is it rather that we can't say for sure whether we are measuring a proper mixed state (or pure state) or an improper one?

And what is the difference between a proper mixed state and a pure state anyway?

Sorry for coming into the discussion with far less knowledge than all of you have. As I see it there are 2 kinds of interactions, some which produce information which has the potential to influence the future and others which do not produce such information. In the classical double slit experiment, both results (generating which-path information and not generating it) are observations, both are measurements. And yet, the setup where we do not attempt to generate which-path information is not considered as a "wavefunction collapse". We have observed the system, undoubtedly so, we have a definite persistent record of the result of the observation (an interference pattern) and yet this observation did not cause the electron's probability wave to collapse, it just maintained its superposition of 50% going through slit A and 50% going through slit B, causing the interference result.

So, observation or measurement 'per se' does not cause collapse. It is only when the process generates irreversible information which could potentially alter the future that we get collapse.

The fundamental issue is consistency. Take the quantum Zeno experiment by Itano et al in 1990. They put some 5000 beryllium atoms in a magnetic trap. By applying a shower of microwaves they could bring the beryllium atoms to an excited energy state. Then by shining a laser beam they could tell how many atoms where in their ground state or in the excited state (because non-excited atoms absorbed some of the laser energy while excited ones did not).

Long story short (if anyone wants the details I will post them), the beryllium atoms were clearly inferred to be in a true superposition of both states non-excited and excited when not observed. But if we shine the laser to generate information about the beryllium atoms' state, their superposition is destroyed. Simply because of consistency, we can not have information about the atoms' state in the laser beam and a superposition of states in the atoms. As simple as that. No definite state information existing = all the possible states consistent with the information coexist in superposition. Definite state information exists = the physical system is in that state, "collapsed". It's just consistency.

What makes the difference between processes which do not require generating definite information, such as our double slit experiment without which-path detectors, and those which demand information to be generated, such as our double slit with which-path detectors switched on? Simply the physical arrangement of the system. It is not necessary that we observe any results, we may as well send them to outer space without looking at them, it is just the physical arrangement of the system which defines if definite state information must be generated or not.

So making a measurement can't be done in a closed system, and that is why decoherence comes in play, right?

It's the other way around, decoherence is what provides the measurement record, and since decoherence requires an open system, measurement requires an open system. (Beware that the notion of a "closed system" here is different from that in classical physics, which usually means that the system doesn't give or take mass with other system. "Closed system" here means that it doesn't give or take information in a sense that can be made precise. Then it's a tautology to say that I can't measure a closed system, because I can't get any information out of it!)

The notion of a proper mixed state, IMO, is introduced only to highlight the conceptual tension between them and mixed states in QM and prevent the interpretation of collapse as a redistribution of probabilities (unless some additional assumption is made e.g. hidden variables).

vanhees71
Gold Member
2021 Award
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.
No! Even if you believe in collapse and take wave mechanics (i.e., non-relativistic (and only non-relativistic!) QT) of a single particle and even if you believe in collapse, in no way a particle can take a fully determined exact position. If so the probability distribution whould be a Dirac-##\delta## distribution, i.e., $$|\psi(\vec{x})|^2=\delta^{(3)}(\vec{x}-\vec{x}_0).$$
Since you cannot take a square root of the ##\delta## distribution, this ##\psi## doesn't exist.

On the other hand you may insist that there is a eigenstate of the position operator. This becomes more clear in the momentum representation, in which
$$\hat{\vec{x}}=\mathrm{i} \vec{\nabla}_{p},$$
and the putative eigenstates are
$$\tilde{u}_{\vec{x}_0}(\vec{p})=\frac{1}{(2 \pi)^{3/2}} \exp[-\mathrm{i}(\vec{x}-\vec{x}_0)],$$
but that's not a square-integrable function and thus cannot describe a state. Fourier-transforming to the position representation you get
$$u_{\vec{x}_0}(\vec{x})=\delta^{(3)}(\vec{x}-\vec{x}_0),$$
which confirms this result, i.e., you cannot even square the ##\delta## distribution!

Here you have generalized eigenvectors of unbound operators. This always happens if you have operators with a continuous spectrum (or part of whose spectrum is continuous). Then you have generalized eigenfunctions (in the sense of distributions of functional analysis) rather than Hilbert-space eigenfunctions, and these distributions can never describe states of the particle.

In other words a true pure state of a particle within Schrödinger's formulation of non-relativistic QT of a single particle is always represented by a square-integrable wave function, which may very narrowly peak around some value ##\vec{x}_0## of the position of the particle, but it's always of finite width and thus the standard deviation of the position is never 0. This is also underlined by Heisenberg's uncertainty relation ##\Delta x \Delta p_x \geq 1/2##, which shows that neither position nor momentum can ever be exactly determined, and that the more you specify the position the less well specified is the momentum of the particle.

bhobba