I decided that a reason contributing to the people's misunderstanding of the meaning and validity of quantum mechanics is the prevailing focus on Schrödinger's equation as the basic dynamical law.

*Felix Bloch (picture) is, together with Lev Landau and perhaps also John von Neumann, credited with the 1927 discovery or invention of the density matrix.*

This equation helps to mislead the people into thinking that the wave function is analogous to a classical wave - an electromagnetic wave, for example. In fact, Erwin Schrödinger himself has never quite understood the meaning of his own wave function.

In my current opinion, each of the three alternative dynamical equations

- equation for the density matrix
- Heisenberg equations for the operators
- Feynman's path integral

However, first, let me note that Schrödinger is getting way too much credit while Heisenberg who should be credited for dynamics is being screwed. Schrödinger found his equation in 1926. Heisenberg found his equations, that were later shown equivalent to Schrödinger's picture by Dirac, in 1925. So Heisenberg was earlier.

Moreover, Heisenberg understood what quantum mechanics meant, unlike Schrödinger. Clearly, people talk about Schrödinger's equation because they are able to understand a partial differential equation - not too different from Maxwell's equations, right? - they actively want to mislead themselves into thinking that the wave function is a classical wave of a sort. It's not.

Heisenberg's equations for the operator are more direct a quantum counterpart of Newton's equations of motion. In this picture, the only novelty are the nonzero commutators between the observables - the Heisenberg uncertainty principle etc.

**Classical phase space**

Fine. Let me now return to the topic I promised in the title - the density-matrix-centered interpretation of the classical-quantum relationship.

We may formulate classical mechanics by the differential equations for x(t), p(t), or whatever degrees of freedom we have. They obey Newton's equations or their generalizations for other degrees of freedom such as Maxwell's equations, and so on. I hope you're still with me.

But in reality, we can't know all these observables with complete accuracy and certainty. Measurements of positions and velocities have nonzero error margins; the detailed motion of molecules in some gas is chaotic and we don't know it; the evolution brings a lot of extra uncertainty. The uncertainty grows bigger.

For all those and other reasons, it's more realistic to assume that even in classical physics, we don't know the exact x(t), p(t). Instead, we know some probability distribution on the phase space (the space of all initial states, or states defining where the system is at any moment):

rho(x,p)In this text, x and p will be shortcuts for many coordinates on the phase space. The interpretation of the phase space probability distribution is that

dP = rho(xis the infinitesimal probability dP that the system is found around the point given by the x and p coordinates, within a small hypercube of volume given by the measure factor. OK?_{i},p_{i}) d^{N}x d^{N}p, i = 1...N

Of course, if we know how x and p would evolve, we may also determine how the probability distribution would evolve. It would evolve according to Liouville's equation for the Hamiltonian system. It can be written as

∂ rho / ∂t = {H,rho}where the curly brackets are the Poisson brackets. You will find everything on that Wikipedia page if you need to refresh your memory.

Now, I want to emphasize that in theory, the description by rho also contains the case in which x,p are known accurately and with certainty: in that case, rho is a delta function located on the point x(t), p(t) of the phase space. In classical physics, a strictly sharp delta-function will remain strictly sharp. Its location is moving as a function of t just like x,p would.

However, the more general, "widely spread" form of rho is much more realistic and universal in its applications. Even when it's spread, in classical physics, you could always imagine that there existed an actual x(t), p(t) at each moment - you just didn't know what it was. However, this thesis has several problems: you can't really prove it because classical physics that uses rho(x,p) as the fundamental object, and encourages you to calculate the probabilities from it, is totally consistent as well.

Moreover, you can't learn anything - make any new predictions or explanations - out of the purely ideological opinion that sharp values of x(t), p(t) existed - if you don't actually know what those x(t), p(t) exactly were. So the idea that a sharp x(t), p(t) exists in classical physics is pure philosophy - there is no physics behind it. This assumption isn't needed for physics to work (physics is consistent with its negation as well); and it's not useful to learn anything (which is why the physical laws without this assumption of "realism" are complete according to any operational definition of the word "complete").

**Quantum mechanics: density matrix**

Now, the object rho(x,p) is replaced by an operator rho in quantum mechanics. You may write a general operator acting on your quantum mechanical space as a function of x,p - the "generating" operators of your Hilbert space that know about all your degrees of freedom. If you worked with spins, you would have to add spins; if you worked with field theory, you would have to put the fields instead, and so on.

It's obvious how the classical Liouville equation for rho(x,p) is generalized in quantum mechanics: the Poisson bracket is simply replaced by the commutator with the right normalization. The equation for the operator rho becomes:

iħ ∂ rho / ∂t = [H,rho] := H rho - rho HThis is easily obtained from Schrödinger equation for psi if you substitute

rho = |psi> <psi|and use the Leibniz rule for the derivative of the product, the ordinary Schrödinger equation for psi, and the complex conjugate Schrödinger equation for the bra-vector psi*. In general, rho is a combination of such products, as we will mention, but the equation is linear so the equation works for the combinations as well.

If you think about it for a while, we have made a truly minimalistic change, indeed. In classical physics, the probabilities of different states were given by rho(x,p) which was a function of c-numbers x,p, the coordinates on the phase space. In quantum mechanics, rho is still a general operator, but it's a function of the operators x,p that don't commute with each other. But you may still imagine particular functional prescriptions for rho such as

rho = exp(-(x-x0)which is a packet concentrated near x0,p0. Of course, three constants should be added in front of the exponential (to make it normalized) as well as both terms in the exponent (to determine the width and obey the dimensional analysis). But I want to keep the formulae simple and comprehensible. It's important to realize that the definition of rho could work for operators rho,x,p, too. In some sense, the number of operators on the Hilbert space is equal to the number of functions on the classical phase space - there exist natural one-to-one maps based on various "orderings" of the operators.^{2}- (p-p0)^{2})

The trace of rho is equal to one - it's the total probability. This is the quantum counterpart of the fact that the integral of rho(x,p) over the phase space is one. The corresponding equations - Liouville's equation and the rho-Schrödinger equation - preserve this normalization of rho.

The difference between the classical and quantum dynamical equations is just hidden in the fact that rho,x,p are operators in quantum theory and they generally don't commute with each other. But otherwise, the interpretation from classical physics may be pretty much directly extended to the quantum theory!

In particular, just like the values of rho(x,p) determine the probabilities as we announced in the definition of rho at the beginning, the operator rho in quantum mechanics determines the probabilities of any states. How does it work?

Well, if you pick a particular state in the Hilbert space, it has a well-defined probability if it's an eigenstate of the density matrix. This is an unusual operation that's not usually talked about - because the density matrix isn't an "observable" in the usual sense - like positions or momenta etc. But it's still an operator on the Hilbert space. I will formality treat the density matrix rho as the "operator for the probability".

**Pure states**

Imagine that you have a pure state psi. In that case, the density matrix rho is just the tensor product psi.psi* we wrote some time ago. Can you say what is the probability that the system described by this rho is finding itself in a particular state of the Hilbert space such as chi?

Yes, but using the analogy between observables and density matrix, you can say such a thing only if chi is an eigenstate of rho. When is chi an eigenstate of the "tensor square" psi.psi*? Well, it's easy. It is if chi is either proportional to psi, or it's orthogonal to psi. In the former case, the eigenvalue of rho is 1 (Yes) and in the latter case, the eigenvalue of rho is 0 (No).

It's this simple. Of course, it's not hard to see that the eigenvalues of rho=psi.psi* are (1,0,0,0,0...). Just choose am orthonormal basis of the Hilbert space whose first basis vector is psi itself.

So I would like you to adopt the following restricted description: you can only "sharply" say what is the probability that the state described by psi - or by rho = psi.psi* - is finding itself in the state chi either if psi,chi are proportional to each other, in which case the probability is 100%, or if they're orthogonal, in which case the probability is 0%. No other linear superpositions have well-defined probabilities because they're not eigenvalues of rho.

This is different from the conventional treatment that talks about probabilities for any state to be in any other state. But what this conventional treatment really means is the "expectation value of the probability", chi*.rho.chi, or |(psi*.chi)|^2. However, because chi is not an eigenstate of rho for a general chi, you shouldn't say that the probability is sharply defined! ;-)

You may also allow a collapse of the wave function onto any of the eigenvectors of such a rho. Why? Because it doesn't do anything at all! ;-) The state psi or its density matrix rho = psi.psi* may collapse into some eigenstates chi of rho with the prescribed probabilities (the eigenvalues of rho in those states). The probabilities are 100% for chi = psi (up to a normalization) and 0% for the orthogonal choices. So the state psi will collapse to psi with probability 100% and nothing changes!

The collapse to the orthogonal vector has a vanishing probability and the collapse to generic linear superposition isn't allowed because they are not eigenstates of the "probability operator" rho, the density matrix.

**Decoherence**

Now, in general, you don't want rho to describe a pure state psi. In general, rho is a combination of the type

rho = sumwhere the states psi_i don't have to be orthogonal but the density matrix is still required to have its trace equal to one. In general, the density matrix is still a Hermitian operator. It follows that it can be diagonalized._{i=1...N}p_{i}|psi_{i}> < psi_{i}|

Now, the eigenvalues of rho pick a privileged basis of states that have well-defined probabilities - it's the corresponding eigenvalues of rho - and they can be measured. In fact, you may always imagine that at any moment t, someone in the Heavens or elsewhere "made" rho collapse so that it was reduced to the form chi.chi* where chi is an eigenvector of rho before the collapse. The probability that the collapse picked a particular rho is the corresponding eigenvalue of rho in this state chi.

Most people including many physics PhDs are obsessed with the collapse and because they find the spreading wave function too complicated while their brains are inadequate to deal with quantum mechanics, they're impatiently waiting for the right to make the wave function collapse, referring to their consciousness as the ultimate justification of this right to finally kill the spreading wave function. They're also interested in the other creatures who have consciousness and the same glorious rights to collapse "waves" as they do. ;-)

I have good news for you! You're allowed to make the the density matrix rho, the generalized state vector, collapse at any moment you wish, without compromising the predictions (e.g. without destroying any interference) whatsoever! The only condition is that you must collapse it to one of the eigenstates of rho at the moment, and the probability imagined by you that the collapse took you to a particular eigenstate chi of the matrix rho has to be given by the eigenvalue of rho in chi.

Isn't it great? This erases all your psychological problems with subjectivity etc. In fact, all "realist" observers may agree that the collapse is taking place all the time. (Well, because of relativity, "at a given moment" will still mean different things for differently moving observers, but staunch "realists" don't care about relativity much.)

In practice, this may fail to satisfy the emotions of the staunchest anti-quantum zealots. Why? Well, we have already mentioned one reason. If you study the evolution of a pure quantum system that is remaining pure and coherent, the eigenvalues of rho remain (0,0,0...., 0,1,0,. ...0,0,0), which means that the collapse doesn't do anything. You're never allowed to collapse into a wrong basis - a pure state may only be collapsed onto itself. So the wave function continues to spread as time goes - the genuine anti-quantum zealots won't like it.

Even if rho has many nonzero eigenstates, they're not necessarily the states that are easy to imagine for the anti-quantum zealots - the eigenstates of rho are typically different than the "intuitively natural" (wrong) states into which the anti-quantum zealots would like to collapse things.

However, if you describe a subsystem by its density matrix that has been traced over the other, environmental degrees of freedom, decoherence guarantees that rho for this system will have many nonzero entries. The corresponding eigenstates of rho will be close to some intuitively "classical" states and in that case, the collapse does something nontrivial.

**Collapse is just in your mind**

I have formulated my interpretatino of quantum mechanics as allowing you to collapse rho into one of its eigenstates, with probabilities given by the corresponding eigenvalues of rho. The essential property of this rule is that if you insert these collapses with a random outcome obeying this rule to arbitrary moments of your time, your predictions won't change at all.

So the collapse is completely subjective.

In the past, you could have thought that when I say that the collapse is just a subjective change of knowledge, it was a philosophical assumption. However, in this case, it is actually a provable theorem! What I mean is that when you insert the collapse of rho onto the rho eigenstates as described above, you will get identical predictions for anything you calculate. Why?

Well, it's simple. When you decompose rho into the eigenstates

rho = sumbut in this case, you require that you use the right basis of rho eigenstates (which always exist) psi_i, each of the terms will be evolving independently and the different states psi_i which are orthogonal at the beginning will stay orthogonal in the future as well, because of the unitarity of the evolution._{i=1...N}p_{i}|psi_{i}> < psi_{i}|

So if you later, at time T, make a measurement of something, which means that you decompose rho(T) into the eigenstates according to the same formula as the last displayed one, and you will ask about the probabilities of different outcomes, it's clear that each individual term in the decomposition of rho(T) has to come from at most one term in the decomposition of rho(t).

To calculate the probability, you will only need p_i extracted from rho(t) while all the other terms will contribute nothing to this evolution and you may forget them. So it doesn't matter whether you will calculate the particular probability of an outcome encoded in the decomposition of rho(T) from the full evolution, or whether you insert the imagined collapse at time t; in both cases, the evolution up to the time t will just add a factor p_i from the decomposition of rho(t).

The density matrix rho is gaining an ever larger number of nonzero eigenvalues - you may view this process, resulting from decoherence (=the loss of a realistic chance of different outcomes to interfere with each other in the future, i.e. the loss of the information about the relative phase of their amplitudes), as the "splitting of the many worlds". However, it's very clear that you only have one world in the prescription above.

Moreover, the very point of this proof was just the opposite - to show that the collapse, when properly defined, is completely subjective and has no detectable consequences. I guess that MWI bigots want to use some related thinking exactly for the opposite cause - to claim that there objectively exist some thing that don't objectively exist.

In this picture, the procedures used to obtain predictions from quantum mechanics are pretty much identical to the predictions obtained from the classical phase space distribution function in classical physics. The only thing is that things including rho are operators and they only have well-defined quantities (such as probabilities in the case of rho) when acting on their eigenstates. The operators don't commute.

**Collapse to a bigger subspace**

In the text above, the possibility of a nontrivial collapse was only possible because we were tracing over the environment, thus obtaining a density matrix with an increasing number of independent eigenstates with nonzero eigenvalues.

Can we do it without this tracing over? Yes. In this case, the density matrix will stay pure if it was pure. As discussed above, a collapse to a pure state is not doing anything. Instead, you may consider a collapse to a set of projection operators P_1...P_K that sum to one.

Analogously to the collapse into a pure state, a collapse into projectors may be freely inserted at any moment whenever all the operators P_1...P_K are chosen to commute with the density matrix rho. If that's so, you may imagine that after the collapse, the density matrix becomes rho.P_j = P_j.rho with probability - well, whatever the normalization of the "reduced magnitude" density matrix has to be divided by to normalize it again.

The relationship between the projection operator and pure states that would appear in the "pure collapse" in the previous formalism may be imagined so that e.g. the projection operator is the projection operator on all microstates with a fixed state of the observer system and any state of what we used to call the environment.

Again, in this case, one may show that such a collapse - describing a measurement of "a subset of degrees of freedom" - will have no impact on future predictions; it is a purely subjective trick.

Needless to say, when I get to this formulation based on projectors that sum up to one, I am "almost" formulating quantum mechanics via the "consistent histories" approach. It's almost the same thing at this moment. Except that I am using a simpler and more explicit rule for the consistency - vanishing of the commutator between the projection operators and the density matrix rho. Note that the eigenvalues of rho which survive the collapse don't have to be equal to each other - the only condition is that you're collapsing onto a subspace of the Hilbert space spanned by rho eigenstates.

Every observer also has the right to use projection operators whose commutator [P_i,rho] is not exactly zero but a small nonzero number. In that case, he will introduce errors and inconsistencies and it's up to her what is tolerable. That's analogous to small violations of the "consistency condition" in the usual consistent histories approach.

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