Thursday, September 24, 2015

Guest post: "Are classical, deterministic, field theories compatible with the predictions of quantum mechanics?"

Elliptic Composability is a blog dedicated to quantum mechanics and a recent post attracted a lot of attention and comments. In particular one of the readers, Andrei, has argued for classical physics and we have engaged in a lot of back and forth discussions about it. Although I disagree with Andrei, I have found his arguments well constructed and I thought my debate with him would be of interest to a much larger audience. So I have invited Andrei to present his best arguments in a guest post to which I will present my counter-arguments in the next post. 

Without further ado, here is Andrei's guest post:

Thank you, Florin, for inviting me to present on your blog my arguments that classical physics can still play a role in understanding nature.

The following objections are usually raised against the search for classical explanations for quantum phenomena:

Objection 1: Classical, local theories have been ruled out by Bell’s theorem.
Objection 2: Classical theories cannot explain single-particle interference (double slit experiment), quantum tunneling, the stability of atoms or energy quantification in atoms or molecules.
Objection 3: Even if one could elude the previous points, there is no reason to pursue classical theories because quantum mechanics perfectly predicts all observed phenomena.

Objection 1
It is a widespread belief that Bell’s theorem rules out local realism. As classical field theories are both local and realistic they couldn’t possible reproduce the predictions of quantum mechanics. However, a more careful reading of the theorem would be:

Assuming the experimenters are free to choose what measurements to perform, the logical conclusion is that no local and realistic theory can reproduce the predictions of quantum mechanics.

One can easily notice that, if the motion of quantum particles (including the particles the experimenter himself is made of) is described by some classical, deterministic, field theory, this freedom of choice does not make any sense.  Classical determinism implies that certain past configuration uniquely determines the future. Anything else would require a violation of the physical law.

One can still object that, even if the experimenters (say Bob and Alice) follow some deterministic process there is still no reason to assume that their behavior would be correlated. Alice could use the decay of some radioactive material to decide the measurement she performs, while Bob could let his measurement be decided by very complex random number generation software. Any assumed correlation between such unrelated processes would amount to a conspiracy no one should seriously consider.

Let’s analyze however, this experiment from the point of view of a classical field theory. One can make the following two observations:

a.       the experiment reduces, at microscopic scale, to three groups of particles: A (Alice, her radioactive material and the detector she controls), B (Bob, his computer and the detector he controls), S (the source of the entangled particles).
b.      The trajectory of a particle in A would be a function of position/momenta of particles in A,B and C; the trajectory of a particle in B would be a function of position/momenta of particles in A,B and C; The trajectory of a particle in C would be a function of position/momenta of particles in A,B and C

   From (a) and (b) it follows that Alice and Bob (and the source of the entangled particles), cannot be independent of one another. The so-called conspiracy is a direct consequence of the mathematical structure of a classical field theory.

In conclusion I have proven that Bell’s theorem cannot rule out classical deterministic field theories.

Objection 2
I will show here how some “mysterious” quantum effects are not in contradiction with what one would expect if some classical field theory describes the motion of particles at the fundamental level. I will start with the iconic double-slit experiment as it is presented by Feynman here:

Feynman says:
“In this chapter we shall tackle immediately the basic element of the mysterious behavior in its most strange form. We choose to examine a phenomenon which is impossible, absolutely impossible, to explain in any classical way, and which has in it the heart of quantum mechanics. In reality, it contains the only mystery. We cannot make the mystery go away by “explaining” how it works.”

I invite everyone to read the whole description of the experiment there. Feynman chooses to use indestructible bullets as the classic analogs of electrons. The experiment is performed with the first slit open, then with the second slit open, and, in the end, with both slits open. The expected result is:

“The probabilities just add together. The effect with both holes open is the sum of the effects with each hole open alone. We shall call this result an observation of “no interference,” for a reason that you will see later. So much for bullets. They come in lumps, and their probability of arrival shows no interference.”

This is all nice, but classical physics is not the same thing as Newtonian physics of the rigid body. Let’s consider a better classical approximation of the electron, a charged bullet. The slits are made of some material that will necessarily contain a large number of charged “bullets”. As the test bullet travels, its trajectory will be determined by the field generated by the slitted barrier. The field will be a function of position/momenta of the “bullets” in the barrier. But the field produced by a barrier with two slits will be different than the field produced by a barrier with only one slit, so the effect with both holes open is NOT the sum of the effects with each hole open alone.

In this paper:

Yves Couder has provided experimental confirmation for classical single-particle interference. The greatest mystery of quantum mechanics has been solved by the good old classical field theory.

Let’s move to another “classical impossibility”: stable atoms with quantified energy levels. It is claimed that classical electrodynamics cannot account for stable atoms, and I would accept this claim, even if the situation is not as clear as it seems. For a possible counterexample I recommend this:

The author, Gryziński, has published its model in top peer-reviewed  magazines so I guess there is some truth about it.

But, for the sake of the argument, let’s assume that classical electrodynamics cannot explain the atom. Does it prove that no classical theory could do it? I don’t think so. We know that stable groups of charges with well-defined energies are possible. They are called ionic crystals. The charged particles do not move (if thermal motion is ignored), so there is no energy loss by radiation. The geometry of the crystal dictates the energy, which has a well-defined value. The reason for the stability of the crystals stays in the repulsive force generated by the orbiting electrons. One can argue that electrons are not composite particles, but a repulsive force can be provided by some modification of the electric field.

The third example I will discuss here is tunneling. Presumably, classical physics cannot explain this phenomenon because the particle does not have enough energy to overcome the barrier. 

There are two observations I can make for this case:

a.       Neither the energy of the particle that tunnels, nor the value of the potential are accurately known at the moment of tunneling. They are average values, and these can be very different from the instantaneous ones.
b.      The actual force acting on the particle depends on the classical theory used to describe the experiment. A new theory could predict a much stronger force.

Objection 3

In the end I would like to point out a few reasons for investigating classical theories.

a.       If nature is not probabilistic after all, there is much to be discovered. Detailed mechanism behind quantum phenomena should be revealed, bringing out a deeper understanding of our universe, and maybe new physical effects.
b.      Quantum theories are not well equipped to describe the universe as a whole. There is no observer outside the universe, no measurement can be performed on it, not even in principle.
c.       Due to its inability to provide an objective description of reality, quantum mechanics may not be able to solve the cosmological constant problem. A theory that states clearly “what’s there” could provide a much better estimate of the vacuum energy. After all we are not interested in what energy someone could find by performing a measurement on the vacuum, but what the vacuum consists of, when no one is there to pump energy into it.
d.      Quantum mechanics requires an infinitely large instrument to measure a variable with infinite precision. When gravity is taken into account, it follows that local, perfectly defined properties cannot exist, because, beyond a certain mass, the instrument would collapse into a black hole.

For a detailed description of the latter two points, please read this paper by Nima Arkani-Hamed:

Thanks again, Florin, for this opportunity to present my arguments, and I am looking forward to seeing your opinion about them.


Friday, September 18, 2015

Heisenberg's matrix mechanics and the Uncertainty Principle

Continuing the exploration of various quantum mechanics formulations today I want to talk about Heisenberg's matrix mechanics. A major stumbling block for early 20th century physics was the explanation of discrete levels of energy exhibited for example by the atomic spectral lines. Classical mechanics was completely unable to provide any explanation and this crisis led to the exploration of new ideas. 

One profound way of thinking about this was to strip the theory of all unobservable baggage (like the trajectory of an electron in an atom) and only talk about what can be actually tested by experiment. The discrete levels of energy would come out of the non-commutativity as follows:

1. Find n matrices \(Q_k\) and \(P_k\) such that \([P_i, P_j] = [Q_i, Q_j] = 0\) and \([P_i, Q_j] = \frac{\hbar}{i}\delta_{ij} \)
2. Diagonalize the Hamiltonian \(H\) by usage of the following transformation: \(S^{-1}Q_i S\) and \(S^{-1}P_j S\)

The diagonal elements are the energy levels observed in experiments and the theory seems at first to have the same explanation power as the old semiclassical formulation of quantum mechanics due to Bohr and Sommerfeld, but lacking a mechanism to predict intensities. However this is not the case, because lurking about there is the standard Hilbert space formulation and von Neumann brought it to the surface in his classical book: The Mathematical Foundations of Quantum Mechanics. (Historical note: Heisenberg used intensity information and working backwards he derived a non-commutative rule of multiplication which turned out to be the usual matrix multiplication)

The diagonalization step is nothing but the usual eigenvalue-eigenvector problem: \(H x = \lambda x\) where \(x\) is a column of \(S\) and the space of \(x\) is the usual Hilbert space. As expected \(P, Q, H\) are operators acting on this Hilbert space, and the intensities predictions arise out of projecting in the Hilbert space.

Werner Heisenberg

Then as now, to defend a new paradigm one needs to produce a physical justification, and Mr. Heisenberg tried to explain the commutation relationship \([P_i, Q_j] = \frac{\hbar}{i}\delta_{ij} \) in terms of the inability to measure simultaneously the position and momenta of a particle in what is now known as the uncertainty principle. Today this is so commonplace that it is easy to overlook some subtle points about it and I'll present two such points.

Suppose that you have a beam of electrons moving in a straight line and with various speeds. By placing this beam in a magnetic field you can separate the electrons according to their speed. Then if you place a photographic plate in front of them you can also measure the electron's position. The momentum has an uncertainty, the position has its uncertainty as well, but if you compute \(\Delta p \Delta x\) you seem to be able to beat the uncertainty principle!!! (try with the usual values of electron speed and detection resolution in an old cathode ray tube TV). However, this does not disprove the uncertainty principle and moreover Mr. Heisenberg was aware of this.

I can tell you the solution to this problem, but it is much more interesting for the reader to try to wrestle with this puzzle and explain the apparent contradiction (I'll provide Heisenberg's take on it in the next post).

The second subtle point (arising out of a advanced functional analysis) is that non-commutativity is not enough to establish the uncertainty principle. In addition one needs that:
1. the operators should be unbounded
2. the operators should either have no point spectrum or if they have a point spectrum that spectrum should lie outside the domain of the other operator.

Position and momenta respect those conditions, but spin does not and as such there is no uncertainty principle for spin! But wait a minute, this sounds preposterous. The commutator: \([S_x, S_y] = i\hbar S_z\) allow us to compute \(\Delta S_x \Delta S_y\) and this is not zero! So how can we claim that there is no uncertainty principle for spin? There is a subtle difference between the spin and the position-momenta case. Can you spot it? Again' I'll present the solution next time because it is much more interesting for the reader to attempt to solve this connundrum as well.

To summarize, in the matrix mechanics formulation, both the spectra and the intensities arise out of the diagonalization step after the (infinite dimensional) matrices were selected to respect the commutation relationship. What the theory does not provide is a concrete transition mechanism (corresponding to the collapse postulate) and therefore quantum mechanics is intrinsically probabilistic. This left open the possibility to "complete" quantum mechanics and various interpretations of quantum mechanics are nothing but solution to the so-called "measurement problem". More on this in subsequent posts.

Thursday, September 10, 2015

Hamilton-Jacobi and Schrodinger equations

Quantum mechanics is a theory significantly larger than its standard textbook formulation. Quantum mechanics is categorical in nature and we need to go beyond parochial points of view and understand the links between different formulations and the categorical approach. Last time I presented a concrete problem in the phase space formulation, and today I want to arrive at the usual Hilbert space formalism by deriving Schrodinger's equation. There is no better way to do that than the follow the original paper by Schrodinger which I was pleasantly surprised to discover it is very clearly written and understandable.

The prerequisite of Schrodinger's approach is the Hamilton-Jacobi formalism and I will start with a short introduction of it.

The Hamiltonian equations of motion are:

\(\dot p = -\nabla_{x} H \)
\(\dot x = \nabla_{p} H \)

We want to explore changes of variables: \((p,x) \rightarrow (P(p,x), X(p,x))\) such that the Hamiltonian equations of motion remain valid in the new coordinates (those are called canonical transformations). This means that that differential form \(\omega = dp \wedge dx\) is preserved:

\(dp \wedge dx = d(pdx) = d(PdX) = dP \wedge dX \)

which demands: \(d(pdx - P dX) = 0\) or \(pdx - PdX = dR\) So far so good, but we want to have a change of coordinates to simplify things. In particular we want the new Hamiltonian to be zero which will make the new \(X\) and \(P\) constants. So we have to see what will happen when we have time-dependent canonical transformations. By notation we say \(f_t(P,X) = f(P,X,t)\). Then:

\(dS_t = pdx - P_t dX_t\)

\(dR = \partial_t R dt + dR_t\)
\(dX = \partial_t X dt + dX_t\)

we get:

\(pdx - Hdt = PdX - (H + P\partial_t X + \partial_t R)dt + dR\)
which means that \(H+ P\partial_t X + \partial_t R\) is the transformed Hamiltonian for the new coordinates: \(K(P(t), X(t))\) The link between the new and old Hamiltonian is given by the use of a generating function \(S\):

\(K = H + \partial_t S\)

where \(\nabla_x S = p\) and  \(S = S(P,x,t)\)

If we demand \(K = 0\) then \(P\) and \(X\) are constants. \(K = 0\) is known as the Hamilton-Jacobi equation:

\(\partial_t S(x,t) + H(\nabla_x S(x,t), x) = 0\)

Because this is derived from the phase space Hamiltonian formalism, this equation is another formulation of classical mechanics in configuration space. So how does Mr. Schrodinger arrive at his equation which describes quantum mechanics?

Erwin Schrödinger

The original paper can be found here and as I said it is remarkably accessible. The physical idea is that of the optical-mechanical analogy and the Eikonal equation in geometric optics because according to de Broglie the particles are waves (of an unspecified kind). In the short wavelength limit the waves becomes rays (and the particles are tracing a well defined path according to classical mechanics).

If we start from the Hamilton-Jacobi equation:

\(\frac{\partial S}{\partial t} + \frac{1}{2m}{(\frac{\partial S}{\partial x})}^2 + V = 0\)
\(p = \frac{\partial S}{\partial x}\)

we have

\(\frac{\partial S}{\partial t} = -H = -(T+V)\)
\(dS = \frac{\partial S}{\partial x}dx +\frac{\partial S}{\partial t } dt = p dx - H dt = L dt\) - the Lagrangian and
\(S = \int_{t_0}^{t} L dt\) - the action 

With a usual decomposition: \(S = -Et + Q(x)\):
\(\frac{\partial S}{\partial t} = -E\)

and from the H-J equation and our decomposition

\(|\nabla S| = \sqrt{2m(E-V)}\) - which is the Eikonal equation.

Then Schrodinger considered surfaces of equal values for \(S\) to see how this would propagate. The normal on those surfaces is:

\(dn = \frac{dS}{\sqrt{2m(E-V)}}\)

and the velocity:

\(u = \frac{E}{\sqrt{2m(E-V)}}\)

Those surfaces almost respect a wave equation but to take de Broglie's idea seriously Schrodinger forced the solution to be a wave:

\(\psi = A e^{\frac{i S}{\hbar}} = A e^{\frac{-iEt}{\hbar} + \frac{iQ(x)}{\hbar}}\)

respecting a wave equation according to the geometric optics insight (thus modifying the Hamilton-Jacobi equation in the process):

\(\Delta \psi -\frac{\ddot \psi}{u^2} = 0\)

Because \(u^2 = \frac{E^2}{2m(E-V)}\), Schrodinger's equation is:

\((-\frac{\hbar^2}{2m} \Delta + V )\psi = E \psi\)

It is informative to see what we get if we directly plug the ansatz \(\psi = A e^{\frac{i S}{\hbar}}\) into the Hamilton-Jacobi equation. Simple calculations yields:

\((-\frac{\hbar^2}{2m \psi} \Delta + V )\psi = E \psi\)

Conversely, if we plug the ansatz into the Schrodinger equation we recover the Hamilton-Jacobi equation and an additional term proportional with \(\hbar\).

Schrodinger's equation and the Hamilton-Jacobi equations are cousins and they both are equations in the configuration space.

There are typical mistakes people make about them. For example working with a single particle, one gets the temptation to identify the particle with its wavefunction, This mistake is cured when considering the wavefunction of N particles where such identification is impossible because we are working in configuration space. Other mistake is "deriving" Schrodinger equation directly from Hamilton-Jacobi, like in this paper: If that were true, quantum mechanics would be nothing but classical mechanics. In this paper the mistake occurs in Eq. 4.5 because total and partial derivatives cannot be interchanged. Here is a simple counterexample: \(x(t) = e^t\)

\(\frac{\partial \frac{d x}{d t}}{\partial x} = \frac{\partial x}{\partial x} = 1\) but
\(\frac{d \frac{\partial x}{\partial x}}{dt} = \frac{d 1}{dt} = 0\)

The original triumph of Schrodinger's equation (derived from free particle considerations) was its application to the bound states of the atom. There the matrix mechanics of Heisenberg was developed to eliminate unobservable classical concepts like the classical path and later on Schrodinger proved the equivalency of the two seemingly different approaches.

But what is the link between Schrodinger's equation and the categorical formulation of quantum mechanics using the abstract products \(\alpha\) and \(\sigma\)? First we need to pass from the Schrodinger picture where the wavefunction evolves in time and the operators are stationary (recall that in Hamilton-Jacobi \(\dot P = \dot Q = 0\)) to the Heisenberg picture where the states are stationary and the operators change in time. Here the time evolution of the operators is given by the abstract product \(\alpha\):

\(\dot A = H\alpha A\)

The categorical formulation of quantum mechanics is a formulation in a Hilbert (and phase space when possible) in the Heisenberg picture. 

Thursday, September 3, 2015

Quantum harmonic oscillator in phase space

Continuing the physics posts, today I want to show how to solve a standard quantum mechanics problem: the harmonic oscillator, but do it in the unusual phase space formulation of quantum mechanics. It is not until we can solve concrete problems that we truly understand a new formalism. For today's result I will follow a pedagogical paper by Emile Grgin and Guido Sandri from 1996.

The Hamiltonian of the harmonic oscillator is obviously:

\(H=\frac{1}{2m} p^2 + \frac{m\omega^2}{2} x^2\)

and it is advantageous to work with dimensionless variables as follows:

\(p=\sqrt{\hbar \omega m} \eta\)
\(x = \sqrt{\hbar / \omega m} \xi\)
\(H = \hbar \omega \chi\)

If we recall the abstract products \(\alpha\) and \(\sigma\) in phase space:

\(\alpha = \frac{2}{\hbar} sin(\frac{\hbar}{2} \overleftrightarrow{\nabla})\)
\(\sigma = cos(\frac{\hbar}{2} \overleftrightarrow{\nabla}))\)


\(\overleftrightarrow{\nabla} = \frac{\overleftarrow{\partial}}{\partial \eta}\frac{\overrightarrow{\partial}}{\partial \xi} - \frac{\overleftarrow{\partial}}{\partial \xi}\frac{\overrightarrow{\partial}}{\partial \eta}\)
\(f\overleftrightarrow{\nabla}g = \{f,g\} \): the Poisson bracket

with \(\eta\) and \(\xi\) cannonical variables (dimensionless momenta and position) and the funny arrows are used as a reminder of how they act: on the left or on the right arguments.

In phase space if \(u\) is a state, the condition for a pure state is:

\(u=u\sigma u\)

the expectation value for an observable \(f\) and a state \(u\) is:

\(\langle f {\rangle}_{u} = \int f \sigma u ~dx dp\)

and the characteristic equation corresponding to the eigenvalue-eigenvector problem in standard Hilbert space formalism is:

\(f \sigma u_{\lambda} = \lambda u_{\lambda}\)

In the Hilbert space formulation, the time-independent Schrodinger equation in the dimensionless parameters reads:

\(\frac{1}{2}(\xi^2 - {\partial_{\xi}}^2) \psi = \lambda \psi\)

and the solution is well known:

\({|\psi|}^2 = \frac{1}{\sqrt{\pi}2^n n!} H_n^2 (\xi) e^{-\xi^2}\)
\(\lambda = n+1/2\)
with \(H\) the Hermite polynomials.

Now what do we get in the phase space formulation? The Hamiltonian is:

\(\chi = \frac{1}{2}(\xi^2 + \eta^2)\)

and we can compute \(\chi \alpha\) and \(\chi \sigma\):

\(\chi \alpha = \eta \partial_{\xi} - \xi \partial_{\eta}\)
\(\chi \sigma = \frac{1}{2}(\xi^2 - \hbar^2 / 4 \partial^2_{\xi}) + \frac{1}{2}(\eta^2 - \hbar^2 / 4 \partial^2_{\eta})\)

because higher powers of \(\overleftrightarrow{\nabla}\) vanish as they apply on the left side as well.

\(\chi \alpha \) is the same as in classical mechanics and \(\chi \sigma \) is a double copy of the Hamiltonian in the Hilbert space formalism!!! And this is the key to prove the equivalence between the two approaches. What you get from the characteristic equation is a sum of the products of the Hermite polynomials which by some Hermite polynomials identity magic reduces to the desired result. But those are mathematical details relevant only to test one's knowledge of functional analysis. The full mathematical details can be found here and here. We are after the conceptual ideas.

For the problem above the link between the phase space state \(u\) and the Hilbert state \(\psi\) is given by:

\({|\psi(\xi)|}^2 = \frac{1}{2\pi}\int_{-\infty}^{+\infty} u(\xi, \eta) ~d\eta\)

What I presented so far is the quantum mechanics formulation in phase space using the observable product \(\sigma\). But the products \(\sigma\) and \(\alpha\) can be combined to create an associative product \(\star = \sigma + J \frac{\hbar}{2}\alpha\) known as the Moyal, or the star product.

Jose Enrique Moyal
Recalling that \(u\sigma u = u\), similarly in the star product formulation the pure states \(F\) respect:

\((2 \pi \hbar) F\star F = F\)

and the eigenvalue-eigenvector problem is not surprisingly:

\(H\star F = \lambda F\)

For details please see here.

Now we can combine all this and past posts to introduce the high level overview of quantum mechanics in phase space and its relationship with quantum mechanics in Hilbert space.

Phase space:

observables = differentiable functions on phase space
generators = vector fields on phase space
1-to-1 map observable and generators: J with \(J^2 = -1\)

observable product: \(\sigma = cos(\frac{\hbar}{2} \overleftrightarrow{\nabla}) \)
generator product: \(\alpha = \frac{2}{\hbar} sin(\frac{\hbar}{2} \overleftrightarrow{\nabla})\)
associated product: \(\star = e^{J\frac{\hbar}{2}\overleftrightarrow{\nabla}}\)
state space (for the products \(\sigma\) or \(\star\)) = phase space

Hilbert space:

observables = hermitean operators
generators = anti-hermitean operators
1-to-1 map observable and generators: J with \(J^2 = -1\)

observable product: \(A\sigma B = \frac{1}{2}(AB + BA)\) - the Jordan product
generator product: \(A\alpha B = \frac{J}{\hbar}(AB-BA) = \frac{J}{\hbar} [A,B] \)
associated product: \(A\cdot B = A B\) regular operator multiplication
state space (for the products \(\sigma\) or \(\cdot\)) = Hilbert space

The states can be  defined for the observables product or for the associated product. In the first case one encounters the Jordan-GNS construction and in the second case the usual GNS construction.

One goes from the phase space formalism to the Hilbert space formalism by a quantization procedure, and the simplest one is Weyl quantization which directly constructs the operators in the Hilbert space. Many other quantization procedures are known.

The same thing is valid for classical mechanics with the main difference that \(J^2 = 0\)!!!! There the product \(\alpha\) is the Poisson bracket, and the product \(\sigma\) is regular function multiplication on phase space.

Deformation quantization is the process which transforms the products \(\alpha\) and \(\sigma\) from classical to quantum realizations. the dimension of the map \(J\) is \(\hbar\) and this observable to generator map (known as dynamic correspondence) corresponds to Noether's theorem.

So now what? How can we understand quantum mechanics? There are basically two camps, the ontic and the epistemic ones. If you are in the ontic camp you you may like the Bohmian interpretation where particles have a well define position at all times. But what prevents us to attach an ontic interpretation to the phase space formulation as well? After all phase space is well understood and classical mechanics does not have any interpretation issues. Sure, in the phase space formulation one encounters quasi-probabilities in the form of Wigner functions, but you have to compare their strangeness this with the strangeness of the quantum potential. Something has to be different than in the case of classical mechanics otherwise you don't get quantum effects.

If we attach an ontic meaning to quantum mechanics in phase space, now we have two distinct ontic interpretations. But an ontic interpretation must be unique, otherwise it cannot be taken seriously. 

The existence of the phase space formulation of quantum mechanics presents the greatest challenge to the Bohmian interpretation. This formalism has the same limitations as Bohmian quantum mechanics particularly in the treatment of spin which is a pure Hilbert space phenomena with no classical counterpart. Why should non-detectable violations of relativity (in the case of Bohmian quantum mechanics) be better than non-detectable violations of positive probabilities (in the case of quantum mechanics in phase space)?