9
$\begingroup$

In his book on Quantum Mechanics, P.A.M. Dirac talks about the stability of the atom as a means of demonstrating the need for quantum mechanics. He writes:

The necessity for a departure from classical mechanics is clearly shown by experimental results. In the first place the forces known in classical electrodynamics are inadequate for the explanation of the remarkable stability of atoms and molecules, which is necessary in order that materials may have any definite physical and chemical properties at all. The introduction of new hypothetical forces will not save the situation, since there exist general principles of classical mechanics, holding for all kinds of forces, leading to results in direct disagreement with observation. For example, if an atomic system has its equilibrium disturbed in any way and is then left alone, it will be set in oscillation and the oscillations will get impressed on the surrounding electromagnetic field, so that their frequencies may be observed with a spectroscope. Now whatever the laws of force governing the equilibrium, one would expect to be able to include the various frequencies in a scheme comprising certain fundamental frequencies and their harmonics. This is not observed to be the case. Instead, there is observed a new and unexpected connexion between the frequencies, called Ritz's Combination Law of Spectroscopy, according to which all the frequencies can be expressed as differences between certain terms, the number of terms being much less than the number of frequencies. This law is quite unintelligible from the classical standpoint.

Since de-Broglie Bohm theory essentially introduces a quantum force to remedy classical mechanics, does Dirac's argument stand in contradiction to Bohm's theory?

$\endgroup$
  • 9
    $\begingroup$ You will find that there is rather less than universal acceptance of de Broglie Bohm theory. It has it's adherents (some of whom are quite adamant that it is obviously better than other interpretations), but also its detractors (who are often equal are sure that it is the least likely interpretation and was clearly dreamed up under the influence of strong drugs). There are also many, many people who simply don't care about the interpretational battles unless and until there is a discriminating experiment proposed. $\endgroup$ – dmckee Oct 20 '15 at 3:28
9
$\begingroup$

Dirac does not intend classical to mean non-quantum mechanics; he intends classical to mean pre-quantum mechanics1. So no, this says nothing about de Broglie–Bohm theory.

Dirac opens his paragraph with

The necessity for a departure from classical mechanics is clearly shown by experimental results.

Dirac is not talking about any theory that one might call "classical" today. He is talking about the theory of classical mechanics that existed before quantum mechanics was formulated. If you read his paragraph, all his criticisms are directed against that particular theory. For example, he says

the forces known in classical electrodynamics

and

if an atomic system has its equilibrium disturbed in any way and is then left alone, it will be set in oscillation and the oscillations will get impressed on the surrounding electromagnetic field.

These properties are specific to pre-quantum classical mechanics. His argument thus says nothing about de Broglie–Bohm theory. In fact, if I suspect that if you had asked Dirac, he would have said that de Broglie–Bohm theory was not classical mechanics.

EDIT: I want to address Lubos's comments, because some people seem to be agreeing with him and downvoting me. Further, I was asked in the comments to explain why Dirac's criticism doesn't apply to Bohmian mechanics. If you think Lubos is a genius and take everything he says uncritically, go ahead and downvote. If you actually can reason critically and think indepedently, please read my rebuttal to Lubos's comments before downvoting.

Lubos is wrong about almost everything he says about Bohm's theory in his comments. Lubos claims that the pilot wave in Bohm's theory is observable. The pilot wave in Bohm's theory is exactly the same as the Schroedinger wave function, so if the wave function in quantum mechanics is unobservable, so is the pilot wave in Bohmian mechanics. Furthermore, Bohm's theory has exactly the same predictions as quantum mechanics, so it is clear that in Bohm's theory, a system will not "be set in oscillation and the oscillations will get impressed on the surrounding electromagnetic field," simply because it isn't in standard quantum mechanics.

The only way I can see to save Lubos' statements is to realize that Bohm's theory doesn't actually work relativistically2, and then to reason that because photons are intrinsicially relativistic, photons don't actually work in it. But even though Dirac was an incredibly smart man, it is completely ridiculous to claim that

  1. Dirac intended his argument to work for all non-quantum theories and
  2. he foresaw Bohm's theory and was clever enough to realize that his argument would be saved by the fact that Bohm's theory doesn't work relativistically.

Dirac was smart enough to know that you can't prove a theorem like this (i.e., no theory except quantum mechanics can explain observation) without stating the hypotheses. His hypotheses, which I am sure he thought he had clearly stated, were that physics behaves along the lines of pre-1900 physics ... e.g., the dictionary definition of classical as traditional in style and form. In 1930, I don't believe anybody was using "classical physics" to mean "non-quantum". (I'll retract this statement if you can find an example of "classical" clearly meaning "non-quantum" rather than "pre-quantum" before 1940.) They were using "classical" to mean "pre-1900 physics", i.e., physics before quantum mechanics and relativity came along.

1 A standard definition of classical is: traditional in style or form.

2 Bohm's theory's adherents may claim otherwise, but I don't believe anybody has demonstrated that one can have a Bohmian theory with Lorentz-invariant particle trajectories and general relativity, which means that relativistically, the theory has severe drawbacks.

$\endgroup$
  • 3
    $\begingroup$ No, "classical" and "pre-quantum" are exactly synonymous - and not an epsilon has changed about the meaning of these adjectives since the publication of Dirac's book. Dirac's argument is a meritocratic argument that only depends on the intrinsic characteristics of the theory and not on the year in which someone proposed a theory or other sociological irrelevancies. When one goes through the argument, it is spectacularly obvious that it applies to Bohm's theory as well because it's a classical theory according to the relevant, and nearly rigorous, terminology he was very careful to follow. $\endgroup$ – Luboš Motl May 18 '16 at 8:57
  • 1
    $\begingroup$ Of course it does apply to any classical theory including Bohmian mechanics (Bell's was a typo of yours, wasn't it?). It follows from the definition of an equilibrium. Any object in a classical theory including Bohmian mechanics has some pure states that are represented in the equilibrium. And it also has arbitrarily weak deformations of this state. The response of the electromagnetic field is a continuous function of the phase coordinates of the object, so a small deformation will unavoidably lead to weak imprints in the electromagnetic field. $\endgroup$ – Luboš Motl May 18 '16 at 12:39
  • 1
    $\begingroup$ If the state is stable, it follows that the first derivative of the energy with respect to the coordinate is zero, and the second derivative is positive - it's a minimum, not a maximum. A maximum would imply an instability. With the positive second derivative, the degrees of freedom oscillate just like a harmonic oscillator. No way to avoid it. In any classical theory, weak perturbations translate to weak imprints in the radiation, just like he says it. $\endgroup$ – Luboš Motl May 18 '16 at 12:41
  • 1
    $\begingroup$ The only reason why Bohmians don't actively realize that this is obviously the case for a Bohmian theory as well is that they have never even dared to think about how an object at equilibrium could be described by the theory of the Bohmian type. They know that nothing they have discussed is powerful enough to do such things. But if a well-defined theory without the uncertainty principle is written down, of course it can't avoid the general facts of classical physics and Dirac's treatment, and the same conclusions follow. They may be inconvenient but they are unavoidable. $\endgroup$ – Luboš Motl May 18 '16 at 12:42
  • 2
    $\begingroup$ Bohmian mechanics is exactly the type of theories that Dirac immediately excludes at the first page of his book The Principles of Quantum Mechanics. Have you read at least the first page of the book? It doesn't look so. He explicitly discusses why new types of forces won't help in any way - this was exactly meant to deal with forces such as the force from the pilot wave. Just read it, you will see the argument why these theories can't work. $\endgroup$ – Luboš Motl May 18 '16 at 12:45
5
$\begingroup$

Yes, absolutely, Dirac's argument shows that one could never construct a complete theory – which specifies the rules for evolution as well as predictions for the measurement and what happens after the measurement – that would be compatible with the basic facts about the atoms.

This no-go theorem applies to Bohmian mechanics because it is just another classical theory, indeed. Bohmian mechanics is a classical theory composed of a classical field $\psi$, the pilot wave, which is said to obey the same equations as the wave function in quantum mechanics but has a completely different interpretation, plus some extra beables – typically classical particle positions $x_i(t)$. These positions are influenced by the pilot wave but the opposite reaction doesn't exist.

Like any normal enough classical system, the configuration space of Bohmian mechanics is continuous so one may always divide a small perturbation by two and what this perturbation evolves out of this perturbation is the original outcome divided by two, too.

That's in direct contradiction with the facts such as the discrete spectra of atoms etc. (The energy eigenstates of atoms may be present mathematically but Bohmian mechanics won't allow any mechanism that could imprint the spectra to the electromagnetic radiation of emitted light.) The dynamical equation (i.e. Schrödinger's equation in particular) in quantum mechanics is continuous but the meaning of the wave function is probabilistic so a very tiny perturbation of $\psi$ means a very small probability of a new (but finitely strong) new process, not a guaranteed presence of a very weak process. This is a key different of quantum mechanics from any classical theory, including Bohmian mechanics.

Dirac's textbook starts with several other arguments that instantly rule out classical theories including Bohmian mechanics, especially the argument about the observer low heat capacity of all atoms. If Bohmian theory could be extended to a full theory of interacting atoms, they would be described by a huge phase space because they have many classical plus new degrees of freedom. If the thermal equilibrium were possible in such a theory, the entropy of an atom (the logarithm of the volume of the phase space of states that are accessible at the thermal equilibrium) would be huge due to these extra degrees of freedom, in a direct conflict with experiments (which shows that the atom's heat capacity is always comparable to $k_B$).

Quantum mechanics allows these correct predictions of the low heat capacity because the state of the bound states is basically unique or has low degeneracy when the energy is required to be close to the ground state energy. Small perturbations of the ground state don't refer to mutually exclusive states. Instead, quantum mechanics says that a state that is mutually exclusive with the ground state – so that it could contribute to the entropy – has to be orthogonal to the ground state, i.e. very different. This is equivalent to the usual "quantization of the phase space" that is effectively divided to "cells". That's how quantum mechanics manages to produce a "small number of cells" which is needed e.g. for the low heat capacities.

Advocates of Bohmian mechanics never discuss any of these elemenentary, but still advanced relatively to the "Bohmian demos" issues – thermal equilibrium of objects, heat capacity, but also propagation of photons and other bosons, emission of sharp spectral lines, the observed collapse of the atoms to energy eigenstates once the energy of the photon is measured, the existence of fermionic quantum fields, and so on. And the main reason is that all these important parts of physics as we know it are irreconcilable with Bohmian mechanics – and every other "realist" i.e. classical theory.

$\endgroup$
  • 4
    $\begingroup$ You repeatedly write that Bohmians can't actually do any calculations. Then you write a bunch of qualitative statements that Bohmian mechanics gives the wrong heat capacities, but you yourself perform no calculations whatsoever. If Bohm's theory gives the wrong heat capacities, I personally challenge you to calculate the heat capacities in Bohm's theory, and show that they differ from what you get when you calculate them in Quantum mechanics. $\endgroup$ – user7348 May 16 '16 at 15:02
  • 4
    $\begingroup$ It is easy to prove - and I (and before me, Dirac) have proven - that you get huge capacities well exceeding $O(R)$ of quantum mechanics (confirmed by experiments) in any theory based on the classical including Bohmian paradigm. Bohmians probably don't want to see this proof - and they surely never write it or publish it themselves - but that doesn't make the proof go away. When Dirac was teaching quantum mechanics, all students knew these basic things after the first lecture. $\endgroup$ – Luboš Motl May 16 '16 at 17:18
  • 3
    $\begingroup$ The entropy of any classical i.e. realist system is the logarithm of the volume of the phase space of states that macroscopically behave in the same way. Bohmian theory includes the phase space of the wave function misinterpreted as a classical field which already gives huge, infinite values, and on top of that, some extra beables such as particle positions. It's clear that the equilibrium will produce entropy that is infinite and even when any regularization/truncation is added, parameterically higher than $O(R)$. $\endgroup$ – Luboš Motl May 16 '16 at 17:22
  • 3
    $\begingroup$ As I reminded you in the other thread, this choice of fields as beables instantly falsifies Bohm's theory because the point-wise detection of photons may be experimentally measured. A theory only has two choices: the soon-to-be-detected location of the photon may already be calculated from the existing degrees of freedom right before the measurement (which would be the case of Bohm's theory if positions were beables, but you admitted it can't work), or the outcome is produced randomly and then we deal with proper (Copenhagen) quantum mechanics. There's no third option aside from Yes and No. $\endgroup$ – Luboš Motl May 17 '16 at 4:28
  • 3
    $\begingroup$ One can't perform a precise calculation in a non-existing theory. Your request is analogous to the request to calculate the proton-to-electron mass ratio from Genesis. There exists no formulation of Bohmian mechanics where the full configuration space would be known and that would admit thermal equilibrium at all. One can only perform the calculations with the precision at most equal to the precision with which the theory is formulated. $\endgroup$ – Luboš Motl May 17 '16 at 4:37
3
$\begingroup$

Since de-Broglie Bohm theory essentially introduces a quantum force to remedy classical mechanics,

That isn't what dBB theory does. It introduces a state-dependent force. Which means it introduces states (waves) and then the wave exerts a quantum force.

does Dirac's argument stand in contradiction to Bohm's theory?

No. And not for the reasons above. The dBB theory is for non relativistic quantum mechanics. And that means it doesn't really address electromagnetism the same way the Schrödinger equation doesn't address electromagnetism.

When you solve for, say the energy levels of hydrogen, you don't use electromagnetic fields you just use a scalar potential. Even when you solve it as a two particle problem (so a wave in a 6d configuration space) you still just write a scalar potential like $$V(x_e,y_e,z_e,x_p,y_p,z_p)=\frac{-k^2e^2}{\sqrt{(x_e-x_p)^2+(y_e-y_p)^2+(z_e-z_p)^2}}$$ which is just the electrostatic potential. No magnetism for instance, and no electrodynamics. If you have an external classical potential you can put that in, but this isn't the full classical electromagnetism in particular you aren't assigning a state to the electromagnetic field, one a state to the particles.

So dBB is really incomplete, it doesn't have a full proper QFT version.

But all Dirac was saying is that a force doesn't change the nature that a disequilibrium in classical physics imprints into the wider world. But in dBB you essentially have a radical departure from classical physics in that in addition to a configuration you have a state, a wave. And the state dependent wave acts as a pilot wave exert state dependant forces. Which allows you to have different states exert different forces.

That's the change from classical physics and since dBB does it, dBB isn't a classical theory and Dirac is right, you have to do something different to get quantum mechanics.

In you wanted to fully explore how Dirac's comments apply to dBB you should look a deviation from equilibrium, i.e. a state that isn't an energy eigenstate and see how it evolves and look for an imprint. And the naive nonrelativistic Schrödinger equation might have each energy eigenstate evolve just a phase and so by linearity the superposition evolve just as the sum of each.

But when you throw in an external electromagnetic potential you get, for instance, transitions up from some fields and transitions down from other fields. It doesn't just evolve to permanently be in that superposition. And that's because QM becomes QFT as you include the vacuum fields and other fields in a way where those fields themselves can change.

But the dBB theory just doesn't have that part of quantum mechanics because it doesn't have a QFT version. So it really just doesn't apply to the exact situation Dirac was discussing.

$\endgroup$

protected by Qmechanic May 18 '16 at 21:00

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

Not the answer you're looking for? Browse other questions tagged or ask your own question.