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The Copehagen interpretation of QM explains the Stern–Gerlach experiment by asserting that a particle is in a superposition of states and doesn't have a definite spin until measured. However, the de Broglie–Bohm interpretation suggests that there is always a definite position of the particle, with a well-defined initial state (which may not be measurable with enough precision), including its spin. This initial state defines the path the particle will take through the apparatus.

How does one explain the results that the spin of a particle is always measured with its full magnitude along the plane of measurement? I would guess it has something to do with the pilot-wave's effects on the particle as its angular momentum is changed by the magnetic field.

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    $\begingroup$ Briefly, even in the S-G experiment, one is measuring the position of the particle, not the spin. On the BM view, spin is not a property of the particle; the S-G apparatus interaction splits the spinor valued wave function (packet) into two separated packets one of which is empty. See, for example: plato.stanford.edu/entries/qm-bohm/#sp $\endgroup$ – Alfred Centauri Jan 24 '15 at 12:36
  • $\begingroup$ Very interesting link. So spin is a property of the wave function, and not of the particle? $\endgroup$ – B T Jan 24 '15 at 18:47
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How does one explain the results that the spin of a particle is always measured with its full magnitude along the plane of measurement?

In Bohmian mechanics, the spinor valued wave function (packet), under the influence of the apparatus, splits into two discrete packets only one of which guides the Bohmian particle; the other packet is 'empty'.

Which packet the particle ends up in is determined by the initial configuration.

On this view, spin is not a property of the particle but wavefunction

From the article "Bohmian Mechanics" at SEP

Bohmian mechanics makes sense for particles with spin, i.e., for particles whose wave functions are spinor-valued. This occurs because the Stern-Gerlach magnets are so designed and oriented that a wave packet (a localized wave function with reasonably well defined velocity) directed towards the magnet will, by virtue of the Schrödinger evolution, separate into distinct packets — corresponding to the spin components of the wave function and moving in the discrete set of directions. The particle itself, depending upon its initial position, ends up in one of the packets moving in one of the directions.

The probability distribution for the result of such a Stern-Gerlach experiment can be conveniently expressed in terms of the quantum mechanical spin operators — for a spin-1/2 particle given by certain 2 by 2 matrices called the Pauli spin matrices — in the manner alluded to above. From a Bohmian perspective there is no hint of paradox in any of this — unless we assume that the spin operators correspond to genuine properties of the particles.


Very interesting link. So spin is a property of the wave function, and not of the particle?

From the paper "What you always wanted to know about Bohmian mechanics but were afraid to ask"

Thus, in general, 'measurements' do not measure anything in the closer meaning of the term. The only exception being of course position measurements, and, in some sense momentum-measurements. The latter do indeed measure the asymptotic (Bohmian) velocities. Hence, the only properties of a Bohmian particle are its position and its velocity.

Just as $\psi$ is no classical field, the Bohmian particles are no classical particles, i.e. they are no bearers of properties other than position. Therefore a physical object like e.g. an electron should not be confused with the Bohmian particle at position $Q_i$ . It is represented by the pair $(\psi, Q_i )$.

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  • $\begingroup$ Hmm, if a bohmian particle only has position, how does it interact with other bohmian particles to transfer momentum? Surely the bohmian particle itself has momentum as well as position, right? Which would mean that properties of the particle are at least mass, position, and velocity. $\endgroup$ – B T Jan 24 '15 at 19:36
  • $\begingroup$ @BT, not on the Bohmian view. Take a look at the 2nd link in my answer (which I've fixed). If it still isn't clear, why not ask this as another question? There's too much to say about it in a comment and it's reasonable follow-up to this question. I haven't checked to see if it's been asked here before. $\endgroup$ – Alfred Centauri Jan 24 '15 at 22:20
  • $\begingroup$ Hokay: physics.stackexchange.com/questions/161303/… $\endgroup$ – B T Jan 24 '15 at 23:32
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I am not supposed to place an answer here, as I am no more active in this site: however, a comment doesn't offer enough place.

So, you ask:

1) "spin is a property of the wave function, and not of the particle?"

Please pay attention to the following differences between the standard quantum theory (SQT) and the Bohmian interpretation (BI):

SQT doesn't work with the concept of particle, but with the concept of wave-function. When we send the wave-packet onto a detector and we get a click, a recording, we say sometimes "we detected a particle". This is not a rigorous expression from the point of view of SQT, we should say that "we got a detection". There is another concept valid from the point of view of the SQT, "type-of-particle", which comprises rest-mass, spin, charge, and others. I'll elaborate more, in continuation.

To the difference, BI distinguishes between two concepts: the guide-wave and the particle.

Now, some clarifications. By solving the Schrodinger eq. we get a wave-packet. SQT says that all the properties of the type-of-particle are displayed at every point in the wave-packet. This is all that the SQT gives us. (Of course, when we measure some observable, position, linear momentum, spin-projection on some axis, i.e. when we do the classical measurement (i.e. with the macroscopic detector), there come decoherence and collapse, but our talk here is not on this.)

BI also recognizes the type-of-particle properties, and they are also carried by the wave-function which is considered a sort of guiding wave. To your question, spin is considered to be carried by the wave-function. When you split the wave-function according to the spin-projection, each separated wave-packet carries one of the values of the spin-projection.

What then is the Bohmian particle if all the type-of-particle properties are carried by the wave-function? Different followers of Bohm say different things. Some of them say that the Bohmian particle is a special point inside the wave-function, with the property that makes a detector click. When we split the wave-function into space-separated branches, only one of them is supposed to take the so-called particle. However, Basile Hiley (a friend of Bohm that collaborated with him in some articles), told me that he sees the particle rather as some sort of structure.

Anyway, this point, or structure, etc., is supposed to have at any time a position, and a Bohmian velocity. These are all the properties that the Bohmian particle is supposed to possess. The Bohmian velocity multiplied by the mass of the type-of-particle gives a Bohmian linear momentum. If the wave-guide displays a well-defined linear momentum, then the two linear momenta coincide. But if the wave-guide doesn't have a well defined linear momentum, the Bohmian linear momentum is not equal to the average linear momentum displayed by the wave-guide.

I also saw your question

2) "if a bohmian particle only has position, how does it interact with other bohmian particles to transfer momentum? Surely the bohmian particle itself has momentum as well as position, right? Which would mean that properties of the particle are at least mass, position, and velocity."

NO!!! It's not the Bohmian particle that interacts with another Bohmian particle. The wave-guides interact, exactly as requires the Schrodinger equation. We get a two-body wave-function - wave-guide - that guides the two-particles. And the two Bohmian particles have, each one, at any time, position and Bohmian velocity.

FINALLY to your question

3) "Answers to my question here seem to say that the imaginary position of the bohmian particle actually has nothing to do with "where" the detector detects the particle."

That's exactly opposite. Exactly for this purpose was built the BI. It was proposed as a tentative to get rid of the collapse, i.e. that part of the wave-function that comprises the Bohmian particle, that part makes the detector to do a recording. See my explanations above. The detector is supposed to produce a recording iff the Bohmian particle passes through the detector, and at the time when it passes.

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  • $\begingroup$ Thanks for answering my questions, this is very helpful. So I do understand that its not the bohmian particle that directly interacts with the detector (or more precisely, th wave functions of the bohmian particles that comprise the detector). But surely the bohmian particle interacts with the detector indirectly, right? So What I meant to ask is, how does momentum get transferred from one bohmian particle to another? Does the bohmian particle's position/velocity affect its wave-function, which then in turn interacts with the wave functin of the detector? $\endgroup$ – B T Apr 24 '15 at 20:55
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    $\begingroup$ @BT The supporters of the BI are not very clear about the mechanism of impressing a detector. They only say that the detector is impressed if and only if the Bohmian particle impinges on it. Of course it seems inconsistent to claim that the physical properties, mass, charge, etc., are carried by the guide-wave (at every point and point of it), however the detector makes a recording only at the contact with the Bohmian particle, that possesses only position and Bohmian velocity. About transfer of momenta, why don't you read attentively my answer? (see the next comment) $\endgroup$ – Sofia Apr 25 '15 at 10:58
  • $\begingroup$ @BT Please read attentively my answer and comments, I told you, my time is very short. When two quantum objects interact, you write the Schrodinger equation. See what I answered to your question 2). $\endgroup$ – Sofia Apr 25 '15 at 11:07
  • $\begingroup$ @BT About the inconsistency mentioned in my comment, I had a talk with prof. D. Durr, an expert in BI, but what he answered me was that this is what the equations say. I wasn't pleased by this answer, I see this as a conceptual problem in BI. Still, I won't refute BI for this weakness, as the concept of collapse in the SQT is also very unpleasant. $\endgroup$ – Sofia Apr 25 '15 at 11:17
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GENERALITIES : Let's precise some issue: the Bohm interpretation (BI) suggests that particles have from the beginning, i.e. even in the initial wave-function, not only positions but also velocities. The mathematics of the BI, in essence, predicts trajectories. In these calculi, the quantum potential tells through which regions the particle may pass, and through which it may not.

The way BI works, is first of all to solve the Schrodinger equation and find the wave-function. From the wave-function, BI gives us the velocity of a particle for any pair (position, time). Thus, given an initial position, we can calculate the full trajectory. So is that in the Stern-Gerlach experiment, if we start with a particle in a certain initial position, we can calculate if it will end in the spin-up beam, or in the spin-down beam.

REFERENCES :

"A Suggested Interpretation of the Quantum Theory in Terms of "Hidden" Variables. I", D. Bohm, Physical Review, vol. 85, no. 2, January 15, 166, (1952),

"A Suggested Interpretation of the Quantum Theory in Terms of "Hidden" Variables. I", D. Bohm, Physical Review, vol. 85, no. 2, January 15, 180, (1952).

For your questions suits the part I, and you can find the calculi for trajectories in the section 4.

SPECIFICALLY : for the Stern-Gerlach experiment, one writes the Hamiltonian which includes the spin-magnetic field interaction. Then one solves the Schrodinger equation. Once the wave-function is found, it is put in the form

$\psi(\vec r, t, \vec {\sigma}) = R(\vec r, t, \vec {\sigma}) e^{iS(\vec r, t, \vec {\sigma})/ \hbar}, \ \ \ |R(\vec r, t, \vec {\sigma})|^2 = P(\vec r, t, \vec {\sigma}) $ .

The velocity of a particle at any point is

$v^{Bohm} (\vec r, t, \vec {\sigma}) = \frac {\nabla S(\vec r, t, \vec {\sigma})}{m} $.

Thus, for a particle that begins at $t_0$ in a position $\vec r_0$, and in a spin state $\vec {\sigma}$, we can find $v_0$, and as I explained above, we then can calculate all the trajectory. This answers also your last question, why "the spin of a particle is always measured with its full magnitude along the plane of measurement?" Because this is what the wave-functions gives us: for a spin 1/2 fermion it gives us two possible beams, or, if the spin is polarized in the direction of the field, one beam.

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  • $\begingroup$ Most of what you're saying is that "you calculate the wave function, and that's what you get". I understand that, but I'm looking for more of an intuitive explanation of why the equation comes out that way. How is spin interpreted? Is it a property of the pilot wave or the particle? $\endgroup$ – B T Jan 24 '15 at 18:40
  • $\begingroup$ You ask me more than it's written in Bohm's article. The hidden variables of Bohm are positions and velocities, not spin. I saw that you ask why in Bohm's interpretation the spin of a particle is "always measured with its full magnitude along the plane of measurement". Bohm's interpretation didn't get into the internal dynamics of elementary particles. Look also what says D. Durr, in his book "Bohmian Mechanics": "Spin is indeed a truly quantum mechanical attribute." $\endgroup$ – Sofia Jan 24 '15 at 20:31
  • $\begingroup$ Well hey, I can read the bohm article myself. I'm looking for new info here. But sounds like its a property of the wave function. $\endgroup$ – B T Jan 24 '15 at 21:40
  • $\begingroup$ @BT : what you mean by new info? What said Bohm's followers? $\endgroup$ – Sofia Jan 24 '15 at 22:00

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