In many textbooks (J.J. Sakurai for instance) S-G experiment is explained through the idea of a collapse of a state of a system. So here, in this discrete case, we can see quantum behavior more clearly than with wave functions. We can interpret results of the experiment by saying that spin state of a particle is not defined, particle is in spin1 and spin2 state, in a superposition of two states, and its state collapses into one of these two when measured ( meaning when passing through magnetic field). My question is this: What other interpretations can anyone offer, and why would anyone invoke consciousness when it is clear that here magnetic field plays a role and not observer? How would other interpretations of QM interpret this? Bohms or any other?

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    $\begingroup$ Do these textbooks explain what is meant by "collapse", other than it being a fancy name for the Born rule? Serious physics textbooks never talk about consciousness. It's not a measurable quantity and therefor beyond the reach of the scientific method. $\endgroup$
    – CuriousOne
    Jan 4, 2015 at 10:06
  • $\begingroup$ If you think of electron or a system of any kind as being in a superposition of some eigenstates then collapse would be decision to take on a definite eigenstate. Or something like that. $\endgroup$ Jan 4, 2015 at 16:26

1 Answer 1


There are a couple of mistakes in what you say. It may ne that Sakurai is not so clear. This is a book mentioned by other people too.

So, let's fix the things.

1) You say "in this discrete case, we can see quantum behavior more clearly than with wave functions." No, wave-function is not something opposite to the continuous case. Both in the discrete and in the continuous case, the behavior of these microscopic particles is described by a wave-function. The description is good up to the measurement by a macroscopic apparatus. The macroscopic apparatus is not the magnetic field, but the detector. I will return on these things below.

2) Next, you say "We can interpret results of the experiment by saying that spin state of a particle is not defined, particle is in spin1 and spin2 state, in a superposition of two states, ..." Let me use better the standard expressions: the wave-function of a fermion with spin $\hbar /2$ is a superposition of two states of the projection of the spin on the direction of the magnetic field,

(1) $|\psi> = \frac {|\uparrow> + |\downarrow>}{\sqrt{2}}$.

When passing between the two magnetic poles, the non-homogeneity of the field sends particles with spin-up upwards, and particles with spin-down, downwards. But in our case we have one single particle. Then, what happens? The component $|\uparrow>$ of the wave-function goes up, and the component $|\downarrow>$ goes down.

3) Now you say "its state collapses into one of these two when measured (meaning when passing through magnetic field)."

No, as long as the wave-function doesn't meet the macroscopic detector, we have no collapse. A macroscopic apparatus, e.g. detector, is an apparatus in which the wave-function comes into contact with many particles. The joint wave-function of both, studied particle and apparatus' particles, becomes very complicated,

(2) $|\Psi> = \frac {1}{\sqrt{2}}(|\uparrow>|D_1^*> |D_2> + |\downarrow>|D_1> |D_2^*>)$.

Here, $D_1>$ and $D_2>$ denote the wave-functions of all the particles in the detector $D_1$ respectively $D_2$, and the asterisk means that the state of the detector is excited, i.e. the studied particle is inside, and cause excitation of atoms, electron emissions, whatever happens when a detector detects a particle. You also see that if one detector is excited, the other isn't.

Well, in the end we have the collapse, i.e. only one of the detectors make a recording. How so? There are all sort of explanations, you meet words like "decoherence", and others, but none is satisfactory.

4) Let me return on one thing. As long as the detectors weren't met (or other macroscopic apparatuses), we have the wave-function (1). This wave-function can be reversed, i.e. we can place mirrors on both the upward path and downward path, and from the magnet the wave will get out as it entered, one single beam. But the wave-function (2) we cannot undo. Even more, after one detector records a detection, we cannot restore the wave-function (1). The "collapse" cannot be undone, it is irreversible, and even the wave-function (2), which means not yet collapse, cannot be undone.

5) Now, other interpretations. There are many, but I know two, which are very elaborated, very serious, though none of them is flawless.

  • Bohm's interpretation assumes that from the two waves that exit the magnet, one upwards and one downwards, only one "contains the particle", the other wave is just a field, but is devoid of a particle. So Bohm explains why only one of the detectors makes a recording. Unfortunately, Bohm's interpretation has problems, one of them being that it is unable to deal with photons.

  • The GRW (Ghirardi-Rimini-Weber) interpretation, which makes changes in the wave-function for explaining the collapse. Also, there are problems with this.

  • $\begingroup$ I wont be precise here but I believe that you will know what am I talking about. When you take a beam of spin 1/2 particles, and you let them pass through the S-G aparatus which measures sz projection, you get about 50 percent up and 50 percent down. When you choose up particles and let them pass the same aparatus again, you get 100 percent up. But, take them through sx aparatus, and you will get 50 percent one direction and 50 other. Take the same particles again, let them through the first aparatus, and you will get 50 percent up and 50 percent down, so they forgot all about being in up stat $\endgroup$ Jan 4, 2015 at 16:20
  • $\begingroup$ and about discreteness, i meant you have discrete and continuous quantum numbers. $\endgroup$ Jan 4, 2015 at 16:23
  • $\begingroup$ 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: physics.stackexchange.com/questions/161163/… . Is that not correct? $\endgroup$
    – B T
    Apr 23, 2015 at 0:39
  • $\begingroup$ Also, how is Bohm's pilot wave interpretation not able to deal with photons? $\endgroup$
    – B T
    Apr 23, 2015 at 0:40
  • $\begingroup$ @BT The Bohmian velocity contains the factor $\hbar / m$ where $m$ is the rest-mass of the particle. The photon has no rest-mass. I had one or two years ago a talk with Roherich Tumulka, an expert in the Bohmian Mechanics (BM). I asked him whether we cannot take as mass of the photon the value $\hbar \omega /c^2$. Tumulka replied that there are problems with this. Additional variants have been tried, but, as far as I know from Prof. Detlef Dürr, who is a great expert in the BM, they have problems with extending BM to the electromagnetic field. $\endgroup$
    – Sofia
    Apr 24, 2015 at 1:22

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