Suppose we start out by having two entangled electrons. We separate them by some distance and we put one electron inside a thin loop of wire connected to an extremely sensitive voltage measuring device at lab 1 and the second electron at lab 2. At this point, no measurement is made. Both electron's spin's are undetermined.


-The direction of the spin is undetermined.

-We do not know if the electron's spin, $s=\pm\hbar$.

-We do not know its magnetic moment,$ m = (-gu_bS)/\hbar$. $S = \frac{h}{2\pi}\sqrt{s(s+1)}$, $g$ is g factor, $u_b$ is Bohr Magneton.

-We do not know the Magnetization, $M = (N/V)m$, where $N$ is number of magnetic moments and $V$ is the volume of the system in question.

-We do not know the magnetic field, $B = \mu_0(H + M)$, where $\mu_0$ is vacuum permeability, $H = M/X$, where $X$ is the magnetic susceptibility.

The magnetic field of the first electron at lab 1 is undetermined because no measurement has been made, therefore no magnetic field can possibly be present, $B = 0$. (Please correct if wrong) Now, we measure the second electron's spin by sending through a Stern-Gerlach device and having that electron hit a screen to record its spin value, $+$ or $-$, at lab 2.

Regardless of whether or not the second electron's spin is up or down, we know that the first electron's spin is now determined. This means that the magnetic field at lab 1 has been determined and therefore a magnetic field must be present. Since there is a change in magnetic field from 0 to some non-zero value, there must be a change in voltage from the law of induction, $V = -\frac{d}{dt}\left(BNA\cos\theta\right)$, where $A$ is the area the magnetic flux is going through, and $N$ is the number of coiled wire. This suggest that there is a measurable effect, although extremely small, at lab 1 due to the entanglement breaking in lab 2.

My question is, is this theoretically correct? If yes, then I would suggest this as a method of communication by the following.

By creating a large ensemble of these entangled electrons, A-A', B-B', C-C', D-D', where A is an electron at lab 1, entangled to a second electron, A', at lab 2, etc. For example, by choosing to measure A' and C' and leaving B' and D' alone at lab 2, we create a measurable effect at lab 1 for the electrons' A and C, voltage is changed. Thus, this would constitute a sent message as (1 0 1 0). Where 1 would be a voltage change and 0 would be no voltage change. This would of course be a one time messaging system, but it still does not negate the fact that it would be able to send a message via entanglement by this specific scheme. This is true only if my scheme is logically and theoretically correct.

I am open to scrutiny and correction. Please help me determine if my scheme is wrong. Thank you. :D

  • $\begingroup$ I have discussed this question with a graduate student and he told me that the act of putting the entangled electron in a loop will measure the spin of the electron. This would destroy the entire scheme if true. $\endgroup$ – QEntanglement Feb 29 '12 at 23:17
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    $\begingroup$ How is magnetic field undetermined equivalent to magnetic field zero? I would say that magnetic field zero is very determined. $\endgroup$ – Raskolnikov Feb 29 '12 at 23:23

1) Putting an electron in a loop would measure it's spin. You have to get it to the loop and that involves a change of flux

2) Your magnetic field went from $\frac{1}{\sqrt{2}}|B_\uparrow\rangle+\frac{1}{\sqrt{2}}|B_\downarrow\rangle \to |B_\uparrow\rangle$ Now I don't know how Maxwell's laws work on wavefunctions, but it definitely didn't start at 0.

In QM, when we say a state is in a superposition, it's not the classic sort of superposition. You do not vector-add the states (not classically-vector-add, after all, $|B_\uparrow\rangle$ itself is a vector, but it's orthogonal to $|B_\downarrow\rangle$). Rather, you say that it is in both states at the same time.

Picture a particle with x-coordinate +1, y-coordinate -1. It's net coordinate is not 0. There's no such thing as a net coordinate. It's net coordinate-thingy(you know what I mean) can be said to be both at the same time, which is true. So we write it as $(1,-1)$

Edit: I asked @DavidZaslavsky on chat how to apply Maxwell's laws to a superposition. His logic came from the many-worlds interpretation, and basically it turns out that you just apply the laws to each ket separately and then add them (according to their relative weightages). There may be some QFT involved, but this seems correct to both of us. Anyways, using this we get that $V=\frac{1}{\sqrt{2}}|d\phi_\uparrow/dt\rangle+\frac{1}{\sqrt{2}}|d\phi_\downarrow/dt\rangle$. So again, your potential is also a superposition. Tying to measure it will collapse it. So, having a wire loop around the electron with a voltmeter measures the potential and prematurely collapses the wavefunction of B. So we get point (1) from point (2).


The flaw in your question lies in assuming that induction occurs when the entangled electron is remotely detected.

The rule for entanglement is extremely simple: If information that determines the state of the system exists anywhere in the universe, the system is classical, not quantum, at least for the state in question. Your induction premise assumes that the existence of the electron's field was detectable in some fashion prior to the detection of its entangled partner. But simply asserting that makes the system non-quantum, and thus no longer entangled with the other particle.

The no-information-anywhere rule is very powerful, incidentally. It can in fact be used as a definition of whether a system will interfere -- that is, whether it will behave under the rules of quantum mechanics.

  • $\begingroup$ Why would induction not occur? Why isn't there a change in voltage? $\endgroup$ – QEntanglement Feb 29 '12 at 19:08

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