Consider a device consisting of a Stern-Gerlach apparatus (minus the detector screen), Hadamard gate, double-slit, and screen in sequence, tuned to direct electrons with spin up along a predetermined axis towards one slit, and those with spin down towards the other. Without the Hadamard gate, I am sure that an ensemble of electron in a superposition of spin-up and spin-down states would not generate an interference pattern on the screen, because in the total wavefunction $\psi_0(x) \left| x \right> \left| 0 \right> + \psi_1(x) \left| x \right> \left| 1 \right> $, where $\psi_0$ and $\psi_1$ encode the alterered trajectories from the Stern-Gerlach apparatus, we can't simply drop the spin state, and $\psi_0$ and $\psi_1$ will not interfere. However, with the gate inserted, the wavefunction before passing through the double-slit becomes

$$\frac{1}{\sqrt{2}} \left( \psi_0(x) + \psi_1(x) \right) \left| x \right> \left| 0 \right> + \frac{1}{\sqrt{2}} \left( \psi_0(x) - \psi_1(x) \right) \left| x \right> \left| 1 \right> $$

and this seems to open up the possibility of an interference pattern being generated on the screen.

Can the different components of the electron's wavefunction indeed interfere with each other in this manner after passing through the SG apparatus and Hadamard gate, with appropriate design and tuning of the whole system? I can imagine one reason not might be that the interaction with the external magnetic field is perhaps enough to decohere the original state and environment, but I don't see why this has to be true in principle.

Assuming yes, consider a pair of electrons prepared in an EPR state, or rather an ensemble of such pairs prepared one at a time. One member of the pair is moved to Bob's lab, where Bob will measure its spin along the $\hat z$ axis. Meanwhile, the latter is moved to Alice's lab, where Alice will measure its position on a screen after passing it through the aforementioned device.

What should we expect to happen in different scenarios? Edit: And is this experiment essentially a variant of the quantum eraser, that I might have half-remembered and garbled?

  • Scenario 1: Bob measures the electron's spin before (in the causal past of) Alice's measurement. I'd think that then, Alice can be sure that her electron's spin is the opposite of Bob's result, and no interference pattern should develop in her measurement.
  • Scenario 2: Bob loses all his electrons. After measuring the position of each electron in the ensemble, my feeling is that Alice does not see an interference pattern because she is blind to the spin state of her electron (which has doubtless been "measured" though Alice could not record it). the interference pattern generated by all the cases where the electron was in state $\left| 0 \right>$ is proportional to $\left| \psi_0(x) + \psi_1(x) \right|^2$, while that for the cases where the electron was in the state $\left| 1\right>^2$ is $\left| \psi_0(x) - \psi_1(x) \right|^2$ -- opposite interference patterns that add up to $|\psi_0(x)|^2 + |\psi_1(x)|^2$.
  • Scenario 3: This time, Bob remembers to apply an inverse Hadamard gate to his electron, prior to Alice's measurement (in its causal past or at a spacelike interval). The joint spin state is thus still an EPR pair, at least until Alice's electron hits the screen. Afterwards, Bob measures their spins and communicates this to Alice. Sorting the data into the two categories, does Alice see the two opposite interference patterns emerge?
  • Scenario 4: Alice decides Bob is unreliable and defines an even more complicated apparatus that measures the spin and position of her electrons, by putting another Stern-Gerlach apparatus after the double-slit, so that the different interference patterns would be shifted on an orthogonal axis on the screen. What effect can Bob, who has gone rogue, have on Alice's result? If he measures the electron's spin prior to Alice's measurements, as in Scenario 1, surely Alice would not see an interference pattern. What if he measures them at different sorts of intervals relative to Alice, with-and-without applying a Hadamard gate?
  • Are there any other scenarios that are interesting to think about?
  • $\begingroup$ To understand your experiment, how are the SG apparatus and the double-slit related (with or without the Hadamard gate)? In particular, if you have a superposition of spin up and spin down in the beginning and then the SG measures its spin to be either up or down and sends it through one of the slits of the double-slit depending on the outcome of the measurement then why is there supposed to be an interference pattern (even with a Hadamard gate)? [...] $\endgroup$ – Dvij D.C. Mar 28 at 14:51
  • $\begingroup$ [...] Because it looks like the SG is essentially doing the job of measuring which slit the electron passes through and thus, there won't be interference regardless of whether there is the Hadamard gate or not. In my understanding, it is not the spin that is preventing the interference, it is that there isn't a superposition of the paths of the double-slit -- it seems like your setup is equivalent to putting detectors on the slit and destroying the interference from the get-go. I might be missing something crucial. $\endgroup$ – Dvij D.C. Mar 28 at 14:51
  • $\begingroup$ When I say "SG apparatus" I'm not referring to the full apparatus including a detector screen, but instead just the magnetic field used to alter the trajectory of states of different spin. My assumption is that without the detector screen, this does not constitute a measurement, but I'm wasn't sure whether the interaction of the electron with the magnetic fields was enough by itself to entangle the electron spin with the lab environment $\endgroup$ – jwimberley Mar 28 at 14:54
  • $\begingroup$ I've edited the question to clarify that I was slightly abusing the term "SG apparatus" $\endgroup$ – jwimberley Mar 28 at 14:56
  • $\begingroup$ Ah, I see. Yes, it makes sense after reading it again. I suppose I agree that one can, in principle, imagine a decoherence-free entanglement of the SG apparatus with the "two branches" of the spin and in Everettean language, your set-up would amount to the SG apparatus coming into a superposition of having "measured" the spin up and having "measured" the spin down. Correct me if I am wrong. $\endgroup$ – Dvij D.C. Mar 28 at 14:58

If I understand your set-up correctly, Alice would not see an interference pattern in any of the cases. In order for the interference pattern to develop in this set-up, it is essential that the initial state of the particle that gets thrown into the experiment is a coherent superposition of $\vert 0\rangle$ and $\vert 1\rangle$. This condition is not met in any of the Bell pair scenarios. In the case where the incoming state is either $\vert 0\rangle$ or $\vert 1\rangle$, the state after applying the Hadamard gate would read

\begin{align} \vert \phi_{0,1}\rangle = \frac{1}{\sqrt{2}}\psi_{0,1}(x)\vert x\rangle\otimes\big(\vert 0\rangle\pm\vert 1\rangle\big) \end{align}

depending on whether the initial state was $\vert 0\rangle$ or $\vert 1\rangle$ respectively. Thus, in all of the Bell pair scenarios, the state of the particle after the Hadamard gate would be given by a density matrix obtained by considering that each of the two states written above has a $1/2$ probability of being true. Thus, this state would be represented by the density matrix (in the spin-basis) given by

\begin{align} \rho &= \frac{1}{2}\big(\vert \phi_0\rangle\langle\phi_0\vert + \vert \phi_1\rangle\langle \phi_1\vert \big)\\&=\frac{1}{2}\pmatrix{\vert \psi_0(x)\vert^2\vert x\rangle\langle x\vert & 0 \\ 0 & \vert \psi_1(x)\vert^2\vert x\rangle\langle x\vert} \end{align}

Thus, over a number of experiments, the distribution on the screen would look like $\vert \psi_0\vert ^2+\vert\psi_1\vert^2$. However, I would note that if Alice separates the data into two categories based on (reliable) results from Bob, she would still not see an interference pattern but she would see two different diffraction patterns $\vert \psi_0(x)\vert^2$ and $\vert \psi_1(x)\vert^2$ respectively. On the other hand, if she separates the spins by placing an SG after the double-slit, she would simply see $\vert \psi_0(x)\vert^2+\vert\psi_1(x)\vert^2$ as can be read off by looking at the expressions of the $\phi$ states I wrote above.


To me, the more interesting part of your experiment is the original setup without the Bell pairs because the Bell pairing takes away the necessary coherence from the scenario in my understanding.

In the original setup, let's say Alice throws in a coherent superposition $\frac{1}{2}(\vert 0\rangle + \vert 1\rangle)$ to the experiment. The state just before the final screen reads

\begin{align} \frac{1}{\sqrt{2}}\Big[\big(\psi_0(x)+\psi_1(x)\big)\vert x\rangle\otimes\vert 0 \rangle + \big(\psi_0(x) - \psi_1(x)\big)\vert x\rangle\otimes\vert 1 \rangle\Big] \end{align}

Or, in the spin-basis, it is the pure-state density matrix given by

\begin{align} \frac{1}{2}\pmatrix{\vert \psi_0(x)+\psi_1(x)\vert^2 \vert x\rangle\langle x\vert & \big(\psi_0(x)+\psi_1(x)\big)\big(\psi_0^*(x)-\psi_1^*(x)\big) \vert x\rangle\langle x\vert \\ \big(\psi_0^*(x)+\psi_1^*(x)\big)\big(\psi_0(x)-\psi_1(x)\big) \vert x\rangle\langle x\vert & \vert \psi_0(x)-\psi_1(x)\vert^2 \vert x\rangle\langle x\vert } \end{align}

Now, if we only measure the position operator, we have to consider the following reduced density matrix obtained by taking trace over the spin-states

\begin{align} \big(\vert\psi_0(x)+\psi_1(x)\vert^2+\vert\psi_0(x)-\psi_1(x)\vert^2\big)\vert x\rangle\langle x\vert = \big(\vert \psi_0(x)\vert^2 + \vert \psi_1(x)\vert^2\big)\vert x\rangle\langle x\vert \end{align}

So, the interference pattern is still destroyed. The only way for Alice to get an interference pattern is for her to get two interference patterns ;) She would have to introduce an SG after the double-slit to generate two patterns on two different screens -- one for the spins measured to be up at that stage and one for spins measured to be down at that stage. In that case, we would isolate the $\vert \psi_0(x)+\psi_1(x)\vert^2$ interference on the screen where we send the states measured to be $\vert 0\rangle$ in the last SG and we would isolate the $\vert \psi_0(x) - \psi_1(x)\vert^2$ interference on the screen where we send the states measured to be $\vert 1\rangle$.

In case I have misunderstood your set-up completely, kindly let me know.

  • $\begingroup$ Thanks for the answer, I'm going to take some time to digest this. My thinking on the Bell pair was that Alice would apply a Hadamard transform and Bob would apply an inverse Hadamard transform (Scenario 3), which I thought would rescue coherence but I likely erred (I think I made a sign mistake and thought some interference terms cancelled) $\endgroup$ – jwimberley Apr 3 at 14:33

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