# The uncertainty principle and spin

I realize that this may be a very basic question, but I've been unable to find the answer elsewhere so thanks in advance for the help.

Suppose an electron's spin is measured about an axis, and then about an axis perpendicular to the previous one. Both spins cannot be known simultaneously (by the uncertainty principle), so what happens to prevent this?

• Traditionally, the view has been that by observing the spin along one direction, you have collapsed the spin wave function and can no longer identify information about the other spin axis. It is analogous to not being able to know the position and momentum of a particle; the microsecond you observe position of the particle you lost all information about the momentum. – Benjamin Horowitz Feb 23 '15 at 9:08

## 2 Answers

Nothing prevents the electron's spin from being measured along a particular axis, and then subsequently measured along an axis perpendicular to the first. In this situation, however, the spins along the perpendicular axes would not be known simultaneously, so the uncertainty principle would not be violated.

As an example, say that we perform our own Stern-Gerlach Experiment, in which we send a beam of particles through a series of Stern-Gerlach magnets in order to measure their spins in various directions. First, let's measure spin in the $z$ direction, and then only keep those particles which we have found to be spin up. Now, we are in possession of a beam of particles which are all spin up in the $z$ direction.

Next, let's take this beam of $\left| +z\right\rangle$ particles and make a subsequent measurement in the $x$ direction. What we will find is that $50\%$ of the particles will be measured as spin right in the $x$ direction, (in the $\left| +x\right\rangle$ state) and $50\%$ will be found spin left in the $x$ direction (in the $\left|-x\right\rangle$ state). Finally, let's say we keep only the beam which we have measured to be in the $\left|+x\right\rangle$ state.

We may be tempted now to draw a conclusion which, classically, seems reasonable, but in reality is false. If from our first measurement we kept only particles in the $\left|+z\right\rangle$ state, and then out of those $\left|+z\right\rangle$ particles we kept only those which were found in the $\left|+x\right\rangle$ state, we may believe that we are now in possession of a beam of particles which are simultaneously spin up in the $z$ direction and spin right in the $x$ direction. This would violate the uncertainty principle, however, so something must be wrong with this conclusion. To answer what exactly is wrong, let's make one more spin measurement.

Let's take our beam which we first measured as $\left|+z\right\rangle$ and subsequently as $\left|+x\right\rangle$ and make another measurement on it in the $z$ direction. The result will be that $50\%$ of the particles will be found as spin up and $50\%$ will be found as spin down. But if all of the particles in the beam truly had been in the $\left|+z\right\rangle$ state after the measurement in the $x$ direction, we should have found $100\%$ of the particles spin up in the $z$ direction in our third measurement. Because this is not what we observe, we are forced to conclude that our measurement in the $x$ direction destroyed our previous knowledge gained about the system from the first $z$ measurement. Thus, we did not know the spin states of the $x$ and $z$ directions simultaneously, just as, qualitatively, the uncertainty principle states.

• Thanks for the explanation, but there's one thing that I don't understand. Since all of the electrons have spin up in the z direction before the second measurement and only half have spin up after the second measurement, how does this reconcile with the conservation of spin? – Sheepman Feb 23 '15 at 18:40
• It's important to remember that the conserved quantity is total spin, which is $S^2=S_x^2+S_y^2+S_z^2$. The expectation value of this quantity is conserved through all measurements. When the measurement in the x direction is made, thereby changing the expectation value of $S_x$, the expectation values in the z and y directions adjust accordingly such that the total spin angular momentum is conserved. – wgrenard Feb 23 '15 at 19:52
• Thanks, and I have one last question. Suppose that two electrons are entangled such that they have opposite spins. The spin of the first electron is measured about the z axis, and then the spin of the second electron is measured about the x axis. Then by the uncertainty principle, if the spin of the first electron were again measured about the z axis, it would have a 50-50 chance of spinning up or spinning down about that axis. Does this mean that the spin of the first electron was changed be the measurement taken on the second? – Sheepman Feb 23 '15 at 20:25
• @SamuelMontgomery I apologize, I believe I was mistaken in my original comment answering your question involving entanglement. I am not familiar enough with the concept to give you a complete answer, I'd advise you post a new question so someone more experienced can answer. Best of luck. – wgrenard Feb 23 '15 at 21:35

"By the uncertainty principle" is the answer. In more detail, let's say we're talking about x and y axes. The first measurement puts the electron into an eigenstate of the spin X observable (the question of how it does this is the quantum measurement problem). Whichever of the two X eigenstates this "collapse" ends up in, it not an eigenstate of the spin Y observable: both spin Y eigenstates have a nonzero projection onto the either of the spin X eigenstates. So the second measurement can go either way, with probabilities governed by the size of these projections. The spin observables about each axis are the Pauli matrices.

If we were in a situation where the two measurements had common eigenstates, then the second measurement would be wholly determined by the first. The uncertainty principle is simply the quantification of all of this: you can work out the variance of the second measurement from the commutator bracket of the two observables in question. This variance is nought if the two observables have common eigenstates (e.g. momentum and energy eigenstates of the photon).