# Spin angular momentum conservation and entanglement

Suppose we have a system with $S=0$ . Suddenly, this system decays into two particles, A and B, so we know the sum of the spin of these two particles is zero. Suppose we measure the spin of A in the z direction and get $$s_z,a=1/2$$ Then we know B must an eigenstate of $S_{z,b}$, the operator associated with the value of spin of B in the z direction, with eigenvalue $-1/2$, such that the sums of spins still equal 0.

But what happens if, before measuring the spin of B in z direction, I measure it in the x direction? Then B will collapse to an eigenstate of $S_{x,b}$, which is a superposition of states which are eigenstates $S_{z,b}$ with eigenvalues +1/2 and -1/2. So now there´s some probability that we will find $S_z,b=1/2$ and the sum of spins $S\neq0$. Have we changed the sum of spins by just observing it?

• By measuring the spins you have changed the spins and the system is not in an entangled state any longer. Maybe I am missing something? – CuriousOne May 30 '15 at 23:13
• related: physics.stackexchange.com/q/4047 – Mark Eichenlaub May 30 '15 at 23:25

There are many operators you can call spin.

The operator $S^2_A$ that measures the total spin on particle $A$ is a reasonable operator to call spin, however for a spin $1/2$ particle it always gives the number $\hbar^23/4$ so it is kinda boring. Same with the operator $S^2_B$ that measures the total spin on particle $B$ is an equally reasonable operator to call spin, however for a spin $1/2$ particle it is equally boring because it also always gives the number $\hbar^23/4$. So there are the operators $S_{z,A}$ and $S_{z,B}$ which can give $\pm \hbar/2$ depending on which state you give them. But there is also $S_{x,A}$, $S_{x,B}$, $S_{y,A}$, and $S_{y,B}$, and I think you've used these before. And there are their sums. For instance the operator $S_z=S_{z,A}+S_{z,B}$ which could also be written $S_z\otimes I + I\otimes S_z$ which sends the state $|+\rangle_A\otimes|+\rangle_B$ to $\hbar|+\rangle_A\otimes|+\rangle_B$ and sends $|+\rangle_A\otimes|-\rangle_B$ and $|-\rangle_A\otimes|+\rangle_B$ to zero and sends $|-\rangle_A\otimes|-\rangle_B$ to $-\hbar|-\rangle_A\otimes|-\rangle_B$, it measures the combined/total z component. Since any possible state can be written in terms of those four, that actually spells it out completely. However, this isn't a simple situation. For instance $S^2_A=S_{x,A}^2+S_{y,A}^2+S_{z,A}^2$, but if you look at the total spin $$S^2=S_{x}^2+S_{y}^2+S_{z}^2,$$ you get $$S^2=S^2_A+S^2_B+2S_{x,A}S_{x,B}+2S_{y,A}S_{y,B}+2S_{z,A}S_{z,B},$$which is not at all a boring operator, even though $S^2_A$ and $S^2_B$ were both incredibly boring.

So, what I'm trying to get at is that all these things could rightfully be called spin, but your text is rather vague. For instance you said,

Suppose we have a system with $S=0$

If you mean suppose your system is an eigenstate of $S_z=S_{z,A}+S_{z,B}$ with eigenvalue 0, then that is a full two (complex) dimensional subspace. Which includes totally different states, states with different eigenvalues of $S^2=S^2_A+S^2_B+2S_{x,A}S_{x,B}+2S_{y,A}S_{y,B}+2S_{z,A}S_{z,B}.$ For instance the state $|+\rangle_A\otimes|-\rangle_B+|-\rangle_A\otimes|+\rangle_B$ has a different eigenvalue under $S^2=S^2_A+S^2_B+2S_{x,A}S_{x,B}+2S_{y,A}S_{y,B}+2S_{z,A}S_{z,B}$ than $|+\rangle_A\otimes|-\rangle_B-|-\rangle_A\otimes|+\rangle_B$ does.

That said. Your text probably meant $|+\rangle_A\otimes|-\rangle_B-|-\rangle_A\otimes|+\rangle_B$, but that assumes no couplings during the decay and such, and even conservation of angular momentum a thing that needs to be given. Spin isn't conserved alone when there is coupling. The answer so far could lok like I'm merely nit picking, however we've already touched on how spin angular momentum is not always conserved by itself when there is coupling. It is different than conservation of energy, there is a flow of energy and what comes from some place must go somewhere else. For instance rest energy is not conserved. Kinetic energy is not conserved. And similarly spin angular momentum is not conserved even though total angular momentum is conserved. And angular momentum isn't just frame dependent like energy, it can depend on where you take your origin to be so it is origin dependent.

That said, your question has a very direct answer. When you measure something you force it into an eigenstate of the operator in question. if you had something spin up in the z direction and measured in z you'd get up. However if you measured it in the z direction you'd get a random $\pm$ answer and if then measured in the z direction you'd get a random $\pm$ answer. The fact that doing $S_{x,B}$ then $S_{z,B}$ gives different answers than doing $S_{z,B}$ then $S_{x,B}$ is just what is means for two operators to not commute. So we aren't surprised to get different results.

OK. So really the question isn't about the singlet state $|+\rangle_A\otimes|-\rangle_B-|-\rangle_A\otimes|+\rangle_B$, and it is isn't even about whether $S_{z,B}$ and $S_{x,B}$ commute. Really it is about how a state moves from a general state to an eigenstate when you measure it.

And if that's what you want to know, ask an experimentalist how they measure it and then look at what the theory says about how that happens. So for instance you can measure the spin with a Stern-Gerlach device. A detailed and realistic (maybe overly realistic) description is available in Travis Norsen's "The pilot-wave perspective on spin" in the American Journal of Physics 82, 337 (2014); http://dx.doi.org/10.1119/1.4848217 arxiv version. You can watch it become polarized as the beams diverge.

This can still seem weird if you are used to thinking of spin as conserved. But spin precesses in a homogeneous magnetic field. Angular momentum is conserved, not spin, spin is just one kind of angular momentum. Just like kinetic energy is one kind of energy and kinetic energy is not conserved. There is just a small region where the field is inhomogeneous. Note that you need another system, in this case an electromagnetic field, and you have to interact with it in order to separate out a spin component. You actually affect your system too, and the system and the particle whose spin you separate become entangled with each other.

There is a legitimate sense where it is that combine system, particle plus system in its entangled state that is the time evolution of the pre-measurement system and particle. However, by the time you get to the point where you start talking probabilities you act as if just one of those entangled branches (particle $+\hbar/2$ system with it deflected one way) versus (particle $-\hbar/2$ system with it deflected the other way).

And the singlet state and the entanglement of the decay products have nothing really to do with this because you destroy the entanglement between the decay products when you start to measure the separate parts. Destroyed in the sense that as you separate the beam for the particle you are measuring not only does it smoothly develop into one of two eigenstates, but the other particle will smoothly develop into the opposite eigenstate, but thereafter they aren't influencing each other, they've fully realized their entanglement. Their entanglement was just a particular way of being in a superposition, a superposition where any separation of either that polarized either ends up polarizing them both. This is because when one is separated, the joint pair of particles is separated. And they were anti-polarized when the separation happened and the separation polarized one of them so polarized both of them oppositely. And its done its job because the entanglement has been passed to the various states of the system that measured it, so once it gets to the point where you treat one of the states of the system to measure it as having happened, the little particles are now just in a particular spin state, there is no entanglement between them anymore (from the perspective of a particular state of the system used to measure the spin).

Absolutely nothing I've said is controversial, if you just write down the Hamiltonian for the two particles and the system you use to measure one of them and then just solve the Schrödinger equation for that combined system it just evolves, and evolves just like I describe. Expecting anything other than the Schrödinger equation evolution for the actual experimental system under consideration will be redundant at best and possibly wrong.

There might be helpful details in my answer Spin in magnetic field and eigenvalues

Note that the total angular momentum is conserved in this system. After the decay of the particle with total angular momentum of $0$ the total angular momentum of the system is described by $$\mathbf J = \mathbf S_A + \mathbf{S}_B + \mathbf{L}$$

, where $\mathbf L$ is the relative angular momentum between $A$ and $B$. From the addition of angular momentum we have

$$j_1 \otimes j_2 = (j_1+j_2) \oplus (j_1+j_2-1) \oplus \dots \oplus\left|j_1-j_2\right|$$

$$1/2\otimes 1/2 \otimes l = \Big[ 1 \oplus 0 \Big] \otimes l = 1 \otimes l + 0 \otimes l \tag{1}$$
We want to get the a singlet from the equation $(1)$ in order to have a total angular momentum of $0$. If you try various values for $l$ you would see that only for $l=1$ and $l=0$ you get a singlet state and thus obey the conservation of the total angular momentum.
You see directly from equation $(1)$ the result that Serg obtain in his/her answer ie after the measurement the particles $A$ and $B$ are either in a state with total spin $1$ or $0$. However this change in the spin is compensated by the relative angular momentum between the particles $A$ and $B$.