Consider a system of two particles. The system was prepared (e.g., by some prior measurement) to have zero total spin. Another measurement is made, either immediately after the preparation or at a later time if Schrodinger evolution does not disturb the total spin of the system. This other measurement finds the spin of one particle to be "up" in some selected direction. We conclude that the spin of the other particle must be "down" relative to the same selected direction. However, this conclusion can only be valid if the measurement did not disturb the total spin of the two-particle system. I don't doubt that this is correct, but my problem is not knowing how to derive, from basic principles that I have learned, that the spin measurement of a single particle does not disturb the total spin of the two-particle system. Using more mathematical terminology, the problem is to show that an operator associated with single-particle spin (I'm not sure what that means for a two-particle system) commutes with the operator associated with total system spin. How can we show that these operators commute?
The state space (Hilbert space) of the total spin of two distinguishable particles is the tensor product of the state space of the spin of particle 1, say space 1, and the state space of the spin of particle 2, say space 2.
The spaces 1 and 2 describe the spin states of the particles 1 and 2 before the two particles become a system, and since spin is an intrinsic quantum property of particles, the spaces 1 and 2 are independent.
The spin operators S1 and S2 that describe the spin of particle 1 and 2, respectively, act on the space 1 and 2, respectively, and since the two spaces are independent, S1 and S2 commute.
The operator of the total spin S of the two particles is the sum of S1 and S2, and since S1 and S2 commute, and S1 commutes with itself and S2 commutes with itself too, then S commutes with both S1 and S2.
Actually S is not just the sum of S1 and S2, because S acts on a space that is different from the spaces on which S1 and S2 act, so you have to “embed”, in some way, the operators S1 and S2 into the state space of total spin, but this is a mathematical procedure (that is done by taking tensor products of the two operators S1 and S2 with respective unit operators) that does not change the previous result.
If you want to see the details of this, you may refer to https://ocw.mit.edu/courses/physics/8-05-quantum-physics-ii-fall-2013/lecture-notes/MIT8_05F13_Chap_08.pdf
The state space (Hilbert space) of the total spin of two distinguishable particles is the tensor product of the state space of the spin of particle 1, say space 1, and the state space of the spin of particle 2, say space 2.
The spaces 1 and 2 describe the spin states of the particles 1 and 2 before the two particles become a system, and since spin is an intrinsic quantum property of particles, the spaces 1 and 2 are independent.
The spin operators S1 and S2 that describe the spin of particle 1 and 2, respectively, act on the space 1 and 2, respectively, and since the two spaces are independent, S1 and S2 commute.
The operator of the total spin S of the two particles is the sum of S1 and S2, and since S1 and S2 commute, and S1 commutes with itself and S2 commutes with itself too, then S commutes with both S1 and S2.
Actually S is not just the sum of S1 and S2, because S acts on a space that is different from the spaces on which S1 and S2 act, so you have to “embed”, in some way, the operators S1 and S2 into the state space of total spin, but this is a mathematical procedure (that is done by taking tensor products of the two operators S1 and S2 with respective unit operators) that does not change the previous result.
If you want to see the details of this, you may refer to https://ocw.mit.edu/courses/physics/8-05-quantum-physics-ii-fall-2013/lecture-notes/MIT8_05F13_Chap_08.pdf
Thank you Spiros for your answer. You have the qualifier "distinguishable particles" but I think I know how your answer still applies even to identical particles. If the degree of localization of the particles compared to the spatial separation between them is such that we can unambiguously say that the two particles are at different locations, then location can serve as a particle identifier and this makes identical particles treatable in the same way as distinguishable particles, so your answer for distinguishable particles still applies. Also, the notes that you attached are exactly what I needed to fill in the educational gap that made me unable to answer my own question. Thanks again.
The statements are too vague-and, indeed, the answer to the main question is that it need not be possible to show it. It depends on the properties of the Hamiltonian. Nor is it true that, when measuring a spin component of a multiparticle state, this necessarily implies anything particular about the state of the corresponding component of the other spins, that is independent of the properties of H or M. However, for any choice of H and M it is possible to compute any property of the spin system.
The statement about measurements ``disturbing'' spins, by itself doesn't mean anything. ``Disturbing'' means the Hamiltonian H (or the evolution operator U=exp(iH) ) doesn't commute with the measurement operator M; ``not disturbing'' means that it does. Therefore the action of a measurement M, that doesn't commute with the evolution operator leads to a state that isn't an eigenstate of the Hamiltonian, but a linear combination thereof. Furthermore, the statement that the spins interact at all means that the Hamiltonian itself isn't block-diagonal in the individual spins. Everything else is just a consequence of linear algebra.
It's not the Hamiltonian-that enters in the Schrödinger equation, which describes the evolution of the spin system-that ``disturbs'' the system; it's the measurement operator, M, that acts on the state of the system, if it doesn't commute with H (or U); M describes the interaction of the spin system with another system.
This is a standard exercise in quantum mechanics.
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