# I Do spin operators 'appear' to commute for an entangled spin state?

#### muonion

Let's consider Bohm's paradox (explaining as follows). A zero spin particle converts into two half-spin particles which move in the opposite directions. The parent particle had no angular momentum, so total spin of two particles is 0 implying they are in the singlet state.
Suppose we measured Sz of particle 1 which happened to be +1/2. Then automatically Sz of particle 2 would be -1/2.
From Weinberg's 'Lectures on Quantum Mechanics' Ch-12 (P. 394)

... the observer could have measured the x-component of the spin of particle 1 instead of its z-component, and by the same reasoning, if a value h/2 or −h/2 were found for the x-component of the spin of particle 1 then also the x-component of the spin of particle 2 must have been −h/2 or h/2 all along. Likewise for the y-components. So according to this reasoning, all three components of the spin of particle 2 have definite values, which is impossible since these spin components do not commute.
I don't get the bold lines above. Whenever we make a Sz measurement of 1 and then the same of 2 and then Sx measurement of 1- this x measurement destroys the previous info of particle 2. Hence particle 2 shouldn't have 2 definite components at a time, let alone 3 definite values.
What did Weinberg imply there?

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#### PeterDonis

Mentor
Whenever we make a Sz measurement of 1 and then the same of 2 and then Sx measurement of 1- this x measurement destroys the previous info of particle 2.
Weinberg is not talking about successive measurements. He's talking about alternative possibilities for single measurements (one on each particle). He's simply pointing out that, since measuring the same spin component of both particles will always give opposite results, no matter which component is measured (x or y or z), any hidden variable model, i.e., any model that attributes the correlations between these measurements to pre-existing properties of the particles, would have to attribute definite spin components (+1 or -1) in all three directions (x and y and z) to each particle, i.e., the pre-measurement state of each particle would have to have definite values for spin-x, spin-y, and spin-z. But that is not possible because no quantum state can have definite values for multiple non-commuting operators.

#### muonion

Thank you for your answer. I have got another question reading your answer. Why are we relying on established quantum facts (i.e. no 3 definite values for spin simultaneously) than hidden variable model? We are rejecting the idea that hidden variable model could give room to both quantum superposition and entanglement in a new theory. Why is that?

#### PeroK

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Thank you for your answer. I have got another question reading your answer. Why are we relying on established quantum facts (i.e. no 3 definite values for spin simultaneously) than hidden variable model? We are rejecting the idea that hidden variable model could give room to both quantum superposition and entanglement in a new theory. Why is that?
There are the experiments related to Bell's theorem:

In my view, however, the nature of spin for a single electron already makes hidden variables seem very odd. For example, if you have an electron in the z-spin-up state:

1) A measurement of spin in the x-direction will return up or down with 50% probability each. So when an electron is in the z-up state, there must be a binary variable that gets set with equal probability to control a subsequent measurement of spin in the x-direction.

2) A measurement of spin along an axis in the x-z plane returns up-down with a probability that is determined by the angle with the z-axis: the nearer to the z-axis, the more likely will be spin up; and the nearer to the x-axis, the greater the likelihood of spin down (with 50% chance when you reach the x-axis). There must be some complicated hidden variables there to control at what point the spin-down is returned.

It also seems to me that there must still be a probabilistic mechanism inside the electron to decide how the variables are set!

The other interesting thing here is that the probability of getting spin-up or down about an axis in the x-z plane, as described above, depends on the angle in a way that can be calculated using quantum probability amplitudes. Although, you can have rules for the setting of hidden variables that matches these values.

Now, the trick with Bell's inequality was to highlight the difference between probabilistic calculations for hidden variables, which would depend on classical probabilities, and quantum mechanics, which depends on probability amplitudes. This, in the more complicated case of entangled particles and measurements about different axes, actually gives different numerical values for the probabilities, which can be tested by experiment.

The experiments agree with the probabilities from QM and not from local hidden variables.

#### vanhees71

Gold Member
The point is that the total spin has three determined components, which is the outstanding special feature of the $S=0$ state. This implies that if $s_{z1}=1/2$, then $s_{z2}=-1/2$ and vice versa. Though the single-particle spin components are maximally indetermined, i.e., the single-particle spin states are $\hat{\rho}_1=\hat{\rho}_2=1/2 \hat{1}$.

This implies though you have this 100% correspondence between measuring the spin components in the SAME direction, this doesn't imply that after the spin measurement of $\sigma_{1z}$ any spin component of particle 2 in another direction would be determined as well. Knowing the complete two-spin state,
$$\hat{\rho} = |\Psi \rangle \langle \Psi| \quad \text{with} \quad |\Psi \rangle=\frac{1}{\sqrt{2}} (|1/2,-1/2 \rangle-|-1/2,1/2 \rangle),$$
you can calculate the probabilities to get $\pm 1/2$ for spin components of the single particles in any direction.

"Do spin operators 'appear' to commute for an entangled spin state?"

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