What Are the Probabilities of Measuring Spin Components in a Spin Singlet State?

[confuted]

Homework Statement

Consider a system made up of two spin 1/2 particles. Initially the system is prepared in a spin singlet state with total spin $$S_{total}=0$$. We'll define the spin components of one particle as $$(s_{1x},s_{1y},s_{1z})$$ and of the other as $$(s_{2x},s_{2y},s_{2z})$$

a. If no measurements are made on particle 2, what is the probability that a measurement of the:

i. $$s_{1z}$$ component of particle 1 will yield $$+\frac{\hbar}{2}$$?
ii. $$s_{1x}$$ component of particle 1 will yield $$+\frac{\hbar}{2}$$?
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b. Now assume that a measurement of $$s_{2z}$$ component of particle 2 yielded $$+\frac{\hbar}{2}$$. What is the probability that a subsequent measurement of the:

i. $$s_{1z}$$ component of particle 1 will yield $$+\frac{\hbar}{2}$$?
ii. $$s_{1x}$$ component of particle 1 will yield $$+\frac{\hbar}{2}$$?
​

Homework Equations

Spin singlet state: $$|\psi>=\frac{1}{\sqrt{2}}\left(|\uparrow\downarrow>-|\downarrow\uparrow>\right)$$

The Attempt at a Solution

a. i. The state is just random, right? p=1/2
ii. All we're doing is changing the axis we're measuring - nothing physical has set it. p=1/2

b. i. The total spin of the system is 0. Since particle 2 was spin up, particle 1 must be spin down. p=0
ii. The measurements of the x and z components of the spin don't commute, so measuring $$s_{1x}$$ destroys the information we knew about $$s_{2z}$$, and we're back to case a ii. p=1/2

Those are my answers and my reasoning - am I correct in my thinking?

 

[badphysicist]

I agree with you and all your answers check out for me.