A query regarding Rotational Invariance

  • #26
PeroK
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Got it. But if we limit ourselves to a plane (which I was assuming as its in the video example hence the confusion) we can set ##\phi##=0 in #21 then from it we can get the probability of opposite spin. But, it doesn't answer about the probability of ##\uparrow \uparrow## and ##\downarrow \downarrow## each given arbitrary α, β.

PS Sorry for testing your patience.
With ##\phi = 0##, the probability of getting opposite spins is:
$$\frac 1 2 |\alpha - \beta|^2 \sin^2 \theta $$
 
  • #27
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With ##\phi = 0##, the probability of getting opposite spins is:
$$\frac 1 2 |\alpha - \beta|^2 \sin^2 \theta $$

I totally understand.

But how do we know about the same spin probability of each case ↑↑ and ↓↓ given arbitrary α, β?
 
  • #28
PeroK
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I totally understand.

But how do we know about the same spin probability of each case ↑↑ and ↓↓ given arbitrary α, β?
You use the coefficients of the state expressed in the relevant basis. From a few posts ago:
$$\psi = [\alpha \cos^2(\frac \theta 2) + \beta e^{-2i\phi}\sin^2(\frac \theta 2)]\uparrow_n\uparrow_n + [\alpha \sin^2(\frac \theta 2) + \beta e^{-2i\phi}\cos^2(\frac \theta 2)]\downarrow_n\downarrow_n$$ $$ + \frac 1 2\sin\theta(\beta e^{-2i\phi} - \alpha)(\uparrow_n\downarrow_n + \downarrow_n\uparrow_n)$$
 
  • #29
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You use the coefficients of the state expressed in the relevant basis. From a few posts ago:
$$\psi = [\alpha \cos^2(\frac \theta 2) + \beta e^{-2i\phi}\sin^2(\frac \theta 2)]\uparrow_n\uparrow_n + [\alpha \sin^2(\frac \theta 2) + \beta e^{-2i\phi}\cos^2(\frac \theta 2)]\downarrow_n\downarrow_n$$ $$ + \frac 1 2\sin\theta(\beta e^{-2i\phi} - \alpha)(\uparrow_n\downarrow_n + \downarrow_n\uparrow_n)$$

This has been very helpful. Thanks a ton again.
 
  • #30
vanhees71
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So, just to be clear/reconfirm:

if Alice and Bob have 1 particle each and Alice decides to measure its particle (in a direction that has an angle ##\theta## with the original direction ##z##) the probability of both Alice and Bob getting a ##\uparrow \uparrow## and getting a ##\downarrow \downarrow## in new direction is given by the eqns in #9 correctly. right?
Ok, that's a different question.

To answer the question about joint measurements when both are measuring the spin component in direction ##\theta## (using the same states as in my previous posting) it's most simple to first express the old basis by the new, i.e.,
$$|1/2 \rangle=\cos(\theta/2) |\chi_+ \rangle - \sin[\theta/2]|\chi_- \rangle, \quad |-1/2 \rangle = \sin(\theta/2) |1/2 \rangle + \cos(\theta/2) |\chi_- \rangle.$$
The state ket the particle is prepared in is thus given in this basis by
$$|\Psi \rangle = [\alpha \cos^2(\theta/2) + \beta \sin^2(\theta/2)] |\chi_+,\chi_+ \rangle + (\beta-\alpha) \cos(\theta/2) \sin(\theta/2) (|1/2,-1/2 \rangle+|-1/2,1/2 \rangle) + (\alpha \sin^2 \theta/2 + \beta \cos^2 \theta/2) |-1/2,-1/2 \rangle.$$
From this you read off immediately for the probabilities for the 4 possible outcomes of joint measurements of the spin components in ##\theta## direction
$$P_{++}=|\alpha \cos^2 \theta/2 + \beta \sin^2 \theta/2|^2,$$
$$P_{+-}=P_{-+}=|\alpha-\beta|^2 \sin^2 \theta/2 \cos^2 \theta/2,$$
$$P_{--}=|\alpha \sin^2 \theta/2+\beta \cos^2 \theta/2)|^2.$$
You can simplify a bit using some trigonometry:
$$P_{++}=\frac{1}{4} [|\alpha+\beta|^2 + 2 (|\alpha|^2-|\beta|^2)\cos \theta + |\alpha-\beta|^2 \cos^2 \theta],$$
$$P_{+-}=P_{-+}=\frac{1}{4} |\alpha-\beta|^2 \sin^2 \theta,$$
$$P_{--}=\frac{1}{4}[|\alpha+\beta|^2 - 2 (|\alpha|^2-|\beta|^2) \cos \theta + |\alpha-\beta|^2 \cos^2 \theta].$$
As expected, the probabilities in general depend on ##\theta##. That's because any spin-triplet state is not rotation invariant, because the total spin is ##S=+1##. The spin-singlet state, i.e., ##S=0## is the only rotationally invariant state, i.e.,
$$|\Psi ' \rangle=\frac{1}{\sqrt{2}}(|1/2,-1/2 \rangle-|-1/2,1/2 \rangle),$$
as already mentioned in some postings above.
 
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