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## GR0877 Problem 88

PROBLEM STATEMENT: This problem is still being typed.

SOLUTION: (D) Let us first determine the normalization constant $A$: $\chi^{\dag} \chi = |A|^2 ((1-i)(1+i)+4) = 6 |A|^2 = 1$ $\implies$ $A = 1/\sqrt{6}$. Now, $\displaystyle \chi = a\chi_{+} + \chi_{-} = \frac{1+i}{\sqrt{6}} \begin{pmatrix} 1 \\ 0 \end{pmatrix} + \frac{2}{\sqrt{6}} \begin{pmatrix} 0 \\ 1 \end{pmatrix}$, where $\chi_{+}$ represents spin up and $\chi_{-}$ is spin down. When you measure $S_z$ on a particle in the state $\chi$, you could get $+\hbar/2$ with probability $|a|^2$ or $-\hbar/2$ with probability $|b|^2$. Thus, the probability of finding the particle with spin projection $S_z = -\hbar/2$ is $\displaystyle |\frac{2}{\sqrt{6}}|^2 = \frac{2}{3}$. For more on this, see section 4.4 of Griffiths’ “Introduction to quantum mechanics” (2nd ed.) and, in particular, identical Example 4.2.

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Categories: 0877 Tags: , , , ,
1. 03.11.2012 at 02:44

I think you’re missing a ‘b’ term in your spinor equation, should be X= aX(+) + bX(-)

2. 27.09.2014 at 04:59

how you guys get a or b?

• 06.10.2015 at 13:21

I think the a and b are just constants . Their squared value gives the probability for either spin up or spin down. You get the full picture after decomposing the matrix into a + and a – . I am not sure if I made much sense, but if you still have trouble let me know.

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