- #1
Bill Foster
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Homework Statement
A spin ½ system is known to be in an eigenstate of [tex]\textbf{S}\cdot\hat{\textbf{n}}[/tex] with eigenvalue [itex]\frac{\hbar}{2}[/itex] , where [tex]\hat{\textbf{n}}[/tex] is a unit vector lying in the xy-plane that makes an angle γ with the positive z-axis.
a. Suppose [tex]S_x[/tex] is measured. What is the probability of getting [itex]+\frac{\hbar}{2}[/itex]?
The Attempt at a Solution
[tex]|\hat{\textbf{n}},+\rangle = \cos\left(\frac{\beta}{2}\right)|+\rangle + \sin\left(\frac{\beta}{2}\right)|-\rangle[/tex]
[tex]\langle S_x,+|\hat{\textbf{n}},+\rangle = \left(\frac{\langle +|+\langle -|}{\sqrt{2}}\right)\left(\cos\left(\frac{\gamma}{2}\right)|+\rangle + \sin\left(\frac{\gamma}{2}\right)|-\rangle\right)=\frac{1}{\sqrt{2}}\left(\cos\left(\frac{\gamma}{2}\right)+\sin\left(\frac{\gamma}{2}\right)\right)[/tex]
The probability of getting [itex]\frac{\hbar}{2}[/itex] is
[tex]P=|{\langle S_x,+|\hat{\textbf{n}},+\rangle|^2=\frac{1+\sin\gamma}{2}[/tex]
I don't really understand this solution.
Why are they using [tex]\langle S_x,+|\hat{\textbf{n}},+\rangle[/tex] instead of [tex]\langle S_x|\hat{\textbf{n}}\rangle[/tex]?
And how does [itex]\frac{\hbar}{2}[/itex] come into play here? What if we were looking for the probability of getting something other than [itex]\frac{\hbar}{2}[/itex], like [itex]\frac{\hbar}{4}[/itex] for example? How would that change it?