Phy5646/Grp3SpinProb

Worked Spin Problem

Consider a unit vector ${\hat {n}}$, measured an angle $\phi$ from the positive ${\hat {x}}$ axis in the $xy$ plane and angle $\theta$ from the positive $z$ axis.

Let the components of the spin vector ${\overrightarrow {S}}$ along ${\hat {n}}$ be $S_{n}={\hat {n}}\cdot {\overrightarrow {S}}$.

a.) Solve the resulting eigenvalue equation. ($S_{n}\lambda =v\hbar \lambda$)

$S_{n}=S_{x}\sin(\theta )cos(\phi )+S_{y}\sin(\theta )\sin(\phi )+S_{z}\cos(\theta )$

Which, in matrix form (using the definitions of $S_{x},S_{y},S_{z}$) looks like:

$S_{n}={\frac {\hbar }{2}}\left({\begin{array}{cc}cos(\theta )&sin(\theta )e^{-i\phi }\\sin(\theta )e^{i\phi }&-cos(\theta )\end{array}}\right)$

And define $\chi =\left({\begin{array}{c}a\\b\end{array}}\right)$.

So that:

$S_{n}\chi ={\frac {\hbar }{2}}\left({\begin{array}{c}a\cos(\theta )+b\sin(\theta )e^{-i\phi }\\a\sin(\theta )e^{i\phi }-b\cos(\theta )\end{array}}\right)$

And the eigenvalue equation becomes:

$\left({\begin{array}{c}a(\cos(\theta )-2v)+b\sin(\theta )e^{-i\phi }\\a\sin(\theta )e^{i\phi }-b(\cos(\theta )+2v)\end{array}}\right)=\left({\begin{array}{c}0\\0\end{array}}\right)$

Which has nontrivial solutions

$v=\pm {\frac {1}{2}}$

For $v_{+},{\frac {b}{a}}=\tan({\frac {\theta }{2}})e^{i\phi }$

Which means that:

$\chi _{+}=\left({\begin{array}{c}\cos({\frac {\theta }{2}})\\\sin({\frac {\theta }{2}})e^{-i\phi }\end{array}}\right)$

And, for $v_{-},{\frac {b}{a}}=-\cot({\frac {\theta }{2}})e^{i\phi }$

So that $\chi _{-}=\left({\begin{array}{c}\sin({\frac {\theta }{2}})\\-\cos({\frac {\theta }{2}})e^{i\phi }\end{array}}\right)$

b.) Verify that the two resulting eigenvectors are orthogonal.

Show that

$\langle \mathrm {X} _{+}|\mathrm {X} _{-}\rangle =0$

Where, from above,

$\langle \mathrm {X} _{+}|=\left(\cos({\frac {\theta }{2}})\;\;\sin({\frac {\theta }{2}})e^{-i\phi }\right)$

$|\mathrm {X} _{-}\rangle =\left({\begin{array}{c}\sin({\frac {\theta }{2}})\\-\cos({\frac {\theta }{2}})e^{i\phi }\end{array}}\right)$

So that,

$\langle \mathrm {X} _{+}|\mathrm {X} _{-}\rangle =\cos({\frac {\theta }{2}})\sin({\frac {\theta }{2}})-\sin({\frac {\theta }{2}})\cos({\frac {\theta }{2}})e^{i\phi }e^{-i\phi }=0$

As expected.

c.) Show that these vectors satisfy the closure relation.

This amounts to showing that

$\sum _{i}|\mathrm {X} _{i}\rangle \langle \mathrm {X} _{i}|=\left({\begin{array}{cc}1&0\\0&1\end{array}}\right)$

So:

$\langle \mathrm {X} _{+}|=\left(\cos({\frac {\theta }{2}})\;\sin({\frac {\theta }{2}})e^{-i\phi }\right)$

And thus,

$|\mathrm {X} _{+}\rangle \langle \mathrm {X} _{+}|=\left({\begin{array}{c}\cos({\frac {\theta }{2}})\\\sin({\frac {\theta }{2}})e^{-i\phi })\end{array}}\right)\left(\cos({\frac {\theta }{2}})\sin({\frac {\theta }{2}})e^{-i\phi }\right)$

$|\mathrm {X} _{+}\rangle \langle \mathrm {X} _{+}|=\left({\begin{array}{cc}\cos ^{2}({\frac {\theta }{2}})&\sin({\frac {\theta }{2}})\cos({\frac {\theta }{2}})e^{-i\phi }\\\sin({\frac {\theta }{2}})\cos({\frac {\theta }{2}})e^{-i\phi }&\sin ^{2}({\frac {\theta }{2}})\end{array}}\right)$

While

$|\mathrm {X} _{-}\rangle \langle \mathrm {X} _{-}|=\left({\begin{array}{c}\sin({\frac {\theta }{2}})\\-\cos({\frac {\theta }{2}})e^{i\phi })\end{array}}\right)\left(\sin({\frac {\theta }{2}})-\cos({\frac {\theta }{2}})e^{i\phi }\right)$

And $|\mathrm {X} _{-}\rangle \langle \mathrm {X} _{-}|=\left({\begin{array}{cc}\sin ^{2}({\frac {\theta }{2}})&-\sin({\frac {\theta }{2}})\cos({\frac {\theta }{2}})e^{i\phi }\\-\sin({\frac {\theta }{2}})\cos({\frac {\theta }{2}})e^{i\phi }&\cos ^{2}({\frac {\theta }{2}})\end{array}}\right)$

So, clearly, we can get: $|\mathrm {X} _{-}\rangle \langle \mathrm {X} _{-}|+|\mathrm {X} _{+}\rangle \langle \mathrm {X} _{+}|=\left({\begin{array}{cc}1&0\\0&1\end{array}}\right)$