# Calculate spin wave function given probabilities of its alignment along 2 axes

Problem: An $e^{-}$ exists in such a state that the probability of its spin aligning across the $x_{(+)}$ axis is $P_{x+}=1/2$ and across the $y_{(+)}$ axis is $P_{y+}=1/2$ as well. What is the spin wave function of the electron ?

Solution: Let $\psi_s$ be the spin wave function of the ${e^-}$: $\psi_s=\left(\begin{array}{c}a\\b\\\end{array}\right), a,b\in C$. Then: $P_{x+}=\left|\langle\psi_s, X_{x+}\rangle\right|^2$ and $P_{y+}=\left|\langle\psi_s, X_{y+}\rangle\right|^2$. As for the $X_{X_+}$ and $X_{y_+}$ they are calculated via the formula $X_{n_+}=\left(\begin{array}{c}\cos(θ/2)\\\sin(θ/2)e^{i\varphi}\\\end{array}\right)$, where $φ,θ$ are the angles of the unitary vector $\vec{n}$ with axes $x,z$ respectively:

$X_{X_+}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\\\end{array}\right), X_{y_+}=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\i\\\end{array}\right)$

Therefore, we end up with 2 equations and 2 unknowns:

$\left|\langle\left(\begin{array}{c}a\\b\\\end{array}\right),\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\1\\\end{array}\right)\rangle\right|^2=1/2$ and

$\left|\langle\left(\begin{array}{c}a\\b\\\end{array}\right),\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\i\\\end{array}\right)\rangle\right|^2=1/2$

But I'm having some difficulties to solve the system:

$|a^*+b^*|^2=1, |a^*+i b^*|^2=1$

Any hints ? Also the fact that both $a=1,b=0$ and $a=0,b=1$ both satisfy the system worries me.

EDIT: The missing clue was $|a|^2 + |b|^2=1$.

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## migrated from physics.stackexchange.comSep 29 '13 at 7:19

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(A more math-y title might be a good idea for this post. THe core question is a math one) – Manishearth Sep 29 '13 at 7:20

What you've done is correct. You need to also make use of the fact that the state is normalized, so $|a|^2 + |b|^2 = 1$.
So your first equation expands out to be $a^*b + b^*a = 0$ and the second to be $i a b^* - i b^*a =0$ which implies that $a^*b = 0$. So either $a = 0$ or $b = 0$. which are the two solutions you guessed (but turns out they're the only solutions).
Physically it makes sense, since the spin z-up and spin z-down state can be written as $\frac{1}{\sqrt{2}}(|+\rangle_i \pm e^{i\phi_i}|-\rangle_i)$ where $i = x, y$ (i.e. in the spin-x or spin-y bases). you can verify that this will give you what you want.