# 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

## 1 Answer

Why are you worried? Why do you think the solution must be unique?

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.

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