Find the Pauli vector's eigenvalues

Question

Question:

Find the eigenvalues of $\hat{a} \cdot \vec{\underline{\underline{\sigma}}}$.

My answer:

1- First let's note that $ (\hat{a} \cdot \vec{\underline{\underline{\sigma}}})^2 = \hat{a} \cdot \vec{\underline{\underline{\sigma}}} \cdot \hat{a} \cdot \vec{\underline{\underline{\sigma}}} = (a_x \cdot \underline{\underline{\sigma_x}} + a_y \cdot \underline{\underline{\sigma_y}} + a_z \cdot \underline{\underline{\sigma_z}} ) \cdot (a_x \cdot \underline{\underline{\sigma_x}} + a_y \cdot \underline{\underline{\sigma_y}} + a_z \cdot \underline{\underline{\sigma_z}} ) $ is equal to:
$ a_x^2 \underline{\underline{\sigma_x}} \cdot \underline{\underline{\sigma_x}} + a_x a_y \underline{\underline{\sigma_x}} \cdot \underline{\underline{\sigma_y}} + a_x a_z \underline{\underline{\sigma_x}} \cdot \underline{\underline{\sigma_z}} + $ $ a_y^2 \underline{\underline{\sigma_y}} \cdot \underline{\underline{\sigma_y}} + a_y a_x \underline{\underline{\sigma_y}} \cdot \underline{\underline{\sigma_x}} + a_y a_z \underline{\underline{\sigma_y}} \cdot \underline{\underline{\sigma_z}} + $ $a_z^2 \underline{\underline{\sigma_z}} \cdot \underline{\underline{\sigma_z}} + a_z a_x \underline{\underline{\sigma_z}} \cdot \underline{\underline{\sigma_x}} + a_z a_y \underline{\underline{\sigma_z}} \cdot \underline{\underline{\sigma_y}}$
And as $i \neq j \Rightarrow \sigma_i \cdot \sigma_j = - \sigma_j \cdot \sigma_i$ and that $ \sigma_i ^2 = I$ we have that: $(\hat{a} \cdot \vec{\underline{\underline{\sigma}}})^2 = a_x^2 \underline{\underline{I}} + a_y^2 \underline{\underline{I}} + a_z^2 \underline{\underline{I}}$. Thus by definition: $(\hat{a} \cdot \vec{\underline{\underline{\sigma}}})^2 = | \vec{a} |^2 \underline{\underline{I}}$

2- $(\hat{a} \cdot \vec{\underline{\underline{\sigma}}})^2 = | \vec{a} |^2 \Rightarrow (\hat{a} \cdot \vec{\underline{\underline{\sigma}}})^2 - | \vec{a} |^2 = 0 = (\hat{a} \cdot \vec{\underline{\underline{\sigma}}} + | \vec{a} |)(\hat{a} \cdot \vec{\underline{\underline{\sigma}}} - | \vec{a} | ) $

3- Now by definition an eigen value $\lambda $ will verify that $\hat{a} \cdot \vec{\underline{\underline{\sigma}}} | \psi > = \lambda | \psi > \Rightarrow (\hat{a} \cdot \vec{\underline{\underline{\sigma}}}) ( \hat{a} \cdot \vec{\underline{\underline{\sigma}}}) | \psi > = ( \hat{a} \cdot \vec{\underline{\underline{\sigma}}}) \lambda | \psi > = \lambda^2 | \psi > \Rightarrow (( \hat{a} \cdot \vec{\underline{\underline{\sigma}}})^2- \lambda^2) | \psi > = 0 $. Thus we have to solve $(( \hat{a} \cdot \vec{\underline{\underline{\sigma}}})^2- \lambda^2)=0$

4- But we have proved in "2-" that a solution to $(( \hat{a} \cdot \vec{\underline{\underline{\sigma}}})^2- \lambda^2)=0$ is $\lambda^2 = | \vec{a} | ^2 $. From here we can conclude that $ \lambda_{1,2} \in \left \{ | \vec{a} |, -| \vec{a} | \right \} $. More over as we know that by property $Tr \left \{ \hat{a} \cdot \vec{\underline{\underline{\sigma}}} \right \}= \lambda_1 + \lambda_2=0$ thus we can conclude that the two eigen values of $\hat{a} \cdot \vec{\underline{\underline{\sigma}}}$ are $| \vec{a} |, -| \vec{a} |$.

Q.E.D.

Is this correct?

Answer 1

According to https://math.stackexchange.com/users/147873/winther my solution is correct.

In order to add information we can write that the 2 corresponding projector $P_{\pm}$ satisfying $\hat{a} \cdot \vec{ \sigma }= \lambda_+ P_+ + \lambda_- P_-$,are $P_{\pm} = \frac{1 \pm \hat{a} \cdot \vec{ \sigma }}{2} $.

Indeed $\hat{a} \cdot \vec{ \sigma }$ is an Hermitian matrix (easy to prove )so we can write $\hat{a} \cdot \vec{ \sigma }= \lambda_+ P_+ + \lambda_- P_-$. $\hat{a}$ is a unit vector so $|\hat{a}|=1$ so $ \hat{a} \cdot \vec{ \sigma } = P_+ - P_-$ and so from here it is easy to guess that $P_{\pm} = \frac{1 \pm \hat{a} \cdot \vec{ \sigma }}{2}$.