# Eigenstates of a Hamiltonian [closed]

For a particle with a spin of 1/2, which was exposed to both magnetic fields

$$B_{0}=B_{z}e_z$$ and $$B_1=B_xe_x$$

I already found the eigenvalues of its Hamiltonian which is given by $$$$H = -\gamma (B_0+B_1)\cdot S=-\dfrac{\gamma \hbar}{2} (B_x\sigma_x+B_z\sigma_z)$$$$

and they were

$$E_{\pm}=\mp \dfrac{\gamma \hbar}{2} \sqrt{B_z^2+B_x^2}$$

can anyone guide me to find its eigenstates?

## closed as off-topic by John Rennie, Aaron Stevens, By Symmetry, Jon Custer, ZeroTheHeroOct 23 '18 at 8:03

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Hint: Consider a matrix $$\mathbf{A}$$ with eigenvalue $$\lambda$$ and eigenvector $$\mathbf{u}$$.
\begin{align} \mathbf{A} \mathbf{u} \; = \; \lambda \mathbf{u} \end{align} Since your question is for $$2 \times 2$$ matrix, we explicitly write it as \begin{align} \underbrace{\begin{pmatrix} a & b\\ c & d \end{pmatrix}}_{\mathbf{A}} \; \underbrace{ \begin{pmatrix} u_{1}\\ u_{2} \end{pmatrix} }_{\mathbf{u}} \; = \; \lambda \; \underbrace{ \begin{pmatrix} u_{1}\\ u_{2} \end{pmatrix} }_{\mathbf{u}} \end{align} For each eigenvalue $$\lambda$$, we are ready to solve $$u_{1}$$ and $$u_{2}$$ \begin{align} au_{1} + bu_{2} = \lambda u_{1} \tag{1}\\ cu_{1} + du_{2} = \lambda u_{1} \tag{2} \end{align} Notice that Eq. (1) and (2) are linearly dependent in this case. So you actually need one equation only (say Eq. (1)). Thus, we have \begin{align} b u_{2} \; = \; (\lambda - a)u_{1} \end{align} Now, it seems that we don't have unique solution. Yes, remember that eigenvector is in general not unique since the vector norm is not fixed.
We pick $$u_{1} = 1$$, then we have $$u_{2} = (\lambda - a)/b$$. Since we are talking a quantum state, we have to normalize the eigenvector by $$\sqrt{u_{1}^{2} + u_{2}^{2}}$$.