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Consider a two-state system:

The Hamiltonian takes the general form $$ H= \begin{pmatrix} a_1 & c-id \\ c+id & a_2 \\ \end{pmatrix} $$ where $a_1, a_2, c, d \in\mathcal{R}$

$H$ can be written as $$ H=a.\sigma_0+c.\sigma_1+d.\sigma_2+b.\sigma_3 $$ where $\sigma_i$ are the Pauli matrices and $a=\frac{(a_1+a_2)}{2}$, $b=\frac{(a_1-a_2)}{2}$ and $a, b \in \mathcal{R}$

$$ H=a.\sigma_0+\vec{r}.\vec{\sigma} $$ where $\vec{r}=(c,d,b)$ and $\vec{\sigma}=(\sigma_1,\sigma_2,\sigma_3)$

Given the time-independent Hamiltonian $H$ what is the general way of proving that the eigenvalues are:

$E^{\pm}=a\pm|\vec{r}|$ ?

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  • $\begingroup$ These equations are quite confusing $a=\frac{(a+b)}{2}$ and $b=\frac{(a-b)}{2}$. $\endgroup$ – vnd Feb 16 '16 at 8:41
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    $\begingroup$ @vnd i'm really srry...i have edited it.pls check $\endgroup$ – ss1729 Feb 16 '16 at 8:43
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$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ $\newcommand{\ket}[1]{\left\vert\,{#1}\,\right\rangle}$ \begin{align} \pars{a\,\sigma_{0} + \vec{r}\cdot\vec{\sigma}}\ket{\Psi} = E\ket{\Psi} & \implies\quad \vec{r}\cdot\vec{\sigma}\ket{\Psi} = \pars{E - a}\ket{\Psi} \label{1}\tag{1} \\[5mm] \implies \pars{\vec{r}\cdot\vec{\sigma}}^{2}\ket{\Psi} = \pars{E - a}\vec{r}\cdot\vec{\sigma}\ket{\Psi} & \implies\quad \vec{r}\cdot\vec{\sigma}\ket{\Psi} = {r^{2} \over E-a}\ket{\Psi} \label{2}\tag{2} \end{align}


Compare \eqref{1} and \eqref{2}: $$ E - a = {r^{2} \over E - a}\quad\implies\quad \bbox[#ffb,5px,border:2px groove navy]{\ E = a \pm r = a \pm\verts{\vec{r}}\ } $$

Note that $$ \pars{\vec{r}\cdot\vec{\sigma}}^{2} = \vec{r}\cdot\vec{\sigma}\,\,\vec{r}\cdot\vec{\sigma} = \vec{r}\cdot\vec{r} + \ic\,\vec{\sigma}\cdot\underbrace{\vec{r} \times \vec{r}}_{\ds{=\ \vec{0}}} = r^{2} $$

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The basis vectors are $\psi_1 = \begin{bmatrix}1\\0\end{bmatrix}$ and $\psi_2=\begin{bmatrix}0\\1\end{bmatrix}$. A general state is $$\psi=\eta\psi_1 + \mu\psi_2.$$All one needs to do is solve the Eigenvalue Equation $H\psi=E\psi$. If you strictly want to use the pauli matrices for calculation, then one has to work out the action of the pauli matrices on the basis states. It is not very difficult to verify that $\sigma_0 \psi_{1,2}=\psi_{1,2}$, $\sigma_1 \psi_{1,2}=\psi_{2,1}$, $\sigma_2 \psi_{1,2}=\pm i\psi_{2,1}$ and $\sigma_3 \psi_{1,2}=\pm\psi_{1,2}$. Using these relations and the expression of the hamiltonian in terms of the pauli matrices, one can obtain $$H\psi=\left[\eta a+\mu c-i\mu d+\eta b\right]\psi_1+\left[a\mu + \eta c+i\eta d-\mu b\right]\psi_2=E\eta\psi_1+E\mu\psi_2.$$ From the second equality in the above equation, and using the fact that $\psi_1$ and $\psi_2$ are independent, you get the following set of equations in $\eta$ and $\mu$.$$\eta\left[a+b-E\right]+\mu\left[c-id\right]=0$$ $$\eta\left[c+id\right]+\mu\left[a-b-E\right]=0.$$ The condition that the coefficient matrix has vanishing determinant for non trivial values of $\eta$ and $\mu$ gives the required eigenvalues.

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