Question

The Hamiltonian for a certain two-level system is

$\hat{\mathbf{H}}\mathbf{=}\dot{\mathbf{o}}\left(\overline{)1><\overline{)1}-\overline{)2><\overline{)2}+\overline{)1}><\overline{)2}+2><\overline{)1}}}\right)$

where$\overline{)\mathbf{1}\mathbf{>}\mathbf{,}\overline{)\mathbf{2}}\mathbf{>}}$is an orthonormal basis and localid="1658120083298" $\dot{\mathbf{o}}$ is a number with the dimensions of energy. Find its eigenvalues and eigenvectors (as linear combinations oflocalid="1658120145851" $\overline{)\mathbf{1}\mathbf{>}}$ and$\overline{)\mathbf{2}\mathbf{>}}$ . What is the matrix H representing$\hat{\mathbf{H}}$ with respect to this basis?

Short answer

Eigenvalues are ${\lambda }_1$and ${\lambda }_2$, and corresponding eigenvectors:

$$\begin{gathered} {\lambda }_1=-\sqrt{2\epsilon } \\ {\phi }_1={\left(4-2\sqrt{2}\right)}^{-1/2}\left[\left(1-\sqrt{2}\right)\overline{)1>+\overline{)2>}}\right] \\ {\lambda }_2=-\sqrt{2\epsilon } \\ {\phi }_2={\left(4+2\sqrt{2}\right)}^{-1/2}\left[\left(1+\sqrt{2}\right)\overline{)1>+\overline{)2>}}\right] \end{gathered}$$

Matrix of given Hamiltonian in new basis:

$H=\left(\begin{array}{cc}-\sqrt{2\epsilon }& 0\\ 0& \sqrt{2\epsilon }\end{array}\right)$

Step-by-step solution

Concept of Eigenvalue

The term "eigenvalues" refers to the precise collection of scalars connected to the system of linear equations. A characteristic value, a characteristic root, suitable values, or latent roots are other names for eigenvalues. An eigenvector is changed into a scalar by using an eigenvalue.

Find the matrix of the Hamiltonian.

Eigenvalues and corresponding eigenvectors of following Hamiltonian is

$\hat{\mathrm{H}}=\epsilon \left(\overline{)1><\overline{)1}-\overline{)2><\overline{)2}+\overline{)1}><\overline{)2}+2><\overline{)1}}}\right)$

where $\overline{)1>,\overline{)2>}}$is an orthonormal basis, and $\epsilon$is energy.

First, determine the matrix of Hamiltonian as:

$$\begin{gathered} \overline{)1>}<\overline{)1}=\left(\begin{array}{c}1\\ 0\end{array}\right).\left(\begin{array}{cc}1& 0\end{array}\right)=\left(\begin{array}{cc}1& 0\\ 0& 0\end{array}\right) \\ \overline{)2>}<\overline{)2}=\left(\begin{array}{c}0\\ 1\end{array}\right).\left(\begin{array}{cc}0& 1\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 1\end{array}\right) \\ \overline{)1>}<\overline{)2}=\left(\begin{array}{c}1\\ 0\end{array}\right).\left(\begin{array}{cc}0& 1\end{array}\right)=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right) \\ \overline{)2>}<\overline{)1}=\left(\begin{array}{c}0\\ 1\end{array}\right).\left(\begin{array}{cc}1& 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right) \end{gathered}$$

So, the matrix of Hamiltonian is:

$H=\left(\begin{array}{cc}\epsilon & \epsilon \\ \epsilon & \epsilon \end{array}\right)$

Diagonalize H 

To find its eigenvalues ${\lambda }_i$and corresponding eigenvectors ${\phi }_i$diagonalize H:

$$\begin{gathered} H{\phi }_i={\lambda }_i{\phi }_i \\ \left(H-{\lambda }_i.I\right){\phi }_i=0 \\ \left(\begin{array}{cc}\epsilon -{\lambda }_i& \epsilon \\ \epsilon & -\epsilon -{\lambda }_i\end{array}\right).\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0\left(1\right) \end{gathered}$$

To obtain eigenvalues, the determinant of $\left(H-{\lambda }_i.I\right)$must vanish:

$\left|\begin{array}{cc}\epsilon -{\lambda }_i& \epsilon \\ \epsilon & -\epsilon -{\lambda }_i\end{array}\right|=0$

$$\begin{gathered} \left(\epsilon -{\lambda }_i\right)\left(\epsilon -{\lambda }_i\right)-{\epsilon }^2=0 \\ -\left(\epsilon -{\lambda }_i\right)\left(\epsilon +{\lambda }_i\right)-{\epsilon }^2=0 \end{gathered}$$

$$\begin{gathered} -{\epsilon }^2+{\lambda }_i^2-{\epsilon }^2=0 \\ {\lambda }_i^2=2{\epsilon }^2 \end{gathered}$$

Find the first eigenvector φ1  with a lower eigenvalue λ1=-2ε 

To find eigenvectors solve the system of equations (1) for $a_1$and $a_2$:

${\lambda }_1=-\sqrt{2\epsilon }\left(\begin{array}{cc}\epsilon +\sqrt{2\epsilon }& \epsilon \\ \epsilon & -\epsilon +\sqrt{2\epsilon }\end{array}\right).\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0$

$$\begin{gathered} \epsilon (1+\sqrt{2})a_1+\epsilon a_2=0\left(2\right) \\ \epsilon a_1+\epsilon (\sqrt{2}-1=0(3) \end{gathered}$$

Subtract two previous formulas to get:

$$\begin{gathered} \epsilon \sqrt{2}{a}_1+\epsilon \left(2-\sqrt{2}\right)=0 \\ \sqrt{2}a=\left(\sqrt{2}-2\right)a_2 \\ {a}_1=\left(1-\sqrt{2}\right)a_2 \end{gathered}$$

Choose $a_2=1$and eigenvector ${\phi }_1$with this eigenvalue is:

${\phi }_1=\frac{1}{N_1}\left(\begin{array}{c}1-\sqrt{2}\\ 1\end{array}\right)$

where $N_1$is normalization which has to be calculated:

$$\begin{gathered} N_1^2=a_1^2+a_2^2 \\ ={\left(1-\sqrt{2}\right)}^2+1 \\ =1-2\sqrt{2}+2+1 \\ =4-2\sqrt{2}{N}_1 \\ =\sqrt{4-2\sqrt{2}} \end{gathered}$$

Finally, the first eigenvector ${\phi }_1$with a lower eigenvalue ${\lambda }_1=-\sqrt{2}\epsilon$ is:

${\phi }_1={\left(4-2\sqrt{2}\right)}^{-1/2}\left(\begin{array}{c}1-\sqrt{2}\\ 1\end{array}\right)$

In basis${\phi }_1={\left(4-2\sqrt{2}\right)}^{-1/2}\left[\left(1-\sqrt{2}\right)\overline{)1>+\overline{)2>}}\right]$

Find a second eigenvector φ2  with a higher eigenvalue λ2=2ε 

$$\begin{gathered} {\lambda }_2=\sqrt{2\epsilon }\left(\begin{array}{cc}\epsilon -\sqrt{2}\epsilon & \epsilon \\ \epsilon & -\epsilon -\sqrt{2}\epsilon \end{array}\right).\left(\begin{array}{c}{a}_1\\ a_2\end{array}\right)=0 \\ \epsilon \left(1-\sqrt{2}\right)a_1+\epsilon a_2=0\left(4\right) \\ \epsilon a_1-\epsilon \left(\sqrt{2}+1\right)=0\left(5\right) \end{gathered}$$

Subtract two previous formulas and get the result:

$$\begin{gathered} -\epsilon \sqrt{2}{a}_1+\epsilon (2+\sqrt{2})=0 \\ \sqrt{2}{a}_1=\left(\sqrt{2}+2\right)a_2 \\ {a}_1=\left(1+\sqrt{2}\right)a_2 \end{gathered}$$

Again, choose $a_2=1$, and eigenvector ${\phi }_2$ with this eigenvalue is:

${\phi }_2=\frac{1}{N_2}\left(\begin{array}{c}1+\sqrt{2}\\ 1\end{array}\right)$

Here $N_2$ is the normalization which has to be calculated:

$$\begin{gathered} N_2^2=a_1^2+a_2^2 \\ ={\left(1+\sqrt{2}\right)}^2+1 \\ =1+2+\sqrt{2}+2+1 \\ =4+2\sqrt{2}{N}_2 \\ =\sqrt{4+2\sqrt{2}} \end{gathered}$$

Finally, the second eigenvector ${\phi }_2$with a higher eigenvalue ${\lambda }_2=\sqrt{2}\epsilon$ is:

${\phi }_2={\left(4+2\sqrt{2}\right)}^{-1/2}\left(\begin{array}{c}1+\sqrt{2}\\ 1\end{array}\right)$

In basis $\overline{)1>,\overline{)2>:}}$

${\phi }_2={\left(4+2\sqrt{2}\right)}^{-1/2}\left[\left(1+\sqrt{2}\right)\overline{)1>\overline{)2>}}\right]$

Matrix of given Hamiltonian in a new basis

When the orthonormal basis consists of eigenvectors of some matrix, then the matrix is diagonal in that basis. Elements of the matrix are corresponding eigenvalues. So, the matrix of Hamiltonian in the basis of eigenvectors ${\phi }_1,{\phi }_2$ is:

$$\begin{gathered} H=\left(\begin{array}{cc}{\lambda }_1& 0\\ 0& {\lambda }_2\end{array}\right) \\ =\left(\begin{array}{cc}-\sqrt{2\epsilon }& 0\\ 0& \sqrt{2\epsilon }\end{array}\right) \end{gathered}$$

Thus, Eigenvalues are ${\lambda }_1$and ${\lambda }_2$, and corresponding eigenvectors:

$$\begin{gathered} {\lambda }_1=-\sqrt{2}\epsilon \\ {\phi }_1={\left(4-2\sqrt{2}\right)}^{-1/2}\left[\left(1-\sqrt{2}\right)\overline{)1>+\overline{)2>}}\right] \\ {\lambda }_2=-\sqrt{2\epsilon } \\ {\phi }_2={\left(4+2\sqrt{2}\right)}^{-1/2}\left[\left(1+\sqrt{2}\right)\overline{)1>+\overline{)2>}}\right] \end{gathered}$$

Matrix of given Hamiltonian in new basis:

$H=\left(\begin{array}{cc}-\sqrt{2\epsilon }& 0\\ 0& \sqrt{2\epsilon }\end{array}\right)$