- #1

RicardoMP

- 49

- 2

[itex]

\begin{bmatrix}

E_0 &-A \\

-A & E_0

\end{bmatrix}

[/itex]

Eventually I found the matrix eigenvalues [itex]E_I=E_0-A[/itex] and [itex]E_{II}=E_0+A[/itex] and eigenvectors [itex]\left|I\right\rangle = \begin{bmatrix}

\frac{1}{\sqrt{2}}\\

\frac{1}{\sqrt{2}}

\end{bmatrix} and \left|II\right\rangle=\begin{bmatrix}

\frac{1}{\sqrt{2}}\\

-\frac{1}{\sqrt{2}}

\end{bmatrix}[/itex].

I found out in the solutions of further problems that I can write these vectors as [itex]\left|I\right\rangle=\frac{1}{\sqrt{2}}\left|1\right\rangle+\frac{1}{\sqrt{2}}\left|2\right\rangle[/itex] and[itex]\left|II\right\rangle=\frac{1}{\sqrt{2}}\left|1\right\rangle-\frac{1}{\sqrt{2}}\left|2\right\rangle[/itex]

But why do we assume that [itex]\left|1\right\rangle=

\begin{bmatrix}

1 \\

0

\end{bmatrix}

[/itex] and [itex]\left|2\right\rangle=

\begin{bmatrix}

0 \\

1

\end{bmatrix} ?

[/itex]

Is this canonical basis, a basis of every matrix?