- #1
RicardoMP
- 49
- 2
I have the following matrix given by a basis [itex]\left|1\right\rangle[/itex] and [itex]\left|2\right\rangle[/itex]:
[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?
[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?