Undergrad Diagonalization and change of basis

[RicardoMP]

I have the following matrix given by a basis \left|1\right\rangle and \left|2\right\rangle:
<br /> \begin{bmatrix}<br /> E_0 &amp;-A \\<br /> -A &amp; E_0<br /> \end{bmatrix}<br />

Eventually I found the matrix eigenvalues E_I=E_0-A and E_{II}=E_0+A and eigenvectors \left|I\right\rangle = \begin{bmatrix}<br /> \frac{1}{\sqrt{2}}\\<br /> \frac{1}{\sqrt{2}}<br /> \end{bmatrix} and \left|II\right\rangle=\begin{bmatrix}<br /> \frac{1}{\sqrt{2}}\\<br /> -\frac{1}{\sqrt{2}}<br /> \end{bmatrix}.
I found out in the solutions of further problems that I can write these vectors as \left|I\right\rangle=\frac{1}{\sqrt{2}}\left|1\right\rangle+\frac{1}{\sqrt{2}}\left|2\right\rangle and\left|II\right\rangle=\frac{1}{\sqrt{2}}\left|1\right\rangle-\frac{1}{\sqrt{2}}\left|2\right\rangle
But why do we assume that \left|1\right\rangle=<br /> \begin{bmatrix}<br /> 1 \\<br /> 0<br /> \end{bmatrix}<br /> and \left|2\right\rangle=<br /> \begin{bmatrix}<br /> 0 \\<br /> 1<br /> \end{bmatrix} ?<br />
Is this canonical basis, a basis of every matrix?

 

[Orodruin]

The column matrix representation of a state $\newcommand{\ket}[1]{\left| #1 \right\rangle} \ket{\psi}$ in a basis $\ket{1}$, $\ket{2}$ is given by
$$
\newcommand{\braket}[2]{\langle #1 | #2 \rangle}
\begin{pmatrix}
\braket{1}{\psi} \\ \braket{2}{\psi}
\end{pmatrix}
$$
so clearly, if you want to represent the state $\ket{1}$ in this matrix representation, you would get
$$
\begin{pmatrix}
\braket{1}{1} \\ \braket{2}{1}
\end{pmatrix} = \begin{pmatrix} 1 \\ 0 \end{pmatrix}
$$

 

[StoneTemplePython]

to be crystal clear, you have real symmetric matrix, with eigenvectors

$\mathbf v_1 \propto \begin{bmatrix} 1\\ 1 \end{bmatrix}$

and

$\mathbf v_2 \propto \begin{bmatrix} 1\\ -1 \end{bmatrix}$

these are orthogonal to each other. If you choose to scale each by $\frac{1}{\sqrt{2}}$ then you may call them orthonormal -- in general this is extremely desirable and hence that is why it was rescaled.

But why do we assume that \left|1\right\rangle=<br /> \begin{bmatrix}<br /> 1 \\<br /> 0<br /> \end{bmatrix}<br /> and \left|2\right\rangle=<br /> \begin{bmatrix}<br /> 0 \\<br /> 1<br /> \end{bmatrix} ?<br />

This appears to be a question about conventions in the use of Dirac Notation, which is not a math question but something for the QM forums.

In general the use of standard basis vectors' coordinates is a common approach when dealing in finite dimensons. Technically for a relevant canonical form you'd look to the Jordan Canonical Form. I don't think using coordinates of the standard basis is canonical per se in math, but again your questions seems to be more about QM conventions than math.

 

[vanhees71]

No, it's an abuse of notation. A ket is an abstract vector, independent of any basis. Column vectors always refer to a basis. Obviously tacitly you assumed that the basis to use should be $\{|1 \rangle,|2 \rangle \}$. Of course the components of the basis vectors wrt. this basis itself are the "canonical basis" vectors of $\mathbb{C}^n$ (in your case $n=2$).