# Ket vectors. Eigenvalues

1. For any two kets $|ψ>$ and $|χ>$ that have finite norm, show that

$Tr(|ψ><χ|)=<χ|ψ>$

(Hint: chose matrix representation and use an arbitrary basis)

$<χ|ψ>=(χ_1^*, χ_2^*, χ_3^*,….)\begin{pmatrix}ψ_1\\ψ_2\\ψ_3&…\end{pmatrix}=χ_1^* ψ_1+χ_2^* ψ_2+χ_3^* ψ_3+⋯$.

$|χ><ψ|=\begin{pmatrix}χ_1^* ψ_1&χ_1^* ψ_2&χ_1^* ψ_3……\\χ_2^* ψ_1&χ_2^* ψ_2&χ_2^* ψ_3\\χ_3^* ψ_1&χ_3^* ψ_2&χ_3^* ψ_3 )\\……\end{pmatrix}$  so  $Tr(|χ><ψ| )=χ_1^* ψ_1+χ_2^* ψ_2+χ_3^* ψ_3+⋯$.

$<χ|ψ>=Tr(|χ><ψ| )$

2. Consider the matrix $A=\begin{pmatrix}0&i\\-i&0\end{pmatrix}$

a) Find the eigenvalues and the normalized eigenvectors for the matrix $A$

b) Check orthogonality of these vectors.

c) Find the matrices representing the operators $|a_1><a_1|$, $|a_2><a_2|$ where $|a_i>$, $i=1,2$ are eigenvectors found in a)

d) Using found matrices check the closure relation $|a_1><a_1|+|a_2><a_2|=I$ , i.e. verify that the eigenvectors form a basis (orthonormal and complete set of vectors).

e) Rewrite matrices found in (c) using vectors $|a_i>$, $i=1,2$ as a basis.

$A=\begin{pmatrix}0&i\\-i&0\end{pmatrix}$

Eigenvalues $λ$  are

$det⁡(A-Iλ)=0$  or $\begin{vmatrix}–λ&i\\-i&-λ)\end{vmatrix}=0$  or $λ^2-1=0$  so that $λ_1=1$ and $λ_2=-1$

Eigenvectors are $X$ where $(A-λI)*X=0$   so that

$λ=1$:  $\begin{pmatrix}-1&i\\-i&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0$  or $\left\{\begin{matrix}x+iy=0\\-ix-y=0\end{matrix}\right.$  or $x=α$  and $y=-iα$  so that

$X=α*\begin{pmatrix}1\\-i\end{pmatrix}$

$λ=-1$:  $\begin{pmatrix}1&i\\-i&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0$  so that $\left\{\begin{matrix}x+iy=0\\-ix+y=0\end{matrix}\right.$   or $x=α$  and $y=iα$ so that

$X=α\begin{pmatrix}1\\i\end{pmatrix}$

Orthogonality check: the matrix constructed with these two vectors as columns has determinant nonzero.

$\begin{vmatrix}1&1\\-i&i\end{vmatrix}=i+i=2i? 0$

c)

$|a_1>=\begin{pmatrix}1\\-i\end{pmatrix}$  ,$<a_1 |=(1,-i)$   and   $|a_2>=\begin{pmatrix}1\\i\end{pmatrix}$,$<a_2 |=(1,i)$

$|a_1><a_1 |=\begin{pmatrix}1^* 1&1^* (-i)\\(-i)^* 1&(-i)^* (-i)\end{pmatrix}=\begin{pmatrix}1&-i\\i&1\end{pmatrix}$

$|a_2><a_2 |=\begin{pmatrix}1^* 1&1^* i\\(i^*)1&(i^*)i\end{pmatrix}=\begin{pmatrix}1&i\\-i&1\end{pmatrix}$
d)

$|a_1><a_1 |+|a_2><a_2 |=\begin{pmatrix}2&0\\0&2\end{pmatrix}=2\begin{pmatrix}1&0\\0&1\end{pmatrix}=2I$

e)

$|a_1><a_1 |=|a_2 >^(-1)*I*|a_2>$  and $|a_2><a_2 |=|a_1 >^(-1)*I*|a_1>$