# Hamiltonian States (3-325)

Consider a system with a hamiltonian

$H=\frac{1}{\sqrt{2}}\begin{pmatrix}1&-i&0\\1&3&3\\0&3&0\end{pmatrix}$ in the initial state given by $|\psi_0>=\begin{pmatrix}4-1\\-2+5i\\3+2i\end{pmatrix}$

a) If the energy is measured, which values will be obtained, and with which probabilities?

b) Find the state of a system in the latter time.

$H$ is hermitian $H=(H^T )^*$ and therefore it has real eigenvalues. Its eigenvalues and normalized eigenvectors (see question 3) are

$E_1=5$;  $E_2=-2$;  $E_3=0$

$|ϕ_1> =\frac{1}{√35}*\begin{pmatrix}-i\\5\\3\end{pmatrix}$    ;$|ϕ_2> =\frac{1}{√14}\begin{pmatrix}-i\\-2\\3\end{pmatrix}$  ;$|ϕ_3> =\frac{1}{√10}*\begin{pmatrix}3i\\0\\1\end{pmatrix}$
$H$ is hermitic so that its eigenvectors are orthogonal. The initial state $|ψ_0>$ can be written as a linear combination of the eigenvectors

$|ψ_0>=i√35*|ϕ_1>+√14*|ϕ_2> -i√10*|ϕ_3>$

Normalization of the initial state yields

$|Ψ_0>=i\sqrt{35/59}*|ϕ_1>+\sqrt{14/59}*|ϕ_2> -i\sqrt{10/59}*|ϕ_3>$

Therefore the values obtained after measuring the energy are

$E_1=5$;  $4E_2=-2$;  $E_3=0$

With the probabilities:

$P_1=|i\sqrt{35/59}|^2=35/59$   ;$P_2=14/59$;  $P_3=10/59$

b)

At a later time $t>0$ taking $ℏ=1$ we have

$|Ψ_0>=i\sqrt{\frac{35}{59}}|ϕ_1>exp⁡(-iE_1 t) +\sqrt{\frac{14}{59}}|ϕ_2>exp⁡(-iE_2 t) -i\sqrt{\frac{10}{59}}|ϕ_3>exp⁡(-iE_3 t)$

$|Ψ_0>=i\sqrt{\frac{35}{59}}|ϕ_1>exp⁡(-5it) +\sqrt{\frac{14}{59}}|ϕ_2>exp⁡(2it) -i\sqrt{\frac{10}{59}}|ϕ_3>$

or writing $i$ as $\exp(…)$:

$|Ψ_0>=\sqrt{\frac{35}{59}}|ϕ_1>exp⁡(i(-5t+π/2) ) +\sqrt{\frac{14}{59}}|ϕ_2>exp⁡(2it)+\sqrt{\frac{10}{59}}|ϕ_3>exp⁡(i*3π/2)$