# Probability of State (Homework 3-325)

The components of the state of a system $|\varphi_1>$ in some basis $|\phi_1>, |\phi_2>,|\phi_3>$ are given by

$<\phi_1|\varphi_1>=1/\sqrt{3}$, $<\phi_2|\varphi_1>=0$

Find the probability of finding the system in state $|\varphi_2>$ whose components in the same basis are

$<\phi_1|\varphi_1>=(1+i)/\sqrt{3}$, $<\phi_2|\varphi_1>=\sqrt{1/6}$, $<\phi_3|\varphi_1>=\sqrt{1/6}$

The state $|φ_1>$ can be written as

$|φ_1>=i/√3*| ϕ_1>+√(2/3)*|ϕ_2>$

The state $|φ_2>$ can be written as

$|φ_2>=(1+i)/√3*|ϕ_1>+√(1/6)*|ϕ_2>+√(1/6)*|ϕ_3>$

The probability to find the system described by $|φ_1>$ in the state $|φ_2>$ is

$P=\frac{(|<φ_2 | φ_1>|^2)}{(<φ_1 |φ_1>)}=\frac{|i/√3*(1-i)/√3+√(2/3)*√(1/6)+0|^2}{(1/3+2/3)}=|i+1|^2/3^2 +√2^2/√18^2 =$

$=√2^2/9+1/9=1/3$

Consider a system with a Hamiltonian

$H=(1/\sqrt{2})*\begin{pmatrix}0 & -i & 0\\ i & 3 & 3\\ 0 & 3 & 0\end{pmatrix}$

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

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

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

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

$E_1=5$; $E_2=-2$; $E_3=0$

$|ϕ_1> =(1/√35)*(-i, 5,3)$

$|ϕ_2> =(1/√14)*(-i,-2,3)$

$|ϕ_3> =(1/√10)*(3i,0,1)$

$H$ 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

$|Ψ_0>=i√(35/59)*|ϕ_1>+√(14/59)*|ϕ_2> -i√(10/59)*|ϕ_3>$

Therefore the values obtained after measuring the energy are

$E_1=5; E_2=-2; E_3=0$

With the probabilities:

$P_1=|i√(35/59)|^2=35/59 ;P_2=14/59; P_3=10/59$

b)

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

$|Ψ_0>=i√(35/59)*|ϕ_1>*exp⁡(-i E_1 t) +√(14/59)*|ϕ_2>*exp⁡(-i E_2 t) -$

$-i√(10/59)*|ϕ_3>*exp⁡(-i E_3 t)$

$|Ψ_0>=i√(35/59)*|ϕ_1>*exp⁡(-5it) +√(14/59)*|ϕ_2>*exp⁡(2it) -i√(10/59)*|ϕ_3>$

$|Ψ_0>=√(35/59)*|ϕ_1>*exp⁡(i(-5t+π/2) ) +√(14/59)*|ϕ_2>*exp⁡(2it)+$

$+√(10/59)*|ϕ_3>*exp⁡(i*3π/2)$