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
$|Ψ_0>=i√(35/59)*|ϕ_1>*exp(-i E_1 t) +√(14/59)*|ϕ_2>*exp(-i E_2 t) -$
$-i√(10/59)*|ϕ_3>*exp(-i E_3 t)$
$|Ψ_0>=i√(35/59)*|ϕ_1>*exp(-5it) +√(14/59)*|ϕ_2>*exp(2it) -i√(10/59)*|ϕ_3>$
$|Ψ_0>=√(35/59)*|ϕ_1>*exp(i(-5t+π/2) ) +√(14/59)*|ϕ_2>*exp(2it)+$
$+√(10/59)*|ϕ_3>*exp(i*3π/2)$
