Matrix Hamiltonian

Consider a system whose Hamiltonian is given by the matrix

$H=ϵ\begin{pmatrix}0&-i&0\\i&0&2i\\0&-2i&0\end{pmatrix}$

a) If we measure energy, which values we will obtain?

b) If the system starts in the state $(1/√3) \begin{pmatrix}1\\1\\1\end{pmatrix}$ what are the probabilities of getting various values of the energy?

c) Write down the ket vector describing time evolution of this state.

d) if the result of measurement of the energy produced value $E=\sqrt{5}\epsilon_0$, and immediately afterward an observable presented by matrix $a*\begin{pmatrix}0&-i&0\\i&1&1\\0&1&0\end{pmatrix}$ is measured, what is the probability of getting value of this observable equal to zero?

$H=ϵ\begin{pmatrix}0&-i&0\\i&0&2i\\0&-2i&0\end{pmatrix}$  and the state $|ψ>=(1/√3) \begin{pmatrix}1\\1\\1\end{pmatrix}$

$H$ is hermitian because $H=(H^T )^*$  so that is has real eigenvalues.

Eigenvalues and eigenvectors of H are

$E_1=-√5ϵ$   ;       $E_2=√5ϵ$   ;       $E_3=0$
$|ϕ_1>=\sqrt{2/5}\begin{pmatrix}1/2\\-(i√5)/2\\1\end{pmatrix}$   ; $|ϕ_2>=\sqrt{2/5}\begin{pmatrix}1/2\\(i√5)/2\\1\end{pmatrix}$;  $|ϕ_3>=(1/√5)\begin{pmatrix}-2\\0\\1\end{pmatrix}$

a)

Possible energies that are measured are its eigenvalues : $±√5 ϵ$  and $0$

b)

To find the probabilities of different energies we need to write

$|ψ> =c_1|ϕ_1> +c_2|ϕ_2> +c_3|ϕ_3>$

So that

$c_1=<ϕ_1 |ψ>=(3+i√5)/√30$   ; $c_2=<ϕ_2 |ψ> =(3-i√5)/√30$   and $c_3=1/√15$

The probabilities of E1, E2 and E3 are just

$P_1=|c_1 |^2=14/30$;   $P_2=|c_2 |^2=14/30$;  $P_3=|c_3 |^2=1/15$   with $P_1+P_2+P_3=1$
c)

The ket vector for the time evolution is

$|ψ(t)>=c_1 | ϕ_1>\exp⁡(-(i E_1)/ℏt)+c_2 |ϕ_2>\exp⁡(-(i E_2)/ℏt)+c_3 | ϕ_3>\exp⁡(-(i E_3)/ℏt)$

d)

If we measure the energy $E_2=√5ϵ$ , after the measurement the system is left in the state $|ϕ_2>$.

The eigenvalues of

$A=a*\begin{pmatrix}0&-i&0\\i&1&1\\0&1&0\end{pmatrix}$

Are

$a_1=2$;  $a_2=-1$ and $a_3=0$

And the corresponding eigenvectors are

$|a_1>=(1/2)\begin{pmatrix}-i\\2\\1\end{pmatrix}$;   $|a_2>=(1/√3)\begin{pmatrix}-i\\-1\\1\end{pmatrix}$  and $|a_3>=(1/√2)\begin{pmatrix}i\\0\\1\end{pmatrix}$
The probability of getting value $a_3=0$ is just

$P_3’=|<a_3 | ϕ_2>|^2=\left |\frac{1}{√2}(-i ,0 ,1)*\sqrt{\frac{2}{5}}\begin{pmatrix}1/2\\(i√5)/2\\1\end{pmatrix} \right |^2=$

$=|(-i/√20)+(1/√5)|^2=|(2-i)/√20|^2=5/20=1/4$