The Hamiltonian for a certain three-level system is represented by the matrix

$\mathit{H}\mathbf{=}\sout{\mathbf{h}}\mathit{\omega }\left[\begin{array}{ccc}1& 0& 0\\ 0& 2& 0\\ 0& 0& 2\end{array}\right]$ Two other observables, A and B, are represented by the matrices $\mathit{A}\mathbf{=}\mathit{\lambda }\left[\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 2\end{array}\right]\mathbf{,}\mathbf{}\mathit{B}\mathbf{=}\mathit{\mu }\left[\begin{array}{ccc}2& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right]$,where ω, , and μ are positive real numbers.

(a)Find the Eigen values and (normalized) eigenvectors of H, A and B.

(b) Suppose the system starts out in the generic staterole="math" localid="1656040462996" $\mathbf{|}\mathit{S}\left(0\right)\mathbf{>}\mathbf{=}\mathbf{\left(}\begin{array}{c}{\mathbf{c}}_{\mathbf{1}}\\ {\mathbf{c}}_{\mathbf{2}}\\ {\mathbf{c}}_{\mathbf{3}}\end{array}\mathbf{\right)}$

with ${\left|c_1\right|}^{\mathbf{2}}\mathbf{+}{\left|c_2\right|}^{\mathbf{2}}\mathbf{+}{\left|c_3\right|}^{\mathbf{2}}\mathbf{=}\mathbf{1}$. Find the expectation values (at t=0) of H, A, and B.

(c) What is $\mathbf{|}\mathbf{}\mathit{S}\left(t\right)\mathbf{>}$? If you measured the energy of this state (at time t), what values might you get, and what is the probability of each? Answer the same questions for A and for B.

H:

$E_1=\sout{h}\omega ,E_2=E_3=2\sout{h}\omega ;|h_1>=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right),|h_2>=\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right),|h_3>=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right).$

A:

localid="1656041551772" $$\begin{gathered} \left|\begin{array}{ccc}-a& \lambda & 0\\ \lambda & -a& 0\\ 0& 0& \left(2\lambda -a\right)\end{array}\right|=a^2\left(2\lambda -a\right)-\left(2\lambda -a\right){\lambda }^2=0\Rightarrow a_1=2\lambda ,a_2=\lambda ,a_3=-\lambda \\ \lambda \left(\begin{array}{ccc}0& 1& 0\\ 1& 0& 0\\ 0& 0& 2\end{array}\right)\left(\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right)=a\left(\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right)\Rightarrow \left\{\begin{array}{l}\lambda \beta =a\alpha \\ \lambda \alpha =a\beta \\ 2\lambda \gamma =a\gamma \end{array}\right.. \end{gathered}$$

(1)

$\begin{array}{c}\lambda \beta \\ \lambda \alpha \\ 2\lambda \gamma \end{array}\left.\begin{array}{c}=2\lambda \alpha \Rightarrow \beta =2\alpha \\ 2\lambda \beta \\ 2\lambda \gamma \end{array}\right\}\alpha =\beta =0;|a_1>=\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right).$

(2)

localid="1656042243144" $\left.\begin{array}{r}\begin{array}{c}\lambda \beta =\lambda \alpha \Rightarrow \beta =\alpha \\ \lambda \alpha =\lambda \beta \Rightarrow \alpha =\beta \\ 2\lambda \gamma =\lambda \gamma ;\Rightarrow \gamma =0\end{array}\end{array}\right\};|a_2>=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\\ 0\end{array}\right).$

(3)

$\left.\begin{array}{r}\begin{array}{c}\lambda \beta =-\lambda \alpha \Rightarrow \beta =-\alpha \\ \lambda \alpha =-\lambda \beta \Rightarrow \alpha =-\beta \\ 2\lambda \gamma =-\lambda \gamma ;\Rightarrow \gamma =0\end{array}\end{array}\right\};|a_3>=\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ -1\\ 0\end{array}\right).$

B:

$$\begin{gathered} \left|\begin{array}{ccc}\left(2\mu -b\right)& 0& 0\\ 0& -b& \mu \\ 0& \mu & -b\end{array}\right|=b^2\left(2\mu -b\right)-\left(2\mu -b\right){\mu }^2=0\Rightarrow b_1=2\mu ,b_2=\mu ,b_2=-\mu . \\ \mu \left(\begin{array}{ccc}2& 0& 0\\ 0& 0& 1\\ 0& 1& 0\end{array}\right)\left(\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right)=b\left(\begin{array}{c}\alpha \\ \beta \\ \gamma \end{array}\right)\Rightarrow \left\{\begin{array}{l}2\mu \alpha =b\alpha \\ \mu \gamma =b\beta \\ \mu \beta =b\gamma \end{array}\right.. \end{gathered}$$

(1)

role="math" localid="1656043183873" $\left.\begin{array}{r}2\mathrm{\mu \alpha }=2\mathrm{\mu \alpha }\\ \mathrm{\mu \gamma }=2\mathrm{\mu \beta }\Rightarrow \mathrm{\gamma }=2\mathrm{\beta }\\ \mathrm{\mu \beta }=2\mathrm{\mu \gamma }\Rightarrow \mathrm{\beta }=2\mathrm{\gamma }\end{array}\right\}\mathrm{\beta }=\mathrm{\gamma }=0;|{\mathrm{b}}_1>=\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right).$

(2)

$\left.\begin{array}{r}2\mathrm{\mu \alpha }=\mathrm{\mu \alpha }\Rightarrow \mathrm{\alpha }=0\\ \mathrm{\mu \gamma }=\mathrm{\mu \beta }\Rightarrow \mathrm{\gamma }=\mathrm{\beta }\\ \mathrm{\mu \beta }=\mathrm{\mu \gamma };\Rightarrow \mathrm{\beta }=\mathrm{\gamma }\end{array}\right\};|b_1>=\frac{1}{\sqrt{2}}\left(\begin{array}{c}0\\ 1\\ 1\end{array}\right).$

(3)

$\left.\begin{array}{r}2\mathrm{\mu \alpha }=-\mathrm{\mu \alpha }\Rightarrow \mathrm{\alpha }=0\\ \mathrm{\mu \gamma }=-\mathrm{\mu \beta }\Rightarrow \mathrm{\gamma }=-\mathrm{\beta }\\ \mathrm{\mu \beta }=-\mathrm{\mu \gamma };\Rightarrow \mathrm{\beta }=-\mathrm{\gamma }\end{array}\right\};|b_1>=\frac{1}{\sqrt{2}}\left(\begin{array}{c}0\\ 1\\ -1\end{array}\right).$

$$\begin{gathered} |S\left(0\right)>=c_1|h_1>+c_2|h_2>+c_3|h_3>\Rightarrow |S\left(t\right)>=c_1{e}^{-i{E}_{1}tI\sout{h}}|h_1>c_2{e}^{-i{E}_{2}tI\sout{h}}+|h_2>c_3{e}^{-i{E}_{3}tI\sout{h}}+|h_3> \\ =c_1{e}^{-i\omega t}|h_1>c_2{e}^{-2i\omega t}+|h_2>c_3{e}^{-2i\omega t}+|h_3>. \end{gathered}$$

$=e^{-2i\omega t}\left[c_1{e}^{i\omega t}\left(\begin{array}{c}1\\ 0\\ 0\end{array}\right)+c_2\left(\begin{array}{c}0\\ 1\\ 0\end{array}\right)+c_3\left(\begin{array}{c}0\\ 0\\ 1\end{array}\right)\right]=e^{-2i\omega t}\left(\begin{array}{c}{c}_1{e}^{i\omega t}\\ c_2\\ c_3\end{array}\right).$

$\mathrm{H}::h_1=\sout{h}\omega ,$probability ${\left|c_1\right|}^2;h_2=h_3=2\sout{h}\omega$, probability $\left({\left|c_2\right|}^2+{\left|c_3\right|}^2\right).$

localid="1656045665650"

Note that the sum of the probabilities is 1.

B:

Again, the sum of the probabilities is 1.