Question 3.P.7: Consider a system whose Hamiltonian H and an operator A are ...

Consider a system whose Hamiltonian H and an operator A are given by the matrices

H=\varepsilon _{0}\left(\begin{matrix} 1 & -1 & 0 \\ -1 & 1 & 0 \\ 0 & 0 & -1 \end{matrix} \right),      A=a \left(\begin{matrix} 0 & 4 & 0 \\ 4 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right),

where \varepsilon _{0} has the dimensions of energy.

(a) If we measure the energy, what values will we obtain?

(b) Suppose that when we measure the energy, we obtain a value of -\varepsilon _{0}. Immediately
afterwards, we measure A. What values will we obtain for A and what are the probabilities corresponding to each value?
(c) Calculate the uncertainty Δ A.

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(a) The possible energies are given by the eigenvalues of H. A diagonalization of H yields three nondegenerate eigenenergies E_{1}=0,E_{2}=-\varepsilon _{0},and E_{3}=2\varepsilon . The respective eigenvectors are

|\phi _{1}〉=\frac{1}{\sqrt{2}}\left(\begin{matrix} 1 \\ 1 \\ 0 \end{matrix} \right),       |\phi _{2}〉=\left(\begin {matrix} 0 \\ 0 \\ 1 \end{matrix} \right),       |\phi _{3}〉=\frac{1}{\sqrt{2}}\left(\begin{matrix} -1 \\ 1 \\ 0 \end{matrix} \right);      (3.183)

these eigenvectors are orthonormal.

(b) If a measurement of the energy yields -\varepsilon _{0}, his means that the system is left in the state |\phi _{2}〉. When we measure the next observable, A, the system is in the state |\phi _{2}〉. The result we obtain for A is given by any of the eigenvalues of A. A diagonalization of A yields three nondegenerate values: a_{1}=-\sqrt{17}a, a_{2}=0, and a_{3} = \sqrt{17}a;their respective eigenvectors are given by

|a_{1}〉=\frac{1}{\sqrt{34}}\left(\begin{matrix} 4 \\ -\sqrt{17} \\ 1 \end{matrix} \right),       |a_{2}〉=\frac{1}{\sqrt{17}} \left(\begin{matrix} 1 \\ 0 \\ -4 \end{matrix} \right),          |a_{3}〉=\frac{1}{\sqrt{2}}\left(\begin{matrix} 4 \\ \sqrt{17} \\ 1 \end{matrix} \right).              (3.184)

Thus, when measuring A on a system which is in the state |\phi _{2}〉, the probability of finding -\sqrt{17}a is given by

P_{1}(a_{1})=\left|〈a_{1}|\phi _{2}〉\right| ^{2}=\left|\frac{1}{\sqrt{34}}\left(\begin{matrix} 4 & -\sqrt{17} & 1\end{matrix} \right) \left(\begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right) \right| ^{2}=\frac {1}{34}.                (3.185)

Similarly, the probabilities of measuring 0 and \sqrt{17}a are

P_{2}(a_{2})=\left|〈a_{2}|\phi _{2}〉\right| ^{2}=\left|\frac{1}{\sqrt{17}}\left(\begin{matrix} 1 & 0 & -4\end{matrix} \right)\left (\begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right) \right| ^{2}=\frac{16}{17},              (3.186)

 

P_{3}(a_{3})=\left|〈a_{3}|\phi _{2}〉\right| ^{2}=\left|\frac{1}{\sqrt{34}}\left(\begin{matrix} 4 & \sqrt{17} & 1\end{matrix} \right)\left(\begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right) \right| ^{2}=\frac{1}{34}.              (3.187)

(c) Since the system, when measuring A is in the state |\phi _{2}〉, the uncertainty Δ A is given by \Delta A =\sqrt{\left|〈\phi _{2}|A^{2}|\phi _{2}〉\right| ^{2}-\left|〈\phi _{2}|A|\phi _{2}〉\right| ^{2}} , where

〈\phi _{2}|A|\phi _{2}〉=a\left(\begin{matrix} 0 & 0 & 1\end{matrix} \right)\left(\begin{matrix} 0 & 4 & 0 \\ 4 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right)\left(\begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right)=0,                 (3.188)

 

〈\phi _{2}|A^{2}|\phi _{2}〉=a^{2}\left(\begin{matrix} 0 & 0 & 1\end{matrix} \right)\left(\begin{matrix} 0 & 4 & 0 \\ 4 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right)\left(\begin{matrix} 0 & 4 & 0 \\ 4 & 0 & 1 \\ 0 & 1 & 0 \end{matrix} \right)\left(\begin{matrix} 0 \\ 0 \\ 1 \end{matrix} \right)=a^{2}.                      (3.189)

Thus we have Δ A = a.

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