Question 3.P.4: A system is initially in the state |ψ0〉 =[√2|Φ1〉 +√3|Φ2〉 ...

(a) A measurement of the energy yields E_{n}= 〈\phi _{n}|\hat {H}|\phi _{n}〉=n^{2}\varepsilon _{0}, that is

E_{1}=\varepsilon _{0},    E_{2}=4\varepsilon _{0},    E_{3}=9\varepsilon _{0},    E_{4}=16\varepsilon _{0}       (3.167)

Since |\psi _{0}〉 is normalized, 〈\psi _{0}|\psi _{0}〉= (2+3+1+1)/7=1, and using (3.2),

P_{n}(a_{n} )=\frac{\left|〈\psi _{n}|\psi〉\right| ^{2} }{〈\psi|\psi〉} =\frac{\left|a_{n}\right| ^{2}}{〈\psi|\psi〉} ,

we can write the probabilities corresponding to (3.167) as P (E_{n})= \left|〈\phi _{n}|\psi _{0}〉\right| ^{2}/〈\psi _{0}|\psi _{0}〉=\left|〈\phi _{n}|\psi _{0}〉\right| ^{2}; hence,

using the fact that 〈\phi _{n}|\phi _{m}〉=\delta _{nm}, we have

P(E_{1})= \left|\sqrt{\frac{2}{7}}〈\phi _{1}|\phi _{1}〉 \right| ^{2}=\frac{2}{7},      P(E_{2})= \left|\sqrt{\frac{3}{7}}〈\phi _{2}|\phi _{2}〉 \right| ^{2}=\frac{3}{7},              (3.168)

P(E_{3})= \left|\frac{1}{\sqrt{7} } 〈\phi _{3}|\phi _{3}〉 \right| ^{2}=\frac{1}{7},      P(E_{4})= \left|\frac{1}{\sqrt{7} } 〈\phi _{4}|\phi _{4}〉 \right| ^{2}=\frac{1}{7}.                  (3.169)

(b) Similarly, a measurement of the observable \hat{A} yields a_{n}= 〈\phi _{n}|\hat{A}|\phi _{n}〉=(n+1)a_{0} ; that is,

a_{1}=2a_{0},    a_{2}=3a_{0},    a_{3}=4a_{0},    a_{4}=5a_{0}.              (3.170)

Again, using (3.2) and since |\psi _{0}〉 is normalized, we can ascertain that the probabilities corresponding to the values (3.170) are given by P(a_{n})= \left|〈\phi _{n}|\psi _{0}〉\right| ^{2}/〈\psi _{0}|\psi _{0}〉=\left|〈\phi _{n}|\psi _{0}〉\right| ^{2} , or

P(a_{1})= \left|\sqrt{\frac{2}{7}}〈\phi _{1}|\phi _{1}〉 \right| ^{2}=\frac{2}{7},      P(a_{2})= \left|\sqrt{\frac{3}{7}}〈\phi _{2}|\phi _{2}〉 \right| ^{2}=\frac{3}{7},                 (3.171)

P(a_{3})= \left|\frac{1}{\sqrt{7} } 〈\phi _{3}|\phi _{3}〉 \right| ^{2}=\frac{1}{7},      P(a_{4})= \left|\frac{1}{\sqrt{7} } 〈\phi _{4}|\phi _{4}〉 \right| ^{2}=\frac{1}{7}.                  (3.172)

(c) An energy measurement that yields 4\varepsilon _{0} implies that the system is left in the state |\psi _{2}〉.

A measurement of the observable A immediately afterwards leads to

〈\phi _{2}|\hat{A}|\phi _{2}〉=3a_{0}〈\phi _{2}|\phi _{2}〉=3a_{0}.                       (3.173)