 Solution

ID: past-exam-of-the-mathematical-methods-i-course-of-the-university-of-oxford/2014/7/iv/solution

λ_{2} e_{1}^{†} e_{2} = e_{1}^{†} H e_{2} = λ_{1} e_{1}^{†} e_{2}

. The given normalization then gives

e_{1}^{†} (e_{1} + e_{2}) = 1 + 0 = 1.

 New to topics? Read the docs here!