From eq132 of the previous article, we have
From eq113 where and eq130, we have
Comparing eq134 and eq135, both eigenstates and are associated with the same eigenvalue . Therefore, one must be proportional to the other (e.g. the eigenstates and have the same eigenvalue), i.e.
Similarly, for the lowering operator (using eq114), we have,
where and are scalars; or in summary
If the complete set of eigenstates is normalised, multiplying the above equation on the left by gives the matrix elements of :
From eq116, we have . Using eq132 and eq133,
From eq136 and eq137
Comparing eq140 and eq141
Question
Show that .
Answer
From eq139, where we let
Substituting eq109 in the above equation
Since and are Hermitian,
Substituting eq108 in the above equation
Substituting eq139 in the above equation
Therefore, .
Substituting in eq142
Substituting the above equation in eq136,
To find , we repeat the above steps for eq139 onwards and using eq115 to give
From eq137,
Comparing eq145 and eq146,
Substituting eq143, where , in the above equation
Substituting the above equation in eq137,
Combining eq144 and eq147