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