Question

Show that 2|ψ1||ψ2| cos (φ1 − φ2) = 2 (A1A2 + B1B2)
given that ψ1 = A1 + iB1 and ψ2 = A2 + iB2.

Answer 1

Given:

Its given that Ф1 = A1 + iB1 and Ф2 = A2 + iB2.

To Prove:

2|Ф1||Ф2| cos (ω1 − ω2) = 2 (A1A2 + B1B2)

Solution:

Given Ф1 = A1 + iB1 and Ф2 = A2 + iB2.

The given pair of complex number are actually points.

Lets convert this into  coordinates form ( x,y):

• Ф1 = ( A1, B1)

• Ф2 = (A2, B2)

In vector form it will look like,

• Ф1 = A1 i + B1 j

• Ф2 = A2 i + B2 j

Taking  the dot product of 2 vectors. Ф1.Ф2 :

• Ф1.Ф2 = |Ф1||Ф2| cos (ω1 − ω2)

where ω1 − ω2 will be the angle vector Ф1 makes with vector Ф2.

Now,

• Ф1.Ф2 =  (A1i + B1j).(A2i + B2j) = A1A2 + B1B2 .

We have proved that 2|Ф1||Ф2| cos (ω1 − ω2) = 2 (A1A2 + B1B2).