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Show that ψ(x)=A'eikx+B'e-ikxis equivalent to ψ(x)=Asinkx+Bcoskx, provided that A'=12(B-iA)B'=12(B+iA).

Short Answer

Expert verified

Hence, the proof for the equation is obtained.

Step by step solution

01

Concept involved

According to theEuler’s formulain complex numbers, it can be written that:

eiφ=cosφ+isinφ……(1)

02

Given/known parameters

Consider the given function:

ψx=A'eikx+B'e-ikx

Consider the equations:

A'=12B-iA……(2)

B'=12B+iA……(3)

03

Solution

Apply Euler’s formula from equation (1) and solve:

ψx=A'coskx+isinkx+B'coskx-isinkx

Rewrite the above equation as,

ψx=A'+B'coskx+A'-B'sinkx…..(4)

Now, by using equation (2) and (3) in equation (4) solve as:

ψx=12B-iA+12B+iAcoskx-12B-iA-12B+iAsinkxψx=Bcoskx+Asinkx

Thus, you can say that: ψx=A'eikx+B'e-ikxis equivalent to

ψx=Bcoskx+Asinkxprovided that A'=12B-iA andB'=12B+iA

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