Momentum operator eigenfunction problem

[interdinghy]

Homework Statement

Given f(x)1 = cos(kx) and f(x)2 = sin(kx), find a linear combination which is an eigenfunction of the momentum operator.

Homework Equations

The momentum operator:

-iħ d\dx(Acos(kx)+Bsin(kx) = λ[Acos(kx) + Bsin(kx)] where λ is some constant.

The Attempt at a Solution

I've tried a few different A's and B's, mainly permutations of -1 and 1, but I honestly can't see how this is doable. No matter what you do the derivatives' signs aren't going to behave the same way so I just don't see how you can get the same function back out after taking the derivative, no matter what constants you multiply them by.Edit: Nevermind, figured it out.

Edit 2: Occurs to me that just in case someone is looking for this question on google or something in the future, I should post the solution I got. Answer was A = 1 B = i.

 

[mark726]

Solution:

To find a linear combination that is an eigenfunction of the momentum operator, we need to solve the following equation:

-iħ d/dx(Acos(kx)+Bsin(kx)) = λ(Acos(kx) + Bsin(kx))

We can rewrite this equation as:

-iħ(Akcos(kx)+Bksin(kx)) = λ(Acos(kx) + Bsin(kx))

We can then equate the coefficients of cos(kx) and sin(kx) on both sides of the equation:

-Akħ = λA and Bkħ = λB

Solving for A and B, we get:

A = -iħ/λk and B = ħ/λk

Therefore, a linear combination that is an eigenfunction of the momentum operator is:

f(x) = (-iħ/λk)cos(kx) + (ħ/λk)sin(kx)