Wavefunction boundary condition solve for k

[zhillyz]

Homework Statement

A wave function is given by:
$\Psi (x) = a cos(2\pi x) + b sin (2\pi x) for\: x<0 \\ and\\ \Psi (x) = Ce^{-kx} for\: x>0 \\$

Determine the constant k in terms of a, b and c using the boundary conditions and discuss the case a >> b.

Homework Equations

Wavefunctions and their first order derivatives are continuous at the boundaries. So at x = 0 they will equal each other.

The Attempt at a Solution

$acos (2\pi x) + b sin(2\pi x) = Ce^{-kx}$

sin 0 = 0 and cos 0 = 1 and exp 0 = 1 therefore;

$a = C \\ -(2\pi x)asin(2\pi x) + (2\pi x)bcos(2\pi x) = -ake^{-kx}$

again sin 0 = 0 cos 0 = 1 exp 0 = 1 and x = 0 therefore;

$-ak = 0$

Soo pretty sure this is correct so far but not sure on the final step?

 

[TSny]

Check your expressions for the derivatives of the sine and cosine functions. Did you use the chain rule properly?

 

[zhillyz]

Ahh forgot to get rid of the x so that would mean k = 2pi*b/a. And for cases where a<<b then it is just 2pi*b?

 

[TSny]

I get a different sign for k. The initial problem statement says to consider the case a >> b.

The question seems a little odd to me. The constants a, b, c need not be positive numbers (or even real for that matter.) Anyway, I guess you could make a conclusion about the size of k under the assumption that |a| >> |b|.

 

[Nickie]

The final step in solving for k would be to divide both sides by -a and take the natural logarithm of both sides, giving us:

k = -ln(b/a)

This solution holds true for all cases, including the case where a >> b. In this case, the wave function will be dominated by the cosine term and the exponential term will become negligible. Therefore, the value of k will be very small, approaching 0 as a >> b. This makes intuitive sense, as a larger value of a will result in a higher amplitude for the cosine term and a slower decay for the exponential term.