- #1

a1234

- 78

- 6

- Homework Statement
- I am asked to solve the Schrodinger Equation for the "particle in a box" problem (with box width going from 0 to a), but using complex exponentials instead of sin and cos.

- Relevant Equations
- sin(x) = (exp(ix) - exp(-ix))/(2i);

psi(x) = A(exp(ikx) - exp(ikx))

psi(x) = A*2i*sin(kx)

psi(x) = A*sin(kx)

Using the boundary conditions where psi is 0, I found that k = n*pi/a, since sin(x) is zero when k*a = 0.

I set up my normalization integral as follows:

A^2 * integral from 0 to a of (((exp(ikx) - exp(-ikx))*(exp(-ikx) - exp(ikx)) dx) = 1

After simplifying, and accounting for the fact that sin(2*n*pi) = 0, I arrived at the following expression:

2*a*A^2 = 1

A = 1/√(2a)

However, once we substitute the above expression for A in the original wave function, we get psi(x) = i√(2/a)*sin(kx).

I would like to know if there is a way to avoid the extra factor of i in the wavefunction. I've read online that the extra i is often absorbed into the constant A, but is there a way around this? I've been told by my instructor that the constant A is supposed to be complex, and that the final wavefunction is supposed to have real components.

I set up my normalization integral as follows:

A^2 * integral from 0 to a of (((exp(ikx) - exp(-ikx))*(exp(-ikx) - exp(ikx)) dx) = 1

After simplifying, and accounting for the fact that sin(2*n*pi) = 0, I arrived at the following expression:

2*a*A^2 = 1

A = 1/√(2a)

However, once we substitute the above expression for A in the original wave function, we get psi(x) = i√(2/a)*sin(kx).

I would like to know if there is a way to avoid the extra factor of i in the wavefunction. I've read online that the extra i is often absorbed into the constant A, but is there a way around this? I've been told by my instructor that the constant A is supposed to be complex, and that the final wavefunction is supposed to have real components.