Solution to the Schrodinger Equation

[ehrenfest]

I am somewhat confused about how general the solution $$\psi_E(x) = Aexp(ikx)+Bexp(-ikx)$$

is?

Can someone complete one or both of these sentences:

$$\psi_E(x) = Aexp(ikx)+Bexp(-ikx)$$

is the solution to all equations of the form ...

or

$$\psi_E(x) = Aexp(ikx)+Bexp(-ikx)$$

is a solution to the Schrodinger equation when...

 

[Dick]

They are the solutions when the potential is constant, and if you mean k to be real, then when the energy is greater than the potential. Haven't we just been working on such problems? These are the free particle solutions.

 

[maverick280857]

Perhaps not required, but I will add to what Dick said:

The 1D Time Independent Schrodinger Equation,

$$-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x)$$

for a free particle (under a constant potential $V_{x} = V_{0}$) becomes

$$\frac{d^2\psi(x)}{dx^2} = \frac{2m}{\hbar^2}(V_{0} - E) \psi(x)$$

For $E < V_{0}$, solutions to this equation have the form stated in your first post. This happens, as Dick as pointed out, for real values of the constant k.

If you are asking how the general solution is a sum of the two solutions, then, well, that is a basic theorem for linear differential equations which says that if you can find two linearly independent solutions, then their linear combination is also a solution (as may be verified easily by substitution). The general solution is a linear combination of all linearly independent solutions (which in this case will be energy eigenfunctions for the free particle).

 

[Gokul43201]

For $E < V_{0}$, solutions to this equation have the form stated in your first post. This happens, as Dick as pointed out, for real values of the constant k.

Did you mean to write $E > V_{0}$?

 

[maverick280857]

Did you mean to write $E > V_{0}$?

Yes, I apologize for the typo. Please read it as $E > V_{0}$.