Schrödinger Equation for a freely falling body near the surface of Earth

Question

Near Earth's surface the Schrödinger equation of a freely falling particle takes the form, $$ \frac {-\hbar^2}{2m} \frac {d^2 \psi (y)}{dy^2} + mgy\psi (y) = E \psi (y). $$ Putting $k=\frac {\sqrt {2mE}}{\hbar}$ and $\kappa = \frac {mg}{E}$ we have, $$ \psi ''(y)+k^2(1-\kappa y)\psi (y) = 0. $$ This is an airy differential equation with general solutions in terms of Airy functions, $$ c_1 \text{Ai} \left[ \left( \frac {k}{\kappa} \right)^{\frac {2}{3}} (\kappa y - 1) \right] + c_2 \text{Bi} \left[ \left( \frac {k}{\kappa} \right)^{\frac {2}{3}} (\kappa y - 1) \right]. $$ The problem lies here. In order to find relations for the coefficients $c_1$ and $c_2$ , I need appropriate boundary conditions which I don't know. For instance, will the wavefunction collapse at the surface (that is, at $\psi (0)=0$)? Is it continuous and differentiable at the the surface? Or are there other boundary conditions? Moreover, what is the normalization condition? Can I even evaluate the normalization integral? Any help is appreciated.

Edit

Okay, the term containing $\text{Bi}$ function can be ignored , as it is divergent for infinitly large arguments. So, the problem remains to normalise the wavefunction, $$ \psi (y)=c_1 \text{Ai} \left[ \left( \frac {k}{\kappa} \right)^{\frac {2}{3}} (\kappa y - 1) \right] $$

Answer 1

It is better if you begin to solve the time dependent Schrödinger equation, that is:

$\frac{-\hbar^2}{2m}\frac{\partial^2\Psi(y,t)}{\partial y^2} + V(y)\Psi(y,t) = i\hbar\frac{\partial\Psi(y,t)}{\partial t}$

Then, you can apply the method of separation of variables to reduce this partial differential equation to an ordinary differential equation:

$\Psi(y,t) = \psi(y)\phi(t)$

This gives us:

$\frac{-\hbar^2}{2m}\phi\psi" + V(y)\phi\psi = i\hbar\phi'\psi$

Dividing by $\phi\psi$ gives then:

$\frac{-\hbar^2}{2m}\frac{\psi"}{\psi} + V(y) = i\hbar\frac{\phi'}{\phi}$

The left side of this equation is a function of y alone, the right side is a function of t alone. This equality is true if both sides are equal to a constant E, as you use it in your equation. Then, we have 2 ODE's. The first one is easy to solve:

$i \hbar \frac{\phi'}{\phi} = E$

$\phi (t) = Aexp(\frac{-iEt}{\hbar})$

We are now at the point to solve the other side of the equation, namely:

$\frac{-\hbar^2}{2m}\psi" + mgy\psi = E\psi$

$\psi" = \frac{-2m(E - mgy)}{\hbar^2}\psi$

So, as you can see, the solution of this second equation depends on E. You can have bound state ($E<0$) or scattering state ($E>0$). Once you know the state of the system for which you solve the Schrödinger equation with the well defined potential V, in this case the gravitational potential energy, you can get your solution and define the required boundry conditions.

I hope this can help you...