3 Bound States

Stationary States

From last time:

$$ \bbox[silver]{i\hbar\frac{\partial\Psi(x,t)}{\partial t}=\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+U\right)\Psi(x,t).} $$

To solve this equation, a standard technique is separation of variables. Let assume

$$ \Psi(x,t)=\psi(x)\phi(t). $$

Then we have

$$ -\frac{\hbar^2}{2m}\frac{1}{\psi(x)}\frac{\partial^2}{\partial x^2}\psi(x)+U(x)=i\hbar\frac{1}{\phi(t)}\frac{\partial}{\partial t}\phi(t)=C. $$

$$ i\hbar\frac{1}{\phi(t)}\frac{\partial}{\partial t}\phi(t)=C. $$

General solution:

$$ \phi(t)=\mathrm{e}^{-i(C/\hbar)t}. $$

It turned out that $C=E$, so

$$ \Psi(x,t)=\psi(x)\mathrm{e}^{-i(E/\hbar)t}. $$

Notice:

$$ \Psi^*(x,t)\Psi(x,t)=\psi^*(x)\mathrm{e}^{i(E/\hbar)t}\psi(x)\mathrm{e}^{-i(E/\hbar)t}=\psi^*(x)\psi(x). $$

The probably density function is time independent. That is why the phrase "stationary state" is used.

Since $C=E$, the spatial part reduces to the time-independent Schrödinger Equation,

$$ \bbox[silver]{E\Psi(x,t)=\left(-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+U\right)\Psi(x,t).} $$

$$ \int_{\text{all space}}|\Psi(x,t)|^2dx=1. $$