Study goodness

I just had one of my most fruitful study sessions ever today, studying for quantum mechanics. Got some tough problems done, and understood some things a lot better than I had done previously. Now for anyone who's interested, and because I want to play around with LaTeX in this blog, I'm going to show how to formulate the time independent Schrödinger equation.

We'll start with the time dependant version:

i \hbar \frac{\partial \Psi}{\partial t} = - \frac{\hbar^2}{2m} \frac{\partial^2 \Psi}{\partial x^2} + V \Psi

Where \Psi = \Psi \left( x,t \right) i.e. a function of x and t, and likewise for V.

Now we shall assume \Psi \left( x,t \right) = \psi \left( x \right) \phi \left( t \right) i.e. it is separable.

Then the equation becomes:

i \hbar \psi \frac{\partial \phi}{\partial t} = - \frac{\hbar^2}{2m} \phi \frac{\partial^2 \psi}{\partial x^2} + V \psi \phi

and dividing through by \psi \phi gives:

i \hbar \frac{1}{\phi} \frac{\partial \phi}{\partial t} = - \frac{\hbar^2}{2m} \frac{1}{\psi} \frac{\partial^2 \psi}{\partial x^2} + V

If V is only a function of x, then the left hand side is dependant on t alone, and the right hand side is only dependant on x, therefore they must be equal to some constant. Lets call this E.

Now, it is easy to solve for \phi:

\begin{align*}
i \hbar \frac{1}{\phi} \frac{\partial \phi}{\partial t} &= E \\
\int \frac{1}{\phi} \frac{\partial \phi} dt &= \int \frac{E}{i \hbar} dt \\
\ln \left| \phi \right| &= - \frac{i E t}{\hbar} \\
\phi &= e^{- \frac{i E t}{\hbar}}
\end{align*}

Note that the left side is always positive, so the absolute value works out alright.

The other part of the earlier equation becomes the time independent Schrödinger equation:

- \frac{\hbar^2}{2m} \frac{\partial^2 \psi}{\partial x^2} + V \psi = E \psi

Which can only be solved if the potential, V, is defined.

Now I hope that is accurate, but is not unlikely that I made a mistake somewhere.