Hamiltonian & Schrodinger Equation

The Hamiltonian is an operator which gives the total energy of a system by adding together the system's kinetic energy and potential energy. Schrodinger's time-independent equation is a simple mathematical equivocation of this relation between Hamiltonians and total energy. Schrodinger's time-independent equation, more difficultly derived, shows how the the total energy of a system can also be found using operations which rely on the time evolution of wave functions. We'll first derive the Hamiltonian, then piece together both the time-independent and time-dependent Schrodinger equations.

Hamiltonian

To derive the Hamiltonian, recall that classically the total energy, $E$, can be related to in this way:

\begin{equation}
E= \frac{\mathbf{p}^2}{2m} + V(\mathbf{r}, t)
\end{equation}

where $\mathbf{p}$ is the momentum vector of the system, $m$ is the mass, and $V(\mathbf{r}, t)$ is the potential energy that the system has at its position vector, $\mathbf{r}$, and time, $t$. Notice that on the right side of the above equation, the first term gives the kinetic energy of the system and the second gives the potential energy. To write the quantum mechanical equivalent of this formula, all that needs to be done is the substitution of the momentum operator, $\hat{p} = -i\hbar\nabla$, for $\mathbf{p}$. Therefore, the Hamilitonian, $\hat{H}$, can be derived to be:

\begin{equation}
\hat{H} = \frac{-\hbar^2}{2m} \nabla^2 + V(\mathbf{r}, t)
\end{equation}

Or, if we wanted to use the Laplacian, $\Delta=\nabla^2$,

\begin{equation}
\hat{H} = \frac{-\hbar^2}{2m} \Delta + V(\mathbf{r}, t)
\end{equation}

Time-Independent Schrodinger Equation

Schrodinger's time-independent equation simply states that:

\begin{equation}
E\psi = \hat{H}\psi
\end{equation}

That doesn't look all that impressive, so to satisfy your geeky ego, here's a fancier way of writing it:

\begin{equation}
E\psi(\mathbf{r}) = \frac{-\hbar^2}{2m} \Delta\psi(\mathbf{r}) + V(\mathbf{r})\psi(\mathbf{r})
\end{equation}

Note that the potential energy, $V(\mathbf{r})$, is assumed to not change with time.

Time-Dependent Schrodinger Equation

Recall that the energy of a photon is given by:

\begin{equation}
E = \hbar\omega
\end{equation}

where $\hbar$ is the reduced Planck constant and $\omega$ is the angular frequency (unit of angle of oscillation per unit of time) of the photon. Wave functions, much like electromagnetic waves, also define the energies of their systems using the same relation, so $\hat{H} \phi \left(\mathbf{k}, \omega\right) = \hbar\omega \phi\left( \mathbf{k},\omega \right)$ where $\mathbf{k}$ is the three-dimensional wavenumber vector of the system and $\phi \left(\mathbf{k}, \omega\right)$ is the wave function in $\mathbf{k}$-space and $\omega$-domain. But what if we had a wave function in position, $\mathbf{r}$, space and time, $t$, domain, $\psi \left(\mathbf{r}, t\right)$? If you've read my derivation of the momentum operator, you probably know where I'm taking this. Consider the Fourier transform equation for transformations from position space and time domain to momentum space ($k$-space) and angular frequency domain ($\omega$-domain):

\begin{equation}
\phi \left( \mathbf{k}, \omega\right) = \frac{1}{4\pi^2} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \psi\left(\mathbf{r}, t\right)e^{-i \left(\mathbf{k} \mathbf{r}-\omega t\right)} dtd \mathbf{r}
\end{equation}

(If you're wondering why it's $1/(4\pi^2)$, the above equation needs $1/\sqrt{2\pi}$ to normalize the integration over time domain, then another $1/ \sqrt{2\pi}^3$ to normalize the three-dimensional integration over position space.) Notice how $\omega t$ is subtracted from the term $\mathbf{k} \mathbf{r}$. This is what adds time dependence by translating the wave $\omega$ angles of oscillation in the positive direction for every $t$ time. Now, we know that applying the Hamiltonian to both sides of the equation gives this:

\begin{equation}
\hat{H}\phi \left(\mathbf{k}, \omega\right)= \frac{1}{4\pi^2} \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \hbar{\omega}\psi \left( \mathbf{r}, t\right) e^{i\left(\omega t-\mathbf{k} \mathbf{r} \right)}dtd \mathbf{r}
\end{equation}

Integrating by parts,

\begin{equation}
\hat{H}\phi \left(\mathbf{k}, \omega\right) = \frac{1}{4\pi^2} \int_{-\infty}^{\infty} \left[\frac{\hbar\omega}{i \omega}\psi \left( \mathbf{r}, t\right) e^{i\left(\omega t-\mathbf{k} \mathbf{r} \right)} - \int\frac{\hbar \omega}{i \omega}\frac{ \partial\psi \left(\mathbf{r}, t\right)}{\partial t}e^{i(\omega t-\mathbf{k} \mathbf{r})} dt \right]_{t=-\infty}^{t=\infty} d \mathbf{r}
\end{equation}

Because $\lim_{t\to\pm\infty} \psi\left(\mathbf{r}, t\right)=0$,

\begin{equation}
\hat{H}\phi \left(\mathbf{k}, \omega\right)= \frac{1}{4\pi^2} \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{\hbar}{-i}\frac{\partial \psi(\mathbf{r}, t)}{\partial t}e^{i(\omega t-\mathbf{k}\mathbf{r})} dtd \mathbf{r}
\end{equation}

Comparing the above to equation $(8)$, we can complete the derivation of the time-dependent Schrodinger equation:

\begin{equation}
i\hbar \frac{\partial}{\partial t}\psi= \hat{H}\psi
\end{equation}

or if you want to be really fancy about it:

\begin{equation}
i\hbar \frac{\partial}{\partial t} \psi(\mathbf{r}, t)= \frac{-\hbar^2}{2m} \Delta\psi(\mathbf{r}, t) + V(\mathbf{r}, t) \psi(\mathbf{r}, t)
\end{equation}