## The Derivation of the Schrödinger Equation

The image at the top of each page on this site is a photo of some notes I had made a while back on the derivation of the Schrödinger Equation. Here is a transcription of those notes.

Consistent with the results of the Davisson-Germer experiment, let us assume that the free-particle solution is that of a plane wave with a complex phase factor: $\displaystyle \Psi(\vec{r},t) = Ae^{i(\vec{k}\cdot\vec{r} - \omega t)} = Ae^{\frac{i}{\hbar}(\vec{p}\cdot\vec{r} - E t)} \qquad\qquad(Eq .1)$

Here we have used the Planck-Einstein relation $E=\hbar\omega$ and the de Broglie relation $\vec{p}=\hbar\vec{k}$.

Taking the the first order partial derivative with respect to space: $\displaystyle \nabla\Psi = \frac{i}{\hbar}\vec{p}Ae^{\frac{i}{\hbar}(\vec{p}\cdot\vec{r} - Et)} = \frac{i}{\hbar}\vec{p}\Psi \qquad\qquad(Eq. 2)$

We can also take the derivative of (1) with respect to time: $\displaystyle \frac{\partial\Psi}{\partial t} = \frac{-iE}{\hbar}Ae^{\frac{i}{\hbar}(\vec{p}\cdot\vec{r} - Et)} = \frac{-iE}{\hbar}\Psi \qquad\qquad(Eq. 3)$

Now we have something to work with. First of all, let us multiply both sides of (2) by $-i\hbar$ in order to rearrange and isolate what we shall define as our momentum operator: $\displaystyle -i\hbar\nabla\Psi = \vec{p}\Psi \Rightarrow \hat{p}\equiv -i\hbar\nabla \qquad\qquad(Eq. 4)$

Similarly, we can multiply both sides of (3) by $i\hbar$ to obtain a definition for our energy operator: $\displaystyle i\hbar\frac{\partial\Psi}{\partial t} = E\Psi \Rightarrow \hat{E} \equiv i\hbar\frac{\partial}{\partial t} \qquad\qquad(Eq. 5)$

Now, let us take the energy equation from classical mechanics, and substitute in our quantum mechanical equivalents for the various terms: $\displaystyle E = \frac{\vec{p}\cdot\vec{p}}{2m} + V \Rightarrow \hat{E} \Psi = \left(\frac{\hat{p}^2}{2m} + V\right)\Psi \qquad\qquad(Eq. 6)$

Substituting in our operator definitions, we get the Schrödinger equation: $\displaystyle i\hbar\frac{\partial\Psi}{\partial t} = \left(\frac{-\hbar^2}{2m}\nabla^2 + V\right)\Psi = \hat{H}\Psi \qquad\qquad(Eq. 7)$

Here, of course, $\hat{H}$ is the Hamiltonian, $\hat{H}=\frac{-\hbar^2}{2m}\nabla^2 + V$

# Update [3/28/2013]

As I was showering this morning, some observation about this derivation struck me, and I felt I should share them. (Yep, that’s the sort of thing I think about while I’m showering.)

Note that we obtained the energy operator by taking the derivative with respect to time, and that we obtained the momentum operator by taking the derivative with respect to space. This is not accidental. There are fundamental relationships between these pairs of quantities.

• By Noether’s Theorem, conservation of energy arises due to the symmetry known as  time-translation invariance. Likewise, conservation of momentum arises due to spatial-translation invariance.
• The product of energy and time, and the product of momentum and displacement, are both in units of action (and thus in the same units as Planck’s constant). Keep in mind the fundamental role that action plays in physics due to the Action Principle, both in classical physics (Hamiltonian and Lagrangian mechanics) and in quantum theory (the path-integral formulation, QED). The Action Principle, in plain English, basically states that nature is lazy, and that physical processes will unfold in whatever manner will minimize the quantity known as action. I am, of course, grotesquely over-simplifying here, but that is the gist.
• Energy and time, as well as momentum and position, are canonical conjugates of each other, which means that their commutators have a value of $i\hbar$. In other words, their operators do not commute with each other in quantum mechanics. (In the classical limit, as Planck’s constant goes to 0, they DO commute with each other, with the commutator relations becoming equivalent to the classical Poisson brackets describing their commutative nature.)  Another way of saying this is that they are Fourier transforms of one another.
• As a consequence of the previous item, these pairs are both bound by the Heisenberg Uncertainty Principle: $\Delta E \Delta t \geq \frac{\hbar}{2}$ and $\Delta p \Delta x \geq \frac{\hbar}{2}$ 