# Joan Lindsay Orr

$$\newcommand{\CC}{\mathbb{C}} \newcommand{\NN}{\mathbb{N}} \newcommand{\RR}{\mathbb{R}} \newcommand{\TT}{\mathbb{T}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\H}{\mathcal{H}} \newcommand{\e}{\epsilon} \newcommand{\x}{\mathbf{x}} \newcommand{\y}{\mathbf{y}}$$

# Notes on the Schrödinger Equation

## Derivation of the Time-Dependent Schrödinger Equation

The two key ingredients are the Planck equation for the energy of a photon, and the de Broglie wavelength of a matter wave.

The Planck equation for the energy, $$E$$ of a photon with frequency $$f$$ is:

$E = h f$

The de Broglie wavelength, $$\lambda$$, of the matter wave asoociated with a particle with momentum $$p$$ is:

$\lambda = \frac{h}{p}$

It's convenient to recast this in terms of the wave number $$k = 2\pi / \lambda$$ and the normalized frequency $$\omega = 2\pi f$$. (Conceptually, $$k$$ represents the number of wave cycles in an interval of length $$2\pi$$, and $$\omega$$ represents the numbers of cycles in a time interval of length $$2\pi$$.) In these terms, the Planck equations and the de Broglie wavelength formulas become respectively $E = \hbar \omega \qquad\text{and}\qquad p = \frac{h}{\lambda} = \hbar\frac{2\pi}{\lambda} = \hbar k$ where $$\hbar = h/2\pi$$ is the adjusted Planck constant, approximately $$1.05 \times 10^{-34} J\cdot s$$.

The Schrödinger Equation is motivated by looking for a wave-type equation which has as a solution the simplest possible candidate for a matter wave, i.e., a plane wave of the form $\Psi(x,t) = e^{i(kx - \omega t)}$

Then compute $\frac{\partial\Psi}{\partial t} = -i\omega \Psi = \frac{-iE}{\hbar}\Psi$ and so $i\hbar\frac{\partial\Psi}{\partial t} = E\Psi$

Next we calculate $\frac{\partial^2\Psi}{\partial x^2} = -k^2 \Psi$ and then, using the classical (non-relativistic) relationship $E = \frac{p^2}{2m}$ together with the de Broglie formula, we obtain $\frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2} = \frac{\hbar^2 k^2}{2m} \Psi = \frac{p^2}{2m} \Psi = E\Psi$

Putting together the two formulas for $$E\Psi$$ we obtain the free Schrödinger Equation: $i\hbar\frac{\partial\Psi}{\partial t} = E\Psi = \frac{-\hbar^2}{2m}\frac{\partial^2\Psi}{\partial x^2}$

Finally, since we are interpreting $\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2}$ as the energy operator for a free particle, it makes sense that the energy operator for a particle in potential $$V(x, t)$$ is $\frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t)$ and so we reassemble the equations for a particle in a potential as $i\hbar\frac{\partial\Psi}{\partial t} = \left[ \frac{-\hbar^2}{2m}\frac{\partial^2}{\partial x^2} + V(x,t) \right] \Psi$

## The Time-Independent Schrödinger Equation

Suppose that $$V=V(x)$$ does not depend on $$t$$. Then the Schrödinger Equation becomes a separable PDE. Write $$\Psi(x,t) = \phi(t)\psi(x)$$ and substitute: $i\hbar\frac{d\phi(t)}{dt}\psi(x) = - \phi(t)\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + V(x)\phi(t)\psi(x)$ Rearraging this gives $i\hbar\frac{1}{\phi(t)}\frac{d\phi(t)}{dt} = \left(-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + V(x)\psi(x)\right) \frac{1}{\psi(x)}$ The lhs doesn't depend on $$x$$ and the rhs doesn't depend on $$t$$ and so both must equal a common constant value, $$E$$. From this $\frac{d\phi(t)}{dt} = \frac{-iE}{\hbar}\phi(t)$ and so $\phi(t) = e^{-iEt/\hbar}$ (the multiplicative constant of integration is absorbed into $$\psi$$ for convenience). This leaves the time-independent Schrödinger equation: $-\frac{\hbar^2}{2m}\frac{d^2\psi(x)}{dx^2} + V(x)\psi(x) = E\psi(x)$