Materials Science and Engineering/Derivations/Quantum Mechanics

Schrodinger Equation
Schrödinger's equation follows very naturally from earlier developments:

In 1905, by considering the photoelectric effect, Albert Einstein had published his
 * $$E = h f\;$$

formula for the relation between the energy E and frequency f of the quanta of radiation (photons), where h is Planck's constant.

In 1924 Louis de Broglie presented his de Broglie hypothesis which states that all particles (not just photons) have an associated wavefunction $$\Psi\;$$ with properties:
 * $$p=h / \lambda\;$$, where $$\lambda\,$$ is the wavelength of the wave and p the momentum of the particle.

De Broglie showed that this was consistent with Einstein's formula and special relativity so that
 * $$E = h f\;$$

still holds, but now this is hypothesized to hold for all particles, not just photons anymore.

Expressed in terms of angular frequency $$\omega = 2\pi f\;$$ and wavenumber $$k = 2\pi / \lambda\;$$, with $$\hbar = h / 2 \pi\;$$ we get:
 * $$E=\hbar \omega$$

and
 * $$\mathbf{p}=\hbar \mathbf{k}\;$$

where we have expressed p and k as vectors.

Schrödinger's great insight, late in 1925, was to express the phase of a plane wave as a complex phase factor:
 * $$\psi \approx e^{i(\mathbf{k}\cdot\mathbf{x}- \omega t)}$$

and to realize that since
 * $$ \frac{\partial}{\partial t} \psi = -i\omega \psi $$

then
 * $$ E \psi = \hbar \omega \psi = i\hbar\frac{\partial}{\partial t} \psi $$

and similarly since:
 * $$ \frac{\partial}{\partial x} \psi = i k_x \psi $$

then
 * $$ p_x \psi = \hbar k_x \psi = -i\hbar\frac{\partial}{\partial x} \psi $$

and hence:
 * $$ p_x^2 \psi = -\hbar^2\frac{\partial^2}{\partial x^2} \psi $$

so that, again for a plane wave, he got:
 * $$ p^2 \psi = (p_x^2 + p_y^2 + p_z^2) \psi = -\hbar^2\left(\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}\right) \psi = -\hbar^2\nabla^2 \psi $$

And by inserting these expressions into the Newtonian formula for a particle with total energy E, mass m, moving in a potential V:
 * $$E=\frac{p^2}{2m}+V$$ (simply the sum of the kinetic energy and potential energy; the plane wave model assumed V = 0)

he got his famed equation for a single particle in the 3-dimensional case in the presence of a potential:
 * $$i\hbar\frac{\partial}{\partial t}\Psi=-\frac{\hbar^2}{2m}\nabla^2\Psi + V\Psi$$

Solution of the Time-Dependent Schrodinger Equation
On inserting a solution of the time-independent Schrödinger equation into the full Schrödinger equation, we get


 * $$\mathrm{i} \hbar \frac{\partial}{\partial t} \left| \psi_n \left(t\right) \right\rangle = E_n \left|\psi_n\left(t\right)\right\rang. $$

It is relatively easy to solve this equation. One finds that the energy eigenstates (i.e., solutions of the time-independent Schrödinger equation) change as a function of time only trivially, namely, only by a complex phase (waves)|phase:


 * $$ \left| \psi \left(t\right) \right\rangle = \mathrm{e}^{-\mathrm{i} Et / \hbar} \left|\psi\left(0\right)\right\rang. $$

It immediately follows that the probability amplitude,
 * $$\psi(t)^*\psi(t) = \mathrm{e}^{\mathrm{i} Et / \hbar}\mathrm{e}^{-\mathrm{i} Et / \hbar}

\psi(0)^*\psi(0) = |\psi(0)|^2, $$ is time-independent. Because of a similar cancellation of phase factors in bra and ket, all average (expectation) values of time-independent observables (physical quantities) computed from $$\psi(t)\,$$ are  time-independent.

Energy eigenstates are convenient to work with because they form a complete set of states. That is, the eigenvectors $$ \left\{\left|n\right\rang\right\} $$ form a basis (linear algebra)|basis for the state space. We introduced here the short-hand notation $$|\,n\,\rang = \psi_n$$. Then any state vector that is a solution of the time-dependent Schrödinger equation (with a time-independent $$H$$) $$ \left|\psi\left(t\right)\right\rang $$ can be written as a linear superposition of energy eigenstates:


 * $$\left|\psi\left(t\right)\right\rang = \sum_n c_n(t) \left|n\right\rang \quad,\quad H \left|n\right\rang = E_n \left|n\right\rang \quad,\quad \sum_n \left|c_n\left(t\right)\right|^2 = 1.$$

(The last equation enforces the requirement that $$ \left|\psi\left(t\right)\right\rang $$, like all state vectors, may be normalized to a unit vector.) Applying the Hamiltonian operator to each side of the first equation, the time-dependent Schrödinger equation in the left-hand side and using the fact that the energy basis vectors are by definition linearly independent, we readily obtain


 * $$\mathrm{i}\hbar \frac{\partial c_n}{\partial t} = E_n c_n\left(t\right).$$

Therefore, if we know the decomposition of $$ \left|\psi\left(t\right)\right\rang $$ into the energy basis at time $$t = 0$$, its value at any subsequent time is given simply by


 * $$\left|\psi\left(t\right)\right\rang = \sum_n \mathrm{e}^{-\mathrm{i}E_nt/\hbar} c_n\left(0\right) \left|n\right\rang. $$

Note that when some values $$c_n(0)\,$$ are not equal to zero for differing energy values $$E_n\,$$, the left-hand side is not an eigenvector of the energy operator $$H$$. The left-hand is an eigenvector when the only $$c_n(0)\,$$-values not equal to zero belong the same energy, so that $$\mathrm{e}^{-\mathrm{i}E_nt/\hbar}$$ can be factored out. In many real-world application this is the case and the state vector $$\psi(t)\,$$ (containing time only in its phase factor) is then a solution of the time-independent Schrödinger equation.