# 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
       $E=hf\;$
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
       $E=hf\;$
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:
       ${\frac {\partial }{\partial x}}\psi =ik_{x}\psi$
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
    $\mathrm {i} \hbar {\frac {\partial }{\partial t}}\left|\psi _{n}\left(t\right)\right\rangle =E_{n}\left|\psi _{n}\left(t\right)\right\rangle .$
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:
    $\left|\psi \left(t\right)\right\rangle =\mathrm {e} ^{-\mathrm {i} Et/\hbar }\left|\psi \left(0\right)\right\rangle .$
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 $\left\{\left|n\right\rangle \right\}$ form a basis (linear algebra)|basis for the state space. We introduced here the short-hand notation $|\,n\,\rangle =\psi _{n}$. Then any state vector that is a solution of the time-dependent Schrödinger equation (with a time-independent $H$)
$\left|\psi \left(t\right)\right\rangle$ can be written as a linear superposition of energy eigenstates:
    $\left|\psi \left(t\right)\right\rangle =\sum _{n}c_{n}(t)\left|n\right\rangle \quad ,\quad H\left|n\right\rangle =E_{n}\left|n\right\rangle \quad ,\quad \sum _{n}\left|c_{n}\left(t\right)\right|^{2}=1.$
(The last equation enforces the requirement that $\left|\psi \left(t\right)\right\rangle$,
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 $\left|\psi \left(t\right)\right\rangle$ into the energy basis at time $t=0$, its value at any subsequent time is given simply by
    $\left|\psi \left(t\right)\right\rangle =\sum _{n}\mathrm {e} ^{-\mathrm {i} E_{n}t/\hbar }c_{n}\left(0\right)\left|n\right\rangle .$
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 $\mathrm {e} ^{-\mathrm {i} E_{n}t/\hbar }$ 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.