User:Maschen/Schrödinger equation (exact solutions)

This article summarizes the exact solutions to the Schrödinger equation:
 * $$i\hbar \dfrac{\partial \psi}{\partial t}=\hat{H}\psi$$

in 1d and 3d.

Free particle
For no potential, V = 0, so the particle is free and the equation reads:


 * $$ - E \psi = \frac{\hbar^2}{2m}{d^2 \psi \over d x^2}\,$$

which has oscillatory solutions for E > 0 (the Cn are arbitrary constants):


 * $$\psi_E(x) = C_1 e^{i\sqrt{2mE/\hbar^2}\,x} + C_2 e^{-i\sqrt{2mE/\hbar^2}\,x}\,$$

and exponential solutions for E < 0


 * $$\psi_{-|E|}(x) = C_1 e^{\sqrt{2m|E|/\hbar^2}\,x} + C_2 e^{-\sqrt{2m|E|/\hbar^2}\,x}.\,$$

The exponentially growing solutions have an infinite norm, and are not physical. They are not allowed in a finite volume with periodic or fixed boundary conditions.

Constant potential


For a constant potential, V = V0, the solution is oscillatory for E > V0 and exponential for E < V0, corresponding to energies that are allowed or disallowed in classical mechanics. Oscillatory solutions have a classically allowed energy and correspond to actual classical motions, while the exponential solutions have a disallowed energy and describe a small amount of quantum bleeding into the classically disallowed region, due to quantum tunneling. If the potential V0 grows at infinity, the motion is classically confined to a finite region, which means that in quantum mechanics every solution becomes an exponential far enough away. The condition that the exponential is decreasing restricts the energy levels to a discrete set, called the allowed energies.

Harmonic oscillator


The Schrödinger equation for this situation is


 * $$ E\psi = -\frac{\hbar^2}{2m}\frac{d^2}{d x^2}\psi + \frac{1}{2}m\omega^2x^2\psi $$

It is a notable quantum system to solve for; since the solutions are exact (but complicated - in terms of Hermite polynomials), and it can describe or at least approximate a wide variety of other systems, including vibrating atoms, molecules, and atoms or ions in lattices, and approximating other potentials near equilibrium points. It is also the basis of perturbation methods in quantum mechanics.

There is a family of solutions - in the position basis they are


 * $$ \psi_n(x) = \sqrt{\frac{1}{2^n\,n!}} \cdot \left(\frac{m\omega}{\pi \hbar}\right)^{1/4} \cdot e^{

- \frac{m\omega x^2}{2 \hbar}} \cdot H_n\left(\sqrt{\frac{m\omega}{\hbar}} x \right) $$

where n = 0,1,2..., and the functions Hn are the Hermite polynomials.

Hydrogen atom
This form of the Schrödinger equation can be applied to the Hydrogen atom:


 * $$ E \psi = -\frac{\hbar^2}{2\mu}\nabla^2\psi - \frac{e^2}{4\pi\epsilon_0 r}\psi $$

where e is the electron charge, r is the position of the electron (r = |r| is the magnitude of the position), the potential term is due to the coloumb interaction, wherein ε0 is the electric constant (permittivity of free space) and


 * $$ \mu = \frac{m_em_p}{m_e+m_p} $$

is the 2-body reduced mass of the Hydrogen nucleus (just a proton) of mass mp and the electron of mass me. The negative sign arises in the potential term since the proton and electron are oppositely charged. The reduced mass in place of the electron mass is used since the electron and proton together orbit each other about a common centre of mass, and constitute a two-body problem to solve. The motion of the electron is of principle interest here, so the equivalent one-body problem is the motion of the electron using the reduced mass.

The wavefunction for hydrogen is a function of the electron's coordinates, and in fact can be separated into functions of each coordinate. Usually this is done in spherical polar coordinates:


 * $$ \psi(r,\theta,\phi) = R(r)Y_\ell^m(\theta, \phi) = R(r)\Theta(\theta)\Phi(\phi)$$

where R are radial functions and $$\scriptstyle Y_{\ell}^{m}(\theta, \phi ) \,$$ are spherical harmonics of degree ℓ and order m. This is the only atom for which the Schrödinger equation has been solved for exactly. Multi-electron atoms require approximative methods. The family of solutions are:


 * $$ \psi_{n\ell m}(r,\theta,\phi) = \sqrt {{\left ( \frac{2}{n a_0} \right )}^3\frac{(n-\ell-1)!}{2n[(n+\ell)!]^3} } e^{- r/na_0} \left(\frac{2r}{na_0}\right)^{\ell} L_{n-\ell-1}^{2\ell+1}\left(\frac{2r}{na_0}\right) \cdot Y_{\ell}^{m}(\theta, \phi ) $$

where:
 * $$ a_0 = \frac{4 \pi \varepsilon_0 \hbar^2}{m_e e^2} $$ is the Bohr radius,
 * $$ L_{n-\ell-1}^{2\ell+1}(\cdots) $$ are the generalized Laguerre polynomials of degree n − ℓ − 1.
 * n, ℓ, m are the principal, azimuthal, and magnetic quantum numbers respectively: which take the values:
 * $$ \begin{align} n & = 1,2,3 \cdots \\

\ell & = 0,1,2 \cdots n-1 \\ m & = -\ell\cdots\ell \end{align}$$

NB: generalized Laguerre polynomials are defined differently by different authors - see main article on them and the Hydrogen atom.

Two-electron atoms or ions
The equation for any two-electron system, such as the neutral Helium atom (He, Z = 2), the negative Hydrogen ion (H–, Z = 1), or the positive Lithium ion (Li+, Z = 3) is:


 * $$ E\psi = -\hbar^2\left[\frac{1}{2\mu}\left(\nabla_1^2 +\nabla_2^2 \right) + \frac{1}{M}\nabla_1\cdot\nabla_2\right] \psi + \frac{e^2}{4\pi\epsilon_0}\left[ \frac{1}{r_{12}} -Z\left( \frac{1}{r_1}+\frac{1}{r_2} \right) \right] \psi $$

where r1 is the position of one electron (r1 = |r1| is its magnitude), r2 is the position of the other electron (r2 = |r2| is the magnitude), r12 = |r12| is the magnitude of the separation between them given by


 * $$ |\mathbf{r}_{12}| = |\mathbf{r}_2 - \mathbf{r}_1 | \,\!$$

μ is again the two-body reduced mass of an electron with respect to the nucleus of mass M, so this time


 * $$ \mu = \frac{m_e M}{m_e+M} \,\!$$

and Z is the atomic number for the element (not a quantum number).

The cross-term of two laplacians


 * $$\frac{1}{M}\nabla_1\cdot\nabla_2\,\!$$

is known as the mass polarization term, which arises due to the motion of atomic nuclei. The wavefunction is a function of the two electron's positions:


 * $$ \psi = \psi(\mathbf{r}_1,\mathbf{r}_2). $$

There is no closed form solution for this equation.