*QUANTUM
MECHANICS: a synopsis*

*Wave
function and Schrodinger equation for a free particle*

**Schrodinger's equation** for a particle
in one dimension:

i hbar psi_t = - (hbar^2/2m) psi_xx

where hbar is **Planck's constant**
h divided by 2Pi, m is the mass of the particle, and psi is the **wave function.
**Because of the factor of i on the left hand side, all solutions to the
Schrodinger equation must be **complex**. Numerically, hbar ~= 2/3
eV-fs = (6.63/2Pi ) x 10^(-34) J-s. For macroscopic systems hbar is
a TINY number, but for atomic systems it is of order unity. More specifically,
the binding energy of an electron in an atom is of the order of eV (electron
volt) and the time to complete one orbit around the nucleus is of the
order of one fs (1 fs = femtosecond = 10^(-15) s).

*Physical
meaning*

The physical meaning of psi is a "probability
density amplitude", that is, the probability of finding the particle between
x=a and x=b at time t is the integral from a to b of |psi(x,t)|^2,
the squared modulus of psi(x,t). Since the particle must be SOMEWHERE the
integral of |psi(x,t)|^2 over all space must be 1. This condition is automatically
preserved in time for solutions of the Schrodinger equation. A particle doesn't
have a definite position, it has only a **probability** of being found
in any given region.

*Energy
and momentum*

The above equation describes a force-free particle,
so it is to quantum mechanics what Newton's first law is to particle mechanics.
For a wave function of the form Aexp(-iwt)exp(ikx), Schrodinger's equation
implies the dispersion relation hbar w = (hbar k)^2/2m. This is the
same as the relation E = p^2/2m between energy and momentum for a free particle,
if one adopts the correspondence E = hbar w and p = hbar k. Equivalently,
E = h nu and p = h/lambda, where nu and lambda are the frequency and wavelength.
Lambda is called the **de Broglie wavelength** of the particle. In terms
of E and p the above free particle Schrodinger wave is A exp(- iEt/hbar)
exp(ipx/hbar). Note that although the Schrodinger equation does not at first
look like a wave equation, since the time derivative is only first order
instead of second order, its solutions are nevertheless like those of the
wave equation. This is due to the magic of i. Replace the i by 1 on the left
hand side and you get a completely different beast: the diffusion equation.

*Wave
packets, the uncertainty relation, and velocity*

A particle with a definite momentum is evenly
spread out in space, equally likely to be found anywhere, since |Aexp(-iwt)exp(ikx)|=|A|.
(Actually this is a bit of an embarassment, since the total probability of
being anywhere must be unity. No value of the amplitude A will produce this
normalization. Therefore a particle with a precisely defined momentum is
not physically meaningful. There must always be at least a very small spread
in momentum.) A more localized particle is described by a wave packet
containing a spread of momenta. A particle localized perfectly at one point
is described by the Dirac delta function, which consists of ALL momenta.
(This extreme case is also not physically meaningful.) In general , the spread
in position is at least inversely proportional to the spread in momentum:
(Delta x) (Delta p) >= hbar/2 (it turns out the exact minimum is hbar
/2 if Delta x and Delta p are defined as the standard deviations from the
mean). This is called the **Heisenberg uncertainty relation**.
As a wave packet peaked at the wave number k evolves, it spreads out,
but its center moves with the group velocity dw/dk evaluated at k. This velocity
is hbar k/m , swhich corresponds to p/m , the particle velocity. So the velocity
of a classical particle corresponds in fact to the group velocity of the
quantum wave packet for the particle! (The phase velocity is half of this.)

*External
forces*

If a force is acting on the particle, the Schrodinger
equation must be modified, If the force is -dV/dx for some potential energy
function V(x), the term V psi must be added to the right hand side of the
Schrodinger equation, i hbar psi_t = - (hbar^2/2m) psi_xx + V(x)
psi. This equation is to quantum mechanics what Newton' *second *law
(F = ma) is to particle mechanics.

*Energy
levels*

Although a quantum particle can not have a definite
position, it can have a definite energy. For example in an atom, the allowed
definite energies form a discrete set and are called the **quantizedenergy
levels **of the atom. Since energy corresponds to frequency, the energy
levels correspond to solutions of the Schrodinger equation with a definite
frequency, together with some spatial dependence: psi(x,t) = exp(-iEt/hbar)
f(x). The spatial part f(x) must satisfy the **time-independent Schrodinger
equation** - (hbar^2/2m) f_xx + V(x) f = E f, written here in one
dimension.Only for particular values of the energy E are there solutions to
the spatial equation for f satisfying the boundary conditions. For a quantum
particle, the "boundary condition" on f is that the total probability is
one, i.e. \int |f|^2 dx = 1. These definite energy level solutions are in
perfect analogy with the **normal modes** of a continuous vibrating system,
like a string for example. So the energy levels of a quantum particle are
the normal modes of the Schrodinger wave.