Concepts and formulas

One-dimensional eigenvalue problems in quantum mechanics

The operator representing the energy of a system is H.  The eigenvalues of H are E.  If the potential energy function U(x) is independent of time, then separation of variables is possible, and we can write ψ(x,t) = Φ(x)Χ(t).  If the wave function is of this form, then Φ(x) = ΦE(x) is an eigenfunction of the operator H, and the energy of the system is certain.  We find the eigenfunction of H by solving HΦE(x) = EΦE(x).

Regions that do not contain a well
There exists an eigenfunction for every E > Umin.  These eigenfunctions, however, are plane waves and are not square integrable.  They cannot represent a single particle, but can represent a constant flux of particles.  We calculate transmission and reflection coefficients by comparing fluxes.  (Flux ∝ k|Φ(x)|2.)

Regions that do contain a well
For E < Erim eigenfunctions exist only for selected eigenvalues.  These eigenfunctions are square integrable.

Confinement leads to energy quantization.

"Square" potentials

We can solve HΦE(x) = EΦE(x) in regions of piecewise constant potentials.
Assume U(x) = U = constant in certain regions of space.  In such a region the Schroedinger equation yields
(∂2/∂x2)Φ(x) + (2m/ħ2)(E - U)Φ(x) = 0.

i.  Let E > U: (∂2/∂x2)Φ(x) + k2Φ(x) = 0.   E - U = ħ2k2/(2m).
The most general solution is Φ(x) = Aexp(ikx) + A'exp(-ikx), with A and A' complex constants.
ii.  Let E < U: (∂2/∂x2)Φ(x) - ρ2Φ(x) = 0.   U - E = ħ2ρ2/(2m).
The most general solution is Φ(x) = Bexp(ρx) + B'exp(-ρx), with B and B' complex constants.
iii.  Let E = U: (∂2/∂x2)Φ(x) = 0.  Φ(x) = Cx + C', with C and C' complex constants.

(Note: A solution exists in the classically forbidden regions.)

How does the wave function behave at a point where U is discontinuous, i.e. at a step?
(a)  At a finite step the boundary conditions are that Φ(x) and (∂/∂x)Φ(x) are continuous.
(b)  At an infinite step (∂/∂x)Φε(x)|ε-->0 is discontinuous, but it has a finite discontinuity.  Therefore Φε(x) remains continuous as ε --> 0.

Square Potentials (utk.edu)

We can solve for the bound states in a square-well potential using a graphical solution.


Angular momentum in Quantum Mechanics

The operator J, whose Cartesian components satisfy the commutation relations
[Ji,Jj] = εijkiħJk  
is defined as an angular momentum operator.  
For such an operator we have [Ji,J2] = 0, i.e. the operator J2 = Jx2 + Jy2 + Jz2 commutes with each Cartesian component of J.  We can therefore find an orthonormal basis of eigenfunctions common to J2 and Jz.  We denote this basis by {|k,j,m>}.  
We have J2|k,j,m> = j(j + 1)ħ2|k,j,m>,  Jz|k,j,m> = mħ|k,j,m>.  
The index j can take on only integral and half integral positive values.  For a given j the index m can take on one of 2j + 1 possible values,  m = -j, -j + 1, ... , j - 1, j.  

We define the ladder operators J+ = Jx + iJy and J- = Jx - iJy.  
We then have Jx = ½(J+ + J-) and Jy = (-i/2)(J+ - J-) .
The operators J± operating on the basis states {|k,j,m>} yield 
J±|k,j,m> = [j(j+1) - m(m±1)]½ħ|k,j,m±1>.

Orbital angular momentum

The operator L = R × P  satisfies the commutation relations [Li,Lj] = εijkħLk and is called the orbital angular momentum operator.  We denote the  common eigenstates of L2 and Lz by {|k,l,m>}.  In coordinate representation we have
Lz = (ħ/i)∂/∂φ
and
L2 = -ħ2[(1/sinθ)∂(sinθ ∂/∂θ)/∂θ + (1/sin2θ)∂2/∂φ2].
The normalized common eigenfunctions of L2 and Lz are called the spherical harmonics.

Properties of the spherical harmonics
Ylm(θ,φ) = [(-1)l/(2l l!)][(2l+1) (l+m)!/(4π (l-m)!)]½eimφ(sinθ)-mdl-m(sinθ)2l/d(cosθ)l-m.
We have
Y00 = (4π),  Y1±1 = ∓(3/8π)½sinθ exp(±iφ),  Y10 = (3/4π)½cosθ,
Y2±2 = (15/32π)½sin2θ exp(±i2φ),  Y2±1 = ∓(15/8π)½sinθ cosθ exp(±iφ),
Y20 = (5/16π)½(3cos2θ - 1).

The Ylm(θ,φ) form a complete set of functions of angle on the unit sphere.  Orthonormality is expressed through
0πsinθ dθ∫0dφ Y*l'm'(θ,φ)Ylm(θ,φ) = δl'lδm'm.
and completeness is expressed through
l=0m=-ll Y*lm(θ,φ)Ylm(θ',φ') = δ(cosθ - cosθ')δ(φ - φ') = δ(θ - θ')δ(φ - φ')/sinθ.

Complex conjugation
Y*lm(θ,φ) = (-1)mYl(-m)(θ,φ).

Parity
PYlm(θ,φ) = Ylm(π - θ,π + φ) = (-1)lYlm(θ,φ).
The parity of the spherical harmonics is well defined and depends only on l.

Spin ½

The state space of a spin ½ particle is two-dimensional.
The common orthonormal eigenbasis of S2 and Sz is {|+>, |->}.
Below are matrices of spin operators in this basis.

image   image   image   image

We write S = (ħ/2)σ.  The matrices

image

are the Pauli matrices.

Properties of σx and σy and σz
det(σi) = -1,  Tr{σi} = 0,  σi2 = I,  σxσy = -σyσx = iσz.
In general, σiσj =  iεijkσk + δijI,   [σij] = i2εijkσk.
Therefore 
[Sx,Sy] = iħSz,   [Sy,Sz] = iħSx,    [Sz,Sx] = iħSy,
since S = (ħ/2)σ.

The operator Su is defined through Su = Sxsinθcosφ + Sysinθsinφ + Szcosθ.
The matrix of Su is (Su) = (Sx)sinθcosφ + (Sy)sinθsinφ + (Sz)cosθ.

The eigenvectors of Su are
|+>u = cos(θ/2)exp(-iφ/2)|+> + sin(θ/2)exp(iφ/2)|->,
|->u = -sin(θ/2)exp(-iφ/2)|+> + cos(θ/2)exp(iφ/2)|->.
Therefore |+> = cos(θ/2)|+>u - sin(θ/2)|->u,  |-> = sin(θ/2)|+>u + cos(θ/2)|->u