Archive for December, 2008

Monday Math 52

December 29, 2008

Last week, we looked at solving Laplace’s equation, in three dimensions, specifically Cartesian and cylindrical coordinates. Now, let us consider spherical coordinates (ρ,φ,θ) (we use the physicist’s convention of polar angle θ and azimuthal angle φ).

In spherical coordinates, Laplace’s equation is

Trying a solution of the form f(ρ,φ,θ)=R(ρ)Φ(φ)Θ(θ), we have:

The first bracket depends only on ρ, while the second depends only on the angular coordinates. We choose a constant to which the first bracket equals:


The angular portion is then:

and we have separated φ and θ terms. Expecting solutions that are continuous and periodic in 0≤φ<2π, we set


where m is an integer.
This gives us the remaining differential equation

.

To examine the solutions to this last equation, we first consider the m=0 case. Here, we have:
.
Let us make the change of variables . Then we have


This is amenable to solution by the method of Frobenius. One finds that the resulting power series converges for all -1<x<1 (0<θ<π), but that it diverges at x=±1 (blows up at the polar axis) unless the series terminates. As the recursive formula for the coefficients of the power series is , the polynomial terminates only if k=l(l+1), where l is a non-negative integer. Then the solution is a polynomial with degree l, which is an even polynomial for even l and an odd polynomial for odd l. The initial coefficient, a0 or a1, is arbitrary, and is usually chosen so that . These polynomials are known as Legendre polynomials.
They can be generated via Rodrigues’ formula:
.
The first few polynomials are:




.

Now, for m≠0, we have, using k=l(l+1) (here l is not necessarily an integer),

,
called the “general Legendre equation.” It has solutions called the associated Legendre functions, written , where l is called the degree and m the order of the function. We already have that
, the legendre polynomial of degree l, when l is a non-negative integer.
It can be shown that the associated Legendre functions are well behaved on -1≤x≤1 for l and m integers with |m|≤l, l≥0, with the solutions for non-negative m m in terms of Legendre polynomials being

,
and the negative m solutions given by
.
(I have used here the Condon-Shortley phase.)
So our solutions will generally have a polar angle component
.
If we combine these with the azimuthal angular component written in terms of complex exponentials, we then have:

where l and m are integers with l≥0 and |m|≤l. Choosing our constant to produce the othonormal relation:

(where the integration is over all solid angles, * indicates the complex conjugate, and is the Kronecker Delta),
we get
.
These are called spherical harmonics.
The first few are:







Now, for our radial component, we have ; with k=l(l+1), we get

Trying a solution of the form , we find



or
So our radial component is of the form , and our solutions to Laplace’s equation in spherical coordinates will be of the form
.

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Physics Friday 52

December 26, 2008

Quantum Mechanics and Momentum
Part 9: Angular Momentum as an Operator


Classically, the angular momentum L about the origin for a mass with linear momentum p at displacement r is given by . The Cartesian components are thus:
,
,
.
To turn this into a quantum operator, we replace the position and momentum r and p with the corresponding quantum operators. The components of the operator L are thus
,
,
.
The commutator of the x and y components is:
.
Similar relations hold for the cyclic permutation of the coordinates:

.
And thus we also have


.
From our previous discussion of commutators, we then see that two different components of the angular momentum cannot be simultaneously fixed for a quantum wavefunction.

Now, we examine the square norm of the angular momentum operator:
.
Finding it’s commutator with one of the components, and using several of the commutator relations we noted previously:
.
Similarly,
.
Thus, we can simultaneously specify the magnitude of the angular momentum and (any) one of it’s individual components, but no more components than that one. Usually, we choose to label the fixed component as our z.

Monday Math 51

December 22, 2008

Last week, we looked at solving Laplace’s equation, in two dimensions. In particular, I demonstrated the technique of separation of variables for both Cartesian and polar coordinates.

Now, let us consider Laplace’s equation in three dimensions. In Cartesian coordinates, this is . If we use separation of variables by trying f(x,y,z)=X(x)Y(y)Z(z), then we get

As before, each of the three terms on the left hand side is a function of a single, different independent variable, each of these three terms is equal to a constant, the sum of which is zero:
, , ,
and so
, ,
Note that we have either two of the three constants positive, and the third negative, or we have one positive and two negative. In the first case, the resulting function is periodic in one dimension and exponential in the other two; in the second, it is periodic in two dimensions and exponential in the third.

Consider the function over the cube 0≤x≤1, 0≤y≤1, 0≤z≤1 with (Dirichlet) boundary conditions , , , . Seeing the x and y boundary conditions, it must be periodic in x and y: the x=0 and y=0 conditions give , , and our x=1 and y=1 conditions give and , where m and n are positive integers. The original equation tells us then that . The condition that tells us that our Am,n are all zero, and we obtain:
.
For z=1, we get

Using the orthogonality relationships of circular functions, we get:
,
and so
.

Now, let us instead consider cylindrical coordinates (r,φ,z). Drawing upon our plane polar coordinate laplacian, we get
. Trying a solution of the form f(r,φ,z)=R(r)Φ(φ)Z(z), and then dividing the resulting equation by R(r)Φ(φ)Z(z), we get

We see that the third term depends only on z, while the other terms depend only on the other two coordinates; thus we have that the third term must be a constant:

,

and also

Here, we have that the last term on the left hand side is a function of φ only, while the remaining terms depend on r only. As we usually want periodic solutions in φ, we set its term at –m2, so


.
To have a continuous function on all φ, 0≤φ<2π, we require that m be a non-negative integer (it can be zero for the constant case ).
The differential equation for R is now


Making the change in variables to the dimensionless variable , this becomes:

which is known as Bessel’s differential equation.
It’s solutions are known as the Bessel functions of the first and second kind, and . Bessel functions of the first kind, , are finite at u=0 for non-negative integer m, while the are singular at u=0. Thus, most problems will have solutions involving Bessel functions of the first kind.

Physics Friday 51

December 19, 2008

Quantum Mechanics and Momentum

Part 8: Commutators

An important mathematical entity in quantum mechanics is the commutator of two operators. For two operators A and B, their commutator, written with the symbol [A,B], is given by [A,B]=AB-BA. Note that commutator is itself an operator. The name “commutator” arises because two operators commute if and only if their commutator is zero.
As demonstrated in this post, if two operators have a non-zero commutator, then they obey an uncertanty relation: . In fact, it can be shown that a wavefunction can simultaneously be an eigenstate for two distinct observables if and only if the two operators commute, and thus have a commutator of zero.

Consider the one-dimensional position and momentum operators in position space, as discussed in earlier posts. Then their commutator is


The commutator is useful in examining the quantum operators for angular momentum. There are some important properties of commutators which we will note here. First, we see [B,A]=-[A,B] by immediate inspection.
Next, for (linear) operators A and B and scalar c,
[A,cB]=A(cB)-cBA=c(AB-BA)=c[A,B] and [cA,B]=cAB-B(cA)=c(AB-BA)=c[A,B];
and for operators A, B, and C:
[A,(B+C)]=A(B+C)-(B+C)A=AB+AC-BA-CA=AB-BA+AC-CA=[A,B]+[A,C]
[(A+B),C]=(A+B)C-C(A+B)=AC-CA+BC-CB=[A,C]+[B,C];
so the commutator is bilinear (linear in both arguements).
We see also [A,A]=0, as any operator commutes with itself. In addition, [A,An]=0, and in fact, for an operator f(A) that is a function of the operator A only, then [A,f(A)]=0.
One more useful relation is
[A2,B]=A2B-BA2
       =A2B-ABA+ABA-BA2
       =A(AB-BA)+(AB-BA)A
[A2,B]=A[A,B]+[A,B]A

Monday Math 50

December 15, 2008

Let us consider Laplace’s equation, in two dimensions. In Cartesian coordinates, this is . We can see by immediate inspection that linear functions f(x,y)=ax+by+c satisfy this. We also see that terms of the form axy satisfy the equation.

A useful method for finding general solutions is the separation of variables for partial differential equations technique. Here, we substitute in a function of the form f(x,y)=X(x)Y(y). Then Laplace’s equation becomes:

Dividing both sides by f(x,y)=X(x)Y(y), we have:

Now, we note that is a function of x only, while is a function of y only. By varying x while holding y fixed, and vice versa, we see that the sum of these two can be zero for all x and y if and only if both functions are constants, which sum to zero. Thus, we have , for some constant λ.
Both of these are second order differential equations: has solution , while has solution .
Combining these, we have solutions of the form , and we can sum any of these to produce another solution.

For a more specific example, let us choose the region 0≤x,y≤1 with boundary conditions (Dirichlet) , , , .
Then we have, from the above, solutions of the form , , . To get a solution , we want
,
for a function . For a nonzero solution which obeys , we need to be periodic in x (complex exponential, rather than real), so we need λ<0. Thus, let λ=-ω2. We then have
,
and tells us that , and thus for positive integer n,
Thus, we have λ=-ω2=-n2π2, and . tells us , which means
,
and so
,
and our solutions for these three parts of the boundary are .
Consider the Fourier series for the square wave function with period 2π , which is

Compare to
With for odd n and for even n, we have when y=1 the terms of the Fourier series for . Thus, our last boundary is met, and our solution found, by
.

Here is a graph of the partial series with the first 25 terms:
laplacegraph

Now, let us consider polar coordinates, useful for solving Laplace’s equation on the disk. In polar coordinates, the laplacian is
.
Trying a function of the form in Laplace’s equation, we get:

Multiplying that equation by r2 and then dividing by , we have:

As the terms in the first set of parentheses is a function of r only, and the second set holds a function of θ only, these two functions are constants, and so we have , .
The latter equation is easily solved: has general solution . Meanwhile, the second order linear differential equation
is a second order Cauchy-Euler equation, which we solve by substituting a trial solution of the form rm:

for λ≠0. With λ=0, we have the equation: , which, with , gives the first order equation , which is separable:

So we have for λ≠0, and for λ=0.
Note that if our region covers all angles θ, then we have the requirement that must be periodic, with . This tells us first that λ≥0; with λ=ω2, we have . Second, the requirement that means that ω=n, n an integer:
.

Examining our radial solution for λ=ω2=n2, we have for positive n the radial function . The latter term has a singularity at the origin, so if the region over which we wish to solve contains the origin, we must have A2=0 for n>0. For n=0, we have radial function , and to avoid a singularity at the origin, we again set A2=0. Thus, the solutions for Laplace’s equation for a domain in polar coordinates containing the origin (and all angles θ) are of the form , n=0,1,2,3,…
For example, n=0 is the constant function; n=1 gives
; these confirm our early result of linear functions in x and y being solutions. n=2 gives us and , this latter which we also noted earlier.

Suppose we want to solve Laplace’s equation on the unit disk with (Dirichlet) boundary condition . Then we see that we want to combine solutions of the form . As our boundary condition is an odd function of θ, the cosine terms disappear, and we have , plugging in r=1, and noting that our boundary condition is the square wave function we defined earlier: , we put , giving solution


Here is a graph of the partial series with the first 25 terms:
laplaceroundgraph

Physics Friday 50

December 12, 2008

Quantum Mechanics and Momentum
Part 7: Plane Waves

Last week, we introduced the time-dependent Schrödinger equation , and we noted that for an energy eigenstate ψn(x) with energy En (), we have time-dependent solution , with .
Now, consider the case of a free particle of mass m. Here, the potential is everywhere zero, and the energy is given entirely by the kinetic energy. Recalling the (non-relativistic) formula for kinetic energy in terms of momentum, we have . Working in three dimensions, and replacing the classical momentum with the quantum momentum operator , we have , and thus the free particle Hamiltonian:
.

Note that a momentum eigenstate with momentum will also be an energy eigenstate with energy . This can be confirmed by applying to the momentum eigenfunction , where . Now, examining the corresponding solution to the time-dependent Schrödinger equation, we have
.
This, we see, is a wave propagating in the direction of , with (angular) frequency ω and (angular) wavenumber . The wavefronts of this wave, the surfaces of constant phase, can be seen to be planes perpendicular to . Thus, this kind of solution is known as a plane wave.

We have .
Now, from this we find that the group velocity of our plane wave is

which is just the classical velocity of the particle of mass m and momentum p (from p=mv).

Evolving the Mona Lisa

December 10, 2008

This is really cool: Genetic Programming: Evolution of Mona Lisa.
Roger Alsing used a genetic algorithm program to produce a replica of the Mona Lisa that uses just 50 semi-transparent polygons. His post has pictures at several steps along the process, from the initial black background (generation 1), to the final product at generation 904314.
[Via bioephemera]

No Beanstalks

December 9, 2008

According to this New Scientist article, a recent study of the full mechanics of a “space elevator” indicates that such might be prohibitively unstable, with coriolis forces converting the mere motion of a payload along the cable into wobbling of the cable. Add in lunar and solar tidal effects, and the solar wind, and it appears the needed stabilizing thrusters and slow payload speeds would remove any benefits from having the elevator in the first place.

The 40-Year-Old Mouse

December 9, 2008

…the computer mouse turned 40 years old today.

A Quote…

December 9, 2008

Stress is when you wake up screaming and you realize you haven’t fallen asleep yet
–(Unknown)