Laplace's Equation in Spherical Coordinates: Legendre Polynomials

In this post we will be solving Laplace's Equation in Spherical coordinates.  The solution to this is the Legendre Polynomials.

Laplace's Equation in Spherical coordinates is...

We now take this equation and employ the separation of variables technique.

In order for three functions of three different variables to equal a constant, they must each themselves be equal to a constant.

Getting back to the main differential equation, we will do some re-arranging and use n(n+1) as the separation constant.

Let us now solve the equation for U.  This is a Cauchy-Euler Equation.

We now return to the main differential equation.

This is the generalized Legendre equation.

We will assume azimuthal symmetry (aka, no dependence on r, aka axial symmetry)  this means that m=0 and we are left with the ordinary Legendre Equation.

We will now solve this differential equation using a series solution.

We now have a recursive relation to generate the other coefficients.

We now plug these coefficients into the series solution..

These are the Legendre Polynomials.

We need to normalize each polynomial so that its value is unity for x=1.

Let us calculate a couple of the polynomials. You will notice that for even values of n, the first term terminates while the other term doesn't. (Likewise for odd values of n and the other series.)  We only need one of them for a solution so we will take the one that terminates.

The Legendre Polynomials can be put into closed form..

But don't forget! We still need to solve for V in our original equation by plugging everything in. Remember we are still assuming azimuthal symmetry. (m=0)

Quantum Mechanics: One Dimensional Potentials

(The following are my notes on Chapter 4 of "Quantum Physics" by Stephen Gasiorowicz)

This post will deal with one dimensional potential functions in the Schrodinger equation.  The first one we will look at is the Potential Step.

The Potential Step:

This problem has the following potential function.

We will start with the time independent Schrodinger Equation..

Before x=0 and after, the solution will be different. (Slightly, only by the magnitude of E-V compared to just E.) When x<0 we have..

To see a post that includes what the flux, or probability current is, go here.

We can view e^(ikx) as an incoming wave with probability current (h-bar)k/m.  If there was no potential function, then this alone would suffice for the solution to the time independent Schrodinger equation.  Therefore we can attribute the second term to the term caused by the potential ie, the reflected wave with flux [(h-bar)kR^2]/m.  R is the reflection coefficient.

We now need to investigate the behavior of the wave function after the point x=0.

The second term must be equal to zero because in our "experiment" wave is coming from the minus x direction and this term would only yield zero (at infinity) if it was coming from the +x direction. Thus our solution is..

Where T is the transmission coefficient.  

We now calculate the flux or probability current of this solution..

It is important to note that even though the wave function is discontinuous at x=0, the wave function is not.  The slope (or rate of change) of the wave function is also continuous.  We show this by integrating the time independent Schrodinger equation over a small distance -e to e. (e is epsilon)

Which implies continuity of the wave function's derivative.  

We now take the two equation that we have above, and solve for R and T.

We can now use these results to calculate the fluxes.

An interesting thing to note is if E is less than V, then q becomes imaginary, and we get an exponential decay "through" the barrier.  It can be shown (not that hard) that R^2 in this case is equal to 1, thus we would have total reflection, however there is still partial "penetration through the barrier" this allows, for a "thin" enough barrier for the wave to "tunnel" through.

The Potential Well:

The next situation we will investigate is the potential well.  It has the following potential function.

We can immediately write down the solutions based upon the logic from the previous discussion. (Where k and q have the same values as before.)

We now impose the same conditions for continuity.  However, we will combine the conditions into one expression that must be true for each solution.  Remember, both the wave function and its derivative must be continuous at the discontinuities.

We now use this condition for each discontinuity. 

We now solve these two equations to eliminate A and B.

The Potential Barrier:

We now consider a similar situation, the potential barrier.  It has the following potential function.

We will examine what happens inside the barrier.  We again start with the time independent Schrodinger equation.

If (kappa)*a >> 1, then this simplifies to...

This, even though the energy is below the potential barrier, there is indeed transmission through the barrier.  This transmission is extremely sensitive to the width of the barrier.

The Harmonic Oscillator:

Now we get to the meat and potatoes of this post.  The quantum harmonic oscillator. This problem deals with a spring potential energy function.  We will start out with the Hamiltonian for this problem and construct the Schrodinger equation for this situation.

For any eigenvalue of epsilon, as y^2 approaches infinity, the term involving epsilon is negligible, thus we require that u(y) asymptotically follow the equation....

We will now find the homogeneous equation. (We are solving this similar to the variation of parameters approach.)

The way we will solve this is by using a series solution.

When this is inserted into the equation for u(x), we get a solution which does not terminate at infinity.  The way to solve this is to have the solution terminate at certain values of epsilon, namely epsilon=2n+1.

For these values of epsilon, the series terminates and the recursion relation yields..

The function h(y) is known as the Hermite Polynomials.

Quantum Mechanics: Eigenvalue Problems

(The following are my notes on Chapter 3 of "Quantum Physics" by Stephen Gasiorowitcz)

The Time-Independent Schrodinger Equation:

We will start with the Schrodinger equation, with a potential function that only depends on the position.

Where the H operator on the right hand side is called the Hamiltonian, which is essentially the kinetic energy plus the potential energy operators acting on the wave function Psi.  Because we are imposing a potential function that does not depend on time, the solution to this equation can be solved by using separation of variables.

The only way two functions of different variables can be equal to each other for all values of x and t, is if they equal a constant value.  Otherwise t would equal x and it wouldn't really be a function of two variables.

This differential equation is called the time-independent Schrodinger equation.  You see now why we chose the letter "E" as the constant that both functions were equal to.  This is because the left side of the T-I Schrodinger equation is the Hamiltonian acting on u(x).

In classical mechanics, F=-dV/dx meaning if we added a constant V_0 to the potential function, it does nothing to affect the force.  The same holds true for the above equation.  Replacing V(x) by V(x)+V_0 and re-doing all the math we get...

As per your differential equations class, the solution to this will be a Fourier series for all values of n.  We can make a term for all the discrete values of n (summation) and all the continuous values of n (integral). (This isn't terribly important at this level!)

Particle in a Box:

Next we deal with an actual problem.  This problem doesn't necessarily hold any physical significance, but rather is used to show principles.  This problem deals with a "potential well" with an infinite amount of energy required to escape the walls.  This is a one dimensional problem. The problem is described by the following conditions and picture.

From this figure we see that. u(x) = 0 when x is less than zero or greater than a.  Inside the box the potential function is zero, thus the time independent Schrodinger equation becomes.. (Which I will also solve)

So, we have found the energy levels associated with this system. Remember, n is an integer! Normalize the eigenfunctions since they represent a probability, and we know that the particle is in there so the total probability must be 1. (There is a different "A" for each value of n.)

An technically, the sum of all the solutions for every "n" is also a solution so, the formal solution would be a summation over all values of n.

We can extract a lot of information from the solutions above.

The state of lowest energy, (aka the ground state) is given when n=1, thus the energy is..

In a classical example, the least energy would be when the particle is at rest, hence E=0.  Here we have a minimum non-zero energy of the particle.

We can calculate the expected value of the momentum of the particle. (Remember the solutions are real!)

Thus, the expected value of the momentum is zero.  This is to be expected since the particle is confined to a box and is going in one direction exactly as much as it is going in the opposite direction. (Remember this is 1D!)

The expected values of the momentum squared does NOT vanish.

But how does the kinetic energy act?

So the kinetic energy increases as u(x) oscillates more and more.

There is a pattern shown by these solutions.  For the odd values of n, are unaltered when they are reflected from the walls, while the even values of n change sign upon reflection.  We will make the following shift.

  

From our previous solution and from Fourier's Theorem, we know that..

We will now find an interpretation of the Fourier coefficients. 

To interpret |A_n|^2, we will note that an energy measurement can only yield one of the eigenvalues.  Thus, a measurement of the average energy of a system will give..

So the coefficients are interpreted as the probability that a measurement of the energy for the state of Psi(x) yields the eigenvalue E_n.

Momentum Eigenfunction

The energy operator H is not the only one that has eigenfunctions and eigenvalues.  We will now solve the Eigenvalue problem for the momentum operator.

The constant C must be determined by normalization.  Since these eigenfunctions are not quantized, (meaning p is not restricted to integers) the operator for momentum is said to have a continuous spectrum.  But we might, by analogy assume it obeys the orthonormallity condition that the hamiltonian follows.

We can, just like with the Hamiltonian write the wave function as a sum of of eigenfunctions.  It is a little different however since we are dealing with a continuous amount of eigenfunctions instead of a discrete amount.

We now have the wave function in momentum space.

For a particle in a box, we found that for odd values of n, we have an even function for the solution and the wave remains unchanged upon reflection.

Thus, the wave function remains even over time.  So evenness and oddness are constants of motion that do not change.

We now introduce the Parity Operator, which job is to simply replace x by -x.

These are both eigenfunctions with eigenvalues +1 and -1 respectively.

Just like in the other cases, the wave function can be written as a sum of the eigenfunctions.

We will define the following function..