Thursday, September 13, 2012

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..












2 comments:

  1. A differentiable function is a function whose derivative exists at each point in its domain. The graph of a differentiable function must have a non-vertical tangent line at each point in its domain.continuous and differentiable

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    1. Yes that's true. But does it pertain to the above post?

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