Quantum harmonic oscillator
The Hamiltonian for the system is the following:
This Hamiltonian is a one dimensional Hamiltonian. Here is what each of the parts of the Hamiltonian mean:
- m is the mass of the particle
- The first term, is the usual kinetic energy term.
- The second term, is the potential.
The potential term is very frequently written as . This is because the spring constant k is related to the oscillator frequency via the relationship . When this is done, the Hamiltonian reads
Time independent Schrödinger equation edit
The time independent Schrödinger equation is
and if we project onto the position basis, we get
Substituting our Hamiltonian into the equation, we get
The constants can be pulled out in front of the bra so the Schrödinger equation now reads
Now, consider the term Recall that so
For the other term , recall that so that Putting all the pieces together, the Schrödinger equation reads
Since we are working in the position basis, we have so we finally get
which is a differential equation which can be solved for . This is of course, the wavefunction of the system in the position basis.
Solutions to the quantum harmonic oscillator edit
There are different approaches to solving the quantum harmonic oscillator. One of them, involves directly solving the differential equation which was obtained in the previous section. We will do this first. Afterwards, we will solve this same system with the "operator factorization method" as a way to motivate the introduction of boson operators into our quantum mechanical theory.
First, let's define a characteristic length for the quantum harmonic oscillator. We can do this heuristically by looking at the units involved in our expression.
- has units of
- has units of
- has units of
Hence, the quantity has the units of length. We will call this length the "characteristic length". If we substitute into the differential equation we will get
Note also, that the units of are also energy units. We can define as the characteristic energy of the system. (In fact, later, we will find that this energy happens to be the ground state zero point energy of the quantum harmonic oscillator.) So, putting we get
To solve this equation, first consider a simpler equation which describes the behaviour of the original wavefunction in some asymptotic limit. In the regime where the energy is very low, , the wavefunction should then satisfy the differential equation
The form of this equation suggests that . Substituting
This differential equation can be solved in many different ways. One approach is to take
The derivatives are (note carefully the summation limits)
Substituting all of these terms into the above yields
and vanishes term by term provided that
The most straightforward way to enforce these relationships is to set the numerator to zero. This leads to
It's a very interesting result since the energy is now constrained to take on certain discrete values.
This is still work in progress. Your feedback is appreciated! You're invited to leave comments on the talk page.