# Quantum harmonic oscillator

The quantum harmonic oscillator is the quantum mechanical analogue of the classical harmonic oscillator. It is one of the most important model systems in quantum mechanics.

## Hamiltonian

The Hamiltonian for the system is the following:

${\displaystyle {\hat {H}}={\frac {{\hat {p}}^{2}}{2m}}+{\frac {1}{2}}k{\hat {x}}^{2}}$

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, ${\displaystyle {\hat {T}}={\frac {{\hat {p}}^{2}}{2m}}}$ is the usual kinetic energy term.
• The second term, ${\displaystyle {\hat {V}}={\frac {1}{2}}k{\hat {x}}^{2}}$ is the potential.

The potential term is very frequently written as ${\displaystyle {\hat {V}}={\frac {1}{2}}m\omega ^{2}{\hat {x}}^{2}}$. This is because the spring constant k is related to the oscillator frequency via the relationship ${\displaystyle k=m\omega ^{2}}$. When this is done, the Hamiltonian reads

${\displaystyle {\hat {H}}={\frac {{\hat {p}}^{2}}{2m}}+{\frac {1}{2}}m\omega ^{2}{\hat {x}}^{2}.}$

## Time independent Schrödinger equation

The time independent Schrödinger equation is

${\displaystyle {\hat {H}}\left|\psi \right\rangle =E\left|\psi \right\rangle }$

and if we project onto the position basis, we get

${\displaystyle \left\langle x\right|{\hat {H}}\left|\psi \right\rangle =\left\langle x\right|E\left|\psi \right\rangle =E\left\langle x|\psi \right\rangle .}$

Substituting our Hamiltonian into the equation, we get

${\displaystyle \left\langle x\right|{\frac {{\hat {p}}^{2}}{2m}}+{\frac {1}{2}}m\omega ^{2}{\hat {x}}^{2}\left|\psi \right\rangle =E\left\langle x|\psi \right\rangle .}$

The constants can be pulled out in front of the bra ${\displaystyle \left\langle x\right|}$ so the Schrödinger equation now reads

${\displaystyle {\frac {1}{2m}}\langle x|{\hat {p}}^{2}|\psi \rangle +{\frac {m\omega ^{2}}{2}}\langle x|{\hat {x}}^{2}|\psi \rangle =E\langle x|\psi \rangle .}$

Now, consider the term ${\displaystyle \langle x|{\hat {x}}^{2}|\psi \rangle .}$ Recall that ${\displaystyle \langle x|{\hat {x}}=\langle x|x=x\langle x|}$ so ${\displaystyle \langle x|{\hat {x}}^{2}|\psi \rangle =x^{2}\langle x|\psi \rangle .}$

For the other term ${\displaystyle \langle x|{\hat {p}}^{2}|\psi \rangle }$, recall that ${\displaystyle \langle x|{\hat {p}}=\langle x|-i\hbar {\frac {d}{dx}}=-i\hbar {\frac {d}{dx}}\langle x|}$ so that ${\displaystyle \langle x|{\hat {p}}^{2}|\psi \rangle =-\hbar ^{2}{\frac {d^{2}}{dx^{2}}}\langle x|\psi \rangle .}$ Putting all the pieces together, the Schrödinger equation reads

${\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}}{dx^{2}}}\langle x|\psi \rangle +{\frac {1}{2}}m\omega ^{2}x^{2}\langle x|\psi \rangle =E\langle x|\psi \rangle .}$

Since we are working in the position basis, we have ${\displaystyle \langle x|\psi \rangle =\psi (x)}$ so we finally get

${\displaystyle -{\frac {\hbar ^{2}}{2m}}{\frac {d^{2}}{dx^{2}}}\psi (x)+{\frac {1}{2}}m\omega ^{2}x^{2}\psi (x)=E\psi (x).}$

which is a differential equation which can be solved for ${\displaystyle \psi (x)}$. This ${\displaystyle \psi (x)}$ is of course, the wavefunction of the system in the position basis.

## Solutions to the quantum harmonic oscillator

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.

• ${\displaystyle \hbar }$ has units of ${\displaystyle {\textrm {J}}\cdot {\textrm {s}}={\frac {{\textrm {kg}}\cdot {\textrm {m}}^{2}}{\textrm {s}}}}$
• ${\displaystyle m}$ has units of ${\displaystyle {\textrm {kg}}}$
• ${\displaystyle \omega }$ has units of ${\displaystyle {\frac {1}{\textrm {s}}}}$

Hence, the quantity ${\displaystyle l={\sqrt {\frac {\hbar }{m\omega }}}}$ has the units of length. We will call this length the "characteristic length". If we substitute into the differential equation ${\displaystyle x\rightarrow {\frac {x}{l}}}$ we will get

${\displaystyle {\frac {\hbar \omega }{2}}\left(-{\frac {d^{2}}{dx^{2}}}+x^{2}\right)\psi (x)=E\psi (x)}$

Note also, that the units of ${\displaystyle \hbar \omega }$ are also energy units. We can define ${\displaystyle E_{0}=\hbar \omega /2}$ 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 ${\displaystyle E\rightarrow {\frac {E}{E_{0}}}}$ we get

${\displaystyle \left(-{\frac {d^{2}}{dx^{2}}}+x^{2}\right)\psi (x)=E\psi (x)}$

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, ${\displaystyle E\approx 0}$, the wavefunction should then satisfy the differential equation

${\displaystyle {\frac {d^{2}\psi (x)}{dx^{2}}}=x^{2}\psi (x)}$

The form of this equation suggests that ${\displaystyle \psi (x)=e^{-{\frac {x^{2}}{2}}}\phi (x)}$. Substituting

${\displaystyle -{\frac {d^{2}}{dx^{2}}}\left(e^{-{\frac {x^{2}}{2}}}\phi (x)\right)+x^{2}e^{-{\frac {x^{2}}{2}}}\phi (x)=Ee^{-{\frac {x^{2}}{2}}}\phi (x)}$
${\displaystyle -e^{-{\frac {x^{2}}{2}}}\phi ''(x)+2xe^{-{\frac {x^{2}}{2}}}\phi '(x)+e^{-{\frac {x^{2}}{2}}}\phi (x)-x^{2}e^{-{\frac {x^{2}}{2}}}\phi (x)+x^{2}e^{-{\frac {x^{2}}{2}}}\phi (x)=Ee^{-{\frac {x^{2}}{2}}}\phi (x)}$

or

${\displaystyle \phi ''(x)-2x\phi '(x)+(E-1)\phi (x)=0\,\;}$

This differential equation can be solved in many different ways. One approach is to take

${\displaystyle \phi (x)=\sum _{n=0}^{\infty }a_{n}x^{n}}$

The derivatives are (note carefully the summation limits)

${\displaystyle \phi '(x)=\sum _{n=1}^{\infty }na_{n}x^{n-1}\Rightarrow x\phi '(x)=\sum _{n=1}^{\infty }na_{n}x^{n}=\sum _{n=0}^{\infty }na_{n}x^{n}}$

and similarily

${\displaystyle \phi ''(x)=\sum _{n=2}^{\infty }n(n-1)a_{n}x^{n-2}=\sum _{n=0}^{\infty }(n+2)(n+1)a_{n+2}x^{n}}$

Substituting all of these terms into the above yields

${\displaystyle \sum _{n=0}^{\infty }x^{n}\left[(n+2)(n+1)a_{n+2}-2na_{n}+(E-1)a_{n}\right]=0}$

and vanishes term by term provided that

${\displaystyle a_{n+2}=a_{n}{\frac {2n+1-E}{(n+2)(n+1)}}}$

The most straightforward way to enforce these relationships is to set the numerator to zero. This leads to

${\displaystyle E=2n+1\Rightarrow E_{n}=\hbar \omega \left(n+{\frac {1}{2}}\right)}$

It's a very interesting result since the energy is now constrained to take on certain discrete values.

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