3.4 Solving the QHO with Ladder Operators

In this section we are going to introduce a technique to solve for the energy eigenvalues and eigenfunctions without solving the complicated differential equation 3.9. This method will be to factorise the Hamiltonian using two new operators, which we will call ladder operators for reasons that will become clear, and then see what their effect on an unknown eigenfunction is.

3.4.1 Factorising the Hamiltonian

Notice that the QHO Hamiltonian (equation 3.6) is a sum of squares of operators. We would like to write it in a more symmetric form and then factorise.

We can do this by removing the units. Define the dimensionless position operator by

\hat{X}=\frac{\hat{x}}{x_{0}},

(3.59)

where $x_{0}=\sqrt{\frac{\hbar}{m\omega}}$ is the natural length scale created using the constants in the Hamiltonian. We can define a dimensionless momentum operator by creating a nautral momentum scale, which is created by dividing $\hbar$ by $x_{0}$ :

p_{0}=\frac{\hbar}{x_{0}}=\sqrt{\hbar m\omega}.

(3.60)

Then we can define

\hat{P}=\frac{\hat{x}}{p_{0}},

(3.61)

as the dimensionless momentum operator.

Substituting these into equation 3.6, we get

$\displaystyle\hat{H}$	$\displaystyle=\frac{p_{0}^{2}\hat{P}^{2}}{2m}+\frac{1}{2}m\omega^{2}x_{0}^{2}% \hat{X}^{2}$	(3.62)
	$\displaystyle=\frac{\hbar m\omega}{2m}\hat{P}^{2}+\frac{m\omega^{2}\hbar}{m% \omega}\hat{X}^{2}$	(3.63)
	$\displaystyle=\frac{\hbar\omega}{2}(\hat{P}^{2}+\hat{X}^{2}).$	(3.64)

Recall that an expression $u^{2}+v^{2}$ can be factored using complex numbers as $(u+iv)(u-iv)$ . When using the same idea with operators, we need to keep in mind that two operators may not necessarily commute. The two parentheses expand to

	$\displaystyle(\hat{U}-i\hat{V})(\hat{U}+i\hat{V})$	$\displaystyle=\hat{U}^{2}+\hat{V}^{2}+i\hat{U}\hat{V}-i\hat{V}\hat{U}$		(3.65)
		$\displaystyle=\hat{U}^{2}+\hat{V}^{2}+i[\hat{U},\hat{V}],$		(3.66)

therefore for the QHO Hamiltonian, we get

\hat{H}=\frac{\hbar\omega}{2}((\hat{X}-i\hat{P})(\hat{X}+i\hat{P})-i[\hat{X},% \hat{P}]),

(3.67)

where we have to subtract of $i$ times the commutator of $\hat{X}$ and $\hat{P}$ to account for the cross terms. The commutator between dimensionless position and momentum is given by

$\displaystyle[\hat{X},\hat{P}]$	$\displaystyle=\left[\frac{\hat{x}}{x_{0}},\frac{\hat{p}}{p_{0}}\right]$	(3.68)
	$\displaystyle=\frac{1}{x_{0}p_{0}}[\hat{x},\hat{p}]$	(3.69)
	$\displaystyle=\sqrt{\frac{m\omega}{\hbar}}\frac{1}{\sqrt{\hbar m\omega}}i\hbar$	(3.70)
	$\displaystyle=i,$	(3.71)

So the Hamiltonian simplifies to

\hat{H}=\frac{\hbar\omega}{2}((\hat{X}-i\hat{P})(\hat{X}+i\hat{P})+1).

(3.72)

Now we will define new operators to be the combination in parentheses:

	$\displaystyle\hat{A}$	$\displaystyle=\frac{1}{\sqrt{2}}(\hat{X}+i\hat{P})$		(3.73)
	$\displaystyle\hat{A}^{\dagger}$	$\displaystyle=\frac{1}{\sqrt{2}}(\hat{X}-i\hat{P}).$		(3.74)

These operators are called the ladder operators for reasons that will become very clear in the next section. If we substitute these into the Hamiltonian, we get

\hat{H}=\hbar\omega\left(\hat{A}^{\dagger}\hat{A}+\frac{1}{2}\right).

(3.75)

This equation looks suspiciously like equation 3.31 for the energy eigenvalues but with operators! This implies that the product $\hat{A}^{\dagger}\hat{A}$ is a “number operator”, it has the energy eigenstates as its eigenfunctions with their corresponding quantum number $n$ as its eigenvalues.

But what do the operators $\hat{A}$ and $\hat{A}^{\dagger}$ actually do? Note that $\hat{A}^{\dagger}$ is the adjoint of $\hat{A}$ , as we can tell by the notation and because it is defined as the “complex conjugate” of $\hat{A}$ . Because $\hat{A}\neq\hat{A}^{\dagger}$ , neither of them are Hermitian and therefore they do not represent observables. We will look at the action of these operators on an energy eigenstate in the next section.