3.5 Ladder Operators

To work out the action of the ladder operators on an energy eigenstate $u_{n}$ , we will first calculate some commutators involving them and then do some algebra to find the result.

3.5.1 Properties of Ladder Operators

We have already seen that since $\hat{A}$ and $\hat{A}^{\dagger}$ are different operators, they cannot be Hermitian.

The first commutator we will need is the commutator of $\hat{A}$ with its adjoint $\hat{A}^{\dagger}$ , which is

$\displaystyle[\hat{A},\hat{A}^{\dagger}]$	$\displaystyle=\frac{1}{2}[\hat{X}+i\hat{P},\hat{X}-i\hat{P}]$	(3.76)
	$\displaystyle=\frac{1}{2}([\hat{X},\hat{X}]+i[\hat{P},\hat{X}]-i[\hat{X},\hat{% P}]-i^{2}[\hat{P},\hat{P}])$	(3.77)
	$\displaystyle=\frac{1}{2}(0-i^{2}-i^{2}+0)$	(3.78)
	$\displaystyle=1,$	(3.79)

So $\hat{A}$ and $\hat{A}^{\dagger}$ do not commute.

The next commutator to calculate is that of the ladder operators with the Hamiltonian. Looking at $\hat{A}$ first, this is

$\displaystyle[\hat{H},\hat{A}]$	$\displaystyle=\hbar\omega\left[\left(\hat{A}^{\dagger}\hat{A}+\frac{1}{2}% \right),\hat{A}\right]$	(3.80)
	$\displaystyle=\hbar\omega\left([\hat{A}^{\dagger}\hat{A},\hat{A}]+\left[\frac{% 1}{2},\hat{A}\right]\right)$	(3.81)
	$\displaystyle=\hbar\omega(\hat{A}^{\dagger}[\hat{A},\hat{A}]+[\hat{A}^{\dagger% },\hat{A}]\hat{A})$	(3.82)
	$\displaystyle=-\hbar\omega\hat{A}.$	(3.83)

By a similar calculation, it can be shown that $[\hat{H},\hat{A}^{\dagger}]=+\hbar\omega\hat{A}^{\dagger}$ .

3.5.2 The Lowering Operator

Suppose we have the an energy eigenstate $u_{i}$ , so $\hat{H}u_{i}=E_{i}u_{i}$ . We will now calculate what $\hat{A}$ when it acts on $u_{i}$ by acting the Hamiltonian on $\hat{A}u_{i}$ . When doing this calculation, we will deliberately “add zero” (add and subtract the same term) in order to introduce the commutators we have calculated above.

$\displaystyle\hat{H}(\hat{A}u_{i})$	$\displaystyle=(\hat{H}\hat{A}-\hat{A}\hat{H}+\hat{A}\hat{H})u_{i}$	(3.84)
	$\displaystyle=[\hat{H},\hat{A}]u_{i}+\hat{A}(\hat{H}u_{i})$	(3.85)
	$\displaystyle=-\hbar\omega\hat{A}u_{i}+E_{i}\hat{A}u_{i}$	(3.86)
	$\displaystyle=(E_{i}-\hbar\omega)\hat{A}u_{i}.$	(3.87)

This does not really look like we have made much progress, but notice that the last line is an eigenvalue equation, which implies that $\hat{A}u_{i}$ is an energy eigenfunction with energy $E_{i}-\hbar\omega$ . Of course, since we know what the energy eigenvalues are, we can see that $\hat{A}u_{i}$ must be proportional to $u_{i-1}$ because $E_{i}-\hbar\omega=E_{i-1}$ !

Let us apply $\hat{A}$ twice to $u_{i}$ and see what happens. Using the same trick, we get

$\displaystyle\hat{H}(\hat{A}^{2}u_{i})$	$\displaystyle=\hat{H}\hat{A}(\hat{A}u_{i})$	(3.88)
	$\displaystyle=(\hat{H}\hat{A}-\hat{A}\hat{H}+\hat{A}\hat{H})(\hat{A}u_{i})$	(3.89)
	$\displaystyle=[\hat{H},\hat{A}](\hat{A}u_{i})+\hat{A}\hat{H}(\hat{A}u_{i})$	(3.90)
	$\displaystyle=-\hbar\omega\hat{A}^{2}u_{i}+(E_{i}-\hbar\omega)\hat{A}^{2}u_{i}$	(3.91)
	$\displaystyle=(E_{i}-2\hbar\omega)\hat{A}^{2}u_{i},$	(3.92)

so we see, in the same way, that $\hat{A}^{2}u_{i}$ is proportional to $u_{i-2}$ since $E_{i}-2\hbar\omega=E_{i-2}$ .

One can show that in general,

\hat{H}(\hat{A}^{n}u_{i})=(E_{i}-n\hbar\omega)\hat{A}^{n}u_{i}.

(3.93)

So the action of $\hat{A}$ is to lower an eigenstate $u_{i}$ by one energy unit of $\hbar\omega$ . For this reason, $\hat{A}$ is called the lowering operator.

3.5.3 The Raising Operator

If $\hat{A}$ lowers an energy eigenstate by one unit of $\hbar\omega$ , one would expect that its adjoint $\hat{A}^{\dagger}$ raises it up by one. Indeed, we find that this is the case, so $\hat{A}^{\dagger}$ is called the raising operator.

We can show that this is the case by doing a similar calculation to above.

$\displaystyle\hat{H}(\hat{A}^{\dagger}u_{i})$	$\displaystyle=(\hat{H}\hat{A}^{\dagger}-\hat{A}^{\dagger}\hat{H}+\hat{A}^{% \dagger}\hat{H})u_{i}$	(3.94)
	$\displaystyle=[\hat{H},\hat{A}^{\dagger}]u_{i}+\hat{A}^{\dagger}(\hat{H}u_{i})$	(3.95)
	$\displaystyle=\hbar\omega\hat{A}^{\dagger}u_{i}+E_{i}\hat{A}^{\dagger}u_{i}$	(3.96)
	$\displaystyle=(E_{i}+\hbar\omega)\hat{A}^{\dagger}u_{i},$	(3.97)

as predicted.

Again, one can show in general that

\hat{H}(\hat{A}^{\dagger\,n}u_{i})=(E_{i}+n\hbar\omega)\hat{A}^{\dagger\,n}u_{% i}.

(3.98)

One easy way to remember which operator raises and which one lowers is by imagining the Hermitian dagger $\dagger$ as a little “+” symbol, which denotes raising.

3.5.4 The Ladder of Energy Levels

The raising and lowering operators imply that we have a “ladder” of energy levels, equally spaced apart by $\hbar\omega$ . However, there is one problem. It appears that there is nothing stopping us from applying the lowering operator $\hat{A}$ indefinitely, implying that there are infinitely many rungs in both directions! We know that this is impossible, since the potential energy $V(x)$ has a minimum value of $V=0$ , which means there are no bound states with $E_{i}<0$ .

We can also show directly that the energy must be greater than or equal to zero. Recall that the expectation value of the Hamiltonian for an energy eigenstate $u_{n}$ is $E_{n}$ :

\langle u_{n}\mathclose{}|\mathopen{}\hat{H}\mathclose{}|\mathopen{}u_{n}% \rangle=\int_{-\infty}^{\infty}u_{n}^{\ast}\hat{H}u_{n}\operatorname{\mathrm{d% }}\!x=E_{n}\int_{-\infty}^{\infty}u_{n}^{\ast}u_{n}\operatorname{\mathrm{d}}\!% x=E_{n}.

(3.99)

Then if we insert equation 3.75, we get

$\displaystyle E_{n}$	$\displaystyle=\hbar\omega\int_{-\infty}^{\infty}u_{n}^{\ast}\left(\hat{A}^{% \dagger}\hat{A}+\frac{1}{2}\right)u_{n}\operatorname{\mathrm{d}}\!x$	(3.100)
	$\displaystyle=\hbar\omega\left[\int_{-\infty}^{\infty}u_{n}^{\ast}\hat{A}^{% \dagger}\hat{A}u_{n}\operatorname{\mathrm{d}}\!x+\frac{1}{2}\int_{-\infty}^{% \infty}u_{n}^{\ast}u_{n}\operatorname{\mathrm{d}}\!x\right]$	(3.101)
	$\displaystyle=\hbar\omega\left[\int_{-\infty}^{\infty}(\hat{A}u_{n})^{\ast}% \hat{A}u_{n}\operatorname{\mathrm{d}}\!x+\frac{1}{2}\right]$	(3.102)
	$\displaystyle=\hbar\omega\left[\int_{-\infty}^{\infty}\lvert\hat{A}u_{n}\rvert% ^{2}\operatorname{\mathrm{d}}\!x+\frac{1}{2}\right]$	(3.103)
	$\displaystyle\geq 0.$	(3.104)

Note that in the third-last line, we used the fact that $u_{n}^{\ast}\hat{A}^{\dagger}$ is the complex conjugate of $\hat{A}u_{n}$ , and to go from the second-last line to the last we recall that probability density must be non-negative everywhere.

All this is to say that we need to impose that there exists a ground state with a minimum energy. To do this, we define the lowering operator to completely annihilate the ground state wavefunction

\hat{A}u_{0}=0.

(3.105)

This is known as the ladder termination condition. It implies that any repeated application of the lowering operator will eventually end up returning zero (since $\hat{A}^{n}u_{0}=0$ for any $n$ ). This actually gives us the value of the ground state energy. If we apply $\hat{H}$ to $u_{0}$ , we get

$\displaystyle\hat{H}u_{0}$	$\displaystyle=\hbar\omega\left(\hat{A}^{\dagger}\hat{A}+\frac{1}{2}\right)u_{0}$	(3.106)
	$\displaystyle=\hbar\omega\hat{A}^{\dagger}(\hat{A}u_{0})+\frac{1}{2}\hbar% \omega u_{0}$	(3.107)
	$\displaystyle=\frac{1}{2}\hbar\omega u_{0},$	(3.108)

since the first term becomes zero. So the ground state energy is $\frac{1}{2}\hbar\omega$ , which is what we found before by solving the Schrodinger equation.

This, along with equation 3.97, gives us the full spectrum of energy levels

E_{n}=\hbar\omega\left(n+\frac{1}{2}\right),

(3.109)

as found previously.

3.5.5 Finding the Eigenfunctions Using Ladder Operators

The only thing we haven’t recovered yet from our knowledge of the QHO using this method is the form of the energy eigenstates in the position basis. According to equation 3.97, $\hat{A}^{\dagger}u_{n}$ is proportional to $u_{n+1}$ . We can calculate the constant of proportionality by setting $\hat{A}^{\dagger}u_{n}=Cu_{n+1}$ and using the fact that the energy eigenstates are normalised.

$\displaystyle 1$	$\displaystyle=\int_{-\infty}^{\infty}u_{n+1}^{\ast}u_{n+1}\operatorname{% \mathrm{d}}\!x$	(3.110)
	$\displaystyle=\frac{1}{C^{2}}\int_{-\infty}^{\infty}(\hat{A}^{\dagger}u_{n})^{% \ast}\hat{A}^{\dagger}u_{n}\operatorname{\mathrm{d}}\!x$	(3.111)
	$\displaystyle=\frac{1}{C^{2}}\int_{-\infty}^{\infty}u_{n}^{\ast}\hat{A}\hat{A}% ^{\dagger}u_{n}\operatorname{\mathrm{d}}\!x.$	(3.112)

We will now use a nifty trick that comes in handy when manipulating expressions with ladder operators. With the commutator $[\hat{A},\hat{A}^{\dagger}]=1$ , we can rewrite

\hat{A}\hat{A}^{\dagger}=\hat{A}\hat{A}^{\dagger}-\hat{A}^{\dagger}\hat{A}+% \hat{A}^{\dagger}\hat{A}=[\hat{A},\hat{A}^{\dagger}]+\hat{A}^{\dagger}\hat{A}=% 1+\hat{A}^{\dagger}\hat{A}.

(3.113)

Using this and substituting the Hamiltonian, we get

$\displaystyle 1$	$\displaystyle=\frac{1}{C^{2}}\int_{-\infty}^{\infty}u_{n}^{\ast}(1+\hat{A}^{% \dagger}\hat{A})u_{n}\operatorname{\mathrm{d}}\!x$	(3.114)
	$\displaystyle=\frac{1}{C^{2}}\int_{-\infty}^{\infty}u_{n}^{\ast}\left(1+\frac{% \hat{H}}{\hbar\omega}-\frac{1}{2}\right)u_{n}\operatorname{\mathrm{d}}\!x$	(3.115)
	$\displaystyle=\frac{1}{C^{2}}\int_{-\infty}^{\infty}u_{n}^{\ast}\left(\frac{% \hat{H}}{\hbar\omega}u_{n}+\frac{1}{2}u_{n}\right)\operatorname{\mathrm{d}}\!x$	(3.116)
	$\displaystyle=\frac{1}{C^{2}}\int_{-\infty}^{\infty}u_{n}^{\ast}\left(\left(n+% \frac{1}{2}\right)u_{n}+\frac{1}{2}u_{n}\right)\operatorname{\mathrm{d}}\!x$	(3.117)
	$\displaystyle=\frac{n+1}{C^{2}}\int_{-\infty}^{\infty}u_{n}^{\ast}u_{n}% \operatorname{\mathrm{d}}\!x$	(3.118)
	$\displaystyle=\frac{n+1}{C^{2}},$	(3.119)

which implies $C=\sqrt{n+1}$ . Thus, we have

\hat{A}^{\dagger}u_{n}=\sqrt{n+1}u_{n+1}.

(3.120)

By a similar calculation, one can show that

\hat{A}u_{n}=\sqrt{n}u_{n-1}.

(3.121)

Using the ladder termination condition and expanding $\hat{A}$ in the position basis, we can find the form of the ground state wavefunction in the position basis.

	$\displaystyle\hat{A}u_{0}=0$		(3.122)
	$\displaystyle\frac{1}{\sqrt{2}}(\hat{X}+i\hat{P})u_{0}=0$		(3.123)
	$\displaystyle\left(\frac{\hat{x}}{x_{0}}+i\frac{\hat{p}}{p_{0}}\right)u_{0}=0$		(3.124)
	$\displaystyle\sqrt{\frac{m\omega}{\hbar}}u_{0}+\frac{i}{\sqrt{\hbar m\omega}}% \left(-i\hbar\frac{\operatorname{\mathrm{d}}\!u_{0}}{\operatorname{\mathrm{d}}% \!x}\right)=0$		(3.125)

This gives us a first-order differential equation which we can solve by integrating:

$\displaystyle\frac{\operatorname{\mathrm{d}}\!u_{0}}{\operatorname{\mathrm{d}}% \!x}$	$\displaystyle=-\frac{m\omega}{\hbar}xu_{0}$	(3.126)
$\displaystyle\int\frac{\operatorname{\mathrm{d}}\!u_{0}}{u_{0}}$	$\displaystyle=-\frac{m\omega}{\hbar}\int x\operatorname{\mathrm{d}}\!x$	(3.127)
$\displaystyle\ln u_{0}$	$\displaystyle=-\frac{m\omega}{2\hbar}x^{2}+c$	(3.128)
$\displaystyle u_{0}$	$\displaystyle=N_{0}e^{-\frac{m\omega}{2\hbar}x^{2}},$	(3.129)

which is exactly the wavefunction we found by solving the Schrodinger equation. Note that if we were solving this problem from scratch using ladder operators, this would be the only differential equation we have to solve, and also the only time that we have to refer back to the definition of the ladder operators in terms of $\hat{X}$ and $\hat{P}$ !

The rest of the eigenstates can be found using equation 3.120 and by expanding the definition of $\hat{A}^{\dagger}$ . For example, for the first excited state we have $\hat{A}^{\dagger}u_{0}=\sqrt{0+1}u_{1}=u_{1}$ , so we get

$\displaystyle u_{1}$	$\displaystyle=\frac{1}{\sqrt{2}}(\hat{X}-i\hat{P})u_{0}$	(3.130)
	$\displaystyle=\frac{1}{\sqrt{2}}\left(\sqrt{\frac{m\omega}{\hbar}}x-\sqrt{% \frac{\hbar}{m\omega}}\frac{\operatorname{\mathrm{d}}\!}{\operatorname{\mathrm% {d}}\!x}\right)N_{0}e^{-\frac{m\omega}{2\hbar}x^{2}}$	(3.131)
	$\displaystyle=N_{0}\sqrt{\frac{2m\omega}{\hbar}}xe^{-\frac{m\omega}{2\hbar}x^{% 2}},$	(3.132)

which is the same as equation 3.46. Note that when using this method for generating the eigenfunctions, they are already normalised!