Lecture 11 Feb 8

3D Harmonic Oscillator

The 3D harmonic oscillator refers to a particle moving in a potential \(V(r) = (1/2) m\omega^2 r^2\), i.e. it experiences a force always directed towards the origin and proportional to the distance to the origin.

Solution in Cartesian coordinates. The simplest way to solve this is to recognize that we can split the Hamiltonian into three separate 1D harmonic oscillators if we work in Cartesian coordinates:

\[\hat{H} = {\hat{p_x}^2+\hat{p_y}^2+\hat{p_z}^2\over 2m} + {1\over 2}m\omega^2 (\hat{x}^2+\hat{y}^2+\hat{z}^2)\]

\[= {\hat{p_x}^2\over 2m} + {1\over 2}m\omega^2\hat{x}^2 + {\hat{p_y}^2\over 2m} + {1\over 2}m\omega^2\hat{y}^2 + {\hat{p_z}^2\over 2m} + {1\over 2}m\omega^2\hat{z}^2.\]

We can write the stationary states as \(\ket{n_x, n_y, n_z}=\ket{n_x}\ket{n_y}\ket{n_z}\), i.e. a separable solution \(\psi(x,y,z)=X(x)Y(y)Z(z)\) which will have an energy

\[E = \hbar\omega\left(n_x + n_y + n_z+{3\over 2}\right) = \hbar\omega\left(n+{3\over 2}\right)\]

for \(n_x,n_y,n_z,n=0,1,2,\dots\). The ground state has energy \((3/2)\hbar\omega\) with \(n_x=n_y=n_z=0\). The first excited state involves increasing one of the \(n_i\)’s to 1, with three possible ways to do it. The second excited state has either two \(n_i\)’s equal to 1 or one of the \(n_i\)’s equal to 2, with 6 possible states, etc.

Spherical coordinates. Now consider solving this problem in spherical coordinates. From the rotational symmetry of the Hamiltonian, we can write the stationary states as \(\ket{E,\ell}\ket{\ell,m}\), ie. the angular part is given by the angular momentum eigenstates. The radial part then satisfies

\[\hat{H}_\ell \ket{E,\ell} = \left[{\hat{p}_r^2\over 2m} + {\ell(\ell+1)\hbar^2\over 2m \hat{r}^2} + {1\over 2}m\omega^2\hat{r}^2\right]\ket{E,\ell} = E\ket{E,\ell}\]

for energy \(E\). Here, we write the radial part of the kinetic energy with a radial momentum operator \(\hat{p}_r\). In position representation, this is

\[\hat{p}_r\rightarrow {\hbar\over i} \left({\partial\over \partial r}+{1\over r}\right)\]

which gives the correct form for the radial part of the Laplacian since

\[\hat{p}_r^2\ket{\Psi}\rightarrow -\hbar^2 \left({\partial\over \partial r}+{1\over r}\right)\left({\partial\over \partial r}+{1\over r}\right)\psi= -\hbar^2 \left({\partial^2\over\partial r^2} + {2\over r}{\partial\over\partial r}\right)\psi = -\hbar^2 \nabla_r^2\psi\]

where \(\nabla^2_r\) is the radial part of the Laplacian. This operator is introduced in Townsend section 9.6. Two useful commutation relations that you can prove are

\[[\hat{r},\hat{p}_r]=i\hbar;\hspace{1cm}\left[{1\over\hat{r}},\hat{p}_r\right]=-{i\hbar\over \hat{r}^2}.\]

We’ll use these later.

Solution to the radial equation. One approach is to solve the radial equation

\[\left[-{\hbar^2\over 2m}\nabla_r^2 + {\ell(\ell+1)\hbar^2\over 2m r^2} + {1\over 2}m\omega^2 r^2\right]R(r) = E R(r)\]

just as we did for the hydrogen atom. You will work through this in HW5. Just as with the hydrogen atom, there is a quantization condition that is needed to truncate the polynomial solution,

\[\boxed{n = 2n_r + \ell}\]

where \(n\) is the principal quantum number and \(n_r\) is the number of radial nodes. The energy is

\[\boxed{E = \hbar\omega \left(n+{3\over 2}\right) = \hbar\omega \left(2n_r +\ell +{3\over 2}\right)}\]

There’s an interesting difference from hydrogen, which is that the radial nodes contribute twice as much energy compared to increasing \(\ell\). So the ground state has \(n_r=0, \ell=0\), same as hydrogen, but there is only one option for the first excited state: \(n_r=0, \ell=1\). For the second excited state, there is \(n_r=1, \ell=0\) or \(n_r=0, \ell=2\). You can check that the number of states at each \(n\) matches what we found in Cartesian coordinates.

An operator approach to the 3D harmonic oscillator

Operator approach in 1D. First a reminder about what happens in 1D: we try to complete the square in the Hamiltonian

\[\hat{H} = {\hat{p}^2\over 2m} + {1\over 2}m\omega^2\hat{x}^2\]

by defining operators

\[\hat{a} = {1\over\sqrt{2m\hbar\omega}} \left(i\hat{p} + m\omega\hat{x}\right); \hspace{1cm} \hat{a}^\dagger = {1\over\sqrt{2m\hbar\omega}} \left(-i\hat{p} + m\omega\hat{x}\right).\]

It is straightforward to show that

\[\hat{H} = \hbar\omega\left(\hat{a}^\dagger\hat{a} + {1\over 2}\right),\]

and also that \([\hat{a}, \hat{a}^\dagger]=1\) (you will need to use \([\hat{x},\hat{p}]=i\hbar\)).

Then we can investigate the effect of \(\hat{a}^\dagger\) and \(\hat{a}\) on the eigenstates of \(\hat{H}\), or equivalently the number operator \(\hat{N}=\hat{a}^\dagger\hat{a}\). First, the commutator between \(\hat{a}\) and \(\hat{N}\) is

\[[\hat{a},\hat{N}] = \hat{a}\hat{a}^\dagger\hat{a} - \hat{a}^\dagger\hat{a}\hat{a} = [\hat{a},\hat{a}^\dagger]\hat{a} = \hat{a}.\]

Write the eigenstates of \(\hat{N}\) as \(\ket{n}\) with corresponding eigenvalues \(n\). Then

\[\hat{a}n\ket{n} = \hat{a}\hat{N}\ket{n} = \hat{N}\hat{a}\ket{n} + \hat{a}\ket{n}\Rightarrow \hat{N}\left(\hat{a}\ket{n}\right)=(n-1)\left(\hat{a}\ket{n}\right)\]

so we see that \(\hat{a}\ket{n}\) must be \(\propto \ket{n-1}\), i.e. \(\hat{a}\) is a lowering operator. A similar argument shows that \(\hat{a}^\dagger\) is a raising operator, since \([\hat{a}^\dagger,\hat{N}]=-\hat{a}^\dagger\). The lowest value of \(n\) is zero since the fact that the ground state is annihilated by the lower operator \(\hat{a}\ket{0}=0\) is then consistent with \(\hat{N}\ket{0} = 0\). We see then that the states have integer values of \(n=0,1,2,\dots\), generated by successively applying \(\hat{a}^\dagger\) to the ground state to generate the excited states. The normalizations of the raising and lowering operators are such that \(\hat{a}\ket{n}=\sqrt{n}\ket{n-1}\) and \(\hat{a}^\dagger\ket{n} = \sqrt{n+1}\ket{n+1}\).

Operator approach in 3D. Let’s see if we can use a similar approach in 3D. Let’s find operators that can complete the square in the radial Hamiltonian

\[\hat{H}_\ell = \left[{\hat{p}_r^2\over 2m} + {\ell(\ell+1)\hbar^2\over 2m \hat{r}^2} + {1\over 2}m\omega^2\hat{r}^2\right].\]

Let’s try

\[\hat{a}_\ell = {1\over \sqrt{2m\hbar\omega}} \left(i\hat{p_r} - {(\ell+1)\hbar\over \hat{r}} + m\omega\hat{r}\right).\]

Then

\[\hbar\omega \hat{a}_\ell^\dagger\hat{a}_\ell = \hat{H}_\ell + {i\omega\over 2}[\hat{r},\hat{p}_r] - i{(\ell+1)\hbar\over 2m}\left[{1\over \hat{r}}, \hat{p}_r\right] + {(\ell+1)\hbar^2\over 2mr^2}-(\ell+1)\hbar\omega\]

or

\[\hat{H}_\ell = \hbar\omega\left(\hat{a}^\dagger_\ell \hat{a}_\ell + \ell + {3\over 2} \right).\]

A natural next step is to interpret \(\hat{a}_\ell^\dagger\hat{a}_\ell=\hat{N}_\ell\) as a number operator that counts the radial excitations. If we label the eigenstates as \(\ket{k,\ell}\), where \(\hat{N}_\ell\ket{k,\ell} = k\ket{k,\ell}\), then we can investigate the effect of \(\hat{a}_\ell\) on \(\ket{k,\ell}\). Just as in 1D, we will need the commutator \([\hat{a}_\ell^\dagger, \hat{a}_\ell]\). Using the definition above gives

\[[\hat{a}_\ell, \hat{a}_\ell^\dagger] = 1 + {\hbar(\ell+1)\over m\omega r^2} = 1 + {\hat{H}_{\ell+1}-\hat{H}_\ell\over\hbar\omega} = 2 + \hat{N}_{\ell+1} - \hat{N}_\ell,\]

so that

\[[\hat{a}_\ell, \hat{N}_\ell] = \hat{a}_\ell\hat{a}_\ell^\dagger\hat{a}_\ell - \hat{a}_\ell^\dagger\hat{a}_\ell\hat{a}_\ell = [\hat{a}_\ell,\hat{a}_\ell^\dagger]\hat{a}_\ell = (2 + \hat{N}_{\ell+1} - \hat{N}_\ell)\hat{a}_\ell\]

\[\Rightarrow \hat{a}_\ell \hat{N}_\ell = 2\hat{a}_\ell + \hat{N}_{\ell+1}\hat{a}_\ell.\]

Now apply this to the state \(\ket{k, \ell}\):

\[\hat{a}_\ell \hat{N}_\ell\ket{k, \ell} = \hat{a}_\ell k\ket{k, \ell} = 2\hat{a}_\ell\ket{k, \ell} + \hat{N}_{\ell+1}\hat{a}_\ell\ket{k, \ell}\]

\[\Rightarrow \hat{N}_{\ell+1} \left(\hat{a}_\ell\ket{k, \ell}\right) = (k-2)\left(\hat{a}_\ell\ket{k, \ell}\right)\]

which tells us that

\[\hat{a}_\ell\ket{k, \ell} \propto \ket{k-2, \ell+1}\]

an eigenstate of \(\hat{N}_{\ell+1}\) with eigenvalue \(k-2\). So we see that \(\hat{a}_\ell\) lowers the radial quantum number by 2 while simultaneously increasing \(\ell\) by 1. You can show in a similar way that \(\hat{a}_\ell^\dagger\) increases the radial quantum number by 2 while decreasing \(\ell\) by 1. Starting at the ground state with \(k=0\) we can apply \(\hat{a}^\dagger\) to add a radial excitation (while also decreasing \(\ell\) by 1), which will increase \(k\) by 2. We can therefore see that the energies will be \(E = \hbar\omega \left(2n_r +\ell +{3/2}\right)\) with \(n_r=0,1,2,\dots\) and \(\ell=0,1,2,\dots\), in agreement with the solution above from the radial equation.

Further reading

• Townsend section 10.2.