2D

From TheTangentBundle

The Harmonic oscillator in 2 dimensions arises from a potential of the form $V(\vec{r}) = \frac{1}{2}k_x x^2 + \frac{1}{2}k_y y^2\,$ , and the Hamiltonian is

$H = \frac{p_x^2}{2m} + \frac{p_y^2}{2m} + \frac{1}{2}k_x x^2 + \frac{1}{2}k_y y^2$ .

If $k_x\,$ and $k_y\,$ are equal, then the oscillator is isotropic, and both dimensions involve a classical frequency of $\omega = \sqrt{\frac{k}{m}}$ . If they are not, then the oscillator is anisotropic and involves two uncoupled modes $\omega_x = \sqrt{\frac{k_x}{m}}$ and $\omega_y = \sqrt{\frac{k_y}{m}}$ . In the general case, we can introduce the following operators in analogy with the 1D case:

$a_x\,$	$=\frac{1}{\sqrt{2\hbar m\omega_x}}\left( m\omega_x x + i p_x \right)$ ,
$a_y\,$	$=\frac{1}{\sqrt{2\hbar m\omega_y}}\left( m\omega_y y + i p_y \right)$ .

The Hamiltonian is then

$H = \hbar \omega_x \left(a_x^\dagger a_x + \frac{1}{2}\right) + \hbar \omega_y \left(a_y^\dagger a_y + \frac{1}{2}\right)$ .

Similarly, we can label states as

$\left|n_x n_y \right\rangle = \frac{{a_x^\dagger}^{n_x} {a_y^\dagger}^{n_y}}{\sqrt{n_x! n_y!}}\left|0, 0\right\rangle$ .

[edit] Degeneracy

In the case of the anisotropic harmonic oscillator, none of the states are degenerate, except for possible accidental degeneracies. However, if the oscillator is isotropic then only the ground state is non-degenerate. For any eigenstate $\left|n_x n_y \right\rangle$ , the energy is $\hbar \omega (n_x + n_y + 1)$ . For a particular value of $n = n_x + n_y\,$ , $n_x\,$ can take on values between $0\,$ and $n\,$ , while $n_y = n - n_x\,$ . Therefore the degeneracy is $n+1\,$ .

[edit] Alternative Bases

Any operator that depends on $x\,$ , $p_x\,$ , $y\,$ , and $p_y\,$ can be represented in terms of $a_x\,$ , $a_x^\dagger\,$ , $a_y\,$ and $a_y^\dagger$ . For simplicity, let us restrict to the isotropic case. Introduce the vector $\mathbf{a}^\dagger = [a_x^\dagger, a_x, a_y^\dagger, a_y]$ . We can write the Hamiltonian in matrix notation as

$H = \mathbf{a}^\dagger \mathbf{H} \mathbf{a}$ ,

where

$\mathbf{H} = \hbar \omega \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ \end{bmatrix}$ .

In the isotropic case, symmetry about the $z\,$ axis means that the $z\,$ component of angular momentum is conserved. We may be interested in the operator $L_z = x p_y - y p_x = -i\hbar(a_x^\dagger a_y - a_y^\dagger a_x)\,$ . We can write $L_z\,$ as

$L_z = \mathbf{a}^\dagger \mathbf{L_z} \mathbf{a}$ ,

where

$\mathbf{L_z} = \frac{-i \hbar}{2} \begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & -1 \\ -1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ \end{bmatrix}$ .

Consider a perturbation on the Hamiltonian that involves $L_z\,$ , e.g. a Zeeman term of the form $H^\prime = B L_z\,$ . Because $\mathbf{H}$ is just a multiple of the identity matrix, it is possible to diagonalize $\mathbf{L_z}$ via a unitary transformation without affecting the form of $\mathbf{H}$ . This is because a unitary matrix $\mathbf{U}$ that diagonalizes $\mathbf{L_z}$ , i.e., $\mathbf{U}^{\dagger}\mathbf{L_z}\mathbf{U}$ is diagonal, has the property that

$\mathbf{U}^{\dagger}\mathbf{H}\mathbf{U} = \hbar \omega \mathbf{U}^{\dagger}\mathbf{I}\mathbf{U} = \hbar \omega \mathbf{U}^{\dagger}\mathbf{U} = \hbar \omega \mathbf{U}^{-1}\mathbf{U} = \hbar \omega \mathbf{I} = \mathbf{H}$ .

A particular diagonalizing matrix is

$\mathbf{U} = \frac{1}{\sqrt{2}} \begin{bmatrix} 0 & 1 & 0 & 1 \\ 1 & 0 & 1 & 0 \\ 0 & i & 0 & -i\\ -i & 0 & i & 0 \\ \end{bmatrix}$ .

In the new basis, the components of $\mathbf{a}^\prime = \mathbf{U}^\dagger \mathbf{a}$ are:

$a_x^\prime$	$= \frac{1}{\sqrt{2}}\left(a_x^\dagger + i a_y^\dagger\right)$ ,
$a_x^{\prime\dagger}$	$= \frac{1}{\sqrt{2}}\left(a_x - i a_y\right)$ ,
$a_y^\prime$	$= \frac{1}{\sqrt{2}}\left(a_x^\dagger - i a_y^\dagger\right)$ ,
$a_y^{\prime\dagger}$	$= \frac{1}{\sqrt{2}}\left(a_x + i a_y\right)$ .

The matrix $\mathbf{L_z}$ is now diagonal:

$\mathbf{L_z} = \frac{\hbar}{2}\begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 0 & -1 \\ \end{bmatrix}$ .

It is customary to use the following labeling for the new basis:

$a_{\pm}^\dagger = \frac{1}{\sqrt{2}}\left(a_x^\dagger \pm i a_y^\dagger\right)$ .

It is important to check the commutation relations for this new basis:

$\left[a_{\pm}, a_{\pm}^\dagger\right]$	$= \frac{1}{2}\left[a_x \mp i a_y, a_x^\dagger \pm i a_y^\dagger\right]$ ,
	$= \frac{1}{2}\left( \left[a_x , a_x^\dagger \right] + \left[a_y, a_y^\dagger\right]\right)$ ,
	$= 1\,$ .