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Suppose a two dimensional isotropic harmonic oscillator. We define the angular momentum operator as $L = XP_y - YP_x$, where $X,Y$ are the position operators and $P_x,P_y$ are the momentum operators.

We define two new operators $$ a_d = \frac{1}{\sqrt{2}}\left(a_x-ia_y\right) $$ $$ a_g = \frac{1}{\sqrt{2}}\left(a_x+ia_y\right) $$ where $a_i$ is the annihilation operator in the $i$-th direction.

I have first shown that $L$ and the Hamiltonian commute, and then showed that they can be written as $$ L = \hbar (a_d^{\dagger}a_d - a_g^{\dagger}a_g),\ H = \hbar\omega(a_d^{\dagger}a_d + a_g^{\dagger}a_g + 1) $$

How can I find the common eigenbasis of the two? I have seen solutions that tell us to to treat $a_d^{\dagger}a_d$ and $a_g^{\dagger}a_g$ as some kind of number operators, and then represent the product space as the product of the two spaces spanned by them, but I can't seem to understand why this product space is the same product space as $X\otimes Y$?

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The "cartesian" basis for the 2D Harmonic oscillator is given in terms of the $x$ and $y$ eigenstates of the corresponding number operators $\hat n_x$ and $\hat n_y$, written as $|n_x\rangle \otimes|n_y\rangle$. These number operators have to been understood as $\hat n_x \equiv \hat n_x \otimes 1_y$ and $\hat n_y \equiv 1_y\otimes \hat n_y $, with $1_x$ and $1_y$ the identity matrix in the $x$ and $y$ space.

One can also solve the same problem using another basis, spanned by the $g$ and $d$ operators $$\hat a_d=\frac{1}{\sqrt{2}}\left(\hat a_x \otimes 1_y - i1_x \otimes \hat a_y\right),\\ \hat a_g=\frac{1}{\sqrt{2}}\left(\hat a_x \otimes 1_y + i1_x \otimes \hat a_y\right),$$ which act on both $x$ and $y$ spaces at the same time. The effect of the $g$ operator is $$\hat a_g |n_x\rangle \otimes|n_y\rangle = \frac{1}{\sqrt{2}}\left(\sqrt{n_x}|n_x-1\rangle \otimes|n_y\rangle+\sqrt{n_y}|n_x\rangle \otimes|n_y-1\rangle\right).$$

One can then show that the Hamiltonian can be expressed in terms of the $g$ and $d$ number operators (in particular $\hat n_x\otimes 1_y+1_x \otimes\hat n_y= \hat n_d + \hat n_g$).

In principle, one can express that eigenstates of $\hat n_d$ and $\hat n_g$, $|n_g,n_d\rangle$ in the basis $|n_x\rangle \otimes|n_y\rangle$.

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  • $\begingroup$ How can I find the representation of the desired basis in the old one? That is, express $\left|n_g,n_d\right>$ with $\left|n_x\right>\otimes\left|n_y\right>$ $\endgroup$ – JonTrav1 Jun 12 '16 at 16:40
  • $\begingroup$ @JonTrav1: write down what is the action of $\hat n_{g,d}$ in the $n_x,n_x$ basis, and try to diagonalise these operators. $\endgroup$ – Adam Jun 12 '16 at 19:21

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