# How to derive the angular momentum operator for 3D harmonic oscillator?

In Angular momentum for 3D harmonic oscillator in two different bases Robin Ekman comes with the expression to $L_i$. I can't see how $$\epsilon_{ijk}(a_j^\dagger a_k^\dagger - a_j a_k) = 0$$ when developing the $L_i$ for isotropic 3D harmonic oscillator $$L_i = i \frac{\hbar}{2} \epsilon_{ijk}(a_j + a_j^\dagger) (a_k^\dagger - a_k) = i \frac{\hbar}{2} \epsilon_{ijk}( a_j a_k^\dagger - a_j a_k + a_j^\dagger a_k^\dagger - a_j^\dagger a_k) = -i\hbar\epsilon_{ijk} a^\dagger_j a_k.$$

I see that $$\left(a_j^\dagger a_k^\dagger\right)^\dagger = a_k a_j = a_j a_k,$$ which means $a_ja_k$ isn't hermitian for $i \ne j$ how does $$\epsilon_{ijk}(a_j^\dagger a_k^\dagger - a_j a_k)$$ go to $0$?

## 2 Answers

It's not so much that $\epsilon_{ijk}(a_j^\dagger a_k^\dagger - a_j a_k)$ is zero, but that each of those terms vanishes on its own: $$\text{both}\quad \epsilon_{ijk}a_j^\dagger a_k^\dagger = 0 \quad\text{and}\quad \epsilon_{ijk}a_j a_k = 0,$$ because those operators commute, so $a_ja_k=a_ka_j$ and ditto for the daggers, and for each pair with $j\neq k$ you have a corresponding term with the opposite sign in the Levi-Civita tensor. Thus, say, for $i=1$, the double-annihilation term reads $$\epsilon_{1jk}a_j a_k = a_2a_3-a_3a_2 = 0,$$ and similarly for the other components and the double-creation term.

The same thing happens with the other two terms, because you're so far left with $$L_i=\frac{\hbar}{2i}\varepsilon_{ijk}(a_j^\dagger a_k^\phantom{\dagger}-a_j^\phantom{\dagger}a_k^\dagger),$$ but the two terms are essentially identical. To see this, take the second term and flip the $j$ and $k$ labels: \begin{align} \frac{\hbar}{2i}\varepsilon_{ijk}a_j^\phantom{\dagger}a_k^\dagger & = \frac{\hbar}{2i}\varepsilon_{ikj}a_k^\phantom{\dagger}a_j^\dagger \\& = \frac{\hbar}{2i}\varepsilon_{ikj}(a_j^\dagger a_k^\phantom{\dagger}+i\delta_{jk}) \\& = -\frac{\hbar}{2i}\varepsilon_{ijk}a_j^\dagger a_k^\phantom{\dagger} + \frac{\hbar}{2}\varepsilon_{ikj}\delta_{jk} \\& = -\frac{\hbar}{2i}\varepsilon_{ijk}a_j^\dagger a_k^\phantom{\dagger} , \end{align} where $\varepsilon_{ikj}\delta_{jk}=1-1=0$ vanishes. Putting this back into the full expression, you get $$L_i=\frac{\hbar}{2i}\varepsilon_{ijk}(a_j^\dagger a_k^\phantom{\dagger}+a_j^\dagger a_k^\phantom{\dagger}) =-i\hbar\varepsilon_{ijk}a_j^\dagger a_k^\phantom{\dagger}$$ as claimed earlier.

• So for isotopic harmonic oscillator in N there won't be necessarily a generalized angular momentum invariant such as $x \wedge p$ it's only true for N=3
– 0x90
Commented Mar 10, 2017 at 18:57
• For an isotropic harmonic oscillator in $N$ dimensions, $H=\frac12\sum_{j=1}^N(p_j^2+x_j^2)$ commutes with all rotations, so it will commute with the generalized angular momentum of that space just as much as any $H= \frac12 \mathbf p^2 +V(r)$ hamiltonian will, though in $N>3$ the form of the angular momentum will obviously change; for more on that, see this question. Commented Mar 10, 2017 at 19:06
• To complement @EmilioPisanty 's answer: in general a basis for the traceless hermitian matrices can be obtained from the symmetric and antisymmetric real matrices. The antisymmetric matrices close on $so(N)$ while the symmetric matrices define (generalized) quadrupole moments but do not close under commutation (they actually commute to an generalized angular momentum). Commented Mar 10, 2017 at 19:27

First, you need to understand $a_i$ is a bosonic operator satisfying the bosonic commutation relation $[a_i,a_j]=0$, meaning that $a_ia_j=a_ja_i$. Now we show that $\epsilon_{ijk}a_ja_k=0$. Because if you fix $i=1$, then $$\epsilon_{1jk}a_ja_k=\epsilon_{123}a_2a_3+\epsilon_{132}a_3a_2=a_2a_3-a_3a_2=0.$$ For other choice of $i$, the proof is similar. Then $\epsilon_{ijk}a_ja_k=0$ implies $\epsilon_{ijk}a_j^\dagger a_k^\dagger=0$ by Hermitian conjugating both sides of the equation. Because both $\epsilon_{ijk}a_ja_k$ and $\epsilon_{ijk}a_j^\dagger a_k^\dagger$ are zero, so their difference is also zero.