quantum mechanics - 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 $0$?

Answer

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.