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 $0$?
 A: 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.
A: 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.
