2
$\begingroup$

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 $0$?

$\endgroup$
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 $\endgroup$ – 0x90 Mar 10 '17 at 18:57
  • 2
    $\begingroup$ 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. $\endgroup$ – Emilio Pisanty Mar 10 '17 at 19:06
  • 1
    $\begingroup$ 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). $\endgroup$ – ZeroTheHero Mar 10 '17 at 19:27
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.