I showed that the ladder operators: $ \hat{\overrightarrow{a}}=(a_x, a_y , a_z)$ and $\hat{\overrightarrow{a}}^{\dagger} = (a_x^{\dagger}, a_y^{\dagger} , a_z^{\dagger})$ can form a vector operator by proving:

$$ [J_k, a_l] = i \hbar \varepsilon_{klm} a_m \hspace{1,5cm} [J_k, a_l^{\dagger}] = i \hbar \varepsilon_{klm} a_m^{\dagger}$$

I also know that to construct spherical components it should look something like this:

$$ V_1 = - \frac{1}{\sqrt{2}} (V_x + i V_y) \hspace{0,8cm}V_0 = V_z \hspace{0,8cm}V_{-1} = \frac{1}{\sqrt{2}} (V_x - i V_y) \hspace{0,8cm}(1) $$

I would simply plug them in the expressions $V_1$ , $V_0$ and $V_{-1}$ in order to get the spherical components for $\hat{\overrightarrow{a}}$ and $\hat{\overrightarrow{a}}^{\dagger}$ but now I'm wondering, how do I arrive at these general expressions for the spherical components of a vector operator $(1)$?


1 Answer 1


You would first find a linear combination of your operators so that $$ [\hat J_+,\hat{T}^\ell_m]=0\, , $$ and once you have that you can ladder down using $$ [\hat J_-,\hat T^\ell_m]=\sqrt{(\ell+m)(\ell-m+1)}\,\hat{T}^{\ell}_{m-1}\, . $$
This does not fix the “norm” of the operator, i.e. $A\hat T^{\ell}_m$ also has the right transformation properties for any constant $A$. $A$ can be considered a normalization factor.

In practice, one can often “guess” at the form of the operator $\hat T^{k}_k$ by comparing with the spherical harmonics in Cartesian coordinates: since $(x+iy)\sim \hat{T}^{1}_1\sim Y_1^1(\theta,\varphi)$, then $(x+iy)^k\sim \hat T^{k}_k$ and ladder down from there. Indeed if you compare your $V_k$ with the spherical harmonics $Y_{1}^m(\theta,\phi)$ in Cartesian form you can immediately see how the combination $V_x\pm i V_y$ occur.

  • $\begingroup$ But, what would I use in this case? By the definition: $\hat{T}_m^l = \sum_{q, q'} C_{q,q',Q}^{j_1,j_2,J} V_qU_{q'}$ should I use: $V_q = \overrightarrow{a}$ and $U_{q'}= \overrightarrow{a}^{\dagger}$ and then show that the commutator you have is zero? $\endgroup$ Commented Nov 28, 2021 at 19:11
  • $\begingroup$ for which case? $\endgroup$ Commented Nov 28, 2021 at 19:17
  • $\begingroup$ assuming that my components $a_1 , a_0$ and $a_{-1}$ have the form in $(1)$, should I have $V_q= U_{q'}= \overrightarrow{a}$, plug it in $\hat{T}_Q^J$ and finally verify that $[\hat{J}_+ , \hat{T}_Q^J] =0$ ? Im confused in what I should use as $V_q$ and $U_{q'}$ in order to prove your commutation relation $\endgroup$ Commented Nov 29, 2021 at 17:13
  • $\begingroup$ In my first comment the indicies $(l,m) $ should actually be $(J,Q)$ $\endgroup$ Commented Nov 29, 2021 at 17:14

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