1
$\begingroup$

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 $(1)$?

$\endgroup$

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$ Nov 28, 2021 at 19:11
  • $\begingroup$ for which case? $\endgroup$ 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 $\endgroup$ Nov 29, 2021 at 17:13
  • $\begingroup$ In my first comment the indicies $(l,m) $ should actually be $(J,Q)$ $\endgroup$ Nov 29, 2021 at 17:14

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.