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This question is about the components of a vector operator in the spherical basis. In 3D real Euclidean space, a vector $\mathbf{v}$ can be expanded in the standard Cartesian components as $$ \mathbf{v} = v^x \,\mathbf{e}_x + v^y \,\mathbf{e}_y + v^z \,\mathbf{e}_z = v^i \,\mathbf{e}_i $$ where the summation convention is assumed. If one now instead considers a complex space, one may introduce a different basis (spherical basis), which I always see defined as $$ \mathbf{e}_{\pm 1} = \mp \frac{1}{\sqrt{2}} \,(\mathbf{e}_x \pm i \mathbf{e}_y) \\ \mathbf{e}_0 = \mathbf{e}_z \tag{1} \label{1} $$ The vector $\mathbf{v}$ can now be expanded in this spherical basis as $\mathbf{v} = v^q \,\mathbf{e}_q$. By substituting the relations between spherical and Cartesian basis vectors, one finds that these components are given by $$ v^{\pm 1} = \frac{1}{\sqrt{2}} \,(\mp v^x + i v^y) \\ v^0 = v^z \tag{2} \label{2} $$ In quantum mechanics, a vector operator $\mathbf{V}$ is defined as a set of operators that satisfy the commutation relation $[J_i, V_j] = i\hbar \epsilon_{ijk} V_k$ with the angular momentum operator $\mathbf{J}$. It would therefore seem reasonable to define the spherical components of such a vector operator analogously to those of a 3D vector via \eqref{2}. However, with this definition one would find that these components, call them $V^q$, would not form a spherical tensor operator $V_q^1$, defined by the relations $$ [J_z, V_q^k] = \hbar \,q \,V_q^k \\ [J_{\pm}, V_q^k] = \hbar \sqrt{k(k+1) - q(q\pm 1)} \,V_{q\pm 1}^k $$ Instead, the spherical components of a vector operator are usually defined as (see Biedenharn and Louck "Angular momentum in quantum physics" or Sobel'man "Introduction to the theory of atomic spectra", for example) $$ V_{\pm 1}^1 = \mp \frac{1}{\sqrt{2}} \,(V_x \pm i V_y) \\ V_0^1 = V_z \tag{3} \label{3} $$ and they form a spherical tensor operator with rank k=1.

So why is the spherical basis chosen as in \eqref{1} if the components resulting from it in \eqref{2} are not taken as "the spherical components of a vector operator"? Rather, the components defined via \eqref{3} are used.

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It allows you to write the commutation relations using Clebsch-Gordan technology $$ [\hat J_\mu,{\cal M}_{JM}]= \sqrt{J(J+1)} C^{J,M+\mu}_{JM 1\mu} {\cal M}_{JM+\mu}\, , $$ construct composite tensors using Clebsch-Gordan technology, and factor matrix elements into a reduced part and a Clebsch-Gordan coefficient. The additional signs can be deduced by looking at the different spherical harmonics expressed in terms of Cartesian coordinates, v.g. \begin{align} Y^{-1}_1(\theta,\varphi)&=+\frac{1}{2}\sqrt{\frac{3}{2\pi}} \frac{x-iy}{r}\, ,\\ Y^{1}_1(\theta,\varphi)&=-\frac{1}{2}\sqrt{\frac{3}{2\pi}} \frac{x+iy}{r}\, . \end{align}

(Indeed the normalizationfactor $\sqrt{J(J+1)}$ is just the reduced matrix element of the tensor operator.)

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  • $\begingroup$ How are your $J_{\mu}$ defined in terms of Cartesian components? $\endgroup$
    – Quantum
    Mar 28, 2020 at 16:11
  • $\begingroup$ What does the "1" in the Clebsch-Gordan coefficient refer to? $\endgroup$
    – Quantum
    Mar 28, 2020 at 16:20
  • $\begingroup$ The angular momentum operators are $J=1$ tensors. The $\mu$ refers to spherical components. Check all this in a multitude of books on angular momentum. I recommend Varshalovich. $\endgroup$ Mar 28, 2020 at 16:54
  • $\begingroup$ I have read a few books on angular momentum, that's why this question arose. I have edited to clarify, but am happy to change it further. I only asked you to explicitly define the $J_{\mu}$ in terms of Cartesian components to make sure we are starting from the same definition. I assume you take the definition that seems to be standard, which is (3) (applied to angular momentum operator) in my question above. But that is exactly what my question is about. Given the definition (1) of a spherical basis, I would expect the spherical components to be defined via (2), rather than (3). $\endgroup$
    – Quantum
    Mar 29, 2020 at 10:52
  • $\begingroup$ Also, the question about the "1" in the Clebsch-Gordan coefficient was because I had not seen that notation before, only $C_{m_1 m_2 m_3}^{j_1 j_2 j_3}$, but I have now searched for it and there seems to be a multitude of different notations for CG coefficients. $\endgroup$
    – Quantum
    Mar 29, 2020 at 10:56

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