# Wigner-Eckart projection theorem

I'm following the proof of Wigner-Eckart projection theorem which states that:

$$\langle \bf{A} \rangle ~=~ \frac{\langle \bf{A} \cdot \bf{J} \rangle}{\langle {\bf{J}}^2 \rangle} \langle \bf{J} \rangle$$

if ${\bf J}$ is conserved.

There is an equation:

$$\langle k'jm | {\bf J} \cdot {\bf A} | k j m\rangle = m \hbar \langle k'jm | A_0 | k j m\rangle + \frac{\hbar}{2} \sqrt{j(j+1) - m(m-1)} \langle k'j(m-1) | A_{-1} | k j m\rangle - \frac{\hbar}{2} \sqrt{j(j+1) - m(m+1)} \langle k'j(m+1) | A_{+1} | k j m\rangle ~\stackrel{(1)}{=}~ c_{jm} \langle k'j \| {\bf A} \| k j \rangle$$

Now, why on the left of (1) there are components of the vector while on the right it is vector itself? I suppose it may be due to:

$$\alpha \langle j \| A_0 \| j \rangle + \beta \langle j \| A_{-1} \| j \rangle + \gamma \langle j \| A_{+1} \| j \rangle ~\stackrel{(1)}{\equiv}~ c \langle j \| {\bf A} \| j \rangle$$ but what justifies this simplified notation?

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The right-hand side is a scalar as well: it is the reduced matrix element which you can see on the $||$ (double vertical lines) notation. This is not a coincidence and it is not a typo: the double lines can't be replaced by a single line. This matrix element with the double lines represents matrix elements in such a way that the object is independent of the $j_z=m$ components or, equivalently, the matrix element with a "vector" operator written in between actually boils down to a scalar; it only depends on the magnitudes of the angular momenta $j_{\rm something}$ and quantum numbers unrelated to the angular momentum $k$. The individual components may be reconstructed because the dependence on the quantum numbers $j_z=m$ is dictated by the rotational symmetry. That's really the point of such theorems.
Hi, thank you for your answer. I am familiar with the reduced matrix elements. What I don't quite understand is the way in which I should interpret reduced matrix element of whole vector, not one of its components like $A^1_m$. – qoqosz May 23 '12 at 22:03
$<jm| A_q^1 | jm > ~ < j ||A^1|| j>=<j||{\bf A}|| j>$ oh, that's how it works. – qoqosz May 23 '12 at 23:03