Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
up vote 1 down vote accepted

All these things have a good reason but I find the way you phrase your question perhaps intentionally confusing. There are different things on the left hand side of (1) and the right hand side of (1) because (1) is an equation and if there were the same things on both sides of (1), an equation, then (1) would be an uninteresting tautology.

However, none of the two sides of the equations (1) is a vector if this was your concern (a possible discrepancy). Both are scalars so they can be equal and they actually are equal.

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.

I think that you should learn what the double-line notation and the concept of "reduced matrix elements" means and then you should actually try to follow the proof by checking "why the next line follows from the previous ones". I am afraid that as you formulated it, your analysis of the proof hasn't reached this sensible attitude yet: it seems that you are asking why some equations have some properties that you may find counterintuitive because of your incomplete knowledge of the objects in the equations and/or for some heuristic, emotional reasons. That's not the way to check mathematical proofs.

share|cite|improve this answer
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
Right, it's usual to write a vector in the middle of the reduced matrix element but because it doesn't matter on the components, it's the same as if you write any component over there. – Luboš Motl May 24 '12 at 5:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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