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In Sakurai's and Napolitano's book "Modern quantum mechanics" there's a nice proof of the theorem. This can be found also almost identical on Wikipedia's Wigner–Eckart theorem - Proof.

The thing that is bugging me about this proof is this part

We therefore have two sets of linear homogeneous equations \begin{align}\sum_c a_{bc} x_c &= 0 & \sum_c a_{bc} y_c &= 0\end{align} one for the Clebsch–Gordan coefficients $x_c$ and one for the matrix elements $y_c$. It is not possible to exactly solve for $x_c$. We can only say that the ratios are equal, that is

$$\frac{x_c}{x_d} = \frac{y_c}{y_d} \quad\text{or}\quad x_c=ky_c.$$

While the rest is clear, I don't get how we conclude the proportionality. In general, this is clearly not true. This is true only if the kernel of the matrix $a_{bc}$ is one-dimensional. So there must be some other property of the coefficients $a_{bc}$ that allows us to conclude that the kernel is one-dimensional, and thus the proportionality of different $x_c$ satisfying the equation.

Qmechanic
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pp.ch.te
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2 Answers2

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You know that the recurrence relation for the Clebsch's determines them up to normalization, so you know that the kernel of the system of linear equations is indeed one-dimensional.

mike stone
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I got the same problem these days and I tried to investigate to get a right explanation. It is not a phase problem. You have to focus on the set of linear equations given by the recurrence relations. For fixed j, j' and j" (sorry for bad notation, but I hope you can get it!) you can "easily" see that you always have N Clebsch-Gordan and N-1 equations that can link those coefficients. Thus you can show that the set of linear equations always gives a one dimensional solution... if and only if the rank of the associated matrix is N-1. You can show that when you realize that the associated matrix of the system is in row echelon form.

LDG
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