How can I show that $(1+2\cos 2\theta)^3=7+2(6\cos 2\theta+3\cos4\theta+\cos6\theta)$ using the tensor product and the Clebsch Gordan Theorem?

Question

How can I show that $$(1+2\cos2\theta)^3=7+2(6\cos2\theta+3\cos4\theta+\cos6\theta)$$ using the tensor product and the Clebsch Gordan Theorem?

Answer 1

We have $$(1+2\cos(2\theta))^3=1+3\cdot 2\cos(2\theta)+3(2\cos(2\theta))^2+(2\cos(2\theta))^3$$ and this is (simplified) $$-1+12\, \left( \cos \left( \theta \right) \right) ^{2}-48\, \left( \cos \left( \theta \right) \right) ^{4}+64\, \left( \cos \left( \theta \right) \right) ^{6} $$ expanding the right-Hand side we get $$-1+12\, \left( \cos \left( \theta \right) \right) ^{2}-48\, \left( \cos \left( \theta \right) \right) ^{4}+64\, \left( \cos \left( \theta \right) \right) ^{6} $$ this is the same! $$\cos(2x)=2\cos(x)^2-1$$ $$\cos(4x)=8\cos(x)^4-8\cos(x)^2+1$$ $$\cos(6x)=32\cos(x)^6-48\cos(x)^4+18\cos(x)^2-1$$

Answer 2

The Clebsch Gordan formula for characters of $SU(2)$ is the equality $$\chi_m \cdot \chi_n = \chi_{m+n} + \chi_{m+n-2} + \cdots + \chi_{|m-n|}$$ where $$\chi_m(\theta)= e^{im\theta} + e^{i(m-2)\theta} + \cdots e^{-i m \theta}$$

So you want to express $\chi_2^3$ as a combination of $\chi_n$'s. We have $$\chi_2^2 = \chi_4 +\chi_2+ \chi_0$$ so $$\chi_2^3 = \chi_2(\chi_4 +\chi_2+ \chi_0)= (\chi_6 + \chi_4 + \chi_2)+ (\chi_4 + \chi_2 + \chi_0)+\chi_2 = \chi_6 + 2 \chi_4+ 3\chi_2+ \chi_0$$

Now we plug in the expressions for $\chi_0$, $\chi_2$, $\chi_4$, $\chi_6$ and get the result.