I have troubles to understand an equation, which was stated in a lecture. Consider the spin-j representation $V_{j}$ of $\mathrm{SU}(2)$ with its standard basis

$$\{\vert j,m\rangle\}_{-j\leq m\leq j}$$

Now, consider the fundamental representation $V_{\frac{1}{2}}$.It was then stated that

$$\vert\uparrow\rangle^{\otimes 2j}=\vert j,j\rangle$$ where $\vert\uparrow\rangle=\vert 1/2,1/2\rangle$ is one of the two basis vectors of $V_{1/2}$. I cannot see why this is the case. I know that $V_{1/2}^{\otimes 2j}$ is a submodule of $V_{j}$, as a consequence of the Clebsch-Gordan decomposition, but I do not see how to obtain the equality from above. I guess, the result has something to do with the Clebsch-Gordan coefficients.


1 Answer 1


You have it backwards: $V_j$ (dimension $2j+1$) is a submodule of $V^{\otimes 2j}_{1/2}$ (dimension $2^{2j}$).

The state $|\uparrow, \ldots, \uparrow\rangle$ is the highest weight state of both the $V_j$ module and the $V^{\otimes 2j}_{1/2}$. In other words it is the unique state with the highest $m$ spin value $m=j$ in both the larger and the smaller space. Applying step-down ladder opertors keeps the resulting states in $V_j$ so $|\uparrow, \ldots, \uparrow\rangle$ must coincide with $|j,j\rangle$.


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