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So, how do we know $J_{+}|j,(m=j)\rangle =|0\rangle$?

I.e. that m is bounded by j.

We know that $J_{+}|j,(m=j)\rangle =C|j, j+1\rangle$, but how do I know that gives zero? Is it by looking at its norm-square?

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You can prove it like this. Apply $J_+$ n times to an eigenket of $\mathbf{J}^2$ and $J_z$. Thus, you get another eigenket of $\mathrm{J}^2$ and $J_z$ where the eigenvalue of $\mathbf{J}^2$ is unchanged and $J_z$ eigenvalue increases by $n\hbar$. As you will see, you can't repeat this operation indefinitely and there is an upper limit to $\beta$, $J_z$ eigenvalue, for a give eigenvalue $\alpha$ of $\mathbf{J}^2$. So this gives you $\alpha\geq\beta$. To see this, you do the following


From this $J_+J_{+}^{\dagger}$ and $J_{+}^{\dagger}J_+$ must have nonnegative expectation values. This leads us to


Thus, there must exist a $\beta_{max}$ s.t. $J_+|\alpha,\beta_{max}\rangle=0$. This implies that $J_-J_+|\alpha,\beta_{max}\rangle=0$. However, you can rewrite $J_-J_+=\mathbf{J}^2-J_{z}^2-\hbar J_z$. Applying this to $|\alpha,\beta_{max}\rangle$ you get the following relation for the eigenvalues $\alpha=\beta_{max}(\beta_{max}+\hbar)$. In a similar fashion you can prove that there must also exist a $b_{min}$ s.t $J_-=|\alpha,\beta_{min}\rangle=0$. By the same steps you find $\alpha=\beta_{min}(\beta_{min}-\hbar)$. Comparing the two equalities for the eigenvalues you find that $\beta_{max}=-\beta_{min}$. So, applying $J_+$ to $|\alpha,\beta_{min}\rangle$ a finite number of times we must find $|\alpha,\beta_{max}\rangle$. This leads you to $\beta_{max}=\beta_{min}+n\hbar=\frac{n\hbar}{2}$

Here we define $j$ as $\frac{\beta_{max}}{\hbar}$ s.t $j=\frac{n}{2}$ and define $m$ as $\beta=m\hbar$. From this you see the $m$ values for a given $j$; $m=-j,-j+1,\dots,j-1,j$ (a number of $2j+1$ states).

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Excellent, thanks. Can you explicitly show that my first equation is true, by calculation? – Alex Feb 28 '13 at 7:08
Yes. When you want to determine the matrix elements of the angular momentum operators, lets say in your case for $J_+$ you first take $\langle j,m|J_{+}^{\dagger}J_{+}|j,m\rangle=\hbar^{2}[j(j+1)-m^{2}-1]$. But $J_{+}|j,m\rangle$ must be equal to $|j,m+1\rangle$ up to a multiplicative constant. Hence $J_{+}|j,m\rangle=c_{jm}^{+}|j,m+1\rangle$. From this you find that $|c_{jm}^{+}|^{2}=\hbar^{2}(j-m)(j+m+1)$. Thus, $J_{+}|j,m\rangle=\sqrt{(j-m)(j+m+1)}\hbar|j,m+1\rangle$. Making j=m, you get zero. – nijankowski Feb 28 '13 at 7:21

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