Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question

1 Answer 1

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

$\mathbf{J}^2$-$J_{z}^2$=$\frac{1}{2}(J_+J_-+J_-J_+)=\frac{1}{2}(J_+J_{+}^{\dagger}+J_{+}^{\dagger}J_+)$

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

$\langle\alpha,\beta|(\mathbf{J}^2$-$J_{z}^2)|\alpha,\beta\rangle\geq0$

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).

share|improve this answer
    
Excellent, thanks. Can you explicitly show that my first equation is true, by calculation? –  Alex Feb 28 '13 at 7:08
1  
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. –  Leonida Feb 28 '13 at 7:21

Your Answer

 
discard

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.