1. Oct 11, 2007

### masudr

I thought that I had angular momentum very well understood, but something has been giving me problems recently.

It is often stated in textbooks and webpages alike, that the angular momentum ladder operators defined as

$$J_{\pm} \equiv J_x \pm i J_y$$

Then the texts often go on to say that these operators satisfy the following crucial commutation relation:

$$\left[ J_z , J_\pm ] = \pm J_\pm$$

The problem I have is that if the above commutation relation holds perfectly, then the Clebsch Gordan coefficients would never arise. Applying the $J_z$ operator to a raised/lowered eigenstate should perfectly give m-1 or m+1, assuming the above commutation relation to be correct. Instead, we are told that there is some factor that creeps in.

To be honest, there is a similar thing with SHO ladder operators, we normally get factors of the form $\sqrt{n}, \sqrt{n+1}$.

If there is something obvious I have missed, then can someone let me know. Also, if anyone knows the derivation of the above commutation rule, or a link to it, that'd be great. I tried to derive it, but had some trouble.

2. Oct 11, 2007

### dextercioby

Assume an invariant common dense everywhere domain (for the angular momentum algebra representation) on which the ang mom ops are essentially self adj. Then (missing the vector for simplicity and taking hbar=1)

$$\left[ J_{z},J_{\pm }\right] =\left[ J_{z},J_{x}\pm iJ_{y}\right] =iJ_{y}\pm J_{x}=\pm \left( J_{x}\pm iJ_{y}\right) =\pm J_{\pm }$$.

Can you justify that "The problem I have is that if the above commutation relation holds perfectly, then the Clebsch Gordan coefficients would never arise. Applying the LaTeX graphic is being generated. Reload this page in a moment. operator to a raised/lowered eigenstate should perfectly give m-1 or m+1, assuming the above commutation relation to be correct. Instead, we are told that there is some factor that creeps in." ?

3. Oct 11, 2007

### Jim Kata

What you are missing is that while $${\mathbf{J}}_ \pm \left| {j,m} \right\rangle$$ is an eigenvector of $${\mathbf{J}}_z$$, it is not a normalized eigenvector, and this is where this extra scale factor comes from. For example, say you have $$\left| {j,m} \right\rangle$$ normalized so that $$\left\langle {{j,m}} \mathrel{\left | {\vphantom {{j,m} {j,m}}} \right. \kern-\nulldelimiterspace} {{j,m}} \right\rangle = 1$$, now $${\mathbf{J}}_ \pm \left| {j,m} \right\rangle = N_{j,m \pm 1} \left| {j,m \pm 1} \right\rangle$$. So, $$\left| {N_{j,m \pm 1} } \right|^2 \left\langle {{j,m \pm 1}} \mathrel{\left | {\vphantom {{j,m \pm 1} {j,m \pm 1}}} \right. \kern-\nulldelimiterspace} {{j,m \pm 1}} \right\rangle = \left\langle {j,m} \right|{\mathbf{J}}_ \mp {\mathbf{J}}_ \pm \left| {j,m} \right\rangle = \left\langle {j,m} \right|{\mathbf{J}}^2 - {\mathbf{J}}_z^2 \mp {\mathbf{J}}_z \left| {j,m} \right\rangle = j(j + 1) - m(m \pm 1)$$. Therefore, $$N_{j,m \pm 1} = \sqrt {(j + {1 \mathord{\left/ {\vphantom {1 {2)^2 - (m \pm 1/2)^2 }}} \right. \kern-\nulldelimiterspace} {2)^2 - (m \pm 1/2)^2 }}} = \sqrt {(j \mp m)(j \pm m + 1)}$$. Finally you get $${\mathbf{J}}_ \pm \left| {j,m} \right\rangle = \sqrt {(j \mp m)(j \pm m + 1)} \left| {j,m \pm 1} \right\rangle$$. The same logic works for angular momentum $${\mathbf{a}}^\dag \left| n \right\rangle = N_{n + 1} \left| n \right\rangle$$ needs to be normalized just like the angular momentum did.