Angular momentum operator identity J²= J-J+ + J_3 + h*J_3 intermediate step

[xyver]

Homework Statement

I do not understand equal signs 2 and 3 the following Angular momentum operator identity:

Homework Equations

$$\hat{J}^2 = \hat{J}_1^2+\hat{J}_2^2 +\hat{J}_3^2$$

$$<br /> = \left(\hat{J}_1 +i\hat{J}_2 \right)\left(\hat{J}_1 -i\hat{J}_2 \right) +\hat{J}_3^2 + i \left[ \hat{J}_1, \hat{J}_2 \right]$$

$$= \hat{J}_+\hat{J}_- + \hat{J}_3^2 - \hbar \cdot \hat{J}_3$$

$$<br /> <br /> = \hat{J}_-\hat{J}_++ \hat{J}_3^2 + \hbar \cdot \hat{J}_3$$


$$\hat{J}_+ = \hat{J}_1 + i\hat{J}_2$$

$$\hat{J}_- = \hat{J}_1 - i\hat{J}_2$$


$$[\hat{J}_i,\hat{J}_j] = i\hbar\epsilon_{ijk}\hat{J}_k$$

The Attempt at a Solution

$$\hat{J}^2= \hat{J}_1^2+\hat{J}_2^2 +\hat{J}_3^2$$


$$= \left(\hat{J}_+ -i \hat{J}_2 \right)^2 \left( \frac{\hat{J}_+ -i \hat{J}_1 }{i}\right)^2 +\hat{J}_3^2$$

$$= \hat{J}_+^2 -i\hat{J}_+ \hat{J}_2 -i\hat{J}_2 \hat{J}_+ +i^2 \hat{J}_2^2 +\hat{J}_3^2 + \frac{\hat{J}_+^2 -i\hat{J}_+ \hat{J}_1 -i\hat{J}_1 \hat{J}_+ + \hat{J}_1^2 } {i^2} +\hat{J}_3^2$$




$$= \hat{J}_+^2 -i\hat{J}_+ \hat{J}_2 -i\hat{J}_2 \hat{J}_+ - \hat{J}_2^2 - \hat{J}_+^2 +\hat{J}_+ \hat{J}_1 +\hat{J}_1 \hat{J}_+ - \hat{J}_1^2 +\hat{J}_3^2$$

$$= -i\hat{J}_+ \hat{J}_2 -i\hat{J}_2 \hat{J}_+ - \hat{J}_2^2 +\hat{J}_+ \hat{J}_1 +\hat{J}_1 \hat{J}_+ - \hat{J}_1^2 +\hat{J}_3^2$$

Unfortunately, this does not lead to the right way. Who can help?

 

[bigplanet401]

Try working backwards. The 1- and 2-components of J are being factored into ladder operators and a commutator is being added on to cancel the extra terms:

$$\begin{align*}<br /> (J_1 + iJ_2)(J_1 - iJ_2) &= J^2_1 + J^2_2 + iJ_2 J_1 - iJ_1 J_2\\<br /> &= J^2_1 + J^2_2 -i [J_1, J_2]<br /> \end{align*}$$

To get to the third expression from the second, replace the ladder operators with their synonyms (J +/-) and use a commutator identity.