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

  • Thread starter xyver
  • Start date
  • #1
6
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Homework Statement


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



Homework Equations


[tex]\hat{J}^2 = \hat{J}_1^2+\hat{J}_2^2 +\hat{J}_3^2[/tex]

[tex]

= \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] [/tex]

[tex]
= \hat{J}_+\hat{J}_- + \hat{J}_3^2 - \hbar \cdot \hat{J}_3 [/tex]

[tex]


= \hat{J}_-\hat{J}_++ \hat{J}_3^2 + \hbar \cdot \hat{J}_3 [/tex]


[tex] \hat{J}_+ = \hat{J}_1 + i\hat{J}_2 [/tex]

[tex]\hat{J}_- = \hat{J}_1 - i\hat{J}_2 [/tex]


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


The Attempt at a Solution


[tex]\hat{J}^2= \hat{J}_1^2+\hat{J}_2^2 +\hat{J}_3^2 [/tex]


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

[tex] = \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 [/tex]




[tex] = \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 [/tex]

[tex] = -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 [/tex]

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

Answers and Replies

  • #2
104
0
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:

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

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

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