Phy5646/AddAngularMomentumProb: Difference between revisions

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<math>\ S^2|m_1m_2\rangle = (S_1^2 + S_2^2 + 2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+})|m_1m_2\rangle </math>
<math>\ S^2|m_1m_2\rangle = (S_1^2 + S_2^2 + 2S_{1z}S_{2z}+S_{1+}S_{2-}+S_{1-}S_{2+})|m_1m_2\rangle </math>
<math>\ = S_1^2|m_1m_2\rangle + S_2^2|m_1m_2\rangle + 2S_{1z}S_{2z}|m_1m_2\rangle+S_{1+}S_{2-}|m_1m_2\rangle+S_{1-}S_{2+}|m_1m_2\rangle </math>
<math>\ = S_1^2|m_1m_2\rangle + S_2^2|m_1m_2\rangle + 2S_{1z}S_{2z}|m_1m_2\rangle+S_{1+}S_{2-}|m_1m_2\rangle+S_{1-}S_{2+}|m_1m_2\rangle </math>
<math>\ = \hbar^2 \frac{1}{2}\left(\frac{1}{2}+1 \right)|m_1m_2\rangle + \hbar^2\frac{1}{2}\left(\frac{1}{2}+1\right)|m_1m_2\rangle
 
<math>\ = \hbar^2 \frac{1}{2}\left(\frac{1}{2}+1 \right)|m_1m_2\rangle + \hbar^2\frac{1}{2}\left(\frac{1}{2}+1\right)|m_1m_2\rangle + \hbar^2\sqrt{\frac{1}{2} \left(\frac{1}{2}+1 \right)-m_1 \left(m_1+1 \right)}\sqrt{\frac{1}{2} \left(\frac{1}{2}+1 \right)-m_2 \left(m_1-1 \right)}|m_1+1;m_2-1\rangle + \hbar^2\sqrt{\frac{1}{2} \left(\frac{1}{2}+1 \right)-m_1 \left(m_1-1 \right)}\sqrt{\frac{1}{2} \left(\frac{1}{2}+1 \right)-m_2 \left(m_1+1 \right)}|m_1-1;m_2+1\rangle





Revision as of 19:44, 25 April 2010

Based on exercise 15.1.1. from Principles of Quantum Mechanics, 2nd ed. by Shankar:

Express as a matrix for two spin-1/2 particles in the direct product basis.

1.) First express in terms of , , , , and :

2.) Then act with this on direct product state :