The NMR Selection Rules: Understanding Allowed Spin State Transitions

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etotheipi
For simplicity I only take a system of two interacting spin-##1/2## nuclei. If the spins have quantum numbers ##m_1## and ##m_2## respectively when in a certain state, then the energy of that particular state is$$E_{m_1m_2} = m_1 v_{0,1} + m_2 v_{0,2} + m_1 m_2 J_{12}$$where ##v_{0,1}## and ##v_{0,2}## are the Larmor frequencies of the first and second spins respectively (and ##J_{12}## is the coupling constant). A quantum number for the system defined by ##M = m_1 + m_2## takes values

spin statesM
##\alpha \alpha##1
##\alpha \beta##0
##\beta \alpha##0
##\beta \beta##-1

The notes say that only transitions with ##\Delta M = \pm 1## are allowed. That is, the only allowed transitions here are ##\alpha \alpha \rightarrow \alpha \beta##, ##\alpha \alpha \rightarrow \beta \alpha##, ##\alpha \beta \rightarrow \beta \beta##, ##\beta \alpha \rightarrow \beta \beta##.

My question is, why aren't transitions where both individual spin states change allowed [i.e. ##\alpha \alpha \rightarrow \beta \beta## and ##\alpha \beta \rightarrow \beta \alpha##]?
 
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Conservation of angular momentum. The photon required for the transition has a spin of 1.
 
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TeethWhitener said:
Conservation of angular momentum. The photon required for the transition has a spin of 1.

Ah, of course! I hadn't considered that at all. Thanks!
 
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