The NMR Selection Rules: Understanding Allowed Spin State Transitions

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The discussion focuses on a system of two interacting spin-1/2 nuclei and the energy states associated with their quantum numbers. The energy of a specific state is determined by the spins' quantum numbers, Larmor frequencies, and a coupling constant. The total quantum number M, which is the sum of the individual spins, can take values based on the combinations of the spin states. Only transitions with a change in M of ±1 are permitted, which includes specific transitions between the states. The key point of inquiry is why transitions where both spins change, such as from αα to ββ, are not allowed. The explanation centers on the conservation of angular momentum, as the photon involved in such transitions has a spin of 1, making these transitions forbidden. The clarification on angular momentum conservation resolves the initial confusion regarding the allowed transitions.
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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I came.across a headline and read some of the article, so I was curious. Scientists discover that gold is a 'reactive metal' by accidentally creating a new material in the lab https://www.earth.com/news/discovery-that-gold-is-reactive-metal-by-creating-gold-hydride-in-lab-experiment/ From SLAC - A SLAC team unexpectedly formed gold hydride in an experiment that could pave the way for studying materials under extreme conditions like those found inside certain planets and stars undergoing...

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