Scalar Hamiltonian and electromagnetic transitions

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Hello! This is probably a silly question (I am sure I am missing something basic), but I am not sure I understand how a Hamiltonian can be a scalar and allow transitions between states with different angular momentum at the same time. Electromagnetic induced transitions are usually represented as a perturbation to some free hamiltonian ##H'(t)##, which is a scalar. However when calculating the matrix element between 2 states ##<J_fM_f|H'|J_iM_i>## we can have non zero values even when ##J_f \neq J_i##. How is the fact that the hamiltonian is a scalar, consistent with the angular momentum conservation? Thank you!
 

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  • #2
vanhees71
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The only thing you need is that ##\hat{H}'## doesn't commute with ##\hat{\vec{J}}##, which indeed implies that it indeed should not be a scalar.

E.g., for atomic transitions you have in the electric-dipole approximation
$$\hat{H}'(t) \propto \vec{E}(t) \cdot \sum_k q_k \hat{\vec{x}}_k,$$
which is not a scalar when considering only the degrees of freedom of the atom. This is intuitively clear, because ##\vec{E}(t)## defines a preferred direction.
 
  • #3
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The only thing you need is that ##\hat{H}'## doesn't commute with ##\hat{\vec{J}}##, which indeed implies that it indeed should not be a scalar.

E.g., for atomic transitions you have in the electric-dipole approximation
$$\hat{H}'(t) \propto \vec{E}(t) \cdot \sum_k q_k \hat{\vec{x}}_k,$$
which is not a scalar when considering only the degrees of freedom of the atom. This is intuitively clear, because ##\vec{E}(t)## defines a preferred direction.
Thank you for this! I am still a bit confused. I thought that the Hamiltonian must always be a scalar (I've seen this statement in many QM readings). So it is not always a scalar?
 
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Thank you for this! I am still a bit confused. I thought that the Hamiltonian must always be a scalar (I've seen this statement in many QM readings). So it is not always a scalar?
Look here:
https://www.physicsforums.com/threads/lagrangian-vs-hamiltonian-in-qft.658142/

I believe you are confusing between the physicist's meaning of the word "scalar" and the mathematicians' use of this word.

For physicists a scalar is: "Formally, a scalar is unchanged by coordinate system transformations", where in maths a scalar is just an element in the field which the vector space is defined over.

In the example vanhees gave you, I assume the Hamiltonian is a pseudo-scalar, and in after a suitable change of coordinates we get a minus sign there.
 
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I believe you are confusing between the physicist's meaning of the word "scalar" and the mathematicians' use of this word.

Really? I've seen both physicists and mathematicians using word 'scalar' in both of this meanings. It just depends on the context.
 
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vanhees71
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Thank you for this! I am still a bit confused. I thought that the Hamiltonian must always be a scalar (I've seen this statement in many QM readings). So it is not always a scalar?
Of course, if you transform both ##\vec{E}## and ##\vec{x}## it's a scalar, but here you consider ##\vec{E}## as an external (classical) field, and thus you break full rotation symmetry to symmetry under rotations of ##\vec{x}## around ##\vec{E}##.

This is not different from classical mechanics!
 

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