Scalar Hamiltonian and electromagnetic transitions

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Discussion Overview

The discussion revolves around the nature of the Hamiltonian in quantum mechanics, particularly in the context of electromagnetic transitions and angular momentum conservation. Participants explore whether a Hamiltonian can be considered a scalar while allowing transitions between states with different angular momentum, and the implications of this for quantum mechanics.

Discussion Character

  • Technical explanation
  • Debate/contested

Main Points Raised

  • One participant questions how a scalar Hamiltonian can permit transitions between states with differing angular momentum, referencing the matrix element calculation.
  • Another participant suggests that the Hamiltonian does not commute with angular momentum operators, implying it should not be a scalar, and provides an example from atomic transitions.
  • A participant expresses confusion about the scalar nature of the Hamiltonian, noting that many quantum mechanics readings state it must always be a scalar.
  • Further clarification is offered regarding the distinction between the physicist's and mathematician's definitions of "scalar," suggesting that the Hamiltonian in certain contexts may be a pseudo-scalar.
  • Another participant argues that the term "scalar" can have different meanings depending on the context, indicating a potential misunderstanding among participants.
  • One participant explains that while the Hamiltonian can be treated as a scalar under certain transformations, the presence of an external field breaks full rotational symmetry, affecting the classification.

Areas of Agreement / Disagreement

Participants express differing views on the scalar nature of the Hamiltonian and its implications for angular momentum conservation. There is no consensus on whether the Hamiltonian can always be considered a scalar, and the discussion remains unresolved.

Contextual Notes

Participants highlight the importance of context in defining the term "scalar" and the implications of external fields on symmetry considerations. There are unresolved distinctions regarding the definitions and properties of the Hamiltonian in quantum mechanics.

kelly0303
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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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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.
 
vanhees71 said:
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?
 
kelly0303 said:
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.
 
MathematicalPhysicist said:
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.
 
kelly0303 said:
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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