Excited States of The Deuteron

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Discussion Overview

The discussion centers on the excited states of the deuteron, particularly whether it possesses any excited states given its classification as a weakly bound system. Participants explore the nature of the deuteron's ground state, its composition as a mixture of different angular momentum states, and the implications of quantum mechanical principles on these states.

Discussion Character

  • Debate/contested
  • Technical explanation
  • Conceptual clarification

Main Points Raised

  • Some participants assert that the deuteron has no excited states, as it is a weakly bound system, while others question the implications of this classification.
  • It is proposed that the deuteron exists as a quantum mechanical superposition of the ##^3 S_1## and ##^3 D_1## states, with neither being an energy eigenstate of the Hamiltonian.
  • Participants discuss the probabilities associated with the deuteron's states, with one claiming a 96% chance of being in the ##^3 S_1## state and a 4% chance in the ##^3 D_1## state, while others argue about the nature of superposition and mixed states.
  • One participant raises a question about the potential for states with different orbital angular momentum to mix, despite being unbound, and challenges the reasoning behind the absence of mixing with P-wave states.
  • Another participant suggests that the states are not eigenstates of the Hamiltonian and that the true eigenstates are linear combinations of the S and D states, with different energy characteristics.
  • It is noted that the deuteron has only one bound state, which is a point of contention in the discussion.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the deuteron's ground state and the existence of excited states. There is no consensus on whether the ground state is a mixed state or a superposition, and the discussion remains unresolved regarding the implications of quantum mechanics on the deuteron's state composition.

Contextual Notes

Participants highlight the complexity of the quantum states involved, including the dependence on definitions of mixed versus pure states and the implications of observable measurements on the state of the deuteron. The discussion also touches on the limitations of the Hamiltonian in describing the system's energy eigenstates.

devd
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All textbooks and material that I've read on the topic state that the deuteron being a weakly bound system, has no excited state. They also go on to state that the deuteron exists as a mixture of ##^3 S_1## and ##^3D_1## states.

So, are these states degenerate in energy? That is, are both of these states the deuteron ground state?
 
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The state is a quantum mechanical superposition of the two. Neither is an energy eigenstate of the Hamiltonian.
 
The ground state, is a mixed state, right? There is 96% chance for the deuteron to be in ##^3 S_1## state and 4% chance to be in ##^3 D_1## state. I don't get why you're saying that the ground state is a superposition! A superposition would be a pure state!
 
devd said:
There is 96% chance for the deuteron to be in ##^3 S_1## state and 4% chance to be in ##^3 D_1## state.
Only if you measure it in the right way. Before that it is in a superposition.
 
devd said:
The ground state, is a mixed state, right? There is 96% chance for the deuteron to be in ##^3 S_1## state and 4% chance to be in ##^3 D_1## state. I don't get why you're saying that the ground state is a superposition! A superposition would be a pure state!
There is absolutely no contradiction in the superposition being a pure state and there being certain probabilities of finding the state to be in the states contained in the superposition. It is just that your observable does not commute with the system Hamiltonian, i.e., the Hamiltonian has matrix elements mixing the two states. The ground state is an eigenstate if the Hamiltonian and therefore a linear combination of the two.
 
I understand why paradeuteron as a spin 0 state is not bound (Opposite spin nucleons experience slightly weaker strong force, barely not enough to bind a spin 0 deuteron). But what forbids a state with orbital angular momentum 1 and spin 0 from mixing with the prevalent state (orbital angular moment 0, spin 1) along with the other minor contributor (orbital angular moment 2, spin -1)? After all, just because a state on its own is unbound does not stop it from mixing - a state of deuteron with orbital angular momentum 2, spin 1, total angular momentum 3 is unbound due to centrifugal force.
 
I think people are overcomplicating things.

You have two pure states, an S-wave and a D-wave. These are not eigenstates of the Hamiltonian, i.e. they don't have a fixed energy. If you ask what states *are* eignestates of the Hamiltonian, one is mostly S with a sprinkling of D and the other is mostly D with a sprinkling of S. One of these has energy below separation threshold and the other is above. Could have had both above (like the di-neutron) or both below (which didn't happen).

The reason that no P-wave state mixes with the S and the D is that it has different quantum numbers.
 
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and the deuteron has only one bound state.
 

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