# Excited States of The Deuteron

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1. Sep 4, 2015

### devd

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?

2. Sep 4, 2015

### Orodruin

Staff Emeritus
The state is a quantum mechanical superposition of the two. Neither is an energy eigenstate of the Hamiltonian.

3. Sep 4, 2015

### devd

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!

4. Sep 4, 2015

### Staff: Mentor

Only if you measure it in the right way. Before that it is in a superposition.

5. Sep 4, 2015

### Orodruin

Staff Emeritus
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.

6. Sep 4, 2015

### snorkack

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.

7. Sep 4, 2015

Staff Emeritus
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.

8. Sep 6, 2015

### vanhees71

and the deuteron has only one bound state.