I have a question on my Quantum Mechanics homework where we consider protons and neutrons to be manifestations of the same particle -- a nucleon. We think of the proton as the "isospin up state", and the neutron as the "isospin down state" are asked to consider a two-particle state as a tensor product of orbital angular momentum $| l m_l \rangle$, spin angular momentum $|S m_S \rangle$, and isospin $|I m_I \rangle$ i.e.

$$|N N ; lm_l, Sm_S, Im_I \rangle = | l m_l \rangle \otimes | S m_S \rangle \otimes | I m_I \rangle $$

We are then asked two questions:

(1). What are the allowed isospin states for two nucleons in terms of the proton/neutron basis?

(2). At low energies, most two-nucleon observables are dominated by the $l = 0$ angular momentum. If we fix $l = 0$, what are the allowed spin and isospin two-nucleon states?

This may be a stupid question, but how does $| l m_l \rangle \otimes | S m_S \rangle \otimes | I m_I \rangle$ describe a two-particle space? Isn't this describing a single nucleon which is a proton if $m_I = \frac{1}{2}$ and a neutron if $m_I = -\frac{1}{2}$? Meaning a two-particle state would look something like

$$(| l m_l \rangle \otimes | S m_S \rangle \otimes | I m_I \rangle)_1 \otimes (| l' m_l' \rangle \otimes | S' m_S' \rangle \otimes | I' m_I' \rangle)_2 $$

I've included my currently line of attack for the question, but am unsure and would greatly appreciate any guidance.

My approach to (1):

The allowed isospin states have the basis $|p \rangle \otimes | p\rangle$, $|p \rangle \otimes |n \rangle$, $|n \rangle \otimes | p \rangle$, and $|n \rangle \otimes |n \rangle$. Since we are adding two spin-1/2 particles, I think we need to use the fact that their total angular momentum $(j)$ ranges between $s_1 + s_2 = 1$ and $|s_1 - s_2| = 0$ to form the various states with different combinations of $l$ and $s$ values.

My approach to (2):

Using the states found in (1), we just restrict to those with $l = 0$ and take tensor products between the


1 Answer 1


Suppose, $$ |N_1 \rangle = | l_1 m_{l_1} \rangle \otimes | S_1 m_{S_1} \rangle \otimes | I_1 m_{I_1} \rangle$$ describes state of first nucleon and $$ |N_2 \rangle = | l_2 m_{l_2} \rangle \otimes | S_2 m_{S_2} \rangle \otimes | I_2 m_{I_2} \rangle$$ describes state of second nucleon.

Then the two nucleon state can be written as $$ |N_1N_2 \rangle = | l m_{l} \rangle \otimes | S m_{S} \rangle \otimes | I m_{I} \rangle, $$where $\vec{l} = \vec{l}_1 + \vec{l}_2 $, $\vec{S} = \vec{S}_1 + \vec{S}_2 $ and $\vec{I} = \vec{I}_1 + \vec{I}_2$. Means it’s a coupled state. If you are not familiar with it, read how angular momentum coupling is done and about CG coefficients. For the 1st question $l$ is arbitrary, so don't worry about its coupling. Couple the spin and isospin states. Write the coupled states in terms of uncoupled states by calculating the CG coefficients.

Depending upon the values of $l$, $ S $ and $I$ you can now determine the symmetry of the states. And then due of Pauli exclusion principle, you have to identify the states which make the total two particle state antisymmetric when we exchange the particles.

For $ l = 0 $ state the space part is always is symmetric. So the product of spin and isospin part has to be antisymmetric.

This will be bit complicated for a beginner, better to follow a nuclear physics book where the deuteron problem (two nucleon problem) is mentioned in detail. Check how deuteron states are formed.

Book recommendation: Nuclear physics by Roy and Nigam

  • 1
    $\begingroup$ Thank you for the detailed answer! $\endgroup$ Commented Mar 13 at 19:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.