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So I'm studying the shell model and I understand where the individual nucleon energy levels come from (Woods-Saxon plus spin-orbit interaction), but I'm stumped on how to find the ground state total quantum number $J$ for the entire nucleus. Take for example oxygen-16 (8 protons and 8 neutrons). Apparently for this case $J = 0$. I know that from filling up nucleon energy levels from the bottom, you'd get a total $M$J value of $0$ by summing over the $m_j$ of the individual nucleons: $$ M_{J,protons} = (-\frac{1}{2} + \frac{1}{2}) + (-\frac{3}{2} - \frac{1}{2} + \frac{1}{2} + \frac{3}{2}) + (-\frac{1}{2} + \frac{1}{2}) = 0$$ and similarly for neutrons. The parentheses separate the sums of $m_j$ over the $1s_{1/2}$, $1p_{3/2}$, and $1p_{1/2}$ energy levels.

How do you get from the above to the conclusion that $J=0$? Intuitively I'd say to find $J$ you'd have to use angular momentum addition theorem for the sum of the total angular momentum quantum numbers $j_i$ of each nucleon $i$, which would give you a range of possible $J$ values in integer steps. How can you just conclude $J=0$? My professor mentioned that due to the arbitrariness of the z-axis, we can take $M_J$ to be equal to $J$ but I have no idea where this is coming from or why it would make any sense.

Any help would be greatlty appreciated! Thanks in advance!

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I would say that for the nuclei with filled shells $J = 0$ follows from the Pauli principle. The question about filled nuclear shells is probably somewhat similar to questions about filled electron shells. At PSE, questions about electron shells were addressed. However, here is my attempt.

Let $|\Psi\rangle$ denote the state vector of the nucleus according to the shell model. To show that $J=0$, it is enough to prove that in the state $|\Psi\rangle$ all components of the total angular momentum are equal to zero: $$ \hat{J^z}|\Psi\rangle = 0,\quad \hat{J^x}|\Psi\rangle = 0,\quad \hat{J^y} |\Psi\rangle = 0. $$ The validity of the first of these equalities was proved by Samuele Fossati, the other two are equivalent to the following $$ \hat{J^+}|\Psi\rangle = 0,\quad \hat{J^-}|\Psi\rangle = 0.\tag{1} $$ Here $\hat{J^\pm} = \hat{J^x}\pm i\hat{J^y}$ are raising and lowering operators of the total angular momentum.

In the shell model of the nucleus, the interaction between nucleons is neglected. Therefore, the $|\Psi\rangle$ state is constructed from one-nucleon states. Nucleons are fermions, so two or more nucleons cannot be in the same one-nucleon state. We can say that there is a set of one-nucleon states, and the basis states in the state space of a system of many nucleons are determined by the sets of occupied one-nucleon states. The vectors $\hat{J^+}|\Psi\rangle$ and $\hat{J^-}|\Psi\rangle$ in the general case are linear combinations of the basis states $\sum_i|\Psi,i\rangle$, where each state $|\Psi,i\rangle$ differs from $\Psi\rangle$ by one occupied one-nucleon state. But in the case of filled shells, $|\Psi\rangle$ in some sense corresponds to the filling of all possible one-nucleon states. This means that there are no basis states $|\Psi,i\rangle$ for constructing non-zero states $\hat{J^+}|\Psi\rangle$ and $\hat{J^-}|\Psi\rangle$. Consequently, in the case of filled shells, equalities (1) are valid and the total angular momentum is equal to zero, $J = 0$.

I admit that the formulations in the last paragraph may not be entirely clear and understandable. I'm afraid that the only way for me to formulate the statement more precisely is to write a lot more formulas and use the method of secondary quantization.

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