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Suppose we have a system of two identical 0-spin bosons, with a certain symmetric, time independent hamiltonian, which involves some general angular momentum for each particle (I am not currently interested in describing a certain effect or anything). Let that hamiltonian be something like $H_0=J_{(1)}^2+J_{(2)}^2$, spin operators not explicitly involved.

Since the states have to be symmetric and all the particles will have the same spin state, the energy levels for that hamiltonian will be ruled by its quantum numbers $j_1$ and $j_2$, something like $E_{j_1j_2}=\hbar^2[j_1(j_1+1)+j_2(j_2+1)]$. Plus, those eigenvalues will be degenerated, in terms of the possible values of $m_1$ and $m_2$.

Now, my question is the following. If we add a term of the form $\overrightarrow{J_{(1)}}\cdot \overrightarrow{J_{(2)}}$, which could be expressed in terms of $J_{(1)}^2$, $J_{(2)}^2$ and $\mathbb{J}^2$, would that eliminate the degeneracies of the spectrum? I've learnt that non-symmetric hamiltonians tend to have less degeneracies, but I am not quite sure whether total angular momentum would be considered as such in a system of two identical particles. For the sake of simplicity, I am considering that the extra term is multiplied by some small constant, such that $H_0+\alpha\overrightarrow{J_{(1)}}\cdot \overrightarrow{J_{(2)}}\approx H_0+\alpha \mathbb{J}^2$.

Thank you in advance.

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Distinguishable bosons

The short answer is: yes, adding a $\vec{J}_1.\vec{J}_2$ term will lift some of the degeneracies from the spectrum. To see this mathematically, let us first agree to drop the parentheses on the $\vec{J}_{(i)}$, and represent them as $\vec{J}_i$ instead. With $$ H_0 = \vec{J}_1^2 + \vec{J}_2^2 $$ $H_0$ is diagonal in $|j_1,m_1\rangle \otimes |j_2,m_2\rangle$. From this we immediately write the eigenvalues as $\hbar^2[j_1(j+1)+j_2(j_2+1)]$. That's a total of $(2j_1+1)(2j_2+1)$ degenerate states for a given pair of $j_1$ and $j_2$. With $\vec{J}_1.\vec{J}_2$ added, the situation get a little better -- the Hamiltonian takes the form $$ H = a \vec{J}^2 + b (\vec{J}_1^2 + \vec{J}_2^2) $$ This time, the Hamiltonian is diagonal in the ''coupled'' $|j_1,j_2;jm\rangle$ basis, with the eigenvalues being $$ \hbar^2 [a j (j+1) + bj_1(j_1+1) +b j_2(j_2+1)] $$ where from the rules of addition of angular momenta, we have $$ |j_1 - j_2| \leq j \leq j_1+j_2 $$ The presence of a $j$-dependent term in the energies breaks some of the degeneracies. From the earlier $(2j_1+1)(2j_2+1)$ degeneracies, we're down to $(2j+1)$ degenerate states for each $j$. As a bonus, we can verify the counting by testing that that $$ \sum_{j=|j_1-j_2|}^{j_1+j_2} (2j+1)= (2j_1+1)(2j_2+1) $$ is indeed satisfied.

Identical bosons

With this understanding, now we can specialize to the more restricted case of identical bosons. Here, for $j_1=j_2$, we can seen in the $|j_1,m_1\rangle \otimes |j_2,m_2\rangle$ basis that the no. of allowed states is $$ n = \frac{1}{2}(2j_1+1)[(2j_1+1)-1] + (2j_1+1) $$ The first term counts the no. of symmetric states with $m_1 \neq m_2$, while the 2nd counts the states with $m_1=m_2$. All of these $n$ states are degenerate under the Hamiltonian $H_0$.

Now, adding the $\vec{J}_1.\vec{J}_2$ term still breaks the degeneracies, but from the smaller set of correspondingly restricted total angular momentum states. This time, the requirement of symmetrization amounts to restricting the total angular momenta to even $j$'s. So you still have the removal of degeneracies.

Example

Taking the example from your comment, for $j_1 = j_2 = 1$, we count $6$ unique, symmetrized states in the $|j_1,m_1\rangle \otimes |j_2,m_2\rangle$ basis. In the total angular momentum basis, we'd have $j \in \{0,1,2\}$ for distinguishable particles, but symmetrization for our identical bosons restricts us to $j \in \{0,2\}$. So again, instead of $6$ degenerate states, the new Hamiltonian $H$ will have $2$ separate degenerate blocks consisting of $2\times 0 +1=1$ and $2\times 2 + 1=5 $ states respectively.

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    $\begingroup$ Right, that was the kind of behaviour I was expecting, though I couldn't really do it through the counting. Thank you for your answer. However, would that degeneracy counting for $H_0$ be correct even if $j_1=j_2$? For example, suppose we have the state $|1,-1⟩\otimes|1,1⟩$, which should be symmetrized since they are bosons. Then, we wouldn't be counting the state $|1,1⟩\otimes|1,-1⟩$, since if we symmetrize the latter, we obtain the exact same state as before. How could this be taken into account? $\endgroup$ – JorgeOvi Mar 25 at 8:31
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    $\begingroup$ Thanks for reminding me that; updating the answer. $\endgroup$ – ConservedCharge Mar 25 at 16:14

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