# Exchange symmetry of spatial wavefunction of system of two particles with $\ell=1$

Consider a system of two identical particles. The combined system of these two particles has a total orbital angular momentum quantum number $$\ell = 1$$.

I am aware that the spatial wavefunction of a system with $$\ell = 1$$ has odd parity, as this can be shown from the spherical harmonics. However, I am not sure if anything can be said about the exchange symmetry of the spatial wavefunction in this case.

I.e. is the spatial wavefunction of the combined system symmetric or antisymmetric under exchange of the particle labels? And could you explain why.

• If you said spin angular momentum of $\ell = 1$, this would be easy. The spin state would be symmetric so the spatial state would be anti-symmetric (assuming they are fermions). Nov 22, 2021 at 16:04
• Yes I deliberately didn't specify whether the particles are fermions or bosons. I want to know whether you can deduce anything about the exchange symmetry of the spatial wavefunction purely from the knowledge that $l=1$. This is to help clear up some confusing claims that my lecturers are making who appear to be mixing up parity and exchange symmetry. Nov 22, 2021 at 16:29
• Duplicate question - physics.stackexchange.com/questions/325502/… Dec 2, 2021 at 15:52

One cannot know in general on the basis of $$L=1$$ alone.

Combining two states with the same $$j$$ will produce $$L=0,1\ldots,2j$$. The resulting states will be of the form \begin{align} \vert LM\rangle =\sum_{m_1m_2} C_{j m_1;j m_2}^{LM} \vert j m_1\rangle \vert j m_2\rangle \end{align} where $$C_{j m_1;j_2 m_2}^{LM}$$ is a Clebsch-Gordan coefficient.

The CGs satisfy the general relation $$C_{j_1 m_1;j_2 m_2}^{LM}=(-1)^{j_1+j_2-L}C_{j_2 m_2;j_1 m_1}^{LM}$$, and applying to your situation where $$j_1=j_2=j$$ yields $$C_{j m_1;j m_2}^{LM}=(-1)^{2j-L} C_{j m_2;j m_1}^{LM}\, . \tag{1}$$ In other words, the CGs of Eq.(1) are even under permutation if $$2j-L$$ is even, and odd if $$2j-L$$ is odd.

Thus, if your states have $$j=1/2$$, the resulting $$L=1$$ states are even while the $$L=0$$ state will be odd. They are respectively the triplet and singlets states.

If, on the other hand, $$j=1$$, then the $$L=2$$ and $$L=0$$ states will be even, while the $$L=1$$ states will be odd.

In fact it is clear that if $$j$$ is half-integer, then the $$L=1$$ states are even under permutation, while if $$j$$ is integer, the $$L=1$$ states are odd.

• Thanks that cleared up a lot of things about CG coefficients. Could you help me apply this to the original context of the question: there are two identical particles (fermions for concreteness) which are orbiting their mutual centre of mass with combined orbital angular momentum quantum number $l = 1$. I want to know the exchange symmetry under permuting these two particles. My current suspicition is that the fact that the particles are orbiting their centre of mass does not affect the exchange symmetry, since it is not clear to me how permuting the particles would affect the orbital state. Dec 1, 2021 at 22:54
• If you are concerned about orbital angular momentum only, then $j$ is necessarily an integer so the $L=1$ states are odd. Dec 1, 2021 at 23:43
• OK I see. Is it correct to assume that the two particles must have $j_1 = j_2$ when they are orbiting their mutual centre of mass (completely ignoring spin)? Because if $j_1 = 0$ and $j_2 = 1$ and $L = 1$, then the exchange symmetry will be even using the rules described in your answer. Dec 2, 2021 at 16:00