Consider a composite particle state $|\psi\rangle$ (like a hadron or a meson) that is an eigenstate of some Hamiltonian (e.g. the QCD Hamiltonian). Since the Hamiltonian is invariant under rotations and parity this particle state is also an eigenstate of the angular momentum and parity operator: $$L^2 |\psi\rangle = l(l+1)|\psi\rangle$$ $$P |\psi\rangle = (-1)^{a}|\psi\rangle$$

where $a$ is an integer number. Why is $a = l$?

For two particles one can use the 'trick' to transform into relative coordinates and then find that in relative coordinates the eigenfunction is $\sim Y_{lm}$. The parity of the spherical harmonics then leads to $(-1)^l$.

I don't see how to extend this to 3 or more particles.

EDIT: I had the following idea how to extend to 3 particles:

For 3 particles the Hamiltonian looks like: $$H = \frac{p_1^2}{2m_1} + \frac{p_2^2}{2m_2} + \frac{p_3^2}{2m_3} + V_1(|x_2-x_1|) + V_2(|x_3-x_1|) + V_3(|x_3-x_2|)$$.

Now choose new coordinates by $$R = \frac{m_1 x_1 + m_2 x_2 + m_3 x_3}{M}$$ $$y = x_2 - x_1$$ $$z = x_3 - x_1$$

The Hamiltonian becomes: $$H = \frac{p_R^2}{2M} + \frac{p_y^2}{2\mu_{12}} + \frac{p_z^2}{2\mu_{13}} + \frac{p_y\cdot p_z}{m_1} + V_1(|y|) + V_2(|z|) + V_3(|z-y|)$$ where $\frac{1}{\mu_{ij}} = \frac{1}{m_i} + \frac{1}{m_j}$ are reduced masses

The total angular momentum is given by $$L = x_1 \times p_1 + x_2 \times p_2 + x_3 \times p_3 = R \times p_R + y \times p_y + z \times p_z$$.

The $l$ in parity $ = (-1)^l$ is given by the internal angular momentum $$L_i = y \times p_y + z \times p_z$$ which commutes with the Hamiltonian.

Therefore an eigenfunction is given by $$|\psi(y,z)\rangle = |f(|y|,|z|)\rangle |L M\rangle_{\hat{y}\hat{z}}$$. Using Clebsch-Gordan coefficients this angular can be written as: $$|L M\rangle_{\hat{y}\hat{z}} = \sum_{m m'} \langle lm,l'm'|LM\rangle Y_{lm}(\hat{y}) Y_{l'm'}(\hat{z})$$ for some $l$ and $l'$

The overall parity is given by $(-1)^{l + l'}$ which does not necessary equals $(-1)^L$. For example $l = l' = 1$ would lead to a (from my point of view valid) solution: $$|10\rangle_{\hat{y}\hat{z}} = \frac{1}{\sqrt{2}} \left(Y_{11}(\hat{y})Y_{1-1}(\hat{z}) - Y_{1-1}(\hat{y})Y_{11}(\hat{z})\right)$$ with parity $(-1)^{l+l'} = (-1)^{1+1} = 1 \neq (-1)^1 = (-1)^L$.

There must be something which excludes such combinations. Why is this solution not valid?


2 Answers 2


$\def\br{\mathbf r} \def\bR{\mathbf R} \def\bl{\mathbf l} \def\bL{\mathbf L}$

Consider a composite particle state $|\psi\rangle$ (like a hadron or a meson)

I don't see where did you take into account particle's compositeness, nor you had to. Actually this a general QM matter - no need to bring up QCD or the like.

For one particle in a fixed potential your argument of spherical harmonics applies. For two particles interacting with each other but otherwise free, the same argument applies to relative coordinates.

For three particles (or more) you follow the same route, only with a somewhat higher complication. Choose (judiciously) two of the particles, introduce their c.o.m. G, then c.o.m. of G and third particle. Thus you have two position vectors: $\br$, going from particle 1 to particle 2, and $\bR$, going from G to particle 3.

It can be shown that kinetic energy splits into two terms, one depending only on $\br$ and the other on $\bR$. Then you may choose a basis of eigenfunctions of two angular momenta, say $\bl^2, l_z$ and $\bL^2, L_z$. You see that total wavefunction has parity $(-1)^{l+L}$. This is rather tricky, as one could erroneously be drawn to believe parity depends on total angular momentum, whereas it's not so: it depends on the sum of separate quantum numbers.

I talked only of kinetic energy, but of course in order parity may be a useful quantum number potential energy (or interaction lagrangian in QFT) are to be invariant under space reflections.

  • 1
    $\begingroup$ I though about this procedure before. But I am not convinced for two reasons: 1. you calculate the angular momentum of the two particle CMS system and then you shift to the three particle CMS system. This includes a velocity shift and therefore changes the angular momentum (operator) for the two particle system. 2. In order for the angular momentum operator to commute with the Hamiltonian you need a potential which only depends on $|\vec{x}_i - \vec{x}_j|$. This is true in the overall CMS frame, but not in relative coordinates if you have more than 2 particles. $\endgroup$
    – toaster
    Nov 12, 2018 at 22:59

$\def\bL{\mathbf L} \def\bl{\mathbf l} \def\bp{\mathbf p} \def\br{\mathbf r} \def\bP{\mathbf P} \def\bR{\mathbf R} \def\frac#1#2{{\textstyle {#1 \over #2}}} \def\half{\frac12}$ In order to reply your objections it's best for me to write some equations. As a general comment: you should not think of a change of reference frame, but only of expressing the original quantities (hamiltonian, angular momentum) in terms of new coordinates. Let's see how.

I'm going to assume all mases are equal, just to make formulas simpler. But you may verify that the argument is quite general. On the other hand, it's a time-honoured approach, known as Jacobi's coordinates and widely used in celestial mechnics more or less since mid-19th century.

Call $\br_1$, $\br_2$, $\br_3$, the position vectors of the three particles. Define $$\bR = {\textstyle {1 \over 3}} (\br_1 + \br_2 + \br_3)$$ $$\br= \half (\br_1 + \br_2) - \br_3.$$ $$\br' = \br_2 - \br_1$$ Kinetic energy: $$K = \half\,m \left(3\,\dot\bR^2 + \frac23 \dot\br^2 + \half {\dot\br'}^2\right).$$ Conjugate momenta: $$\bP = 3\,m\,\dot\bR \qquad \bp = \frac23 m\,\dot\br \qquad \bp = \half m\,\dot\br'$$ $$K = {P^2 \over 6\,m} + {3\,p^2 \over 2\,m} + {{p'}^2 \over m}.$$ Angular momentum: $$\bL_{\mathrm{tot}} = \bR \times \bP + \br \times \bp + \br' \times \bp' = \bL + \bl + \bl'.$$ Commutators are those expected for canonical coordinateses $\bR$ with $\bP$, $\br$ with $\bp$, $\br'$ with $\bp'$.

Parity quantum number refers to the transformation $$\br \to - \br \qquad \br' \to -\br'$$ and consequently $$\bp \to - \bp \qquad \bp' \to -\bp'.$$ Eigenstates of $\bl^2$, $\bl'^2$ have parity $(-1)^{l+l'}$ and I see that you say the same thing, even if with different coordinates (which, incidentally, gives rise to a non-separable hamiltonian).

What I cannot understand is why you are troubled with your example. You built a state of internal angular momentum 1, $z$-component 0, starting from states with $l=l'=1$, then parity +. What's wrong with that? It's a perfectly possible situation.

  • $\begingroup$ Thank you. I'm a bit confused. Is the formula P = (-1)^L only valid for two particle systems? In our lecture we used it for more particle systems. Maybe I got something wrong :D. However I have a question to your result that the parity is (-1)^(l+l') for l,l' some internal angular momentum. In general (with interactions) only the overall angular momentum is a good quantum number. The internal angular momentums are not. So how can one find values for l and l' and therefore for the parity? $\endgroup$
    – toaster
    Nov 16, 2018 at 23:29
  • $\begingroup$ @toaster You wrote "Is the formula $P=(-1)^L$ only valid for two particle systems?" No doubt. You're right that in general we can't write $P=(-1)^{l+l'}$ if those angular momenta are not conserved. There are two answers. The first is that parity is important in studying particle decays, where the final state consists of free particles. Then no problem. Second. It is true that for interacting particles the general stationary state is a superposition of terms $\phi_l(\mathbf{r})\,\chi_{l'}(\mathbf{r'})$. Then only terms with $l+l'$ of equal parities are allowed. $\endgroup$
    – Elio Fabri
    Nov 17, 2018 at 10:08
  • $\begingroup$ Ah, so I cannot calculate the parity of a particle: I can only observe the decays and then compare to the parity of the final state. Thank you $\endgroup$
    – toaster
    Nov 17, 2018 at 17:05

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