Scattering of Identical Particles

Consider two identical particles that scatter off one another. In the center of mass frame, there is no way of distinguishing a deflection of a particle through an angle $\theta$ , and a deflection through an angle $\pi-\theta$ , because momentum conservation demands that if one of the particles is scattered in the direction characterized by the angle $\theta$ then the other is scattered in the direction characterized by $\pi-\theta$ [55]. (Here, for the sake of simplicity, we are assuming that the scattering potential is spherically symmetric, which implies that the motion of the two particles is confined to a fixed plane passing through the origin.)

In classical mechanics, the differential cross-section for scattering is affected by the identity of the particles because the number of particles counted by a detector located at angular position $\theta$ is the sum of the counts due to the two particles, which implies that

$$\frac{d\sigma_{\rm classical}}{d{\mit\Omega}} = \frac{d\sigma(\th... ...left\vert f(\theta)\right\vert^{\,2}+\left\vert f(\pi-\theta)\right\vert^{\,2}.$$ (10.158)

[See Equation (10.28).]

In quantum mechanics, the overall wavefunction must be either symmetric or antisymmetric under interchange of identical particles, depending on whether the particles in question are bosons or fermions, respectively. (See Section 9.3.) If the spatial wavefunction is symmetric then Equation (10.55) is replaced by

$$\psi({\bf x}) = \frac{1}{(2\pi)^{3/2}} \left( {\rm e}^{\,{\rm i}\... ...frac{{\rm e}^{\,{\rm i}\,k\,r}}{r}\left[f(\theta)+f(\pi-\theta)\right] \right),$$ (10.159)

and the associated differential scattering cross-section becomes

$$\frac{d\sigma}{d{\mit\Omega}} = \left\vert f(\theta)+f(\pi-\theta)\right\vert^{\,2}.$$ (10.160)

[See Equation (10.28).] On the other hand, if the spatial wavefunction is antisymmetric then Equation (10.55) is replaced by

$$\psi({\bf x}) = \frac{1}{(2\pi)^{3/2}} \left( {\rm e}^{\,{\rm i}\... ...frac{{\rm e}^{\,{\rm i}\,k\,r}}{r}\left[f(\theta)-f(\pi-\theta)\right] \right),$$ (10.161)

and the associated differential scattering cross-section is written

$$\frac{d\sigma}{d{\mit\Omega}} = \left\vert f(\theta)-f(\pi-\theta)\right\vert^{\,2}.$$ (10.162)

For the case of two identical spin-zero (i.e., boson) particles (e.g., $\alpha$ -particles), the spatial wavefunction is symmetric with respect to particle interchange, which implies that

$$\frac{d\sigma}{d{\mit\Omega}} = \left\vert f(\theta)+f(\pi-\theta)\right\vert^{\,2}$$

$$= \left\vert f(\theta)\right\vert^{\,2}+\left\vert f(\pi-\theta)\... ...t[f^{\,\ast}(\theta)\,f(\pi-\theta) + f(\theta)\,f^{\,\ast}(\pi-\theta)\right].$$ (10.163)

The previous result differs from the classical one because of the interference term (i.e., the final term on the right-hand side), which leads to an enhancement of the differential scattering cross-section at $\theta=\pi/2$ . In fact,

$$\left(\frac{d\sigma}{d{\mit\Omega}} \right)_{\theta=\pi/2}=4\left[f(\pi/2)\right]^{\,2},$$ (10.164)

whereas

$$\left(\frac{d\sigma_{\rm classical}}{d{\mit\Omega}} \right)_{\theta=\pi/2}=2\left[f(\pi/2)\right]^{\,2}.$$ (10.165)

For the case of two identical spin-$1/2$ (i.e., fermion) particles (e.g., electrons or protons), the overall wavefunction is antisymmetric under particle interchange. If the two particles are in the spin singlet state (see Section 7.9) then the spatial wavefunction is symmetric (because the spin wavefunction is antisymmetric), and

$$\frac{d\sigma_{\rm singlet}}{d{\mit\Omega}} = \left\vert f(\theta)+f(\pi-\theta)\right\vert^{\,2}$$

$$= \left\vert f(\theta)\right\vert^{\,2}+\left\vert f(\pi-\theta)\... ...t[f^{\,\ast}(\theta)\,f(\pi-\theta) + f(\theta)\,f^{\,\ast}(\pi-\theta)\right].$$ (10.166)

On the other hand, if the two particles are in the spin triplet state then the spatial wavefunction is antisymmetric (because the spin wavefunction is symmetric), which leads to

$$\frac{d\sigma_{\rm triplet}}{d{\mit\Omega}} = \left\vert f(\theta)-f(\pi-\theta)\right\vert^{\,2}$$

$$= \left\vert f(\theta)\right\vert^{\,2}+\left\vert f(\pi-\theta)\... ...t[f^{\,\ast}(\theta)\,f(\pi-\theta) + f(\theta)\,f^{\,\ast}(\pi-\theta)\right].$$ (10.167)

In the former case, the interference term leads to an enhancement (with respect to the classical case) of the differential scattering cross-section at $\theta=\pi/2$ : that is,

$$\left(\frac{d\sigma_{\rm singlet}}{d{\mit\Omega}} \right)_{\theta=\pi/2}=4\left[f(\pi/2)\right]^{\,2}.$$ (10.168)

In the latter case, the interference term leads to the complete suppression of scattering in the direction $\theta=\pi/2$ : that is,

$$\left(\frac{d\sigma_{\rm triplet}}{d{\mit\Omega}} \right)_{\theta=\pi/2}=0.$$ (10.169)

Consider the mutual scattering of two unpolarized beams of spin-$1/2$ particles. All spin states are equally likely, so the probability of finding a given pair of particles (one from each beam) in the triplet state is three times that of finding it in the singlet state (see Section 7.9), which implies that

$$\left(\frac{d\sigma_{\rm unpolarized}}{d{\mit\Omega}} \right) = \frac{1}{4}\left(\frac{d\sigma_{\rm singlet}}{d{\mit\Omega}} \right)+\frac{3}{4}\left(\frac{d\sigma_{\rm triplet}}{d{\mit\Omega}} \right)$$ (10.170)

$$= \left\vert f(\theta)\right\vert^{\,2}+\left\vert f(\pi-\theta)\... ...t[f^{\,\ast}(\theta)\,f(\pi-\theta) + f(\theta)\,f^{\,\ast}(\pi-\theta)\right].$$

In this case, the interference term leads to incomplete suppression (with respect to the classical case) of the differential scattering cross-section at $\theta=\pi/2$ : that is,

$$\left(\frac{d\sigma_{\rm unpolarized}}{d{\mit\Omega}} \right)_{\theta=\pi/2}=\left[f(\pi/2)\right]^{\,2}.$$ (10.171)