Identical Particles

Abstract

From Cohen-Tannoudji (Cohen-Tannoudji 1977):

“Two particles are said to be identical if all their intrinsic properties (mass, spin, charge, etc.) are exactly the same: no experiment can distinguish one from another. Thus, all electrons in the universe are identical, as are all the protons and all the hydrogen atoms[...]. An important consequence can be deduced from this definition: when a physical system contains two identical particles, there is no change in its properties or its evolution if the roles of these two particles are exchanged.”

Introduction

In classical physics it is possible to keep track of individual particles even though they may look alike. If we have a system consisting of two particles A and B, we can, in principle, follow the trajectory of 1 and that of 2 separately at each instant of time. In quantum mechanics, however, identical particles are truly indistinguishable. This is because we cannot specify more than a complete set of commuting observables for each of the particles. Nor can we follow the trajectory because it would entail a position measurement at each instant of time (Sakurai 1993).

Figure 1. Two different paths, (a) and (b), of a two electron system, for example, in which we cannot assert even in principle which of the paths the electrons pass (Adapted from (Sakurai 1993)).

For simplicity, consider just two particles. Suppose one of the particles, which we call particle 1, is characterized by $$\left|\alpha \right\rangle$$, where $$\alpha$$ is a collective index for a complete set of observables (CSCO). Likewise, we call the ket of the remaining particle $$\left|\beta \right\rangle$$. The state ket for the two particles can be written as the product,

$\left|\alpha \right\rangle\otimes\left|\beta \right\rangle \equiv \left|\alpha \right\rangle\left|\beta \right\rangle \label{eqn:ketab}$

We can also consider

$\left|\beta \right\rangle\left|\alpha \right\rangle \label{eqn:ketba}$

where particle 1 is now characterized by $$\left|\beta \right\rangle$$ and particle 2 by $$\left|\alpha \right\rangle$$. Even though the two particles are indistinguishable, mathematically \eqref{eqn:ketab} and \eqref{eqn:ketba} are distinct kets for $$\alpha \neq \beta$$. In fact, with $$\alpha \neq \beta$$, they are orthogonal to each other (Sakurai 1993).

Suppose we make an experiment with two identical particles and, after measuring them, we obtain that one is at the state $$\alpha$$ and the other is at $$\beta$$. Is the state vector just after the measurement $$\left|\alpha\beta\right\rangle$$ or $$\left|\beta\alpha\right\rangle$$? The answer is, neither. In quantum theory two configurations related by the exchange of identical particles must be viewed as one and the same and be described by the same vector (Shankar 1994). Put in another way, any linear combination of kets of the form

$c_1 \left|\alpha\beta\right\rangle + c_2 \left|\beta\alpha\right\rangle$

lead to an identical set of eigenvalues when measurement is performed. This is known as exchange degeneracy. Exchange degeneracy presents a difficulty since the specification of the eigenvalue of a complete set of observables does not completely determine the state ket (Sakurai 1993).

The difficulties related to exchange degeneracy arise in the study of all systems containing an arbitrary number of identical particles ($$N>1$$).

Consider, for example, a three particle system. We associate a state space, and observables acting on this space with each of the three particles taken separately. Thus, we are led to number the particles, $$\mathcal{E}(1)$$, $$\mathcal{E}(2)$$ and $$\mathcal{E}(3)$$ will denote the three one-particle state states, and the corresponding observables will be labeled by the same indices. The state space of the three-particle system is the tensor product:

$\mathcal{E}=\mathcal{E}(1)\otimes \mathcal{E}(2)\otimes \mathcal{E}(3)$

Now consider an observable $$B(1)$$, initially defined in $$\mathcal{E}(1)$$. We shall assume that $$B(1)$$ alone constitutes a CSCO in $$\mathcal{E}(1)$$ [or that $$B(1)$$ actually denotes several observables which form a CSCO]. The fact that the three particles are identical implies that the observables $$B(2)$$ and $$B(3)$$ exist and that they constitute CSCO’s in $$\mathcal{E}(2)$$ and $$\mathcal{E}(3)$$ respectively. $$B(1)$$, $$B(2)$$ and B(3) have the same spectrum, $$\{ b_n; n=1,2, \ldots \}$$. Using the bases which define these three observables in $$\mathcal{E}(1)$$, $$\mathcal{E}(2)$$ and $$\mathcal{E}(3)$$, we can construct, by taking the tensor product, an orthonormal basis of $$\mathcal{E}$$, which we shall denote by:

$\{ \left| 1: b_i; 2: b_j; 3:b_k\right\rangle; i,j,k= 1,2,\ldots\}$

The kets $$\left| 1: b_i; 2: b_j; 3:b_k\right\rangle$$ are common eigenvectors of the extensions of $$B(1)$$, $$B(2)$$ and $$B(3)$$ in $$\mathcal{E}$$, with respective eigenvalues $$b_i$$, $$b_j$$ and $$b_k$$.

Since the three particles are identical, we cannot measure $$B(1)$$ or $$B(2)$$ or $$B(3)$$, since the numbering has no physical significance. However, we can measure the physical quantity $$B$$ for each of the three particles. Suppose that such a measurement has resulted in three different eigenvalues $$b_n$$, $$b_p$$ and $$b_q$$. Exchange degeneracy then appears, since the state of the system after this measurement can, a priory, be represented by any of the kets on the subspace of $$\mathcal{E}$$ spanned by the six basis vectors:

$\left| 1: b_n; 2: b_p; 3:b_q\right\rangle,\quad \left| 1: b_q; 2: b_n; 3:b_p\right\rangle,\quad \left| 1: b_p; 2: b_q; 3:b_n\right\rangle, \\ \left| 1: b_n; 2: b_q; 3:b_p\right\rangle,\quad \left| 1: b_p; 2: b_n; 3:b_q\right\rangle,\quad \left| 1: b_q; 2: b_p; 3:b_n\right\rangle$

Therefore, a complete measurement on each of the particles does not permit the determination of a unique ket of the state space of the system (Cohen-Tannoudji 1977).