Permutation Symmetry

Consider a quantum system consisting of two identical particles. Suppose that one of the particles--particle 1, say--is characterized by the state ket $\vert k'\rangle$ . Here, $k'$ represents the eigenvalues of the complete set of commuting observables associated with the particle. Suppose that the other particle--particle 2--is characterized by the state ket $\vert k''\rangle$ . The state ket for the whole system can be written in the product form

$$\vert k'\rangle\,\vert k''\rangle,$$ (9.1)

where it is understood that the first ket corresponds to particle 1, and the second to particle 2. We can also consider the ket

$$\vert k''\rangle\,\vert k'\rangle,$$ (9.2)

which corresponds to a state in which particle 1 has the eigenvalues $k''$ , and particle $2$ the eigenvalues $k'$ .

Suppose that we were to measure all of the simultaneously measurable properties of our two-particle system. We might obtain the results $k'$ for one particle, and $k''$ for the other. However, we have no way of knowing whether the corresponding state ket is $\vert k'\rangle\,\vert k''\rangle$ or $\vert k''\rangle\,\vert k'\rangle$ , or any linear combination of these two kets. In other words, all state kets of the form

$$c_1\,\vert k'\rangle\,\vert k''\rangle + c_2\,\vert k''\rangle\,\vert k'\rangle,$$ (9.3)

where $c_1$ and $c_2$ are arbitrary complex numbers, correspond to an identical set of results when the properties of the system are measured. This phenomenon is known as exchange degeneracy. Such degeneracy is problematic because the specification of a complete set of observable eigenvalues in a system of identical particles does not seem to uniquely determine the corresponding state ket. Fortunately, nature has a way of avoiding this difficulty.

Consider the two-particle permutation operator, $P_{12}$ , which is defined such that

$$P_{12}\,\vert k'\rangle\,\vert k''\rangle = \vert k''\rangle\,\vert k'\rangle.$$ (9.4)

In other words, $P_{12}$ swaps the identities of particles $1$ and $2$ . It is easily appreciated that

$$P_{21} = P_{12},$$ (9.5)

$$P_{12}^{\,2} = 1.$$ (9.6)

Now, the Hamiltonian of a system of two identical particles must necessarily be a symmetric function of each particle's observables (because exchange of identical particles could not possibly affect the overall energy of the system). An example of such a Hamiltonian is

$$H = \frac{{\bf p}_1^{\,2}}{2\,m} + \frac{{\bf p}_2^{\,2}}{2\,m} +... ...vert{\bf x}_1-{\bf x}_2\vert) + V_{\rm ext}({\bf x}_1)+ V_{\rm ext}({\bf x}_2).$$ (9.7)

Here, we have separated the mutual interaction of the two particles from their interaction with an external potential. [To be more exact, $V_{\rm pair}(\vert{\bf x}_1-{\bf x}_2\vert)$ is the interaction potential, and $V_{\rm ext}({\bf x})$ the external potential.] It follows that if

$$H\,\vert k'\rangle\,\vert k''\rangle = E\,\vert k'\rangle\,\vert k''\rangle$$ (9.8)

then

$$H\,\vert k''\rangle\,\vert k'\rangle = E\,\vert k''\rangle\,\vert k'\rangle,$$ (9.9)

where $E$ is the total energy. Operating on both sides of Equation (9.8) with $P_{12}$ , and employing Equation (9.6), we obtain

$$P_{12}\,H\,P_{12}^{\,2}\,\vert k'\rangle\,\vert k''\rangle = E\,P_{12}\,\vert k'\rangle\,\vert k''\rangle,$$ (9.10)

or

$$P_{12}\,H\,P_{12}\,\vert k''\rangle\,\vert k'\rangle = E\,\vert k''\rangle\,\vert k'\rangle = H\,\vert k''\rangle\,\vert k'\rangle,$$ (9.11)

where use has been made of Equation (9.9). Because the $\vert k''\rangle\,\vert k'\rangle$ must form a complete set (otherwise, the properties of the system would not be fully observable), we deduce that

$$P_{12}\,H\,P_{12} = H,$$ (9.12)

which implies [from Equation (9.6)] that

$$[H,P_{12}] = 0.$$ (9.13)

In other words, an eigenstate of the Hamiltonian is a simultaneous eigenstate of the two-particle permutation operator, $P_{12}$ . (See Section 1.13.)

Now, according to Equation (9.6), the two-particle permutation operator possesses the eigenvalues $+1$ and $-1$ , respectively. (See Exercise 1.) The corresponding properly normalized (provided that $k'\neq k''$ ) eigenkets are

$$\vert k'\,k''\rangle_+ = \frac{1}{\sqrt{2}}\,\left(\vert k'\rangle\,\vert k''\rangle + \vert k''\rangle\,\vert k'\rangle\right),$$ (9.14)

and

$$\vert k'\,k''\rangle_- = \frac{1}{\sqrt{2}}\,\left(\vert k'\rangle\,\vert k''\rangle - \vert k''\rangle\,\vert k'\rangle\right).$$ (9.15)

(See Exercise 2.) Here, it is assumed that $\langle k'\vert k''\rangle = \delta_{k'\,k''}$ . Note that $\vert k'\,k''\rangle_+$ is symmetric with respect to interchange of particles--that is,

$$\vert k''\,k'\rangle_+ =+ \vert k'\,k''\rangle_+,$$ (9.16)

whereas $\vert k'\,k''\rangle_-$ is antisymmetric--that is,

$$\vert k''\,k'\rangle_- =- \vert k'\,k''\rangle_-.$$ (9.17)

Let us, now, consider a system of three identical particles. We can represent the overall state ket as

$$\vert k'\rangle\,\vert k''\rangle \,\vert k'''\rangle,$$ (9.18)

where $k'$ , $k''$ , and $k'''$ are the eigenvalues of particles 1, 2, and 3, respectively. We can also define two-particle permutation operators:

$$P_{12}\,\vert k'\rangle\,\vert k''\rangle\,\vert k'''\rangle = \vert k''\rangle\,\vert k'\rangle\,\vert k'''\rangle,$$ (9.19)

$$P_{23}\,\vert k'\rangle\,\vert k''\rangle\,\vert k'''\rangle = \vert k'\rangle\,\vert k'''\rangle\,\vert k''\rangle,$$ (9.20)

$$P_{31}\,\vert k'\rangle\,\vert k''\rangle\,\vert k'''\rangle = \vert k'''\rangle\,\vert k''\rangle \,\vert k'\rangle.$$ (9.21)

It is easily demonstrated that

$$P_{21} = P_{12},$$ (9.22)

$$P_{32} = P_{23},$$ (9.23)

$$P_{13} = P_{31},$$ (9.24)

and

$$P_{12}^{\,2} = P_{23}^{\,2} =P_{31}^{\,2} = 1.$$ (9.25)

As before, the Hamiltonian of the system must be a symmetric function of the particle's observables: that is,

$$H\,\vert k'\rangle\,\vert k''\rangle\,\vert k'''\rangle = E\,\vert k'\rangle\,\vert k''\rangle\,\vert k'''\rangle,$$ (9.26)

$$H\,\vert k''\rangle\,\vert k'''\rangle\,\vert k'\rangle = E\,\vert k''\rangle\,\vert k'''\rangle\,\vert k'\rangle,$$ (9.27)

$$H\,\vert k'''\rangle\,\vert k'\rangle\,\vert k''\rangle = E\,\vert k'''\rangle\,\vert k'\rangle\,\vert k''\rangle,$$ (9.28)

$$H\,\vert k''\rangle\,\vert k'\rangle\,\vert k'''\rangle = E\,\vert k''\rangle\,\vert k'\rangle\,\vert k'''\rangle,$$ (9.29)

$$H\,\vert k'\rangle\,\vert k'''\rangle\,\vert k''\rangle = E\,\vert k'\rangle\,\vert k'''\rangle\,\vert k''\rangle,$$ (9.30)

$$H\,\vert k'''\rangle\,\vert k''\rangle\,\vert k'\rangle = E\,\vert k'''\rangle\,\vert k''\rangle\,\vert k'\rangle,$$ (9.31)

where $E$ is the total energy. Using analogous arguments to those employed for the two-particle system, we deduce that

$$[H,P_{12}] = [H,P_{23}] = [H,P_{31}] = 0.$$ (9.32)

Hence, an eigenstate of the Hamiltonian is a simultaneous eigenstate of the three two-particle permutation operators, $P_{12}$ , $P_{23}$ , and $P_{31}$ . (See Section 1.13.) However, according to Equation (9.25), the possible eigenvalues of these operators are $\pm 1$ . (See Exercise 1.)

Let us define the cyclic permutation operator, $P_{123}$ , where

$$P_{123} \,\vert k'\rangle\,\vert k''\rangle\,\vert k'''\rangle =\vert k'''\rangle\,\vert k'\rangle\,\vert k''\rangle.$$ (9.33)

It follows that

$$P_{123} = P_{12}\,P_{31}=P_{23}\,P_{12}=P_{31}\,P_{23}.$$ (9.34)

It is also clear from Equations (9.26) and (9.28) that

$$[H,P_{123}] = 0.$$ (9.35)

(See Exercise 4.) Thus, an eigenstate of the Hamiltonian is a simultaneous eigenstate of the four permutation operators $P_{12}$ , $P_{23}$ , $P_{31}$ , and $P_{123}$ . (See Section 1.13.) Let $\lambda_{12}$ , $\lambda_{23}$ , $\lambda_{31}$ and $\lambda_{123}$ represent the eigenvalues of these operators, respectively. We know that $\lambda_{12}=\pm 1$ , $\lambda_{23}=\pm 1$ , and $\lambda_{31}=\pm 1$ . Moreover, it follows from Equation (9.34) that

$$\lambda_{123} = \lambda_{12}\,\lambda_{31} = \lambda_{23}\,\lambda_{12}=\lambda_{31}\,\lambda_{23}.$$ (9.36)

The previous equations imply that

$$\lambda_{123} = +1,$$ (9.37)

and either

$$\lambda_{12} = \lambda_{23} = \lambda_{31}= +1,$$ (9.38)

or

$$\lambda_{12} = \lambda_{23} = \lambda_{31}= -1.$$ (9.39)

In other words, the multi-particle state ket must be either totally symmetric, or totally antisymmetric, with respect to swapping the identities of any given pair of identical particles. Thus, in terms of properly normalized single-particle kets, the properly normalized (provided that $k'\neq k''\neq k'''$ ) totally symmetric and totally antisymmetric kets are

$$\vert k'\,k''\,k'''\rangle_+ = \frac{1}{\sqrt{3!}}\left(\vert k'\rangle\,\vert k''\rangle\,\ve... ...\,\vert k''\rangle+\vert k''\rangle\,\vert k'''\rangle\,\vert k'\rangle \right.$$

$$\phantom{=}\left.+ \vert k'''\rangle\,\vert k''\rangle\,\vert k'\... ...,\vert k''\rangle+ \vert k''\rangle\,\vert k'\rangle\,\vert k'''\rangle\right),$$ (9.40)

and

$$\vert k'\,k''\,k'''\rangle_- = \frac{1}{\sqrt{3!}}\left(\vert k'\rangle\,\vert k''\rangle\,\ve... ...\,\vert k''\rangle+\vert k''\rangle\,\vert k'''\rangle\,\vert k'\rangle \right.$$

$$\phantom{=}\left.-\vert k'''\rangle\,\vert k''\rangle\,\vert k'\r... ...,\vert k''\rangle- \vert k''\rangle\,\vert k'\rangle\,\vert k'''\rangle\right),$$ (9.41)

respectively.

The previous arguments can be generalized to systems of more than three identical particles in a straightforward manner [103].