4 Identical Particles In this chapter we consider the consequences resulting from the fact that identical par- ticles are indistinguishable in quantum mechanics (as compared to classical mechanics), and the formalism needed to deal with such situations. Some of the material presented in this chapter is taken from Auletta, Fortunato and Parisi, Chap. 7 and Cohen-Tannoudji, Diu and Laloë, Vol. II, Chaps. XIV. 4.1 Exchange Degeneracy In classical mechanics even if two particles forming a system are identical (i.e., they share the same physical properties, such as mass, charge, etc.), there is nothing unusual to the manner with which one would go about analyzing the system’s evolution. That is, the case of two identical particles is simply a special case of the more general situation when the particles are different. In other words, the fact that the particles are identical does not preclude us from following their respective evolutions separately. It is still possible to label the particles (say, particle 1 and particle 2) without any danger of confusion. Things are very different in quantum mechanics since particles do not have definite trajectories, but rather they are described using wave functions. If two identical particles are separated by a distance that is much larger than the extent of their wave functions, such that they do not overlap, then it is presumably easy to label them and keep track of their evolution. But it is possible that their evolution would bring a spatial overlap of their wave functions (e.g., they could be in the process of colliding), and it then becomes impossible to tell them apart. More precisely, if we detect the presence of a particle in a region where both have a sizeable probability of being present (i.e., their wave functions are non-zero overlapping), it then becomes impossible to say which particle was detected. This situation is an example what is referred to exchange degeneracy. Exercise 4.1. Let us consider the case of two spin one-half particles. We assume that one particle is in the spin-up state j+i, where its spin component along the z-axis has the ~=2 eigenvalue, while the other is in the spin-down state |−i. Find the most general ket for this system and determine the probability of finding both particles having their ^ spin component along the x-axis Sx with the ~=2 eigenvalue. Solution. The kets j+; −i and |−; +i both corresponds to a spin-up/spin-down state. The most general form of such state therefore consists of a linear combination of these kets of the form j i = α j+; −i + β |−; +i ; (4.1) 86 4 Identical Particles with jαj2 + jβj2 = 1 to ensure normalization. We therefore find that there exists an infinite number of kets with the same physical state where one spin is up and the other down. This is clearly a case of exchange degeneracy. We know from equations (3.127) and (3.129) of Chapter 3 that the matrix associated to the operator S^x is given by 0 1 S^ = ~ ; (4.2) x 2 1 0 when using the basis 1 0 j+i = ; |−i = : (4.3) 0 1 We could formally diagonalize equation (4.2) and determine its eigenvectors using the material covered in Exercise 1.5 of Chapter 1, but it is straightforward to achieve this by inspection. It will be readily verified that the eigenvectors of S^x are given by 1 |±i = p (j+i ± |−i) ; (4.4) x 2 ^ i.e., Sx |±ix = ±~=2 |±ix. The compound state where both particles are in the j+ix state is thus expressed by the following direct product jχi = j+ix ⊗ j+ix 1 = (j+; +i + j+; −i + |−; +i + |−; −i) : (4.5) 2 The probability of finding the system in that state is 2 P (+; +x) = jhχ j ij 2 1 = (α + β) : (4.6) 2 This result is problematic because it implies that it is not possible to uniquely describe the state of the system. In other words, it implies that physicists performing independent experiments on similar systems would obtain results that are not consistent with each other. This is entirely non-sensical from a physics standpoint. This state of affair is entirely traceable to the exchange degeneracy, which must therefore be lifted in order for such a system to be describable in a way that is to be expected from a physical system. Evidently, the exchange degeneracy is not limited to systems composed of two identical particles, and can be generalized to any arbitrary number N of them. 87 4 Identical Particles 4.2 Hamiltonian Invariance (or Symmetry) If two identical particles are part of a system, then the corresponding Hamiltonian must be invariant (or symmetric) under the permutation of these particles. This could actually be used as a definition for identical particles. We label these two particles as 1 and 2 and define the permutation operator P^21 = P^12 such that given a compound state j' (1) χ (2)i for the two particles we have P^21 j' (1) χ (2)i = j' (2) χ (1)i = jχ (1) ' (2)i : (4.7) The fact that the permutation of two identical particles leaves the Hamiltonian (i.e., the energy) of a system unchanged implies that corresponding operators commute with each other h i P^21; H^ = 0: (4.8) From this we conclude that P^21 (or any permutation operator, for that matter) and H^ share the same set of eigenvectors (see Section 3.2 of Chapter 3). If we denote that basis as fjuj (1) uk (2)ig, we can evaluate the matrix elements of P^21 with D E ^ uj (1) uk (2) P21 um (1) un (2) = huj (1) uk (2) j un (1) um (2)i = δjnδkm: (4.9) ^y Likewise, the matrix elements of P21 are calculated to be D E ^y uj (1) uk (2) P21 um (1) un (2) = huk (1) uj (2) j um (1) un (2)i = δkmδjn (4.10) y ^ ^y since P21 juj (1) uk (2)i = huj (1) uk (2)j P21. A comparison of equations (4.9) and ^y ^ (4.10) reveals that this permutation operator is Hermitian, i.e., P21 = P21. Also, since it is clear that 2 P^21 = 1^; (4.11) ^y ^ ^ we find that P21P21 = 1. The permutation operator is therefore unitary. Exercise 4.2. Use the basis of eigenvectors fjuj (1) uk (2)ig common to H^ and P^21 to find an expression for the latter. Solution. 88 4 Identical Particles P ^ Let us use the closure relation j;k juj (1) uk (2)ihuj (1) uk (2)j = 1 to expand the following (j i is an arbitrary state vector) X P^21 j i = P^21 juj (1) uk (2)i huj (1) uk (2) j i j;k X = P^21 juj (1) uk (2)i huj (1) uk (2) j i j;k X = juk (1) uj (2)i huj (1) uk (2) j i ; (4.12) j;k which implies that X P^21 = juk (1) ; uj (2)ihuj (1) ; uk (2)j j;k X = juj (1) ; uk (2)ihuk (1) ; uj (2)j : (4.13) j;k Although we have so far limited our discussion to a finite basis, the same reasoning applies to infinite, continuous bases or a combination of both. For example, using the compound basis fjr1;"1; r2;"2ig for the positions ^rj and spin components S^jz of two identical particles (j = 1; 2), it is also possible to determine the effect of P^21 on their wave function (r1;"1; r2;"2). That is, if we define j'i = P^21 j i (4.14) ^ ^y we can calculate (since P21 = P21) D E ^ ' (r1;"1; r2;"2) = r1;"1; r2;"2 P21 = hr2;"2; r1;"1 j i = (r2;"2; r1;"1) (4.15) And in a similar manner as with a finite basis only, we find that Z 1 X 3 3 0 P^21 = d xd x jr1;"1; r2;"2ihr2;"2; r1;"1j : (4.16) −∞ "1;"2 4.2.1 Symmetric and Antisymmetric States We already know from equation (4.11) that, for j i one of its eigenvector, 2 P^21 j i = j i ; (4.17) 89 4 Identical Particles and since P^21 is Hermitian and must therefore have real eigenvalues, it follows that P^21 j i = ± j i : (4.18) That is, ±1 are the only possible eigenvalues for P^21. The eigenvectors fj Si ; j Aig such that P^21 j Si = + j Si (4.19) P^21 j Ai = − j Ai (4.20) are respectively called symmetric and antisymmetric states since they are symmetric and antisymmetric to the permutation of two identical particles (i.e., the kets change sign or not after the permutation). We can define the projectors on the spaces to which the symmetric and antisymmetric states belong with 1 S^ = 1^ + P^ (4.21) 2 21 1 A^ = 1^ − P^ ; (4.22) 2 21 since for an arbitrary ket (as always jαj2 + jβj2 = 1) j i = α j Si + β j Ai (4.23) we have 1 S^ j i = 1^ + P^ (α j i + β j i) 2 21 S A = α j Si (4.24) 1 A^ j i = 1^ − P^ (α j i + β j i) 2 21 S A = β j Ai : (4.25) It is also straightforward to verify that S^y = S^ (4.26) A^y = A^ (4.27) S^2 = S^ (4.28) A^2 = A;^ (4.29) and 90 4 Identical Particles h i S;^ A^ = 0^ (4.30) h i S;^ P^21 = 0^ (4.31) h i A;^ P^21 = 0^ (4.32) S^ + A^ = 1^: (4.33) The combinations of equations (4.24) with (4.31) and (4.25) with (4.32), respectively, yield P^21S^ j i = S^P^21 j i = S^ j i (4.34) P^21A^ j i = A^P^21 j i = −A^ j i (4.35) We thus find that S^ and A^ are symmetrizer and antisymmetrizer operators, re- spectively.