Consider two identical particles.  IF EÄE denotes the state space of the two particles and |u_i,u_j> a state vector in EÄE, then P₂₁|u_i,u_j>=|u_j,u_i> must denote the same physical state, since we cannot distinguish the particles.  If |y> is a normalized state vector in EÄE then P₂₁|y>=e|y>.  Since P₂₁²=I, e=±1.  Any physical state of two identical particles must be either symmetric or anti-symmetric with respect to the exchange of the particles.  Any physical state can be as a linear combination of basis vectors of the state space.  The state space of a two particle system therefore is not EÄE, but either the subspace or symmetric or the subspace of anti-symmetric states.  It is restricted.  This extends naturally to an N-particle system.

Postulate:

• Every elementary particle is either a fermion or a boson.  A state of many identical particles is totally anti-symmetric if the particles are fermions, and it is totally symmetric if the particles are bosons.

Problem:

• Solution:

  Classical	1/4	1/2	1/4
  Boson	1/3	1/3	1/3
  Fermion	0	1	0

How do we construct a physical ket that correctly describes the physical state of a system of N identical particle?

• Choose a basis for the single particle space E, for example the set {|u_j>}.  Let the ket |y> in EÄEÄEÄ… be a tensor product vector, |y>=|u_i>Ä|u_j>Ä|u_k>Ä….  Any vector in EÄEÄEÄ… is a linear combination of such tensor product vectors.
• Apply S or A to |y>, depending on whether the particles are bosons or fermions.
• Normalize the ket thus obtained to find |y_S> or |y_A>.

Examples:

• Two bosons, one in a state |f> and one in a state |c>.

  |y>=|f>Ä|c>.
  S|y>=(1/2)(P₁₂+P₂₁)(|f>Ä|c>)=(1/2)(|fc>+|cf>).
  <y|S^TS|y>=(1/4)(<fc|fc>+<cf|cf>)=1/2=1/N².  N=2^1/2.
  |y_S>=(1/2)^1/2(|fc>+|cf>).

• Three bosons, one in a state |f>, one in a state |c> and one in state |w>.

  |y>=|f>Ä|c>Ä|w>=|f,c,w>.
  S|y>=(1/3!) åP_a|y>
  =(1/6)(|f,c,w>+|w,f,c>+|c,w,f>+|f,w,c>+|c,f,w>+|w,c,f>).
  <y|S^TS|y>=(1/6)=1/N².   N=6^1/2.
  |y_S>=(1/6)^1/2(|f,c,w>+|w,f,c>+|c,w,f>+|f,w,c>+|c,f,w>+|w,c,f>).
  If |f>=|c>=|w> then |y>=S|y>=|y_S>=|f,f,f>.

• Three fermions, one in a state |f>, one in a state |c> and one in state |w>.

  |y>=|f>Ä|c>Ä|w>=|f,c,w>.
  A|y>=(1/3!) åe_aP_a|y>
  =(1/6)(|f,c,w>+|w,f,c>+|c,w,f>-|f,w,c>-|c,f,w>-|w,c,f>).
  <y|A^TA|y>=(1/6)=1/N². N=6^1/2.
  |y_A>=(1/6)^1/2(|f,c,w>+|w,f,c>+|c,w,f>-|f,w,c>-|c,f,w>-|w,c,f>).
  If |f>=|c>=|w> then |y>=A|y>=|y_A>=0.

  Look at the definition of a determinant.

  .

  .

  e_ijk=+1 for even permutations, -1 for odd permutations.

  We can therefore write

  1/N!.

  A|y> is called a Slater determinant.  If any of the individual states |f>, |c>, and |w> are the same, then the determinant has two identical columns and therefore is zero.  We have the Pauli exclusion principle.

How do we find a basis for E_S or E_A?

{|y_S>} or {|y_A>} do not form a basis, because not all the vectors are linearly independent.

Example:

• S|u₁,u₁,u₂>=S|u₁,u₂,u₁>= S|u₂,u₁,u₁>.

Applying S to a basis of EÄEÄE we can obtain the same vector in E_S more than once.  We introduce the occupation number n_k of the individual state |u_k>, equal to the number of times that state appears in the tensor product |u_i>Ä|u_j>Ä… .

.

Two different tensor product vectors, which have the same occupation numbers, are equal to each other within a factor of ±1 after applying S or A.  We denote the basis states by |n₁,n₂,n₃,…,n_k,..>, ån_k=N.  For bosons the n_k are arbitrary, subject only to the restriction ån_k=N, for fermions n_k=0, or n_k=1.

Consequences of the indistinguishability on the calculations of physical predictions:

Consider two particles.  We know one is in the state |f> and one is in the state |c>.  The state of the two particle system is

Let {|u_i>} be an orthonormal basis of eigenvectors of some operator B.  B|u_i>=b_i|u_i>.  What is the probability that a measurement of B yields b_n for one particle and b_n' for the other particle?

The eigenstate of B associated with this measurement is

We therefore have for distinguishable particles,

P(b_n,b_n') is the sum of the probabilities in of finding the particles in either of the two distinguishable states.

For indistinguishable particles there exists only one state |u_n,u_n'>,  and we have

(direct term           exchange term)

=|<u_n|f><u_n'|c>|²+|<u_n|c><u_n'|f>|²
+2e(Re(<u_n|f><u_n'|c>)Re(<u_n|c><u_n'|f>)+Im(<u_n|f><u_n'|c>)Im(<u_n|c><u_n'|f>)).
(interference terms)

Indistinguishability causes "interference terms" to appear when we calculate physical predictions.

If |u_n>=|u_n'>, b_n=b_n', then