Quantum statistics of free particles Identical particles Two particles are said to be identical if all their intrinsic properties (e.g. mass, electrical charge, spin, color, . . . ) are exactly the same. Imagine: 2 identical classical objects We can label them because we can keep track of the trajectories 1 2 Heisenberg’s uncertainty principle prevents us from keeping track in qm identical quantum particles are indistinguishable Implications from indistinguishability Consider Hamilton operator H h1 h2 ,e.g., H 2 2m 2 1 2 2m 2 2 With corresponding Schroedinger eq. 2 2 2 2 2m 1 2m 2 ( r 1 , r 2 ) E ( r 1 , r 2 ) For the interaction free situation considered here spin variable look at solutions of hˆ i r , i i r , i labels set of quantum numbers of particular single particle eigenfunction (do not confuse with particle label) for basis functions appropriate to build up ( r 11 , r 2 2 ) Why is a simple product ansatz ( r 1 1 , r 2 2 ) i r1 , 1 i r 2 , 2 not appropriate? 1 2 If we conduct an experiment with indistinguishable particles a correct quantum description cannot allow anything which distinguishes between them. ( r 1 1 , r 2 2 ) i r1 , 1 i r 2 , 2 artificially distinguishes between the 2 particles 1 2 because indistinguishability requires however in general 2 ( r 11 , r 2 2 ) ( r 2 2 , r 11 ) i r1 , 1 i r 2 , 2 i r2 , 2 i r 1 ,1 2 1 2 1 2 2 2 Simple product ansatz introduces unphysical labels to indistinguishable particles What we need is a property like this 2 ( r 11, r 2 2 ) e ( r 2 2 , r 11 ) to fulfill ( r 11 , r 2 2 ) ( r 2 2 , r 11 ) i Nature picks to simple realizations for e ( r 2 2 , r 1 1 ) ( r 1 1 , r 2 2 ) ( r 2 2 , r 1 1 ) ei e0 1, ei 1 2 i bosons fermions These symmetry requirements regarding particle exchange are fulfilled by bosons 1 ( r 11 , r 2 2 ) i1 r1 , 1 i2 r 2 , 2 i1 r2 , 2 i2 r 1 ,1 2 fermions Let’s summarize properties of antisymmetry product ansatz for fermions ( r 1 1 , r 2 2 ) 1 i1 r1 , 1 i2 r 2 , 2 i1 r2 , 2 i2 r 1 , 1 2 1 Solves Schroedinger equations for non-interacting particles 2 Eigenenergies E are given by E i i 1 2 3 Antisymmetry of wave function ( r 11, r 2 2 ) ( r 2 2 , r 11 ) 4 Pauli principle fulfilled: for identical single particle quantum numbers i1 i2 ( r 11 , r 2 2 ) 0 2 identical fermions cannot occupy the same single particle state Antisymmetric wave function for N identical fermions Slater determinant i r 1 , 1 1 i r 2 , 2 ... i r N , N 1 1 1 i2 r 1 , 1 i2 r 2 , 2 ... i2 r N , N i1 ,i2 ,...,iN ( r 1 1 , r 2 2 ,..., r N N ) N! i N r , 1 1 i N r 2 , 2 ... iN r N , N Check N=2 1 i1 r 1 , 1 i1 r 2 , 2 i1 ,i2 ( r 1 1 , r 2 2 ) 2! i2 r 1 , 1 i2 r 2 , 2 1 i1 r 1 , 1 i2 r 2 , 2 i1 r 2 , 2 i2 r 1 , 1 2 Using occupation numbers to characterize N-particle states Let ni be the # indicating how often the single particle state i is occupied within the N-particle state described by fermions ni 0,1 only possibility in accordance with Pauli principle bosons ni 0,1, 2, 3, ... A few examples: N=2 particles bosons 1 2 r 1 , 1 3 r 2 , 2 2 r 2 , 2 3 r 1 , 1 2 fermions n1 0, n2 1, n3 1, n4 0, ... 1 3 r 1 , 1 3 r 2 , 2 3 r 2 , 2 3 r 1 , 1 2 n1 0, n2 0, n3 2, n4 0, ... E 2 3 1 2 r 1 , 1 3 r 2 , 2 2 r 2 , 2 3 r 1 , 1 2 n1 0, n2 1, n3 1, n4 0, ... E 2 3 E 2 3 Summary occupation number representation: 1 N ni i 2 E i ni E (n1 , n2 ,..., ni ,...) i 3 N-particle state characterized by set of occupation numbers of single particle states 4 fermions ni 0,1 bosons ni 0,1, 2, 3, ... (n1 , n2 ,..., ni ,...) i labels set of single particle quantum numbers Partition functions with occupation numbers Partition function of the canonical ensemble Z e E ( n1 ,n2 ,...,ni ,...) e E ( n1 ,n2 ,...,ni ,...) Partition function of the grandcanonical ensemble ZG e N N 0 ni i ni E ( n1 , n2 ,..., ni ,...) N i i Z (N ) e e N 0 ( n1 ,n2 ,...,ni ,...) N 0 ( n1 , n2 ,..., ni ,...) N ni N ni i i We use the grandcanonical ensemble to derive <ni> the average occupation of the single particle state i Let’s consider how the summation works for an example of N=0,1,2,3 fermions N=0 (0,0,0) meaning all single particle states are unoccupied N=1 (1,0,0) (0,1,0) (0,0,1) N=2 (1,1,0) (0,1,1) (1,0,1) N=3 (1,1,1) ni i ni 3 i i ZG e N 0 ( n1 , n2 , n3 ) N ni i 1 e 1 e e 1 2 2 e 1 2 3 3 2 e e 3 1 3 2 e 2 3 2 ni i ni ni i ni 3 i i i i ZG e e N 0 ( n1 , n2 , n3 ) n1 , n2 , n3 N ni i independent summation Next we show Let’s first look at e ni i i i ni over the ni n n n e 1 1 2 2 3 3 and do a summation over n1 e i ni i i ni 0 2 n2 3 n3 2 n2 3 n3 e 1 e 1 n1 n n 2 n2 3 n3 e 2 2 3 3 e 1 Now summation over n1 and n2 e nii ni i i n n 2 n2 3 n3 e 2 2 3 3 e 1 n1 , n2 n2 0 3 n3 2 0 3 n3 3 n3 2 3 n3 e 2 e 1 e 2 e 1 n 3 n3 3 n3 2 3 n3 e 3 3 e 1 e 2 e 1 And finally summation over n1 , n2 and n3 e nii ni i i n1 , n2 , n3 n 3 n3 3 n3 2 3 n3 e 3 3 e 1 e 2 e 1 n3 2 1 e 1 e 2 e 1 Compare with 3 3 2 3 e 3 e 1 e 2 e 1 ni i ni 3 i ZG e i N 0 ( n1 ,n2 ,n3 ) N ni i 1 e 1 e 2 e 3 e 1 2 2 e 1 3 2 e 2 3 2 e 1 2 3 3 ni i ni i i ZG e N 0 ( n1 , n2 ,..., ni ,...) N ni i e 1 n1 2 n2 e ... e e ni i i i ni n1 , n2 ,..., ni ,... i ni ... n1 , n2 ,..., ni ,... 1 n1 2 n2 i ni e ... e ... e n1 n2 ni ZG e i i ni ni Holds for fermions and bosons with the only obvious difference fermions ni 0,1 bosons ni 0,1, 2, 3, ...