Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
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 1 1 , 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 1 1 , r 2 2 ) ( r 2 2 , r 1 1 ) 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 1 1 , r 2 2 ) ( r 2 2 , r 1 1 ) 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 ( r 1 1 , r 2 2 ) 1 r , r , r , r , i1 1 1 i2 2 2 i1 2 2 i2 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 1 1 , r 2 2 ) ( r 2 2 , r 1 1 ) 4 Pauli principle fulfilled: for identical single particle quantum numbers i1 i2 ( r 1 1 , 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 Z G 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 1 3 2 3 e 2 3 2 ni i ni ni i ni 3 i i i i ZG e e n1 ,n2 ,n3 N 0 ( n1 , n2 , n3 ) N ni i independent summation Next we show Let’s first look at e nii 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 ni i i i n1 , n2 ni n n 2 n2 3 n3 e 2 2 3 3 e 1 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 ni i 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, ...