Many-particle wavefunctions: Before we reflect on the problem-statement, consider some math-preliminaries, namely, the notation for a determinant vs. that of a permanent, and their properties,
j = N ± + → Ψ + = A N ∑ p∈S ∏ψ ν (rp ( j ) ) j = N ψ ν (r1 ) ψν (r2 ) ..... ψ ν (r N ) j =1 ± ˆ Ψ ± = A N S± ∏ψ ν (r j ) = AN ± = (1.1) ⋮ ⋮ ⋱ ⋮ j = N j =1 ± − → Ψ = A p r ψ ( ) s i g n − N ∑ p∈S ∏ ν p( j) ψν (r1 ) ψ ν (r1 ) ..... ψ ν (r N ) = 1 j ± S N = [group of the the N! perm permut utat atio ions ns p on set set of N coor coordi dina nate tes] s];; sign sign(( p) ≡ [sig [sign n of the the perm permut utat atio ion n p] = ±1 ψν (r1 ) ψν (r2 )
ψ ν (r N )
.. ...
1
1
1
2
2
2
N
N
N
N
j
N
j
j
In which: the normalization factor, separately for fermions (occupation # either 0 or 1, and 0! = 1 = 1!) is,
∏ =
ν ±′
±
A N
nν ±′ !
+ → A+ = N = − → A N − =
n1± ! n2± !...n∞± !
=
N !
N !
+ n1+ !n2+ !...n∞ ! N !
=
n1 !n2 !...n∞ ! N !
=
0!1!0!0!1!0!...1!0! N !
(1.2)
1 N !
(a) consider an N-boson wave-function (e.g., represented as a Slater “permanent”),
ψ
B ( n1 ...n∞ )
n1 ! n2 !...n∞ !
(x1...x N ) =
N !
N
∑ P∏ φ j =1
P
( x j ); P = [ permutation operator over states {λ1 ...λN }];
λ j
(1.3)
Verify that this (1.3) is invariant under interchange of particles, i i′ . The effect of the permutation operator
is to construct a permutation of the states {λ1...λ j ...λ N } amidst which the N particles at positions {x1...x j ...x N } are partitioned. The sum
∑
P
( )
is defined to sum over all possible permutations. For bosons, both even and
odd permutations are positive in sense (in general, they bear the same phase, eiφ ). Thus: the (1.3), with abbreviations N 0 ≡
x
and φλ j ( x j ) = φ λ j j , has the appearance,
n1 !n2 !...n ! N ! ∞
ψ B( n ...n∞ ) (x1...x N ) = ψ ((nx ......nx∞ )) = N 0 (... + φλx ...φλ x ...φλx′′ ...φλx + ... + φλ x ...φλ x′ ...φλ x ′ ...φλ x + ...) 1
1
1
N
1
i
1
i
i
1
N
i
i
1
N
i
i
i
x
x
N
N
(1.4)
That is, if there is a term in the wavefunction-product containing permutation ...φλ ii ...φ λ i′i′ ... , there must also be a x
x
term with permutation ...φλ i′i ...φ λ ii′ ... , due to there being all possible permutations. Now: suppose there was an operator Π λ ii′ that exchanged particles’ λ -states, i i′ . Apply this to (1.4),
Π ii′ψ ((nx ......nx∞ ) ) = N 0 (... + Π ii′ (φλx ...φλ x ...φλx′ ′ ...φλx ) + ... + Π ii′ (φλx ...φλ x′ ...φλ x ′ ...φλx ) + ...) 1
1
N
1
i
1
i
i
1
N
i
i
1
N
i
i
i
N
N
= N 0 (... + (φλx ...φλx′ ...φλ x ′ ...φλx ) + ... + (φλ x ...φλ x ...φλ x′′ ...φλ x ) + ...) i
1
1
i
i
i
N
i
1
1
N
i
i
(1.5)
N
i
N
Π ii′ψ ((nx ......nx∞ )) = N 0 (... + φλx ...φλx ...φλx′′ ...φλ x + ... + φλx ...φλ x′ ...φλ x ′ ...φλ x + ...) = ψ ((nx ......nx∞ )) 1
1
N
1
1
i
i
i
i
N
N
1
1
i
i
i
i
N
1
N
1
N
N.B.: Fact: the interchange-operator changes even permutations to odd permutations, and odd permutations to even permutations. For fermions, even and odd permutations bear opposite signs. Thus, by the same λ machinations as above, the fermion-wavefunction is anti-invariant (it suffers a phase eiπ ) under action of Π ii′ ,
because the + signs throughout the sum in (1.5) would be replaced by (±)
sgn P
suffer an arbitrary phase eiφ under interchange called “anyons”.
. Finally: there are particles which
∏
Check the orthonormality condition: ψ ( n1 ...n∞ ) | ψ ( n1′ ...n∞′ ) = B
B
∞ i =1
δ n n′ . Quick preliminary: one needs to be aware i i
1
of multinomial coefficients, the general formula for which is, "n" elements to be divided into "r" elements = particles n! → Ω = Ω = ( , ,..., ) ; n n n r 1 distinct piles of sizes n , n ,..., n piles = quantum states ; (1.6) ! !... ! n n n 1 2 r r 1 2 If we consider the “number of piles” to be the number of terms in a summation, and “number of elements” to be the power that summation is raised to, we get the formula, n n n n n n n! (1.7) ( x1 + x2 + ... + xr ) 2 = (n ) x1 1 x2 2 ... xr r = x 1 x2 2 ...xr r ; n + n +...+ n = 2 n1 ,..., nr n + n + ...+ n = 2 n1 !n2 !... nr ! 1
∑
1
2
∑
r
1
2
r
*
Building the inner product from (1.3), then use the formula (1.7) to carry out the bracketed product [ ] [ ] of 2 (1.8), in which only “direct” (rather than cross) terms survive because of orthonormality ,
ψ |ψ ′ = ψ | 11...1 | ψ ′ =
B ( n1 ... n∞ )
N ! n1 ! n2 !...n∞ ! N !
N !
∑
N
P ′, P
N
∫ [∑ P′∏ φ P′
(... + 0 + 0 +
x1 x2 ...x∞ |ψ (Bn1′ ...n∞′ ) d 3 x1 ...d 3 x∞
| x1 x2 ...x∞
n1 ! n2 !...n∞ ! n1 ! n2 !...n∞ !
= ψ |ψ ′ =
∫ψ
N , N
∏
j , j ′=1
j ′ =1
*
λ j ′
( x j′ )] [
∑ P∏ φ j =1
P
( x j )]d 3 x1...d 3 x∞
n1 !n2 !...n∞ !
δ λ ′ ,λ ′ δ j, j ′ + 0 + 0 + ....) = j
λ j
N !
j
(1.8)
N , N
∑ ∏δ (1)
P ′,P
j , j′=1
λ j′ ,λ ′j
δ j , j′
For an N particle system able to be piled into distinct piles of sizes n1 , n2 ,..., n∞ , we have P = P( N , n∞ ) , and the
sum
∑
P , P′
ψ |ψ ′ =
(1) equals the number of distinct permutations, given by (1.6), n !n N !...! n ! , and we get, 1
n1 ! n2 !...n∞ ! N , N
∏δ
N !
j , j ′ =1
N!
δ
λ j′ ,λ ′j j , j ′
n1 ! n2 !...n∞ !
N , N
∏δ
=
j , j′ =1
λ j′ ,λ ′j
∞
2
λ j ≠ λ ′j = [state-index]
N
∏δ
δ j , j′ =
j =1
;
λ j ,λ′j
j = j ′ = [occupation-number index]
(b) Given an N-fermion wave function (e.g., represented as Slater determinant) , φλ1 (x1 ) ... φ λ N ( x1 ) 1 ⋮ ⋱ ⋮ ψ (F n1 ,...,n∞ ) (x1 ,..., x ∞ ) = ; N ! φλ N (x1 ) ... φ λ N ( x N )
; (1.9)
(1.10)
Verify that ψ (F n1 ,..., n∞ ) changes sign under i j . Following the reasoning after (1.5),
Π ii′ψ ((nx ......nx∞ ) ) = N 0 (... ± Π ii′ (φλx ...φλx ...φλx′′ ...φλx ) + ... ∓ Π ii′ (φλx ...φλ x′ ...φλ x ′ ...φλx ) + ...) 1
1
N
1
i
1
i
i
1
N
i
i
1
N
i
i
N
i
N
= N 0 (... ± (φλ x ...φλx′ ...φλ x ′ ...φλx ) + ... ∓ (φλx ...φλ x ...φλ x′′ ...φλ x ) + ...) 1
i
1
i
i
1
N
i
i
1
N
i
i
N
i
N
(1.11)
= N 0 (... ∓ φλx ...φλx ...φλx′′ ...φλx + ... ± φλ x ...φλ x′ ...φλ x ′ ...φλ x + ...) 1
i
1
i
i
1
N
i
1
N
i
i
i
N
i
N
Π ii′ψ ((nx ......nx∞ )) = − N 0 (... ± Π ii′ (φλ x ...φλ x ...φλ x′′ ...φλx ) + ... ∓ Πii′ (φλ x ...φλ x′ ...φλ x ′ ...φλ x ) + ...) = −ψ ((nx ......n x∞ )) 1
N
i
1
1
1
Verify orthonormality, ψ ( n1 ...n∞ ) | ψ ( n1′ ...n∞′ ) = F
F
i
i
∏
N
i
∞ i =1
N
i
1
1
i
i
i
N
N
1
N
1
δ n n′ . Fermions are easier to state-count, requiring none of the i i
machinery of multinomial coefficients,
1
Eight distinct pieces of food; choose three for breakfast, two for lunch, and three for dinner. How many ways to do that?
8! Ω = 3!2!3!
(c.f., Prof. Scott She ffield, math professor). “There is nothing so annoying as a good example.” ~Mark Twain. 2 Note: there is no overlap between (eigen)states with different indices. One does not lose generality due to the completeness of the set of eigenstates (any arbitrary state can be expanded in these).
"n" elements to be divided into "r" elements = particles N ! n n n N → Ω = Ω = = ( , ,..., ) !; r 1 distinct piles of sizes 0,1 only. piles ; (1.12) = quantum states 1! It’s easy to then write, in total analogy with (1.8),
∫ψ 1 ∫ [∑ P′∏φ N !
ψ |ψ ′ = ψ | 11...1 | ψ ′ = N !
ψ |ψ ′ =
1
P0′
0
j ′=1
λ j′
N
*
(x j ′ )] [
∑ (... + 0 + 0 + ∏ δ j , j ′=1
∑ P ∏ φ 0
P0
N , N
N ! P0 , P0′
x1 x2 ... x∞ | ψ (F n1′ ...n∞′ ) d 3 x1...d 3 x∞
| x1x2 ...x∞
N
1
=
F ( n1 ...n∞ )
λ j′ ,λ ′j
j =1
λ j
(x j )]d 3 x1 ...d 3 x∞ ; (P0 , P0′) ≠ (P, P′ );
δ j , j′ + 0 + 0 + ....) =
1
N , N
∑ (1) ∏ δ
N ! P′, P
j , j′=1
λ j′ ,λ ′j
1
δ j , j′ =
(1.13) N
∏δ
N ! N !
j =1
λ j ,λ ′j
=
N
∏δ j =1
λ j ,λ ′j
(c) Write down explicitly the Slater determinant state for two fermions in the zero energy single particle states, F and one in the 2nd excited state. Without spin degeneracy: ψ 2010...0 ( x1 ,x 2 , x3 ) = 0 ; proof of this claim,
1
F Ψ 20100...0 (r1 r2 r3 ) =
=
φ0 (r1 ) φ 0 (r 1 ) φ 2 (r 1 )
3!
φ0 (r2 ) φ2 (r2 )
φ 0 (r 1 )
φ0 (r3 ) φ2 (r3 )
φ0 (r2 ) φ0 (r2 ) φ 2 ( r2 ) =
−
φ0 (r2 ) φ2 (r2 )
φ0 (r2 ) φ0 (r2 )
3!
φ0 (r3 ) φ0 (r3 ) φ 2 (r 3 )
φ 0 (r 1 ) ⋅ 0 + φ2 (r 1 ) (φ0 (r 2 )φ 0 (r 3 ) − φ0 (r3 )φ 0 ( r 2 ) ) 3!
+ φ 2 (r 1 ) φ0 (r3 ) φ2 (r3 ) φ0 (r3 ) φ0 (r3 )
(1.14)
= 0 + φ 2 (r 1 ) ⋅ 0 = 0 ← [ Pauli exclusion]
This (1.14) is an example/demonstration of the Pauli exclusion principle between already-occupied energy states. Contrast this: With spin degeneracy: a spin-space is “grafted onto” the state space (using “ ⊗ ”), and (1.14) is, φ0↑ (r1 ) φ0↓ ( r1 ) φ 2↑ ( r 1 ) 1 F↑ F↓ Ψ nF1...n∞ = Ψ nF1 ...n∞ (r1...rN ) = Ψ101... φ0↑ ( r2 ) φ0↓ ( r2 ) φ 2↑ ( r2 ) (r1r3 ) ⊗ Ψ 010... ( r2 ) = 3! φ0↑ (r3 ) φ0↓ (r3 ) φ 2↑ ( r 3 )
φ0↓ (r2 ) φ2↑ ( r2 ) φ ( r2 ) φ2↑ ( r2 ) φ (r2 ) φ 0↓ ( r 2 ) − φ 0↓ ( r 1 ) 0↑ + φ 2 (r 1 ) 0↑ φ0↑ (r1 ) φ0↓ (r3 ) φ2↑ ( r3 ) φ0↑ ( r3 ) φ2↑ ( r3 ) φ 0↑ (r3 ) φ 0↓ ( r 3 ) 6 1 φ0↑ (r1 )φ0↓ ( r2 )φ2↑ ( r3 ) − 1 φ0↓ ( r1 )φ0↑ ( r2 )φ2↑ ( r3 ) − 1 φ2 ( r1 )φ 0↑ ( r2 )φ0↓ ( r3 ) − = − + 6 φ0↑ (r1 )φ0↓ ( r3 )φ2↑ ( r2 ) 6 φ0↓ ( r1 )φ0↑ ( r3 )φ2↑ ( r2 ) 6 φ2 ( r1 )φ0↑ ( r3 )φ0↓ ( r 2 )
= Ψ nF ...n∞ 1
1
(1.15)
Spin ambiguity: We never said which particles were spin-up, nor spin down. If we were to effect a global change in spin on the state (1.15), the spin-reversed state would be related to the original state by a phase eiπ = −1 , as shown in (1.11). Verify the normalization of this three-fermion state. Fact: Cross terms of the direct product, of which there are 30, are overlap between particles in separate spin or energy states; they are zero,
Ψ
F n1 ...n∞
|Ψ
F n1 ...n∞
=
1 6
φ0↑ (r1 ) φ0↓ ( r1) φ2↑ ( r1 )
†
φ0↑ ( r1 ) φ0↓ ( r1) φ2↑ ( r1 )
φ0↑ (r2 ) φ0↓ (r2 ) φ2↑ (r2 ) φ0↑ ( r2 ) φ 0↓ ( r2 ) φ 2↑
30
6
i =1
i =1
∑ ( 0) + ∑ ( r ) = 2
φ0↑ (r3 ) φ0↓ ( r3 ) φ2↑ ( r3 ) φ0↑ (r3 ) φ0↓ ( r3 ) φ2↑ ( r 3 )
6
(1)
=
6 6
= 1 (1.16)
