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Unformatted text preview: PHY4604 R. D. Field The Exchange Operator
Two Identical Particles in a OneDimensional Box:
particles in a box
Eαβ = ( E1 ) α + ( E 2 ) β = We see that for two h 2π 2 (α 2 + β 2 )
2mL2 where α = 1, 2, 3, ... and β = 1, 2, 3, .... and the wavefunctions are given by
2
ψ αβ ( x1 , x 2 ) = ψ α ( x1 )ψ β ( x 2 ) = sin(απx1 / L ) sin( βπ x 2 / L )
L The Hamiltonian is H op = 2
( p1 ) op 2m + 2
( p 2 ) op 2m and is symmetric under the interchange of the two particles (as it must be since the two particles are
indistinguishable).
Exchange Operator: The exchange operator is the operator that
interchanges the two particles (1↔2) as follows
2
Pexψ αβ ( x1 , x 2 ) = ψ αβ ( x 2 , x1 ) with Pex = 1 The eigenvalue of the Pex operator are +1 and 1 and the with eigenfunctions
given by
1
S
(ψ αβ ( x1 , x2 ) + ψ αβ ( x2 , x1 ) ) (α ≠ β symmetric under 1↔2)
ψ αβ ( x1 , x 2 ) =
2
1
S
ψ αα ( x1 , x 2 ) = (ψ αα ( x1 , x 2 ) + ψ αα ( x 2 , x1 ) ) (α = β symmetric under 1↔2)
2
1
A
(ψ αβ ( x1 , x2 ) −ψ αβ ( x2 , x1 ) ) (antisymmetric under 1↔2)
ψ αβ ( x1 , x 2 ) =
2
For identical particles the Hamiltonian is invariant under 1↔2 and hence
[Pex,Hop] = 0 and Pex is a constant of the motion (it is conserved). Also,
note that
Pexψ αβ ( x1 , x 2 ) = ψ αβ ( x 2 , x1 ) = ψ βα ( x1 , x 2 )
Mixed State: The following superposition is also a solution of
Schrödinger’s equation (for arbitrary angle θ):
mix
A
S
ψ αβ ( x1 , x2 ) = cos θψ αβ ( x1 , x2 ) + sin θψ αβ ( x1 , x2 ) . The probability density for finding one particle at x1 and the other at x2 is ( ∗ ) mix
A
S
A
S
ρ αβ ( x1 , x 2 ) = cos 2 θ  ψ αβ  2 + sin 2 θ  ψ αβ 2 + sin( 2θ ) Re ψ αβψ αβ . It appears that we have lost our ability to predict the probability density
since it depends on the arbitrary angle θ! Department of Physics Chapter5_2.doc University of Florida ...
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 Spring '07
 FIELDS
 mechanics

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