Chapter5_2 - PHY4604 R. D. Field The Exchange Operator Two...

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Unformatted text preview: PHY4604 R. D. Field The Exchange Operator Two Identical Particles in a One-Dimensional 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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