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# Chapter5_all - PHY4604 R D Field Two Particles in a Box One...

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PHY4604 R. D. Field Department of Physics Chapter5_1.doc University of Florida Two Particles in a Box One Particle in a One-Dimensional Box: For one particle in a box ( V(x) = x 0, V(x) = x L and V(x) = 0 0 < x < L ) we found h / ) ( ) , ( t iE n n n e x t x = Ψ ψ with ) / sin( 2 ) ( L x n L x n π = . The energy levels are given by 0 2 2 2 2 2 2 E n mL n E n = = h where 2 2 2 0 2 mL E h = . Two Particles in a One-Dimensional Box: For two ( non-interacting ) particles we look for a solution of the form h h / 2 1 / 2 1 2 1 ) ( ) ( ) , ( ) , , ( iEt iEt e x x e x x t x x = = Ψ with E m p m p x x = + 2 ) ( 2 ) ( 2 2 2 1 . Thus, ) , ( ) , ( 2 ) , ( 2 2 1 2 2 2 1 2 2 2 1 2 1 2 2 x x E dx x x d m dx x x d m = h h ) ( ) ( ) ( ) ( 2 ) ( ) ( 2 2 1 2 2 2 2 1 2 2 1 1 2 2 2 x x E dx x d x m dx x d x m = h h E dx x d x m dx x d x m = 2 2 2 2 2 2 2 1 1 2 1 2 ) ( ) ( 1 2 ) ( ) ( 1 2 h h Hence, E = E 1 + E 2 and ) ( ) ( 2 1 1 2 1 1 2 2 x E dx x d m = h ) ( ) ( 2 2 2 2 2 2 2 2 x E dx x d m = h The total energy is therefore given by 2 2 2 2 2 2 1 2 ) ( ) ( ) ( mL E E E β α αβ + = + = h 2 2 2 2 1 2 ) ( mL E h = 2 2 2 2 2 2 ) ( mL E h = where α = 1, 2, 3, . .. and β = 1, 2, 3, . .. and ) / sin( ) / sin( 2 ) ( ) ( ) , ( 2 1 2 1 2 1 L x L x L x x x x βπ απ = = ) / ( sin ) / ( sin 4 ) ( ) ( ) ( ) ( ) , ( 2 2 1 2 2 2 1 2 * 1 * 2 1 L x L x L x x x x x x ρ = = 1 ) , ( 00 2 1 2 1 = ∫∫ LL dx dx x x 0 L Two Particles in a Box ψ (x 1 ) ψ (x 2 )

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PHY4604 R. D. Field Department of Physics Chapter5_2.doc University of Florida The Exchange Operator Two Identical Particles in a One-Dimensional Box: We see that for two particles in a box 2 2 2 2 2 2 1 2 ) ( ) ( ) ( mL E E E β α π αβ + = + = h where α = 1, 2, 3, . .. and β = 1, 2, 3, . .. . and the wavefunctions are given by ) / sin( ) / sin( 2 ) ( ) ( ) , ( 2 1 2 1 2 1 L x L x L x x x x βπ απ ψ = = The Hamiltonian is m p m p H op op op 2 ) ( 2 ) ( 2 2 2 1 + = 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 ) , ( ) , ( 1 2 2 1 x x x x P ex = with 1 2 = ex P The eigenvalue of the P ex operator are +1 and -1 and the with eigenfunctions given by () ) , ( ) , ( 2 1 ) , ( 1 2 2 1 2 1 x x x x x x S + = ( α β symmetric under 1 2 ) ) , ( ) , ( 2 1 ) , ( 1 2 2 1 2 1 x x x x x x S αα + = ( α = β symmetric under 1 2 ) ) , ( ) , ( 2 1 ) , ( 1 2 2 1 2 1 x x x x x x A = ( antisymmetric under 1 2 ) For identical particles the Hamiltonian is invariant under 1 2 and hence [P ex ,H op ] = 0 and P ex is a constant of the motion ( it is conserved ). Also, note that ) , ( ) , ( ) , ( 2 1 1 2 2 1 x x x x x x P ex βα = = Mixed State: The following superposition is also a solution of Schrödinger’s equation (for arbitrary angle θ ): ) , ( sin ) , ( cos ) , ( 2 1 2 1 2 1 x x x x x x S A mix θψ + = . The probability density for finding one particle at x 1 and the other at x 2 is ( ) + + = S A S A mix x x θ ρ Re ) 2 sin( | | sin | | cos ) , ( 2 2 2 2 2 1 . It appears that we have lost our ability to predict the probability density since it depends on the arbitrary angle θ !
PHY4604 R. D. Field Department of Physics Chapter5_3.doc University of Florida Spin and Statistics Mixed State: We see that the probability density for finding one particle at x 1 and the other at x 2 is ( ) + + = S A S A mix x x αβ ψ θ ρ Re ) 2 sin( | | sin | | cos ) , , ( 2 2 2 2 2 1 .

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## This note was uploaded on 04/19/2008 for the course PHY 4604 taught by Professor Field during the Spring '07 term at University of Florida.

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Chapter5_all - PHY4604 R D Field Two Particles in a Box One...

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