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Unformatted text preview: HOMEWORK ASSIGNMENT 9: Solutions PHYS852 Quantum Mechanics II, Spring 2009 New topics covered: Scattering amplitude, cross-section . 1. Use the Lippman-Schwinger equation, | ψ ) = | ψ ) + GV | ψ ) , (1) to solve the one-dimensional problem of resonant tunneling through two delta-potentials. Take ψ ( x ) = e ikx and V ( x ) = g [ δ ( x ) + δ ( x − L )] . (2) a.) Express eq. (1) as an integral equation for ψ ( x ). Then solve this equation to find the general solution ψ ( x ) for arbitrary k and g . It might be helpful to express your answer in terms of the dimensionless parameter α = Mg/ ( planckover2pi1 2 k ). Answer: Project onto the state ( x | to get: ψ ( x ) = ψ ( x ) + ( x | GV | ψ ) . Insert the projector integraltext dx | x )( x | = 1 between G and V and use V | x ) = | x ) V ( x ) to arrive at ψ ( x ) = e ikx + integraldisplay dx ′ G ( x,x ′ ) V ( x ′ ) ψ ( x ′ ) . Insert the definitions of G ( x,x ′ ) and V ( x ) to get ψ ( x ) = e ikx − iα bracketleftBig e ik | x | ψ (0) + e ik | x − L | ψ ( L ) bracketrightBig , where α = Mg/ ( planckover2pi1 2 k ). Set x = 0 to find ψ (0) = 1 − iα bracketleftBig ψ (0) + e ikL ψ ( L ) bracketrightBig . Set x = L to get ψ ( L ) = e ikL − iα bracketleftBig e ikL ψ (0) + ψ ( L ) bracketrightBig . Solve these two equations for ψ (0) and ψ ( L ) to find ψ (0) = 1 + iα (1 − e 2 ikL ) 1 + 2 iα − α 2 (1 − e 2 ikL ) . ψ ( L ) = e ikL 1 + 2 iα − α 2 (1 − e 2 ikL ) . Plugging this into the equation for ψ ( x ) gives ψ ( x ) = e ikx − iα e ik ( L + | x − L | ) + e ik | x | ( 1 + iα (1 − e 2 ikL ) ) 1 + 2 iα − α 2 (1 − e 2 ikL ) . 1 b.) Compute the transmission probability T = | t | 2 as a function of k , defined via lim x →∞ ψ ( x ) = te ikx . Answer: For x > L this becomes ψ ( x ) = e ikx bracketleftbigg 1 − iα 2 + iα (1 − e 2 ikL ) 1 + 2 α − α 2 (1 − e 2 ikL ) bracketrightbigg . Which leads to t = 1 1 + 2 α − α 2 (1 − e 2 ikL ) . So that the transmission probability is T = | t | 2 = 1 1 + 2( α 2 + α 4 ) − 2( α 4 − α 2 )cos(2 kL ) + 4 α 3 sin(2 kL ) . c.) In the strong-scatterer limit α ≫ 1, at what k values is the transmission maximized? Answer: In the limit α ≫ 1, we can keep only the α 4 terms in the denominator, giving T = 1 1 + 4 α 4 sin 2 ( kL ) which has a maximum of T = 1 at k = nπ/L , where n is any integer. d.) Consider an infinite square-well V ( x ) = 0 from x = 0 to x = L , and V ( x ) = ∞ otherwise. What are the k-values associated with each bound state? Answer: The eigenstates would be φ n ( k ) = radicalbig 2 /L sin( nπx/L ). Thus each bound state has a k-value of k n = πn/L , which correspond to the k ’s of the transmission resonances....
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