851HW7_09Solutions

# 851HW7_09Solutions - PHYS851 Quantum Mechanics I Fall 2009...

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PHYS851 Quantum Mechanics I, Fall 2009 HOMEWORK ASSIGNMENT 7 1. The continuity equation: The probability that a particle of mass m lies on the interval [ a,b ] at time t is P ( t | a,b ) = integraldisplay b a dx | ψ ( x,t ) | 2 (1) Differentiate (1) and use the definition of the probability current, j = i planckover2pi1 2 m ( ψ d dx ψ ψ d dx ψ ) , to show that d dt P ( t | a,b ) = j ( a,t ) j ( b,t ) . (2) Next, take the limit as b a 0 of both (1) and (2), and combine the results to derive the continuity equation: d dx j ( x,t ) = d dt ρ ( x,t ). We start by differentiating Eq. (1), d dt P ( t | a,b ) = integraldisplay b a dx parenleftbigg ψ ( x,t ) d dt ψ ( x,t ) + ψ ( x,t ) d dt ψ ( x,t ) parenrightbigg (3) with d dt ψ ( x,t ) = i planckover2pi1 2 M d 2 dx 2 ψ ( x,t ) V ( x ) planckover2pi1 ψ ( x,t ) this gives d dt P ( t | a,b ) = i planckover2pi1 2 M integraldisplay b z dx parenleftbigg ψ ( x,t ) d 2 dx 2 ψ ( x,t ) ψ ( x,t ) d 2 dx 2 ψ ( x,t ) parenrightbigg (4) Integrating by parts gives d dt P ( t | a,b ) = i planckover2pi1 2 M bracketleftBigg ψ ( x,t ) d dx ψ ( x,t ) vextendsingle vextendsingle vextendsingle vextendsingle b a integraldisplay b a dx d dx ψ ( x,t ) d dx ψ ( x,t ) ψ ( x,t ) d dx ψ ( x,t ) vextendsingle vextendsingle vextendsingle vextendsingle b a + integraldisplay b a dx d dx ψ ( x,t ) d dx ψ ( x,t ) bracketrightBigg = i planckover2pi1 2 M parenleftbigg ψ ( x,t ) d dx ψ ( x,t ) ψ ( x,t ) d dx ψ ( x,t ) parenrightbiggvextendsingle vextendsingle vextendsingle vextendsingle b a = j ( a,t ) j ( b,t ) (5) Taking the limit as b a 0, we can then write the integral (area) as simply the height times the width: lim b a 0 P ( t | a,b ) = | ψ ( x,t ) | 2 ( b a ) = ρ ( x,t )( b a ) (6) where x is taken to be the point onto which a and b converge. This gives d dt ρ ( x,t )( b a ) = j ( a,t ) j ( b,t ) . (7) dividing both sides by b a gives d dt ρ ( x,t ) = lim b a 0 j ( b,t ) j ( a,t ) b a = d dx j ( x,t ) (8) 1

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2. Bound-states of a delta-well: The inverted delta-potential is given by V ( x ) = ( x ) , (9) where g> 0. For a particle of mass m , this potential supports a single bound-state for E = E b < 0. (a) Based on dimensional analysis, estimate the energy, E b , using the only available parameters, planckover2pi1 , m , and g . The only energy scale we can form from g , planckover2pi1 and M is E = mg 2 planckover2pi1 2 = planckover2pi1 2 Ma 2 , where a = planckover2pi1 2 mg is the ‘scattering length’. Thus we should expect the answer to be E b ∼ − planckover2pi1 2 Ma 2 (10) (b) Assume a solution of the form: ψ b ( x ) = ce | x | λ , (11) and use the delta-function boundary conditions at x = 0 to determine λ , as well as the the energy, E b . You can then use normalization to determine c . What is ( X 2 ) for this bound-state? Calling x< 0 region 1, and x> 0 region 2, we have ψ 1 ( x ) = ce x λ (12) ψ 2 ( x ) = ce x λ (13) The first boundary condition is ψ 2 (0) = ψ 1 (0) , (14) which is satisfied by construction. Integrating the energy eigenvalue equation from x = ǫ to x = ǫ gives integraldisplay ǫ ǫ dxEψ ( x ) = planckover2pi1 2 2 M
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