851HW7_09Solutions

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

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 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 | ( x,t ) | 2 (1) Differentiate (1) and use the definition of the probability current, j = i planckover2pi1 2 m ( d dx d dx ) , 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 ( x,t ). We start by differentiating Eq. (1), d dt P ( t | a,b ) = integraldisplay b a dx parenleftbigg ( x,t ) d dt ( x,t ) + ( x,t ) d dt ( x,t ) parenrightbigg (3) with d dt ( x,t ) = i planckover2pi1 2 M d 2 dx 2 ( x,t ) V ( x ) planckover2pi1 ( x,t ) this gives d dt P ( t | a,b ) = i planckover2pi1 2 M integraldisplay b z dx parenleftbigg ( x,t ) d 2 dx 2 ( x,t ) ( x,t ) d 2 dx 2 ( x,t ) parenrightbigg (4) Integrating by parts gives d dt P ( t | a,b ) = i planckover2pi1 2 M bracketleftBigg ( x,t ) d dx ( x,t ) vextendsingle vextendsingle vextendsingle vextendsingle b a integraldisplay b a dx d dx ( x,t ) d dx ( x,t ) ( x,t ) d dx ( x,t ) vextendsingle vextendsingle vextendsingle vextendsingle b a + integraldisplay b a dx d dx ( x,t ) d dx ( x,t ) bracketrightBigg = i planckover2pi1 2 M parenleftbigg ( x,t ) d dx ( x,t ) ( x,t ) d dx ( 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 P ( t | a,b ) = | ( x,t ) | 2 ( b a ) = ( x,t )( b a ) (6) where x is taken to be the point onto which a and b converge. This gives d dt ( x,t )( b a ) = j ( a,t ) j ( b,t ) . (7) dividing both sides by b a gives d dt ( x,t ) = lim b a j ( b,t ) j ( a,t ) b a = d dx j ( x,t ) (8) 1 2. Bound-states of a delta-well: The inverted delta-potential is given by V ( x ) = g ( 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: b ( x ) = ce | x | , (11) and use the delta-function boundary conditions at x = 0 to determine , as well as the the energy, E b . You can then use normalization to determine c . What is ( X 2 ) for this bound-state?...
View Full Document

This note was uploaded on 12/04/2011 for the course PHY 7070 taught by Professor Smith during the Spring '11 term at Wisconsin.

Page1 / 10

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online