Mr. A. Square UnboundWith Apologies to ShelleyThe consequencesNormalizing InfinityThe Free ParticleAssume k>0 & real, and B(k)=0, then Y describes a wave moving from –x to +xAssume k<0 & real, and A(k)=0, then Y describes a wave moving from +x to -xObviouslyThe Wave Packet as a solutionThe Wave Packet cont’dSlide 11The phase velocityThe Group VelocityThe Step PotentialRegion 1Region 2Matching Boundary ConditionsApplying some algebraGraphicallyReflection and Transmission CoefficientsIn terms of Energy,Slide 22The Wave FunctionBoundary ConditionsApply Boundary ConditionsSolvingSlide 27Some ConsequencesFor E<V0What happens if the barrier height is high and the length is long?An Alternate MethodThe Dirac Delta PotentialWavefunctions and Boundary ConditionsFrom the previous lecture, the discontinuity at the singularity is given by:Applying the boundary conditionsSolving for k and EA Matrix Approach to ScatteringWavefunctionsSlide 39ConsequencesMr. A. Square UnboundContinuum States in 1-D Quantum MechanicsWith Apologies to ShelleyIn the previous section, we assumedThat a particle exists in a 1-d spaceThat it experiences a real potential, V(x)That its wavefunction is a solution of the TISE or TDSEThat at infinity, its wavefunction is zero.In this section, those are removedThe consequencesIf the boundary condition at infinity is removed,Then a quantum system is not limited to a discrete set of states butA continuum of energies is allowed.Normalizing InfinityOne problem if (x)∞, how do you normalize it?Well, Postulate 7 (wherein we discuss normalization) is based on the proviso that it mainly applies to bound states.Mathematically, if we have to find a matrix element, we perform the following operation:a x axa a=The Free Particle2 22 222 222222222If V(x)=0 then the TDSE reduces to( , ) ( , )-2m i( , ) ( )Now the TISE:22where 0 's solution is sinusoidal so( ) ( )iEtikx ikxx t x tx tx t x emExkxmEkkxA k e B k eyyyyyyyy--�Y �Y=-� �Y = ��=-��=-�=�+ =�= +hh hhh( ) ( )( , ) ( )( , ) ( ) ( )where iEti kx t i kx tx t x ex t A k e B k eEw wyw-- - +Y = �Y = +=hhAssume k>0 & real, and B(k)=0, then  describes a wave moving from –x to +x( )***** ** *What about p ?p dxa p apa adxdxi xpdxk dx dxp kdx dxp ky yy yy yy yy y y yy y y y�- ��- ��- ��- �� �- � - �� �- � - �= =�� �� ��� �== ==����� �� �hhhhObviously, <p2>=2k2So p=<p2>-<p>2 =0There is no variance in momentum, thus the free particle has mixed momentumThis is in agreement with Newton’s 1st LawAssume k<0 & real, and A(k)=0, then  describes a wave moving from +x to -x( )***** ** *What about p ?p dxa p apa adxdxi xpdxk dx dxp kdx dxp ky yy yy yy yy y y yy y y y�- ��- ��- ��- �� �- � - �� �- � - �= =�� �� ��� �=-= =-=-����� �� �hhhhObviously, <p2>=2k2So p=<p2>-<p>2 =0There is no variance in momentum, thus the free particle has mixed momentumThis is in agreement with Newton’s 1st LawObviouslyeikx represents a particle moving from right to lefte-ikx represents a particle moving from left to rightThe Wave Packet as a solutionAnother solution to the TDSE is a “wave packet”As an example, let B(k)=0 and the solution is in the form of the integral:Note that this is the inverse Fourier transformA complication arises in that  is not really independent of k( )( , ) ( )i kx tx t A k e dkw�-- �Y =�The Wave Packet cont’dTypically, the form of A(k) is chosen to be a Gaussian We also assume that (k) can be expanded in a Taylor series about a specific value of k00220 0 021( ) ( ) ( ) ( )2kkk k k k k kk kw ww w� �= + - + - +� �KThe Wave Packet cont’dThe packet consists of “ripples” contained within an “envelope”“the phase velocity” is the velocity of the ripples“the group velocity” is the velocity of the envelopeIn the earlier expansion, the group velocity is d/dkThe phase velocity22 2 2 2 22222 2 2 2 22222222 2221Classically, E=224 422phasephasecc phasec phased x xtvdt t k kxEEvmEmmvE Ev vm mv vyy w y wyy y�� � -�= = = = =�� � -�= == = � ==hhSo the ripple travels at ½ the speed of the particleAlso, note if <p2>=2k2 then I can find a “quantum velocity”= <p2> /m22k2/m2= E/2m=vqSo vq is the phase velocity or the quantum mechanical wave function travels at the phase speedThe Group Velocity2222 22 222 2222groupgroup cmE mkkmkd dkmd kvdk mk Ev vm mwwww= ==== == = =h hhhhhThe group velocity (the velocity of the envelope) is velocity of the particle and is twice the ripple velocity.BTW the formula for  in terms of k is called the dispersion relationThe Step PotentialRegion 2x=0V(x)=V0Region 10 x>0( )0 x<0VV x�=��Region 11 12211 1 12 212Soik x ik xmEkxAe Beyy yy-�= =�= +h“A” is the amplitude of the incident wave“B” is the amplitude of the reflected waveRegion 2( )220222 2 22 212Soik xm E VkxCeyy yy-�= =�=h“C” is the amplitude of the transmitted waveMatching Boundary Conditions1 21 2 1 1 2Condition 1: (0) (0)Condition 2: '(0) '(0)A B Cik A ik B ik Cy yy y= � + == � - =The problem is that we have 2 equations and 3 unknowns.“A” is controlled by the experimenter so we will always solve ALL equations in terms of the amplitude of the incident waveApplying some algebra( )1 1 2 1 1 21 1 21 2 1 21 21 21 2 1 2 11 2 1 2 1 21121B CA B CA AB Cik A ik B ik C k k kA AB Bk k kA ABk k k kAk kBA k kk k k k kC BA A k k k k k k+ = � + =- = � - =� �- = +� �� �- = +-=++ -= + = + =+ + +If E>V0 then E-V0>0 or “+” Then k2 is real and 2 is an oscillator propagationIf E<V0 Classically, the particle is repelledIn QM, k2 is imaginary and 2 describes an attenuating waveGraphicallyIf E>V0 then E-V0>0 or “+” Then k2 is real and 2 is an oscillator propagationIf E<V0 Classically, the particle is repelledIn QM, k2 is imaginary and 2 describes an attenuating waveRegion 2x=0V(x)=V0Region 1Region 2x=0V(x)=V0Region 1Reflection and Transmission Coefficients2**212122Im( )22Recall2Define 3 currents, , ,Re( )DefineA B CABk xCCBA AJmi x xJ J JkJ AmkJ BmkJ C k emJJR TJ Jy yy y-� �� �= -� �� �� �==-== =hr r rhhhIf k2 is imaginary, T=0If k2 is real, then221kCTk A=2221 21 222Im( )21Re( )BAk xCAJ