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Mr. A. Square Unbound Continuum States in 1-D Quantum Mechanics With 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 removed The 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 y(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: x a x a a a The Free Particle If V(x)=0 then the TDSE reduces to 2 ( x, t ) ( x, t ) 2m x 2 i t 2 2 ( x, t ) y ( x ) e Now the TISE: iEt 2y 2mE y 2 2 x 2y 2 k y 2 x 2mE where k 2 2 2y k 2y 0 's solution is sinusoidal so 2 x y A(k )eikx B (k )e ikx ( x, t ) y ( x ) e iEt ( x, t ) A(k )ei kx t B(k )e i kx t E where Assume k>0 & real, and B(k)=0, then describes a wave moving from –x to +x What about p ? p a p a a a * y py dx * y y dx p * y i x y dx y y dx * y * p * y y dx p k ky dx y y dx * k * y y dx Obviously, <p2>=2k2 So Dp=<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 Law Assume k<0 & real, and A(k)=0, then describes a wave moving from +x to -x What about p ? p a p a a a y * py dx * y y dx p y dx i x * y * y y dx y * p y y dx * p k ky dx y y dx * k Obviously, <p2>=2k2 y y dx * So Dp=<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 Law Obviously eikx represents a particle moving from right to left e-ikx represents a particle moving from left to right The 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 ( x, t ) A(k )ei ( kx t ) dk integral: Note that this is the inverse Fourier transform A complication arises in that is not really independent of k 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 k ( k ) ( k0 ) ( k k0 ) k k0 2 1 2 ( k k0 ) 2 k 2 k0 The 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/dk The phase velocity v 2phase v 2phase 2y 2 2 2 d 2 x x 2 2y y 2 2 2t 2 2 2 y dt y t k y k x 2 E2 2 E 2mE 2m 2 1 Classically, E= mvc2 2 2E E vc2 4 4v 2phase m 2m vc 2v phase 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 speed The Group Velocity k 2 2mE 2 2m k2 2m k d dk 2m d k vgroup dk m 2 2 k 2E 2 vgroup 2 vc2 m m 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 relation The Step Potential V(x)=V0 Region 1 Region 2 x=0 V0 x>0 V ( x) 0 x<0 Region 1 y 1 2mE 2 y k 1 1y1 2 2 x So 2 y 1 Aeik x Beik x 1 1 “A” is the amplitude of the incident wave “B” is the amplitude of the reflected wave Region 2 2y 2 2m E V0 2 y k y2 2 2 2 2 x So y 1 Ceik x 2 “C” is the amplitude of the transmitted wave Matching Boundary Conditions Condition 1: y 1 (0) y 2 (0) A B C Condition 2: y 1 '(0) y 2 '(0) ik1 A ik1 B ik2C 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 wave Applying some algebra A B C 1 B C A A ik1 A ik1 B ik2C k1 k1 B C k2 A A B B k2 1 A A B k1 k2 k1 k2 A B k1 k2 A k1 k2 If E>V0 then E-V0>0 or “+” k1 k1 2k1 C B k k k k 1 1 2 1 2 A A k1 k2 k1 k2 k1 k2 Then k2 is real and y2 is an oscillator propagation If E<V0 Classically, the particle is repelled In QM, k2 is imaginary and y2 describes an attenuating wave Graphically V(x)=V0 Region 1 Region 2 x=0 If E>V0 then E-V0>0 or “+” If E<V0 V(x)=V0 Region 1 Region 2 x=0 Then k2 is real and y2 is an oscillator propagation Classically, the particle is repelled In QM, k2 is imaginary and y2 describes an attenuating wave Reflection and Transmission Coefficients Recall * y y * J y y 2mi x x Define 3 currents, J A , J B , J C k1 2 A m k 2 JB 1 B m k 2 J C 2 C Re(k2 )e 2Im( k2 x ) m Define JA R JB JA T JC JA JB k1 k2 B R JA A k1 k2 2 2 2 J C Re(k2 ) C 2Im( k2 x ) T e JA k1 A If k2 is imaginary, T=0 If k2 is real, then k2 C T k1 A 2 In terms of Energy, If E>V0 then If E<V0 then R=1 and T=0 T 4k1k2 k1 k2 2 E 4 V0 1 2 E 1 V0 12 E E 1 V0 V0 2 The Step Potential V(x)=V0 Region 1 Region 3 Region 2 x=0 0 x<0 V ( x) V0 0<x<a 0 x>a x=a The Wave Function e Re ikx ikx y Ae Be Teik0 x ik0 x ik0 x k 2 0 k 2 2mE 2 2m E V0 2 Boundary Conditions a) y 1 (0) y 2 (0) b) y 2 (a) y 3 (a) c) y 1 '(0) y 2 '(0) d ) y 2 '(a) y 3 '(a) Apply Boundary Conditions a) 1 R A B b) Te ik0 a Ae ika Be ika c) ik0 1 R ik A B d ) ik0Te ik0 a ik Ae ika Be ika Solving Let V k 1 0 k0 E 1 sin ka R 1 sin ka 2i cos ka 2 2 Ae ika i 1 1 sin ka 2i cos ka 2 T e ika Be 2i 1 2 sin ka 2i cos ka ika i 1 1 sin ka 2i cos ka 2 Reflection and Transmission Coefficients R 2 T 1 1 2 2 1 2 sin 2 ka sin ka 2 cos 2 ka 2 2 2 2 2 2 2 2 sin ka 2 cos 2 ka 2 2 Some Consequences R 2 T 1 2 1 2 2 sin 2 ka 2 sin 2 ka 2 cos 2 ka 2 2 2 1 2 2 2 sin 2 ka 2 cos 2 ka 2 When ka=n*p, n=integer, implies T=1 and R=0 This happens because there are 2 edges where reflection occur and these components can add destructively Called “RamsauerTownsend” effect For E<V0 Classically, the particle must always be reflected QM says that there is a nonvanishing T In region 2, k is imaginary Since cos(iz)=cosh(z) sin(iz)=isinh(z) T 2 2 1 2 2 2 sinh 2 ka 2 cosh 2 ka 2 Since cosh2z-sinh2z=1 T cannot be unity so there is no RamsauerTownsend effect What happens if the barrier height is high and the length is long? Consequence: T is very small; barrier is nearly opaque. What if V0<0? Then the problem reduces to the finite box Poles (or infinities) in T correspond to discrete states An Alternate Method We could have skipped over the Mr. A Square Bound and gone straight to Mr. A Square Unbound. We would identify poles in the scattering amplitude as bound states. This approach is difficult to carry out in practice The Dirac Delta Potential The delta barrier can either be treated as a bound state problem or considered as a scattering problem. The potential is given by V(x)=-ad(x-x0) x=x0 Region 1 Region 2 Wavefunctions and Boundary Conditions y 1 Ae Be ikx y 2 Ce y 1 (0) y 2 (0) y 1 '(0) y 2 '(0) ikx ikx From the previous lecture, the discontinuity at the singularity is given by: Dy ' 2ma 2 y (x ) 0 Applying the boundary conditions Aeikx0 Be ikx0 Ceikx0 ikCeikx0 (ikAeikx0 ikBe ikx0 ) 2ma 2 Ceikx0 Elimination of B is straight forward and C ik 2 A ik 2 ma B ma 2 ikx0 e A ik 2 ma 2 C k2 4 T 2 4 A k m 2a 2 B m 2a 2 R 2 4 A k m 2a 2 2 R cannot vanish or only vanishes if k is very large so there is always some reflection Solving for k and E C ik 2 B ma 2 ikx0 e A ik 2 ma A ik 2 ma Both of these quantities become infinite if the divisor goes to zero ik 2 ma 0 k k2 m 2a 2 4 2mE 2 ma i 2 ma 2 E 2 2 This is in agreement with the result of the previous section. If a is negative, then the spike is repulsive and there are no bound states A Matrix Approach to Scattering Consider a general, localized scattering problem V(x) Region 1 Region 2 Region 3 Wavefunctions Region 1 y ( x) Ae Be ikx ikx Region 3 y ( x) Fe Ge ikx ikx (k 2mE (k 2mE 2 2 2 2 ) ) Region 2 y ( x) C f ( x) D g ( x) where f(x) and g(x) are two linearly independent functions which depend on the potential. Boundary Conditions There are four boundary conditions in this problem and we can use them to solve for “B” and “F” in terms of “A” and “G”. B=S11A+S12G F=S21A+S22G Sij are the various coefficients which depend on k. They seem to form a 2 x 2 matrix S11 S S21 S12 S22 Called the scattering matrix (s-matrix for short) Consequences The case of scattering from the left, G=0 so RL=|S11|2 and TL=|S21|2 The case of scattering from the right, F=0 so RR=|S22|2 and TR=|S12|2 The S-matrix tells you everything that you need to know about scattering from a localized potential. It also contains information about the bound states If you have the S-matrix and you want to locate bound states, let kik and look for the energies where the S-matrix blows up.