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Quantum Mechanics I

by: Quinn Larkin

Quantum Mechanics I PHY 851

Quinn Larkin
GPA 3.67


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This 3 page Class Notes was uploaded by Quinn Larkin on Saturday September 19, 2015. The Class Notes belongs to PHY 851 at Michigan State University taught by Staff in Fall. Since its upload, it has received 17 views. For similar materials see /class/207632/phy-851-michigan-state-university in Physics 2 at Michigan State University.

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Date Created: 09/19/15
Lecture 21 February 7 2001 Phase Shifts at Low Energy Scattering Lengths At low energy one can solve the Schrodinger equation for a given partial wave between T 0 and some point 7quot b where b is suf ciently large such that the potential is zero Although the norm of Hg is arbitrary the logarithmic derivative is completely determined by the potential the energy and the point 6 ngk a k b E i 1 A de r Erb We will consider k z 0 one can ignore the energy dependence of 04g For T gt b Rgk 7quot must have the form 1mm 0 mm eZi hgwn 2 which means that the logarithmic derivative at T 6 becomes 7quot h k7quot BZi h k7quot ET Mb 4 gt p A D bk 3 hkb e215hgkb Using the de nition of hg jg W allows one with some algebra to write the phase shift in terms of a jg and 77g Ng T 91 omgkb COW awrva makb 4 Thus by nding the logarithmic derivative at T b one determines the phase shift Note that if the potential is zero Rg would be proportional to jg and the denominator would diverge forcing the phase shift to zero We are now in a position to consider the behavior at low k where jg and 77g have the following behavior T 120197 gt 5 WW gt 6 where 25 1ll E 1 A 3 5 25 1 Inserting these into the expression for the phase shift above i 1 b k b l 04g z m ll lll cot 6gk kb 2i 1 2i g bad 6 I For low k the kinetic term in Schrodinger s equation is negligible compared to the potential and 04g approaches a constant Thus the momentum dependence of the phase shifts at low relative momentum is sin 6gk 0 km 8 Lecture 21 February 7 2001 One can see that all phase shifts tend to an integral multiple of 7r at k 0 and that the cross section is dominated by the s wave contribution at low energy In fact the scattering length a is de ned as the derivative of the i 0 phase shift at k U 3 a E ka50k ku 9 The cross section at very low energy is then 4 a z sin2ka 47mg 10 Example s wave scattering off a square well Consider the repulsive potential Vb T lt b l 1 i 0 rgtb 11 Find the scattering length and the cross section at k z 0 for a particle of mass m We need only consider the 3 wave in this case Using the de nition ukr E t HuUc one knows that the Schrodinger equation for ukr looks exactly like a onedimensional Schrodinger equation Furthermore the solution has the following form in the two regions 2 V E u1k r A sinh my If E 12 a 2mm 739 sink 5 k E 2725 13 a Matching logarithmic derivatives at the boundary gives 1 1 Etanh rib EtanUcbi 14 Solving for 6 for small k 1 k y 6 kb tan tanh Kb 19 1 z k b tanh Kb 16 The scattering length and cross section are thus 1 H a b Etanhrtb 1t 0k 0 47mg 18 Note that in the limit that V0 gt 00 that 6 gt kb 2 Lecture 21 February 7 2001 Levinson s Theorem As can be seen from the previous section all phase shifts begin life at multiples of 7r If the phase shift at k 0 were anything else an in nite cross section would result at small k As k gt 00 phase shifts all tend to zero This asymptotic behavior at large k can be understood by realizing that the in the highenergy limit the phase is changed by VAth where At is the time spent in the potential which goes to zero as the particle moves very quickly Levinson s theorem relates the phase shift at zero energy which is a multiple of 7r to the number of bound states U fVBTF 19 where N1 is the number of bound states of angular momentum i In order to explain the physical motivation of Levinson s theorem we digress to consider the density of states of particles in a large sphere of radius R which feels a short range potential V r with the origin being located at the center of the sphere The wave function at large T is 1M7quot gt 00 x sinUmquot 54 20 so that the boundary conditions restrict the possible values of k to CR Silk mr 21 Thus the density of states in momentum is E 12 1k 7r 7r NC The change of the density of states due to the nonzero potential is 17rd6dk Thus the number of extra states inserted between C 0 and k 00 due to the potential is 22 Y 6kU NVan 23 7r However the net number of states under consideration is not affected by the potential If ANcont states were pushed out of the continuum then they must have become bound states N1 ANcm 0 24 Combining this constraint with the Eq 17 and with the fact that Silk gt 00 0 gives Levinson s theorem Levinson s theorem is important as it gives one an idea of the general behavior to expect from phase shifts Attractive potentials tend to have positive phase shifts If no bound state exists the phase shifts rise near k 0 indicating that the states in the continuum were pulled down to k 0 Then at higher k the phase shifts fall indicating that the density of continuum states was depleted at higher k If a bound state exists the phase shift would generally start at 7r and usually fall as a function of k The falling phase shift denotes a negative correction to the density of states These are the states from which the bound state was formed


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