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Quant Chem & Stat Thermodyn I

by: Ladarius Rohan

Quant Chem & Stat Thermodyn I CEM 991

Marketplace > Michigan State University > Chemistry > CEM 991 > Quant Chem Stat Thermodyn I
Ladarius Rohan
GPA 3.66


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This 18 page Class Notes was uploaded by Ladarius Rohan on Saturday September 19, 2015. The Class Notes belongs to CEM 991 at Michigan State University taught by Staff in Fall. Since its upload, it has received 17 views. For similar materials see /class/207708/cem-991-michigan-state-university in Chemistry at Michigan State University.


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Date Created: 09/19/15
vivaWW1 a constant 3 at in nity P1 W2 0 bound states we must have the constant 0 and hence 215 1P1 9 2 rating once more we have In tpl 1n 1p2ln c ie 12 fl2 which39contradicts the ned linear independence of the two functions Consider two eigenfunctions 30x and tp1x with eigenvalues Equot lt EH1 Since pectrum is supposed to be nondegenerate all eigenfunctions can be taken to be real suitable choice of phase factors From the equations 2m 39 2 39l riI hTEnV39 n 0 wnlEul Vvnl 0 some simple calculation we obtain 9 2m 3 Vn39Pnl 1Pn139 nl F En1 E10 J 39Pn39Pn1 dx s take a and t to be two consecutive zeros of 1pquot n a 3 Then 5 2m 6 V Vnl FEu1 EnJ 1Pn1Pni dx in the interval a 8 0quot does not change sign suppose without loss of generality an gt 0 This means that 1502 gt Oand 7209 lt 0 It follows that Vquot 1x must change n the interval a 8 since otherwise the righthand side of the equality would have the sign as 1p1 while the lefthand side would have the opposite sign Hence between onsecutive zeros of 30quot 11 has at least one zero In this connection note also that functions which are even odd with respect to the re ection x gt x have an even number of zeros and that the ground state is always even Since Hx H x we have HWOC 13100 Hvx Billx 51 x and 1px are eigenfunctions of H with the same eigenvalue E We distinguish asec In this case x CM x and hence x izgvx that is the eigenl correSponding to nondegenerate energy levels are either even or odd Suppose 1 all the energy levels are nondegenerate Then if we write the energy eigenvalues if ing order of magnitude 5 x 83 lt E3 the corresponding eigenfunctions will increasing order of the number of their zeros the function corresponding to 13 22 1 zeros see problem 4 Since the even odd functions have an even odd m zeros it follows that the eigenfunctions will be alternately even and odd the gait wavefunetion being always even 2 The level E is degenerate The degeneracy being twofold the general solution of equation 51 can be it C xwxCnvx Awxwxl3vxv x AwxBx where AB C A B Cz Thus the two eigenfunctions having the same ei can be written in the form of a linear combination of two functions of wellde ne which are themselves ei engtnctions with the same eigenvalue uTi Ufl Y a No 393 Fox 4 ntycuyrgA UWS POML 6 Consrder a poten rel Vx avmg a nite discontinmty at x x0 Fig I the interval an d x9d by replacing Vx by the line segment shown in the W W m v xi m X FIG 1116 one obtains a continuous potential V x The Schrodinger equation then becomes wi X quot2h 5 1013 0 whence xod wixd wixd 2 j V1x Ewxdx xo d Xcid f5 g 4x 2 4 1139 a has all rend o Show 4 90 W M 77L We 6 1395 WMow M QM cum 7kg Wf zf V f x b aD WIf MOQJ 77 MM A 404 mi2Lde 40 A40 7191 V tg Loaf 400w Mm alumr Lfaf v CAVOCJ4 D 91 wr6 1lt MM Kevcwk 0 x0 mink we quot d K L x 1m of QM quotL gx thfof Wx x dx 39 a New dz7w fk r xbu lw39 0M 7amp1 a ma M417 at KO Xofd xa xo d V X ord k COMfo f J a flf Wx 750 ij FLOd k4 QCL quot90 6 090 9 COMLlltuouL 88 Chapter 6 Piecewise Constant Porenrials in One Dimension According to the theorems about Fourier series any function de ned between z 0 and x 2 L may be expanded in terms of these functions From 629 we obtain for it gtgt Art 277252 13 AB 2 En An 01 2n Art An rugEAII n We conclude that owing to the degeneracy the number of energy eigenstates per unit energy interval is approximately 2MM J 135 E E 1 hence is inversely proportional to the square root of the energy The number of states per unit momentum interval is given by 1 In statistical mechanics this last equation is somewhat loosely interpreted by saying The number of states per unit momentum interval and per unit length is lh Conversely a volume it in the phase space of x and p is ascribed to each state 4 The Potential Step Next in order of increasing complexity is the potential step We 2 Vomx as shown in Figure 63 There is no physically acceptable solution for E lt 0 because of the general theorem that E can never be less than the absolute minimum of V x Classically this is obvious But as the examples of the harmonic oscillator and the free particle have already shown us it is also true in quantum mechanics despite the possibility of penetration cask2215 L l I L 10 8 6 4 2 1 Figure 63 Eigenfunction for the potential Step Vex V017x corresponding to an energy E Veil Human e 4 The Potential Step 89 x Figure 64 Shape of the wave function in the nonclassical region into classically inaccessible regions We can prove the theorem by considering the real solutions of the Schrodinger equation see Exercise 47 W wquot Vr Elw 0 2t quot If Vx gt E for all 2 1 has the same sign as 1 Hence if y is positive at 39 some point x the wave function has one of the two shapes shown in Figure 64 depending on whether the slope is positive or negative In Figure 640 1 can never bend down to be nite as 2 gt 00 In Figure 64b 1p diverges as 1 00 Hence there must always be some region where E gt Va and where the particle could be found classically Now we consider the potential step with 0 lt E lt V0 Classically a particle of this energy if it were incident from the left would move freely until re ected at the potential step Conservation of energy requires it to turn around changing the sign of its momentum The Schrodinger equation has the solution ikz ikr Mm Ae Be x lt O 631 Ce x gt 0 Ilere hkJME nKway m The second linearly independent solution for x gt 0 e has been omitted because it is in con ict with the boundary condition that 1 remain nite as quotfquot 3 00 Since one of the two linearly independent solutions is excluded the quotquotQ stationary states for E lt V0 are nondegenerate5 3 By matching the wave function and39 its slope at the discontinuity of the potential 9 O we have ABC m mA mm A See Chapter 5 Footnote 4 wequot 90 39 Chapter 6 Piecewise Constant Potentiois in One Dimension or A g K e at real K 632 g Zik 1 8 A 16 K Substituting these values into 631 we obtain 2 tux2 E Ae cos k2 2 x lt 0 633 2Aeia 2 cos 9quot 2 gt 0 in agreement with the remark made in Section 5 of Chapter 4 that the wave function in the case of no degeneracy is real except for an arbitrary constant factor Hence a graph may be drawn of such a wave function Figure 63 The classical turning point st 0 is a point of in ection of the wave function The oscillatory and the exponential portions can be joined smoothly at 1 O for all values of E between 0 and 10 the energy spectrum is ontinuous W The solution 631 can be given a straightforward interpretation It eprcsents a plane wave incident from the left with an amplitude A and a ected wave which propagates toward the left with an amplitude B ccording to 632 MP 11313 hence the re ection is total Although i has a nite value in the region to the right of the potential step there is no permanent penetration A wave packet which is a superposition of eigenc functions 631 could be constructed to represent a particle incident from the left This packet would move classically heingre ected at the wall and giving again a vanishing probability of nding the particle in the region 01 positive x after the wave packet has receded 5 W These remarks can perhaps be better understood if we observe that for onedimensional motion the conservation of probability leads to particularlj transparent consequences For a stationary state 42 reduces to ajida 0 of y g s 3 0 Hence the current density 3 t merit 39i39 31 634 it quot answer ax LE 4 c has the same value at all points x j when calculated with the wave functio 4quot 633 is seen to vanish as it does for any e ll real wave functior g z w Wat Hence there is no net current anywhere at all To the left of the potential stc the relation Al2 Biz insures that incident and re ected probability currenl cancel one another If there is no current there is no net momentum i Lot c 0 aft weresz w gigging a e 1quot i 4 The Potential Step 3 1 31 91 Apgt state 63 In order to observe the particle in the exponential tail it must be localized within a distance of order A lily Hence its momentum must i i j39 he uncertain by a i we ogrerem A E A gt Ax 392 fur x 2V0 E M i The particle of energy E can thus be located in the neoclassical region only if it is given an energy V8 8 suf cient to raise it into the classically allowed region 4 The case of an in nitely high potential barrier V0 w 00 or K w 00 deserves special attention From 631 it follows that in this limiting case Ms W 0 in the region under the barrier no matter what value the coef cient C may have According to 632 the joining conditions for the wave function at x 0 now reduce formally to limgz 1 iim 0 xrooA K433 for A B m 0 and C O as Va w on These equations show that at a point where the potential makes an infinite jump the wave function must vanish 39Whereas its slope may jump discontinuously from a nite value Zz39kA to quotzero W We next examine the quantum mechanics of a particle which encounters Vt he potential step in one dimension with an energy E gt V0 Classically this f h particle passes the potential step with altered velocity but no change of will direction The particle could be incident either from the right or from the left The solutions of the Schrodinger equation are now oscillatory in both regions r ihence to each value of the energy correspond two linearly independent Aegenerate eigenfunctions For the physical interpretation their explicit in nstruction is best accomplished by specializing the general solution t citsth AikzB ik2 xlt0 quot to e e 635 I Wye Cew oeW a gt e W g V girhere 13k JZnE and his 2 xZntE 0 Two useful particular solutions ii are obtained by setting D 0 or A 0 The first of these represents a wave 5 ncident from the left Re ection occurs at the potential step but there is also transmission to the right The second particular solution represents incidence and transmission from right to left and re ection toward the right We consider here only the rst case D 0 The remaining constants are related by the condition for smooth joining at x 0 4 13 C and kA B klc The discontinuity oi the slope is not in con ict with the condition of smooth joining rived for nite jumps of the potential W58 92 Chapter 6 Piecewise Constant Potentiais in One Dimension from which we solve B k kx C 2k and A kk1 A kicx 636 The current density is again constant but its value is no longer zero Instead j tm ism xlt0 14 file i it2 x gt 0 u The equality of these values is assured by 636 We thus have LEE in 392 637 W2 k W2 In analogy to optics the rst term in this sum is called the re ection coe icieni the second is the transmission coe ciem We have zwzw 633 R W we EEK 4kkls k W a kot Equation 63 aSSures us that R T l R and T depend only on the ratio 8 V0 For a wave packet incident from the left the presence of re ection means that the wave packet may when it arrives at the potential step split into two parts provided that its average energy is close to V0 This splitting up of the wave packet is a distinctly neoclassical effect which affords an argument against the early attempts to interpret the wave function as measuring the matter or charge density of the particle which it accompanies For the splitting up of the wave packet would then imply a physical breakup of the particle and this would be very dif cult to reconcile with the facts of observa tion After all electrons and other particles are always found as complete entities with the same distinct properties On the other hand there is no contradiction between the splitting up of a wave packet and the probability interpretation of the wave function 639 Exercise 65 Show that the coef cients for re ection and transmission at a potential step are the same for a wave incident from the right as for a wave incident from the left 4 2 oneBody Emblems Without Spin Onobimensional Problems Problem 22 Scattering at a symmetric potential barrier A current of particles of energy E strikes a potential barrier Vx re stricted to the region a x 3 The potential is supposed to be an even function of x wnme an The amplitudes of backward and forward scattered waves shall be determined in terms of the logarithmic derivatives of the wave func tion at x in Solution The symmetry condition 221 has the important con sequence that for every energy E there exists a solution of the Schrodin ger equation of even parity 34 ulxluxl ux u3939XL 2242a and a solution of odd parity 9 u mM m m mmnpo emu Since both are of course linearly independent of one another the general solution will be any linear combination of them The two partial solutions 14 and a may be determined inside the interval ag x3 a in the worst case by numerical procedures starting from x0 with u01 u 00 and u00 u 0l in an arbitrary normalization of these two basic solutions Thus we can compute their logarithmic derivatives at xa which we shall write for convenience in a dimensionless form an tn11 a Lr nu aua L A 223 independent of their respective normalizations The logarithmic deriv atives au aui a at x a then follow from 222 to be LJr and L The solution corresponding to a wave of unit amplitude incident from the left can be written eikxBecth mwltxsa ux CuxC2uex axa 224 lFequotquot asxltoo 4 yWW 9601 eVIv a a 1x3gtc2 Czll Problem 22 Scattering at a symmetric potential barrier 43 Continuity of ux and u39x at x j a yields four conditions viz M a edema C1ua c2 ua 225 a ikequotquot Be C1u39aC2u a 225b 1 Fe C ua C2ua 225 c 1 a ikl Fe cuaC2u39a 225d The sum of 225a and 225c gives 2C1ua and the difference of 225 b and 225d 2Cua The ratio of both is e39 1FBe L 39k 226 39g e quotquotlFBequot a By the same procedure with opposite signs we nd ika l 3 Elm Lika e H F 6 226b Solving 226a b with respect to 1FiB and using the abbreviation ka q 227 we nally arrive at the amplitude relations 1 L i Li B 62quot q 4 228a 2 L iq L iq 1 L 39 L 39 HF e 2q q lq 228b 2 L zq L 1q According to the equation ofcontinuity we expect the sum of re ected and transmitted intensities to be equal to the incident one Indeed from 228 a b there follow the formulae lBt2 L 42 2293 LL q22q2L L 2 and 2 L L 2 1F2 q l 2291 L L q22q2L L239 They evidently satisfy the expected relation B21F21 2210 The problem of determining the scattering amplitudes in forward and backward directions has thus been reduced to nding the logarithmic W 44 UneBody Problems without Spin OneDimensional Problems derivatives 223 at xa of the even and the odd parity wave functions This cannot of course be accomplished as long as no special form of the potential 221 has been introduced In contradistinction to Problem 21 we no longer have 8 F Forward scattering prevails if giLr LigtiL L 42li backward scattering in the opposite case Problem 23 Re ection at a rectangular barrier The general formulae derived in Problem 22 shall be applied to a potential barrier with a WVzk in lxlga 23 and V 0 elsewhere The transmittance of the barrier shall be determined Solution Inside the barrier the Schrodinger equation becomes u k2 Iran 0 232 There are therefore solutions of different type for a kinetic energy below kltko and above kgtk0 threshold We begin with the rst case and write k kZLxz u x2u0 233 Then the even solution is nxcoshxx MAO l M39Ol0 2343 and the odd Solution mix 2 sinhxx u00 MAO l 3 234 b Hence I r auauaxatanhxa 2353 L au auaxacothxa 235b The transmittance of the barrier then follows from 229 b by elementary reshaping T211 th mm W3 mmmmmm M 236 Problem 23 Reflection at a rectangular barrier 45 whereas the reflectance according to 2210 is given by RslBl2 l T 237 In classical mechanics the whole flux arriving from the left side would be re ected at the barrier so that B2 1 and 1F20 This happens according to Eq 236 if and only if xa mo i e ifthere is a very great potential mountain above the energy level of the particles the trans parency of the barrier will become very small though still finite tunnel effectquot The transmittance of the barrier may then be written a pproximat ely l6k2x2 k3 its order of magnitude being mainly determined by the exponential factor a 4m 2 fax lZgw E will be generalized to this integral form for any potential Vx below cf Prob lem 116 e4xa 238 s The exponent If on the other hand the kinetic energy exceeds the height of the barrier the quantity x defined by 233 becomes imaginary With the abbreviation K2k2 k x2 239 we may then write instead of236 l 2 Wm T W 2310 17 1lt2k Kgt sin22Ka Though in classical mechanics there should be Tl and RO at these energies the transparency following from 2310 shows maxima of T l only at 2 K a n n n 12 3 Between these there are minima in the neighbourhood of 2Kan7r which lie the closer to Tl the smaller the factor in front of the sine in 2310 ie the higher the energy above threshold The general behaviour of T as a function of the energy in units of the threshold height say U is shown in Fig 5 where T has been drawn as a function of EU for the example 2k0a37t The wave function has been explained in Fig 6 where in2 has been drawn vs x On the right hand side of the barrier we simply have lul21F2 ie constant Munwhom 71 it L wm an 5m coI i WW g 0 NTIquot kiln Ne no M axillcb39h39 K30 46 OneBody Problems without Spin OneDimensional Problems 7 392 whereas on its left there is interfer nce of the incident with the re ected wave Fig 6 shows this feature for k3 2 k and different widths of the barrier The broader the latter the smaller is the intensity transmitted and the more pronounced the interference phenomena become b e 30 15 2 EU Fig5 Transmittanee of potential barrier for E gt U in dependence upon energy a5 Fig 6 Probability density 1113 in the current falling upon the barrier from the left in the ease ltU The two vertical lines mark the width 0 of the barrier The waves on the left are caused by interference between incident and re ected beams Problem 24 Inversion of re ection 4 Problem 24 Inversion of re ection Let a potential threshold Vxgt0 in the interval 0ltxlta form an obstacie to a wave incident from the left x lt 0 It shall be proved that the coef cient of re ection is the same when the wave falls on the obstacle from the right xgtn whatever the potential shape Solution Let ux and vx be two real and independent solutions of the Schrodinger equation in the interval 0ltxlta with the Wronskian un nu 1 241 The wave function in the case of the wave incident from the left is then of the form2 eikxRelkx xlt03 11 AuxBux 0ltxlta 242 Cequot quot altx and the conditions for continuity of if and tb at x0 and xa are 1R Au0 Bn0 ik 1 RAu390Bv 0 Aua Bva 2C Au39iaBv aikC From the last pair of equations making use of 241 we nd A Chin stung B Cu aikua 244 Putting these expressions for A and B into the rst pair of the Eqs 243 we et g 1Rp00 qC 1 RpaovirC 245 243 with the abbreviations pmnmnmnmnm enenmamwnt kntmvnt nmumn eta 1 m 110 v ta v tO u a From 245 we nally nd Oa pao r M 247 nennnn39 2 We write 4 for the space part of the wave function in this problem since u is used in another sense and the re ection coef cient becomes 2 gpotzquotPao2 q quotquot r32 Poaao2 tq r Now take the opposite case with the wave incident from the right hand side The wave function 242 has then to be replaced by 56 xlt0 1 ux Bum 0ltxlta 249 ensux a eiku ai altx39 The conditions of continuity run as follows 11 a slum Beta ltl 4 Ilu a1 v39a 5340 r utot 6 xiu3905 v390 wikC They have the same structure as Eqs243 from which they can be obtained by exchanging the two arguments x0 and Jean and re placing k by k This transformation applied to 246 renders RI 248 240 Poa paoi PanPoai 3 H 3quot 2411 ie the only difference in the nal formulae 247 and 248 occurs by exchanging p00 and pao Therefore since 248 is symmetrical in pea and pas the re ection coefficient flit2 lRl2 2412 is the same for waves incident from both sides as was to be proved This does not however hold for R Eq 247 which written as a ratio Ra8 of two complex numbers a and 3 according to 24 is trans formed into R a Problem 25 Rectangular potential hole For a rectangular potential hole U lxl lto V 2 251 x i O elsewhere the bound state solutions and their eigenvalues shall be determined Solution Since for states of positive energy the general behaviour can be gathered without dif culty from the preceding problem it suf ces to discuss negative energies i e bound states 1 it Lott E11132 39XLU 1 2w 1 u 236th Mt k1u0 utcaslax I lisl i k The potential is invariai so that the solutions have Putting fisz 2m t these solutions are 386 I Acos Mix A cos nitX mix lgAi tone uwxt u x i 2 lA k odd Here the amplitudes inside I adjusted to make ua conti i been calculated frOm the cor Continuity of u at xa ad even k or it odd 4 c Of 3 1W t kt 3 Izzeorltx 9 rtuusem 3 nectangular potennat note 49 The potential is invariant with respect to inversion Vx V x so that the solutions have either even or odd parity cf Problem 20 Putting l I 2 2 thZ i 12 123 U 0 k2k 2 252 2m these solutions are even 1i 2 k 0 i XACOS x QSXSQ stdquot Acosfltaexia x xgta 11 39 m x 253 e i i IAi E kut Stitkacosica cos2ka odd Asinkx ngga M426 Asmkaequot quot xgta uJx ux 2530 1 1 MAE ka smkacoskasm2kn E Here the ampiitudes inside and outside the potential hole have been adjusted to make Ma continuous The normatization constants have i been calculated from the condition d m 1 1 fwdxiu2L 39 2 xiui i jdxmi 00 a A Continuity of u at xa adds another relation viz 56K 739 aquot at I u 153i t even ksmka xcoska 39 Pun1 or 39 4 in tanka E 254e odd kcoska xsinka or i m 3 2540 SO OneBody Problems without Spin One Dimensional Problems Using 254 and 252 we can recast the normalization expressions so that we obtain the same equation IAi a1 255 for both cases 5 in order to find the eigenvalues from 254 we replace it on the righthand side according to 252 and introduce the abbreviation C icon 2553 We thus obtain WWW 2 M 2 even an M Kg l i 257e kn odd tanka 257 o F or a given potential C is a constant determining the size of the hole Czoc U62 and Eqs 257e 0 can be used to nd all values of ice and thus of the energy u 3 239 l ki Koeof hr 12 o Ezwu q i 258 compatible with the size of the hole Fig 7 shows the function tanka vs kc as well as the expressions on the righthand side of 25e and 257 o The solution of these eigenvalue equations is obtained from the intersection of the curves of Fig 7 Graphical solution of Eqs25 e o The line tanka intersects the curves which represent the righthand sides for different values of the size parameter C Curves at positive ordinates for even at negative ones for odd parity 3 w W 6M hm Msi oi o YW my the last t on the si section Cit case A ht with even 72 Fig 83 4 E 3139 value For 2 the point 3 EltE8 b intersection ity for eige ing linearly states The lt they have t are shown it i x


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