SUPPLEMENTARY NOTES ON THE CONNECTION FORMULAE FOR THE SEMICLASSICAL APPROXIMATION


 Thomasine Casey
 9 months ago
 Views:
Transcription
1 Physics 8.06 Apr, 2008 SUPPLEMENTARY NOTES ON THE CONNECTION FORMULAE FOR THE SEMICLASSICAL APPROXIMATION c R. L. Jffe 2002 The WKB connection formuls llow one to continue semiclssicl solutions from n llowe to forbien region n vice vers. However these formuls re subtle n must be use with cre. The purpose of these notes is to lin how to use the them n why it is necessry to be creful. Note tht we re not eriving the connection formuls here. The erivtion is either long n ull (e.g., Griffiths or Merzbcher) or short, elegnt n obscure (e.g., Lnu n Lifschitz). In ny cse, the proper use of the connection formuls is quite inepenent of how they re erive. The semiclssicl pproimtion is vli whenever the rte of chnge of the e Broglie wvelength is smll, λ () where λ = /p() n p() = 2m(E V ()). Eq. () cn be stisfie in clssiclly llowe region, where E > V (), n λ is rel, or in clssiclly forbien region where E < V (), n λ is imginry. As erive in clss the wvefunctions in the semiclssicl limit re given by: ψ() = c + p() i p( ) + c ī p( ) p() h (2) in the clssiclly llowe region, n ψ() = + κ( ) + κ( ) κ() κ() (2b) in the forbien region, where κ() = 2m(V () E). Notice tht ll the integrls hve been written s inefinite integrls. This is becuse chnge in the lower limit mounts to chnge in the vlue of the constnts c ± n ±. In specific pplictions the lower limits n the constnts re
2 chosen to suit the problem. To mke use of the semiclssicl metho it is lmost lwys necessry to continue the wvefunction from the llowe region to forbien region, or vice vers. The trouble is tht these regions re seprte by clssicl turning point, 0, where E = V ( 0 ), so p( 0 ) = 0 n λ/. So the semiclssicl pproimtion breks own t clssicl turning point. The question is, then, how oes one continue solution from n llowe region through clssicl turning point into forbien region, or vice vers? It shoul be possible becuse we re tlking bout the solution to secon orer ifferentil eqution (the Schröinger eqution). Once one hs specifie two constnts of integrtion, the solution is completely etermine, so specifying the solution in one region shoul fi it in the forbien region. In fct, this is not quite true, n tht s the subtlety of the Connection Formuls. Wht re the Connection Formuls? First, let s summrize the formuls n their omin of pplicbility. The formuls epen on whether the clssiclly forbien region lies to the left or right of the clssiclly llowe region. To be complete we give the formuls for both cses. Figure shows the sitution: in () the forbien region is on the right; in (b) it is on the left. () V() V() (b) E E = = b Figure 2
3 I. Forbien region to the right. If the wvefunction is known to be onentilly flling in the forbien region, then it s phse n mplitue re known in the llowe region: κ( ) 2 cos p( ) π κ() p() (3) A wvefunction 90 out of phse in the llowe region continues into growing onentil in the forbien region s follows: cos p( ) + π κ( ) p() κ() () II. Forbien region to the left. If the wvefunction is known to be onentilly flling in the forbien region, then it s phse n mplitue re known in the llowe region: κ() b κ( ) 2 cos p() b p( ) π (5) A wvefunction 90 out of phse in the llowe region continues into growing onentil in the forbien region s follows: cos p( ) + π b κ( ) p() b κ() (6) 2 Are the Connection Formuls equlities? (This section is written by Hong Liu, 2007) Cn we use the connection formuls in the irections opposite to the rrows in (3) (6)? The nswer is yes, but etr cre must be pi in reversing the rrows in equtions () n (6). In contrst, the rrows in (3) n (5) cn be reverse reltively strightforwrly. 3
4 To illustrte the subtleties, let us consier specific emple: Suppose the wvefunction in forbien region to the right is given by ψ() + κ() + κ() κ( ) ( + O()) κ( ) ( + O()). (7) Then using () in the irection opposite to the rrow n (3), we obtin the wve function on the left + cos p( ) + π ( + O()) p() + 2 cos ( + O()). (8) p() p( ) π The bove proceure is correct provie in (7) we know the coefficient before the onentilly flling term κ( ) precisely. In most circumstnces, however, the ccurcy of the WKB pproimtion is not enough for us to know ectly. This is ue to tht the secon term in (7) is so smll compre with the first term, tht it is normlly roppe completely. Note tht for ech term in (7), we hve roppe terms of orer O(), s inicte in the eqution. The O() contribution in the first term, which ws lrey roppe, is much lrger thn the onentilly suppresse secon term in the smll limit. Thus it is completely legitimte to rop the secon term in (7). In such sitution we re simply left with ψ() + κ( ) (9) κ() n it is then incorrect to use eqution () bckwrs to conclue tht the wvefunction on the left is of the form + cos p( ) + π. (0) p() In compring (0) with (8), we see tht (0) misses the secon term in (8), which is of the sme orer s the term in (0). Nevertheless, one oes encounter situtions in which in (7) is known precisely, often ue to symmetry. For emple, consier sitution in which
5 = 0 n both ψ() n κ() re even functions of. Then the symmetry requires tht + =. (We will see such n emple in the problem of ouble well potentil in pset 8). Sometimes it is lso possible to use more sophisticte mthemticl methos to keep trck of the coefficient precisely even without symmetry. In these situtions one cn then use the connection formuls in both irections. 3 An Emple A worke emple will help show how the Connection Formuls re to be pplie. Consier prticle trppe between the origin t = 0 n high potentil V(). Figure 2 gives n illustrtion. For energy E, the clssicl turning point is t = (E). is the point where E = V (). The semiclssicl solution which vnishes t the origin is V() E Figure 2 ψ() = sin 0 p( ). () This form is vli for <. Wht o we o s pproches the turning point t =? We rewrite the sin... s the liner combintion of the two cosines for which we hve connection informtion: ψ() = sin 0 = sin cos p( ) p( ) π 5 + cos cos p( ) + π (2)
6 where = p() π 0. (3) This much is just ppliction of trig ientities. If the coefficient of cos p( ) + π is not zero, then the wve function continues into growing onentil for > ccoring to eq. (). The only thing we cn sy for certin is tht ψ() hs n onentilly growing term in the forbien region: ψ() cos κ() κ( ). () ψ() woul in generl lso contin onentilly flling terms in the forbien region, but we on t hve the ccurcy to compute them. Tiny corrections to the onentilly growing term, which we i not keep trck of in our WKB pproimtion will be much lrger thn the onentilly flling term. However, if cos = 0, then there is no onentilly growing term in the forbien region. Thus, if we know tht ψ() flls onentilly in the forbien region cos must be zero. Since boun stte wvefunction must fll onentilly in the forbien region we lern tht the WKB conition for boun stte is cos = 0, or 0 p() = (n + 3 )π (5) which is the BohrSommerfel Quntiztion conition when there is hr wll on one sie. The problem set contins other problems which require creful use of the Connection Formuls. Deriving the Connection Formule This section c Krishn Rjgopl, 200 Let us consier the cse epicte in the left pnel of Fig., with turning point t = with the llowe region t < n the clssiclly forbien region t >. We know tht in the forbien region ψ() = κ( ) for (6) κ() 6
7 n in the llowe region ψ() = c + i p() p( ) + c ī p() h p( ) for. (7) Our tsk is to relte c + n c to, n we ect to fin the reltion escribe by the connection formul (3). I ll tke you through this erivtion, n leve eriving the other three connection formule to you. In the vicinity of the turning point, the potentil is pproimtely liner n we cn write V E b( ) where b V > 0. (8) = This linerize potentil is goo escription for < L, where L is the length scle over which V curves. The semiclssicl forms for the wve function, (6) n (7), re not vli too close to. For emple, (6) is vli where κ() (9) n we must sk whether there re vlues of ( ) tht re simultneously lrge enough tht (9) is vli, but not so lrge tht (8) breks own. If there is rnge of ( ) in which both (9) n (8) re vli, then within this omin (9) becomes 2mb( ) = We cn therefore conclue tht s long s L 8mb ( ). 8mb, (20) there is region where (9) is vli, mening tht (6) escribes the wve function, n where (8) escribes the potentil. The sme conition (20) implies tht there is rnge of, in the llowe region in which is big enough tht (7) is vli escription of the wve function n smll enough tht (8) still escribes the potentil. The conition (20) must be stisfie by the potentil ner its turning points in orer to ensure the vliity of the nlysis we re pursuing (with semiclssicl wve functions 7
8 fr wy from turning points n connection formule prescribing how they re connecte cross the turning points.) One wy of reing (20) is tht it is lwys stisfie in the 0 limit, n inee this is why the metho is clle semiclssicl. Perhps better wy to re the conition, though, is to leve fie it is fter ll constnt of nture n view (20) s the sttement of how smooth the potentil must be ner its turning points. Henceforth, we ssume tht (20) is stisfie. Within the omin of where both (8) n (9) hol, the wve function is given by ψ() = = 2mb( ) 2mb( ) / / 2mb( ) 2 2mb ( ). (2) 3 Now, we wnt to nlyticlly continue this ression from > 0 to < 0. The trick is to consier ( ) comple vrible, which we shll write s ( ) = ρ iφ, n to strt with φ = 0 n ρ in the rnge such tht ll our pproimtions re vli n then to continuously chnge φ from 0 to π, ll the while keeping ρ fie. In this wy, we en t point in the llowe region (with < 0) where both (8) n (9) hol. Lets see wht hppens to the wve function upon performing this proceure. First, we rewrite the wve function s ψ() = 2 2mb ρ 3iφ (22) (2mbρ) / iφ 3 2 which, for φ = 0, is wht we h before. For φ = π, nmely < 0, the wve function becomes ψ() = +i 2 2mb ρ. (23) (2mbρ) / iπ 3 So, this is the wve function in the llowe region tht we obtin by strting from the wve function in the forbien region n nlyticlly continuing. Note tht the turning point is t ρ = 0, n we never went ner it. By turning into comple vrible, we were ble to strt with > 0, en with < 0, n never go ner = 0. We must now compre 8
9 (23) to the form of the wve function we were ecting to get in the llowe region, nmely (7). Using (8) n, note tht we re in the region where this is vli we cn rewrite the wve function in the llowe region (7) s follows: ψ() = = = c + 2mb( ) i / 2mb( ) c + ī / 2mb( ) h c + +i 2 / 2mb ( ) 2mb( ) 3 c + 2mbρ c + / 2mb( ) +i 2 2mb ρ 3 / c + 2mbρ / i 2 3 i 2 2mb ρ 3 2mb( ) 2mb ( ). (2) We now see tht if we choose c + = iπ (25) then the wve function (23) tht we obtine by nlyticlly continuing the forbienregion wve function to the llowe region is the sme s the c + term in (2)! This looks goo, but wht hs hppene to the c term??? Let s try to figure out why we foun the c + term, but not the c term. To o this, we strt with (2), incluing both the c + n c terms, n try to nlyticlly continue it in the opposite irection to wht we i before, bck to the forbien region. We o the nlyticl continution by chnging φ from π to 0. Note tht the imginry prt of ( ) is positive uring the continution. You cn esily check tht if you strt with (2) n perform this proceure, s you begin the continution into the comple plne (i.e. s you strt reucing φ from π) the mgnitue of the c term becomes onentilly smll compre to the mgnitue of the c + term. As we iscusse in lecture, the semiclssicl pproimtion entils ropping such onentilly smll terms. So, if we strt with (2) n continue bckwr, we 9
10 lose the c term n the c + term turns into the correct wve function in the forbien region once φ is bck to 0. Anlogously, when we strte with the forbienregion wve function n continue forwr, we only obtine the c + term. Now tht we unerstn why we lost the c term, how cn we fin it?? Simple. Strt with the forbienregion wve function (22) gin. This time, chnge φ from 0 to π. As before, we strt in the forbien region n en in the llowe region. This time, though, the imginry prt of is negtive uring the continution. This mens tht ner φ = π, the mgnitue of the c + term is onentilly smller thn tht of the c term, so we ect this time to lose the c + term. An, lo n behol, the wve function in the llowe region tht we obtin by strting from the forbien region n continuing φ from 0 to π is ψ() = (2mbρ) / ( ) i 2 2mb ρ, (26) iπ 3 which is not the sme s (23). Inste, it is precisely the c term in (2), s long s we choose c = ( ). (27) iπ By oing the nlytic continution from φ = 0 to φ = π, we hve lost the c + term n obtine the c term! By performing these two ifferent nlytic continutions, we re ble to strt from the wve function in the forbien region n etermine the complete wve function in the llowe region. Wht we fin is tht in the llowe region, the wve function is given by (2) or, equivlently, (7) with c + n c specifie by (25) n (27). Tht is, in the llowe region ψ() = 2 cos p( ) π (28) p() which is the connection formul (3) we set out to prove. Elegnt, n estce ps? 0
Solutions to Problems in Merzbacher, Quantum Mechanics, Third Edition. Chapter 7
Solutions to Problems in Merzbcher, Quntum Mechnics, Third Edition Homer Reid April 5, 200 Chpter 7 Before strting on these problems I found it useful to review how the WKB pproimtion works in the first
More informationBasic Derivative Properties
Bsic Derivtive Properties Let s strt this section by remining ourselves tht the erivtive is the slope of function Wht is the slope of constnt function? c FACT 2 Let f () =c, where c is constnt Then f 0
More informationf a L Most reasonable functions are continuous, as seen in the following theorem:
Limits Suppose f : R R. To sy lim f(x) = L x mens tht s x gets closer n closer to, then f(x) gets closer n closer to L. This suggests tht the grph of f looks like one of the following three pictures: f
More information221B Lecture Notes WKB Method
Clssicl Limit B Lecture Notes WKB Method Hmilton Jcobi Eqution We strt from the Schrödinger eqution for single prticle in potentil i h t ψ x, t = [ ] h m + V x ψ x, t. We cn rewrite this eqution by using
More informationPolynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230
Polynomil Approimtions for the Nturl Logrithm nd Arctngent Functions Mth 23 You recll from first semester clculus how one cn use the derivtive to find n eqution for the tngent line to function t given
More informationVII. The Integral. 50. Area under a Graph. y = f(x)
VII. The Integrl In this chpter we efine the integrl of function on some intervl [, b]. The most common interprettion of the integrl is in terms of the re uner the grph of the given function, so tht is
More informationConservation Law. Chapter Goal. 6.2 Theory
Chpter 6 Conservtion Lw 6.1 Gol Our long term gol is to unerstn how mthemticl moels re erive. Here, we will stuy how certin quntity chnges with time in given region (sptil omin). We then first erive the
More information221A Lecture Notes WKB Method
A Lecture Notes WKB Method Hmilton Jcobi Eqution We strt from the Schrödinger eqution for single prticle in potentil i h t ψ x, t = [ ] h m + V x ψ x, t. We cn rewrite this eqution by using ψ x, t = e
More informationGoals: Determine how to calculate the area described by a function. Define the definite integral. Explore the relationship between the definite
Unit #8 : The Integrl Gols: Determine how to clculte the re described by function. Define the definite integrl. Eplore the reltionship between the definite integrl nd re. Eplore wys to estimte the definite
More informationMath 211A Homework. Edward Burkard. = tan (2x + z)
Mth A Homework Ewr Burkr Eercises 5C Eercise 8 Show tht the utonomous system: 5 Plne Autonomous Systems = e sin 3y + sin cos + e z, y = sin ( + 3y, z = tn ( + z hs n unstble criticl point t = y = z =
More informationDuality # Second iteration for HW problem. Recall our LP example problem we have been working on, in equality form, is given below.
Dulity #. Second itertion for HW problem Recll our LP emple problem we hve been working on, in equlity form, is given below.,,,, 8 m F which, when written in slightly different form, is 8 F Recll tht we
More informationand that at t = 0 the object is at position 5. Find the position of the object at t = 2.
7.2 The Fundmentl Theorem of Clculus 49 re mny, mny problems tht pper much different on the surfce but tht turn out to be the sme s these problems, in the sense tht when we try to pproimte solutions we
More informationHomework Assignment 5 Solution Set
Homework Assignment 5 Solution Set PHYCS 44 3 Februry, 4 Problem Griffiths 3.8 The first imge chrge gurntees potentil of zero on the surfce. The secon imge chrge won t chnge the contribution to the potentil
More informationlim P(t a,b) = Differentiate (1) and use the definition of the probability current, j = i (
PHYS851 Quntum Mechnics I, Fll 2009 HOMEWORK ASSIGNMENT 7 1. The continuity eqution: The probbility tht prticle of mss m lies on the intervl [,b] t time t is Pt,b b x ψx,t 2 1 Differentite 1 n use the
More informationx dx does exist, what does the answer look like? What does the answer to
Review Guie or MAT Finl Em Prt II. Mony Decemer th 8:.m. 9:5.m. (or the 8:3.m. clss) :.m. :5.m. (or the :3.m. clss) Prt is worth 5% o your Finl Em gre. NO CALCULATORS re llowe on this portion o the Finl
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationLecture 13  Linking E, ϕ, and ρ
Lecture 13  Linking E, ϕ, nd ρ A Puzzle... InnerSurfce Chrge Density A positive point chrge q is locted offcenter inside neutrl conducting sphericl shell. We know from Guss s lw tht the totl chrge on
More informationSturmLiouville Theory
LECTURE 1 SturmLiouville Theory In the two preceing lectures I emonstrte the utility of Fourier series in solving PDE/BVPs. As we ll now see, Fourier series re just the tip of the iceerg of the theory
More information5.3 The Fundamental Theorem of Calculus
CHAPTER 5. THE DEFINITE INTEGRAL 35 5.3 The Funmentl Theorem of Clculus Emple. Let f(t) t +. () Fin the re of the region below f(t), bove the tis, n between t n t. (You my wnt to look up the re formul
More informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More information4.5 THE FUNDAMENTAL THEOREM OF CALCULUS
4.5 The Funmentl Theorem of Clculus Contemporry Clculus 4.5 THE FUNDAMENTAL THEOREM OF CALCULUS This section contins the most importnt n most use theorem of clculus, THE Funmentl Theorem of Clculus. Discovere
More informationThe First Fundamental Theorem of Calculus. If f(x) is continuous on [a, b] and F (x) is any antiderivative. f(x) dx = F (b) F (a).
The Fundmentl Theorems of Clculus Mth 4, Section 0, Spring 009 We now know enough bout definite integrls to give precise formultions of the Fundmentl Theorems of Clculus. We will lso look t some bsic emples
More informationARITHMETIC OPERATIONS. The real numbers have the following properties: a b c ab ac
REVIEW OF ALGEBRA Here we review the bsic rules nd procedures of lgebr tht you need to know in order to be successful in clculus. ARITHMETIC OPERATIONS The rel numbers hve the following properties: b b
More information5.4, 6.1, 6.2 Handout. As we ve discussed, the integral is in some way the opposite of taking a derivative. The exact relationship
5.4, 6.1, 6.2 Hnout As we ve iscusse, the integrl is in some wy the opposite of tking erivtive. The exct reltionship is given by the Funmentl Theorem of Clculus: The Funmentl Theorem of Clculus: If f is
More informationProblem Set 3 Solutions
Chemistry 36 Dr Jen M Stndrd Problem Set 3 Solutions 1 Verify for the prticle in onedimensionl box by explicit integrtion tht the wvefunction ψ ( x) π x is normlized To verify tht ψ ( x) is normlized,
More informationIf we have a function f(x) which is welldefined for some a x b, its integral over those two values is defined as
Y. D. Chong (26) MH28: Complex Methos for the Sciences 2. Integrls If we hve function f(x) which is wellefine for some x, its integrl over those two vlues is efine s N ( ) f(x) = lim x f(x n ) where x
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More informationAntiderivatives Introduction
Antierivtives 0. Introuction So fr much of the term hs been sent fining erivtives or rtes of chnge. But in some circumstnces we lrey know the rte of chnge n we wish to etermine the originl function. For
More informationapproaches as n becomes larger and larger. Since e > 1, the graph of the natural exponential function is as below
. Eponentil nd rithmic functions.1 Eponentil Functions A function of the form f() =, > 0, 1 is clled n eponentil function. Its domin is the set of ll rel f ( 1) numbers. For n eponentil function f we hve.
More informationUsing integration tables
Using integrtion tbles Integrtion tbles re inclue in most mth tetbooks, n vilble on the Internet. Using them is nother wy to evlute integrls. Sometimes the use is strightforwr; sometimes it tkes severl
More information1 nonlinear.mcd Find solution root to nonlinear algebraic equation f(x)=0. Instructor: Nam Sun Wang
nonlinermc Fin solution root to nonliner lgebric eqution ()= Instructor: Nm Sun Wng Bckgroun In science n engineering, we oten encounter lgebric equtions where we wnt to in root(s) tht stisies given eqution
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More informationEnergy Bands Energy Bands and Band Gap. Phys463.nb Phenomenon
Phys463.nb 49 7 Energy Bnds Ref: textbook, Chpter 7 Q: Why re there insultors nd conductors? Q: Wht will hppen when n electron moves in crystl? In the previous chpter, we discussed free electron gses,
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More informationTopic 1 Notes Jeremy Orloff
Topic 1 Notes Jerem Orloff 1 Introduction to differentil equtions 1.1 Gols 1. Know the definition of differentil eqution. 2. Know our first nd second most importnt equtions nd their solutions. 3. Be ble
More informationChapter Five  Eigenvalues, Eigenfunctions, and All That
Chpter Five  Eigenvlues, Eigenfunctions, n All Tht The prtil ifferentil eqution methos escrie in the previous chpter is specil cse of more generl setting in which we hve n eqution of the form L 1 xux,tl
More informationSummary: Method of Separation of Variables
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 1 Summry: Method of Seprtion of Vribles 1. Seprtion of Vribles in Crtesin Coordintes 2. Fourier Series Suggested Reding: Griffiths: Chpter 3, Section
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationAppendix 3, Rises and runs, slopes and sums: tools from calculus
Appendi 3, Rises nd runs, slopes nd sums: tools from clculus Sometimes we will wnt to eplore how quntity chnges s condition is vried. Clculus ws invented to do just this. We certinly do not need the full
More informationNotes on the Eigenfunction Method for solving differential equations
Notes on the Eigenfunction Metho for solving ifferentil equtions Reminer: WereconsieringtheinfiniteimensionlHilbertspceL 2 ([, b] of ll squreintegrble functions over the intervl [, b] (ie, b f(x 2
More informationPhysics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More informationCHAPTER 9 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS
CHAPTER 9 BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS BASIC CONCEPTS OF DIFFERENTIAL AND INTEGRAL CALCULUS LEARNING OBJECTIVES After stuying this chpter, you will be ble to: Unerstn the bsics
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationSection 5.1 #7, 10, 16, 21, 25; Section 5.2 #8, 9, 15, 20, 27, 30; Section 5.3 #4, 6, 9, 13, 16, 28, 31; Section 5.4 #7, 18, 21, 23, 25, 29, 40
Mth B Prof. Audrey Terrs HW # Solutions by Alex Eustis Due Tuesdy, Oct. 9 Section 5. #7,, 6,, 5; Section 5. #8, 9, 5,, 7, 3; Section 5.3 #4, 6, 9, 3, 6, 8, 3; Section 5.4 #7, 8,, 3, 5, 9, 4 5..7 Since
More informationContinuous Quantum Systems
Chpter 8 Continuous Quntum Systems 8.1 The wvefunction So fr, we hve been tlking bout finite dimensionl Hilbert spces: if our system hs k qubits, then our Hilbert spce hs n dimensions, nd is equivlent
More informationMath Lecture 23
Mth 8  Lecture 3 Dyln Zwick Fll 3 In our lst lecture we delt with solutions to the system: x = Ax where A is n n n mtrix with n distinct eigenvlues. As promised, tody we will del with the question of
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More information5.5 The Substitution Rule
5.5 The Substitution Rule Given the usefulness of the Fundmentl Theorem, we wnt some helpful methods for finding ntiderivtives. At the moment, if n ntiderivtive is not esily recognizble, then we re in
More informationpotentials A z, F z TE z Modes We use the e j z z =0 we can simply say that the x dependence of E y (1)
3e. Introduction Lecture 3e Rectngulr wveguide So fr in rectngulr coordintes we hve delt with plne wves propgting in simple nd inhomogeneous medi. The power density of plne wve extends over ll spce. Therefore
More informationBernoulli Numbers Jeff Morton
Bernoulli Numbers Jeff Morton. We re interested in the opertor e t k d k t k, which is to sy k tk. Applying this to some function f E to get e t f d k k tk d k f f + d k k tk dk f, we note tht since f
More informationDo the onedimensional kinetic energy and momentum operators commute? If not, what operator does their commutator represent?
1 Problem 1 Do the onedimensionl kinetic energy nd momentum opertors commute? If not, wht opertor does their commuttor represent? KE ˆ h m d ˆP i h d 1.1 Solution This question requires clculting the
More informationx ) dx dx x sec x over the interval (, ).
Curve on 6 For , () Evlute the integrl, n (b) check your nswer by ifferentiting. ( ). ( ). ( ).. 6. sin cos 7. sec csccot 8. sec (sec tn ) 9. sin csc. Evlute the integrl sin by multiplying the numertor
More informationLecture 3. In this lecture, we will discuss algorithms for solving systems of linear equations.
Lecture 3 3 Solving liner equtions In this lecture we will discuss lgorithms for solving systems of liner equtions Multiplictive identity Let us restrict ourselves to considering squre mtrices since one
More informationMathematics Number: Logarithms
plce of mind F A C U L T Y O F E D U C A T I O N Deprtment of Curriculum nd Pedgogy Mthemtics Numer: Logrithms Science nd Mthemtics Eduction Reserch Group Supported y UBC Teching nd Lerning Enhncement
More informationAPPENDIX. Precalculus Review D.1. Real Numbers and the Real Number Line
APPENDIX D Preclculus Review APPENDIX D.1 Rel Numers n the Rel Numer Line Rel Numers n the Rel Numer Line Orer n Inequlities Asolute Vlue n Distnce Rel Numers n the Rel Numer Line Rel numers cn e represente
More informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationChapters 4 & 5 Integrals & Applications
Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO  Ares Under Functions............................................ 3.2 VIDEO  Applictions
More informationHomework Problem Set 1 Solutions
Chemistry 460 Dr. Jen M. Stnr Homework Problem Set 1 Solutions 1. Determine the outcomes of operting the following opertors on the functions liste. In these functions, is constnt..) opertor: / ; function:
More informationPh2b Quiz  1. Instructions
Ph2b Winter 21718 Quiz  1 Due Dte: Mondy, Jn 29, 218 t 4pm Ph2b Quiz  1 Instructions 1. Your solutions re due by Mondy, Jnury 29th, 218 t 4pm in the quiz box outside 21 E. Bridge. 2. Lte quizzes will
More informationMassachusetts Institute of Technology Quantum Mechanics I (8.04) Spring 2005 Solutions to Problem Set 6
Msschusetts Institute of Technology Quntum Mechnics I (8.) Spring 5 Solutions to Problem Set 6 By Kit Mtn. Prctice with delt functions ( points) The Dirc delt function my be defined s such tht () (b) 3
More informationp(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
More informationSection 6.3 The Fundamental Theorem, Part I
Section 6.3 The Funmentl Theorem, Prt I (3//8) Overview: The Funmentl Theorem of Clculus shows tht ifferentition n integrtion re, in sense, inverse opertions. It is presente in two prts. We previewe Prt
More informationM 106 Integral Calculus and Applications
M 6 Integrl Clculus n Applictions Contents The Inefinite Integrls.................................................... Antierivtives n Inefinite Integrls.. Antierivtives.............................................................
More information7.2 The Definite Integral
7.2 The Definite Integrl the definite integrl In the previous section, it ws found tht if function f is continuous nd nonnegtive, then the re under the grph of f on [, b] is given by F (b) F (), where
More informationf(x) dx, If one of these two conditions is not met, we call the integral improper. Our usual definition for the value for the definite integral
Improper Integrls Every time tht we hve evluted definite integrl such s f(x) dx, we hve mde two implicit ssumptions bout the integrl:. The intervl [, b] is finite, nd. f(x) is continuous on [, b]. If one
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationChapter 3 The Schrödinger Equation and a Particle in a Box
Chpter 3 The Schrödinger Eqution nd Prticle in Bo Bckground: We re finlly ble to introduce the Schrödinger eqution nd the first quntum mechnicl model prticle in bo. This eqution is the bsis of quntum mechnics
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More informationStuff You Need to Know From Calculus
Stuff You Need to Know From Clculus For the first time in the semester, the stuff we re doing is finlly going to look like clculus (with vector slnt, of course). This mens tht in order to succeed, you
More informationPractice Problems Solution
Prctice Problems Solution Problem Consier D Simple Hrmonic Oscilltor escribe by the Hmiltonin Ĥ ˆp m + mwˆx Recll the rte of chnge of the expecttion of quntum mechnicl opertor t A ī A, H] + h A t. Let
More informationChapter 14. Matrix Representations of Linear Transformations
Chpter 4 Mtrix Representtions of Liner Trnsformtions When considering the Het Stte Evolution, we found tht we could describe this process using multipliction by mtrix. This ws nice becuse computers cn
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationIntegration. 148 Chapter 7 Integration
48 Chpter 7 Integrtion 7 Integrtion t ech, by supposing tht during ech tenth of second the object is going t constnt speed Since the object initilly hs speed, we gin suppose it mintins this speed, but
More informationax bx c (2) x a x a x a 1! 2!! gives a useful way of approximating a function near to some specific point x a, giving a powerseries expansion in x
Elementr mthemticl epressions Qurtic equtions b b b The solutions to the generl qurtic eqution re (1) b c () b b 4c (3) Tlor n Mclurin series (powerseries epnsion) The Tlor series n n f f f n 1!! n! f
More informationES.182A Topic 32 Notes Jeremy Orloff
ES.8A Topic 3 Notes Jerem Orloff 3 Polr coordintes nd double integrls 3. Polr Coordintes (, ) = (r cos(θ), r sin(θ)) r θ Stndrd,, r, θ tringle Polr coordintes re just stndrd trigonometric reltions. In
More informationHow do we solve these things, especially when they get complicated? How do we know when a system has a solution, and when is it unique?
XII. LINEAR ALGEBRA: SOLVING SYSTEMS OF EQUATIONS Tody we re going to tlk bout solving systems of liner equtions. These re problems tht give couple of equtions with couple of unknowns, like: 6 2 3 7 4
More informationSection 4: Integration ECO4112F 2011
Reding: Ching Chpter Section : Integrtion ECOF Note: These notes do not fully cover the mteril in Ching, ut re ment to supplement your reding in Ching. Thus fr the optimistion you hve covered hs een sttic
More informationLecture 14: Quadrature
Lecture 14: Qudrture This lecture is concerned with the evlution of integrls fx)dx 1) over finite intervl [, b] The integrnd fx) is ssumed to be relvlues nd smooth The pproximtion of n integrl by numericl
More information5.7 Improper Integrals
458 pplictions of definite integrls 5.7 Improper Integrls In Section 5.4, we computed the work required to lift pylod of mss m from the surfce of moon of mss nd rdius R to height H bove the surfce of the
More informationThe Fundamental Theorem of Calculus Part 2, The Evaluation Part
AP Clculus AB 6.4 Funmentl Theorem of Clculus The Funmentl Theorem of Clculus hs two prts. These two prts tie together the concept of integrtion n ifferentition n is regre by some to by the most importnt
More informationLECTURE 3. Orthogonal Functions. n X. It should be noted, however, that the vectors f i need not be orthogonal nor need they have unit length for
ECTURE 3 Orthogonl Functions 1. Orthogonl Bses The pproprite setting for our iscussion of orthogonl functions is tht of liner lgebr. So let me recll some relevnt fcts bout nite imensionl vector spces.
More informationProblem Set 2 Solutions
Chemistry 362 Dr. Jen M. Stnr Problem Set 2 Solutions 1. Determine the outcomes of operting the following opertors on the functions liste. In these functions, is constnt.).) opertor: /x ; function: x e
More informationInClass Problems 2 and 3: Projectile Motion Solutions. InClass Problem 2: Throwing a Stone Down a Hill
MASSACHUSETTS INSTITUTE OF TECHNOLOGY Deprtment of Physics Physics 8T Fll Term 4 InClss Problems nd 3: Projectile Motion Solutions We would like ech group to pply the problem solving strtegy with the
More informationNew Expansion and Infinite Series
Interntionl Mthemticl Forum, Vol. 9, 204, no. 22, 06073 HIKARI Ltd, www.mhikri.com http://dx.doi.org/0.2988/imf.204.4502 New Expnsion nd Infinite Series Diyun Zhng College of Computer Nnjing University
More informationRecitation 3: More Applications of the Derivative
Mth 1c TA: Pdric Brtlett Recittion 3: More Applictions of the Derivtive Week 3 Cltech 2012 1 Rndom Question Question 1 A grph consists of the following: A set V of vertices. A set E of edges where ech
More informationChapter 5. , r = r 1 r 2 (1) µ = m 1 m 2. r, r 2 = R µ m 2. R(m 1 + m 2 ) + m 2 r = r 1. m 2. r = r 1. R + µ m 1
Tor Kjellsson Stockholm University Chpter 5 5. Strting with the following informtion: R = m r + m r m + m, r = r r we wnt to derive: µ = m m m + m r = R + µ m r, r = R µ m r 3 = µ m R + r, = µ m R r. 4
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More information1 Part II: Numerical Integration
Mth 4 Lb 1 Prt II: Numericl Integrtion This section includes severl techniques for getting pproimte numericl vlues for definite integrls without using ntiderivtives. Mthemticll, ect nswers re preferble
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationState space systems analysis (continued) Stability. A. Definitions A system is said to be Asymptotically Stable (AS) when it satisfies
Stte spce systems nlysis (continued) Stbility A. Definitions A system is sid to be Asymptoticlly Stble (AS) when it stisfies ut () = 0, t > 0 lim xt () 0. t A system is AS if nd only if the impulse response
More informationP 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)
1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this
More informationQuantum Physics I (8.04) Spring 2016 Assignment 8
Quntum Physics I (8.04) Spring 206 Assignment 8 MIT Physics Deprtment Due Fridy, April 22, 206 April 3, 206 2:00 noon Problem Set 8 Reding: Griffiths, pges 7376, 882 (on scttering sttes). Ohnin, Chpter
More information2.4 Linear Inequalities and Interval Notation
.4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn , Less thn or
More informationAdvanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004
Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when
More informationQuantum Physics II (8.05) Fall 2013 Assignment 2
Quntum Physics II (8.05) Fll 2013 Assignment 2 Msschusetts Institute of Technology Physics Deprtment Due Fridy September 20, 2013 September 13, 2013 3:00 pm Suggested Reding Continued from lst week: 1.
More informationIntroduction. Calculus I. Calculus II: The Area Problem
Introuction Clculus I Clculus I h s its theme the slope problem How o we mke sense of the notion of slope for curves when we only know wht the slope of line mens? The nswer, of course, ws the to efine
More informationMain topics for the Second Midterm
Min topics for the Second Midterm The Midterm will cover Sections 5.45.9, Sections 6.16.3, nd Sections 7.17.7 (essentilly ll of the mteril covered in clss from the First Midterm). Be sure to know the
More informationSection 6.1 INTRO to LAPLACE TRANSFORMS
Section 6. INTRO to LAPLACE TRANSFORMS Key terms: Improper Integrl; diverge, converge A A f(t)dt lim f(t)dt Piecewise Continuous Function; jump discontinuity Function of Exponentil Order Lplce Trnsform
More informationNumerical Integration
Chpter 5 Numericl Integrtion Numericl integrtion is the study of how the numericl vlue of n integrl cn be found. Methods of function pproximtion discussed in Chpter??, i.e., function pproximtion vi the
More information