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# Quantum Physics PHYS 4617

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This 99 page Class Notes was uploaded by Iva Cormier on Sunday October 11, 2015. The Class Notes belongs to PHYS 4617 at East Tennessee State University taught by Staff in Fall. Since its upload, it has received 39 views. For similar materials see /class/221413/phys-4617-east-tennessee-state-university in Physics 2 at East Tennessee State University.

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Physics 46175617 Quantum Physics Course Lecture Notes Dr Donald G Luttermoser East Tennessee State University Edition 51 Abstract These Class notes are designed for use of the instructor and students of the course Physics 46175617 Quantum Physics This edition was last modi ed for the Fall 2006 semester II The Wave Function A The Schrodinger Equation 1 As mentioned in l of the notes quantum mechanics approaches the trajectory problem of Newtonian mechanics quite differently On a microscopic level particles do Lot follow trajectories but instead are characterized by their wave function lx t where x is the l dimensional position we will worry about 3 dimensions later of the wave function at time t a The wave function is determined from Schrodinger s Equation Bkl E2 ngl 2 37 372 f i Here 239 1 and h 1054573 x 1034 J s 11 2 W 11 1 ii Whereas Newton s Second Law F ma is the most important equation in all of classical physics Eq ll l is the most important equation in all of quantum physics b Given suitable initial conditions typically llx0 the Schrodinger equation determines lx t for all future times just as in classical mechanics Newton s Second Law de termines 0615 for all future times 2 What exactly is the wave function and what does it do for you once you got it a Whereas a particle is localized at a point in classical me chanics a wave function is spread out in space gt it is a function of x for any given time t 111 Figure ll 1 A hypothetical wave function The particle would be relatively likely to be found near A and unlikely to be found near B The shaded area represents the probability of nding the particle in the range dx b C d Born came up with a statistical interpretation of the wave function which says that ll06t2 gives the proba bility of nding the particle at point x at time t or more precisely 2 i probability of nding the particle lqjm m dx i between 06 and 06 dx at time 15 11 3 The wave function itself is complex but l2 llll where l is the complex conjugate of kl is real and non negative as a probability must be For the hypothetical wave function in Figure ll 1 you would be quite likely to nd the particle in the vicinity of point A and relatively unlikely to nd it near point B 112 From the concept of the wave function it becomes easier to see how the Heisenberg Uncertainty Principle arises in nature The wave function will not allow you to predict with certainty the outcome of a simple experiment to measure a particle s position all quantum mechanics has to offer is statistical information about the possible results B Philosophical Interpretations 1 The Realist Position a We View the microscopic world as probabilistic due to the fact that quantum mechanics is an incomplete theory b The particle really was at a speci c position say point C in Figure 11 1 yet quantum mechanics was unable to tell US SO c To the realist indeterminacy is not a fact of nature but a re ection of our ignorance d If this scenario is in fact the correct one then l1 is not the whole story some additional information known as a hidden variable is needed to provide a complete description of the particle The orthodox position gt the Copenhagen interpretation a The particle isn t really anywhere in space The act of the measurement forces the particle to take a stand though how and why we dare not ask b Observations not only disturb what is to be measured they produce it 113 c Bohr and his followers put forward this interpretation of quantum mechanics d It is the most widely accepted position of the interpreta tion of quantum mechanics in physics The agnostic position a Refuse to answer What sense can there be in making assertions about the status of a particle before a measure ment when the only way of knowing whether you were right is precisely to conduct the measurement in which case what you get is no longer before the measurement b This has been used as a fall back position used by many physicists if one is unable to convince another of the or thodox position In 1964 John Bell astonished the physics community by showing that it makes an observable difference if the particle had a precise although unknown position prior to its measurement a This discovery effectively eliminated the realist position b Bell s Theorem showed that the orthodox position is the correct interpretation of quantum mechanics by proving that any local hidden variable theory is incompatible with quantum mechanics see Bell JS 1964 Physics 1 195 c We won t get into the details of Bell s Theorem at this point in time Su ice it to say that a particle does not have a precise position prior to the measurement any more than ripples in a pond do gt it is the measurement pro cess that insists upon one particular number and thereby in a sense creates the speci c result 114 5 The act of the measurement collapses the wave function to a delta function eg a sharp peak at some position ll soon spreads out again after the measurement in accordance to the Schrodinger equation C Probability and Normalization 1 Because of the statistical interpretation probability P plays a central role in quantum mechanics a b A probability value is the likelihood of a sample point oc curring in a given distribution of points Where a sample point is de ned here as a possible outcome of an experi ment A distribution of points can either be a set Of discrete values or a continuous set of values 2 Discrete Measurements Below are a few de nitions concern ing discrete measurements a b The total number of particles or measurements in a sys tem is 114 where N j is the number of particles or measurements in state 339 The probability of a particle being in state j is N j ll 5 Po N lt gt whereas the sum of all the probabilities is 2P0 l ll 6 J 115 C d e f g h The average or mean of a particle being found in state 339 is EleU a 117 J whereas the most probable value ofj is MAXNj The average of the square of a particle being found in state j is ll 8 In general the average value of some function of j is given by 00 1 Z fJ39PJ39 jl 119 The numerical measure of the amount of spread in a dis tribution with respect to the average is M J39 339 1110 The variance of the distribution is de ned as a 2 Am II11gt where a is called the standard deviation of the mea surement It can be proven that ll 12 Note that ll 13 H6 3 Continuous Measurements For continuous distributions of data or measurements it is Often convenient to de ne the prob ability density px as probability that a random measurement lies between 06 and 06 dx i pm dx 39 11 14 a For these continuous distributions the probability that x lies between a and b a nite interval is given by Pub foamy 11 15 b The following equations are also valid pxdx 1 ll16 x Limpmdx 11 17 x Ofwpxdx 1118 02 A062gtlt062gtlt06gt2 1119 ptx a2 a2 x Figure ll 2 Probability density function for a uniform distribution H7 A uniform distribution as shown in Figure 11 2 is used when there is an equal probability that all of the possible measured values will occur If we set the probability density to a constant value of px A between the limits of a2 to 12 hence the total width of the distribution is a the probability integral can then be used to nd the amplitude A with respect to the width 1 a2 P LaZAdx Aa 1 11 20 hence the amplitude must be 1 A 7 11 21 a lt gt for this probability function to be normalizable Meanwhile a normal distribution re a Gaussian distribu tion can be used to describe the distribution of random events or observations This distribution function is shown in Figure ll 3 and described by the equation 1 2 2 px ie lx ll 22 27w a This distribution is centered around the mean u b Here a is the standard deviation of the distribution c The full width at half marimum FWHM P is related to the standard deviation by r 23540 11 23 1 The probable error of a normalized distribution is de ned to be the absolute value of the deviation x u such that the probability for the deviation of any random observation u to be less is equal to 12 gt that is 118 6c5 4c5 2c5 0 26 46 66 Figure ll 3 Probability density function for a normal distribution half of the observations of an event are expected to fall within the boundaries denoted by u l PE PE 067450 02865 F ll 24 The statistical interpretation of the wave function 69 Eq ll 3 says that ll06t2 is the probability density for nding the particle at point x at time t This dictates the following normal ization f lqxt2dx 1 1125 Without this the statistical interpretation would be nonsense a But is this normalization consistent with Schrodinger s equation ie Eq 11 1 That is is it really valid for all time b Let us take the time derivative of the LHS of Eq ll 25 then d 00 2 i 00 9 2 Lw Mam mi700 Btumm dx 1126 119 C d e f g h Note that the integral is a function only of t since the x terrns will disappear when the limits are applied As such a total derivative ddt is taken for the solution to the integral on the LHS of Eq ll 26 but the integrand e the function inside the integral is a function of x as well as If so a partial derivative 8815 must be used when the derivative is taken inside the integral on the RHS of Eq ll 26 By the product rule we get 8 2 8 6 i Bkl Bkllquot i kl i l Bit at l ll 27 Schrodinger s equation says that Bill 7 E 78 q 3W 1128 8t 7 2m 9x2 h and hence also using Schrodinger s equation for the com pleX conjugate of Eq ll 27 gives 39 2 39 Bkl thkl qu 8t 8562 h mg Using these equations in Eq ll 27 gives 8 lqjlg i ih ngl 8Nquot 87 i 872 8x2 8 ih Bkl Bkllquot a a ax 39 3930 The integral in Eq ll 25 can now be evaluated explicitly d 00 2 2h Bkl Bkllquot 00 m mm dxi w 5 8x 700 ll 31 lx t must go to zero as x gt loo otherwise the wave function would not be norrnalizable It follows that if lqxt2dx 0 1132 dt foo H 10 and hence that the integral on the left is constant inde pendent of time gt if ll is normalized at t 0 it stays normalized for all future time QED Example 1171 Consider the wave function lxt Ae 2z2ge iEth a Find the value of the constant A b If ll 05 at x 0 and t 0 what is the value of Solution a The total probability must equal unity as such f 12dx xingdag 1 The complex conjugate of our wave function is lxt Ae 2IQQeiEth so our normalization equation becomes 00 9 00 7212 71 LOO 1 xt11xtdxA2 LOO e 5 dxAgdgl 2 14 A 7T So our wave function is Z 14 2 2 lxt e75 I e ZEth 7T 01quot Solution b Set ll0 0 05 in the equation above and solve for 4 52 1 o o 1 l 0 0 7 7 lt gt W e e 2 ll 11 01quot D Momentum 1 For a particle in state l the expectation value of x is 00 x LOO lxtxllxtdx The expectation value is the average of repeated measurements on an ensemble of identically prepared systems Lot the average of repeated measurements on one and the same system 1133 The rate of change of this expectation value is dltxgt d 00 00 Bkl 8le 7 7 l l llquot 7 dt dt 00 x dx L00 lt 06 8t l 8t 06 if dx ll 34 a Substituting for Bil8t and my815 from the Schrodinger equation 69 Eq ll l gives dltxgt i 239 00 h ngl 71 c g 700 x ma wq 2 2 9 23 88 Vkllquot xkll dx ll 35 b The terms involving V cancel out and we have dx ih 00 96 ngl ngllquot kl 7 Wm day 1136 x 700 8x2 c Separate out the second term of the integrand and inte grate by parts as follows 00 00 8quot 8V 00 Wquot 958 lf 00267 8x2 dxi 35 Lw 806 1906 dx 39 T T 700 739 du 1137 H 12 d e f g Since ll must be in the form of a group of limited spatial extent in order that the uncertainty in the x coordinate be relatively small both the wave function and its derivatives must go to zero faster than 06 gt loo Consequently the integrand term is equal to zero and we have oo 8quot 008 l 8le Ooxllidx l x 7 8x2 700 806 806 x ll 38 unAT lntegrating by parts again 008xl8l amp f 00 BZMI L00 8x 8x 067 806 ll Loom 9062 dx ll 39 Reducing this again gives the following result for Eq ll 38 oo 8quot 00 9682 Putting this back into Eq ll 36 we have dltxgt 7 t 00 82 82 Consider the bracket in the integrand it can be written 82me any 83 x 8x2 8x2 7 x 9062 806 x 806 i 82KII 82q18q18q1 x 872 872 a Bkl 2 ll 42 Consequently dltxgt ih 00 Bkl 7 7 l7 11 43 dt m 700 lt 806 dx ll 13 h At this point we will postulate that the expectation value of the velocity of the particle is equal to the time derivative of the expectation value of the position of the particle dlt06gt 7 11 44 ltvgt d lt gt i Eq ll 43 tells us then how to calculate v directly from ll 00 Actually it is customary to work with momentum p mv rather than velocity i 00 p 2 mW i Zh LOO kl a dag 1145 a Let s write expressions for and p in a more suggestive way an f llxlfdx ll46 00 6 h 9 p 700 1 g a 1 day 1147 b We say that the operator 06 represents position and the operator represents momentum in quantum mechanics gt to calculate expectation values we sand wich the appropriate operator between l and ll and in tegrate 4 All such dynamic variables can be written in terms of position and momentum Kinetic energy is p h2 00 82 T77 llquot ikld 11 48 lt gt 2m 2m LOO 9062 x E The Uncertainty Principle 1 The uncertainty in the measurement of an event is nothing more than the standard deviation 0 02 of the measurement eg Eq 11 19 ll 14 2 Hence the Heisenberg Uncertainty Principle can be rewrit ten in the form h Ox 0 2 E ll 49 where 01 is the standard deviation in the X position and Up is the standard deviation in the corresponding momentum 3 We shall see from this point forward that wave functions that describe real particles always obey Eq ll 49 We shall prove this relation in lV of the notes Example 1172 Consider the wave function of Example 1 Z 14 x 7 67321 224 ZEth 7 7T Show that this wave function satisfies the Heisenberg Uncertainty Relation ship Solution First calculate the various expectation values 00 52 752 2 24E h 7 52 2 2 39Et h ltxgt 76 IZxeIZdx 00 W 2 00 22 i xe xdx0 7T 700 since we are integrating an odd function over an even interval 00 2 2 2 2 2 ltx2gt 5 5 z 272Eth x2 5 z 2zEth dx 2 00 22 2 00 22 xge fdeEKY mge fdx 2 If 1 7 7T4 2 Z Q Z ll 15 00 g 2 2 h a 2 2 7 73 z 272Eth if 73 z 2zEth p Lwi e iaxgte dx 7 g h 00 752122 2 752122 i A W i L00 6 xe dx 2 00 2 2 i g xe z 1060 7T 2 00 again since we are integrating and odd function over an even interval 2 E267321227iEth 6732122iEthdx foo 7T 2 806 7 g 2 00 752122 92 752122 i LOO 6 87x26 dx 7 g 2 00 752122 9 2 752122 7 h7006 lt xe gtdx i 5 2 2 00 752122 752122 2 752122 Ch e e xlt xgte dx 00 192 32 2 2 073212 22 h Lwe 1 5mm Eh g 00 632 2x2 6329 dx 7T 700 h g 5212 dx gxg 5212 dx F ggg 0072z2 2 002722 Ch pO 65 dx Z O xe xdx 32 1 1 CW 2 2 gf 2 g 32 2 2 7T 1 7T CWNEV Ef i 52 2 2 1 7T 1 2 2 WW2 M2 11 16 Finally using these values in the de nition of the standard deviations Oz lt062gtlt06gt2 i 1 0i 1 i 1 i W W W and 0p p2gtltpgt2 i 1 i 1 ixih iffy of ng 2 So WIFE niLMJh V 2539 2 45 2 H 17 Physics 46175617 Quantum Physics Course Lecture Notes Dr Donald G Luttermoser East Tennessee State University Edition 51 Abstract These Class notes are designed for use of the instructor and students of the course Physics 46175617 Quantum Physics This edition was last modi ed for the Fall 2006 semester VI Angular Momentum and Spin A Angular Momentum 1 In classical mechanics the angular momentum of a particle with respect to the origin is L 739 X p VI 1 Using the rule of cross products this gives three separate equa tions in Cartesian coordinates LI ypz Zpy Ly 2px mp2 and L2 xpy ypz VI 2 Using Eq V Z the corresponding quantum operators are 71 8 8 h 8 8 h 8 8 L2 7 7 7 VI 5 239 x8y y806gt B Eigenvalues of Angular Momentum 1 We rst ask the question do the components of the angular mo mentum operator commute To answer this question let s intro duce a test function fx y 2 that the L1 and Ly operators will act upon Then we can write E MW 22 952 ltZwgt f 8 8 8 8 g ya f l yltz gtyltgt V171 1 y 82f 82f 2 32f mg i a Z 2 282806 yszg Z By x 32f 32f 2 32f 32f lzx ydz ZyBx 82 Z 806 By l 0mg if 82f may xzazay All the terms cancel in pairs by Virtue of the equality of cross derivatives except two Ln Ly f 218 f 272sz and we conclude dropping the test function that L1 Ly z hLz VI6 By cyclic permutation of the indices it follows that 1sz 239th and 122111 ihLy VI 7 Exercise Prove Eq VI 7 by the technique shown in VB1 Since Li Ly and L2 do not commute they are incompatible ob servables Hence according to the generalized uncertainty prin ciple Eq lV lOO we can write 1 2 aim fig 2 WELL Z ltL2gt27 01quot h at at 2 5 ltLzgt VI8 V172 5 It would therefore be futile to look for states that are simultane ously eigenfunctions of LI and of Ly On the other hand the square of the total angular momentum does commute with separately LI Ly and L2 31 The square of the total angular momentum is L2 2 Li Li L3 VI9 b Using the de nition of the commutator Eq lV 42 along with Eqs IV 43 VI 6 VI 7 and the following com mutator relations VI10 VI11 lAaBl 2121 0 21 we can write WM 7 Lin Liam Lilac L1 L1 L1 L1 L1 L1 Ly La LI La LI Ly Lz L2 Lzl L2 Lxl L2 0 0 Ly z39hLz z39hLzLy Lzo39hLy z hLyLz 0 Likewise the same proof can be made for the Ly and L2 components giVing L2LI0 L2Ly0 L2Lzl0 VI12 or more compactly L2 L 0 VI13 V173 Since L2 is compatible with each component of L we don t need to simultaneously nd eigenstates for L2 and each component of L instead only one component will suf ce that is to say that the eigenstates of the other two components will be identical to the one we work on a Let us use L2 as our test component Then we can write the following two eigenfunction equations Lgf Af and sz uf VI 14 b We will now use a ladder operator technique similar to the one used for the harmonic oscillator lllE5 Let Li E LI l iLy VI 15 c lts commutator with L2 is L2 Li 122121 I 239 L2 Ly ihLy l i z hLI ihLI l iLy so L2 Li ihLi VI16 1 Since each component of L is compatible with L2 as per Eq Vl 12 the following commutator results L2 Lil 0 VI17 Exercise Prove Eq VI 17 e We can prove that Li f is an eigenfunction by making use of Eqs VI 14 and VI l7 and substituting Lif for f then l o L2Lilf L2LifLiL2f 0 Lu LAW Lif V174 L2 L2 10 11 and then from the rst equation of Eq VI 14 we get L2Lif Lifa V148 so Li f is an eigenfunction of L2 with the same eigenvalue A Now let s take a look at the second eigenfunction equation of Eq Vl 14 For this equation let s replace f with Li f and write LzLif Lisz Lisz L2 Lil f Liltszgt ihLif LAM M i hXLif so Li f is an eigenfunction of L2 with the new eigenvalues u l h v119 L is called the raising operator because it increases the eigen value of L2 by h L is called the lowering operator because it decreases the eigenvalue of L2 by h We also can investigate various relationships with these angular momentum ladder operators LL LI iLyLI my Li iLILy 2391ny L3 Li L z39LILy VI20 LI iLyLz iLy Li iLzLy 2391ny L3 Li L iLILyl v121 V175 12 and the commutator of L and L results from subtracting Eq VI 21 from Eq VI 20 which gives 4 1 LL LL 2239L1Ly ZhLz VI 22 Exercise Verify the following commutator relations L2 14 hL VI 23 L2 1 hL VI 24 L2 L4 L2 L 0 VI25 If we add Eqs VI 20 and VI 21 we get LL LL 2LZ L3 2L2 L3 VI26 Finally if we use Eq VI 26 along with Eq VI 9 we get L2 Li Li L l LL7 l LL l 14L hLz L LL hLZ L3 v127 VI28 Exercise Prove Eqs VI 27 and VI 28 For a given value of A in Eq VI 14 a ladder of states is obtained with each rung separated from its neighbors by one unit of h in the eigenvalue of L2 see Figure Vl l a To ascend the ladder we apply the raising operatorr b To descend the ladder we apply the lowering operator c This procedure cannot go on forever 7 eventually a state is reached for which the z component exceeds the total angular momentum which is impossible see below V176 7 ft u3h L3f u2h L2f uh Lf u f uh Lf u2h L2f u3h L3f Th fb Figure Vlil The ladder angular momentum states 13 As such there must exist a top rung ft such that Lft 0 VI 29 Let Eh be the eigenvalue of L2 at this toip rung szt 15h ft L219 Aft VI30 a Using our original eigenvalue equations e Eq Vl l4 we can write Li Lin L2 LEM A Mm VI 31 b Since LI Ly L2 and L2 are Hermitian operators with real eigenvalues the square of these eigenvalues must be positive numbers Thus the operator Li L must have a positive eigenvalue and A 2 u 2232 hence supporting the statement made in VliB12ci V177 c Using Eq VI 29 it is obvious that LLft LLft L0 0 VI 32 1 Now using the second equation of Eq VI 30 in conjunc tion with the rst equation of Eq VI 30 along with Eq VI 28 we can write Lgft 111714r l Eh 2h2ft W12 WM W 1W Aft v133 14 Similarly there must exist a bottom rung fb such that 1 2 0 VI 34 Let Wt be the eigenvalue of L2 at this bottom rung then szb thb Lgfb Afb VI 35 a Using Eq VI 34 we see that nub mum Llt0gt 0 VI36gt b Now using the second equation of Eq VI 35 in conjunc tion with the rst equation of Eq VI 35 along with Eq VI 27 we can write L2fb LL hLz Lfb 0 Zh2 Z2hgfb Zhg Z2h2fb a 1h2fb Afb 0137 15 It is clear from the comparison of Eq VI 33 with Eq VI 37 that A g 55 1 1 VI 38 a This implies that either Z E or Z E 1 We can reject the second solution since 6 is de ned to be the top most rung and this second equation would place E above 6 V178 16 b Let N be the number of steps one must make from the bottom 6 state to get to the top 6 state then N E Z E E 01quot e v139 2 c At this point let s note that L2 the angular momentum squared has the same units as h squared remember that E is either an integer or half integer as shown in Eq Vl 39 hence the angular momentum has units of h As such let s rewrite the eigenvalue of the L2 operator as u mh d From these last two points we see that the maximum value of m is 6 see Eq Vl BO and the minimum value of m is E see Eq Vl 35 where E can be an integer or half integer e From this formalism it is clear that there are 26 1 rungs 239e values of m in the angular momentum lad der 65 16 2 10 1 E as shown in Figure VI 2 Since the eigenvalues are composed of 6 s and m s so too must be the eigenfunctions As such we can now write the two angular momentum eigenvalue equations as L72 M 1h2fg L2 2 mhfg v140 where 1 3 60 1 m E1E 1E VI 41 This is the reason Why we chose 6 1 as our differential equation constant for the Schrodinger equation in Eqs V 14 and V 15 V179 Figure Vl72 The allowed projections of the angular momentum for the cases of Z 1 2 and 3 Example V171 The angular momentum operators a Prove that if f is simultaneously an eigenfunction of L2 and L2 the square of the eigenvalue of L2 cannot exceed the eigenvalue of L2 Using Eq VI 14 we have L2 ltfL2fgt ltfL2fgt WV Mflf A and M ltLZ Li LZgt ltLZgt ltL gt 112 But L1 is Hermitian so Li flLi fole llLacfll2 2 0 and likewise Ly 2 0 So b As it turns out see Equations Vl 40 and Vl 41 the square of the eigenvalue of L2 never even equals the eigenvalue of L2 except in the special case 6 m 0 Comment on the implications of this result Show that it is enforced by the uncertainty principle see Eq Vl 8 and explain how the special case gets away with it lmplication There is no state in which the angular momentum points in a speci c direction If there were then we would simultaneously know LI Ly and L2 7 but these are incompatible observables 7 if you know LI precisely you cannot know Ly and L2 precisely The uncertainty principle of Eq VI 8 says that L1 Ly and L2 cannot be simultaneously determined except in the special case LI Ly L2 0 which is to say when E 0 and m 0 C Eigenfunctions of Angular Momentum 1 We need to rewrite LI Ly and L2 see Eqs Vl 3 4 5 in spher ical coordinates a The angular momentum operator in vector notation is L go x V v142 where the gradient operator in spherical coordinates is A18 A 1 8 A8 andrrfso h A A 8 A A 8 A A 1 8 v144 V1711 v A A 1 a L gage 0mg v145 We now need to express the unit vectors and 25 in their Cartesian components cos6 cos b 56 cos6 sin b I sin 639 2 Q3 sin b 02 cos b 7 Using these unit vector relations in Eq VI 45 gives us h 8 L 7 A A 7 Vl 46 Z Sln xcos w80 A A A 1 8 cos6 cosq x cos6 sinq y s1n639z Looking at the Cartesian unit vector components for com parison with L L1 02 Ly L2 2 we see that h 8 8 LI 7 lt sinq aig cosq cot6 VI 47 h 8 19 Ly i lt cosq aig sinq cow VI 48 and h 8 Z 8 We shall also need the raising and lowering operators Li LI l iLy 8 8 sinq iz39cosq Big cosq iz39sinq cot gl but cosq l isinq eat so handling the raising and lowering operators separately we have Lh cosqisinicos isin cot gl 8 8 f 245 39 7 he lt86zcot08 gt V1712 e f and L h cos isin icos isinq cot iqsl coscz isinq icos isin cotOBEA Z 9 8 he lt876 zcot687 gt 8 8 iqu 39 Li ihe 786 l 2 cot 6 78 h SO wwm We are now in a position to determine f e It s an eigenfunction of L2 with an eigenvalue of mh h Bfgn Lzm747 ff 239 Bq i The solution to this simple differential equation is f2 90 emf mhfE wmn ii The function 90 is a constant of integration in b but it can still depend upon 6 iii Remember from Eq V 21 that q 6mm for the b component of the 3 D Schrodinger equation hence f2 90lt1gt VI52 We will now retrieve the functional form of 90 by making use of our second eigenvalue equation for the angular mo mentum operator Using the rst equation of Eq VI 40 in conjunction with Eq VI 27 we get V1713 L21 LL L hLz n he 5 Bf 239 cot 6 he m 8639 Bq 817 Bi 23 112 gtltlt86 zcot6 gas h Bag Z aq aw 1 f2 v153 i At this point let s make use of Eq VI 51 in Eq VI 53 Taking the 6 and b derivatives of Eq VI 51 we get 8f 39 45d 7 m 7 WM 88639 6 d6 gt g imeimdbg VI 55 19 ii Using these equations in Eq VI 53 gives 8 8 Bg 245 7 39 7 2m1 7 e lt86zcot08 gt e gtlt86mgcot6gt mggeim mgeim d dg dg zmdb e idg lt7d6mgcot0gtm 1cot6 d6mgcot6 gt mm Hg EM 1geim iii Canceling emms gives 129 d9 2 d9 W m cot0mg csc 0m 1 NOE mm 1 1 cot2 639 g d2 d T g cot6 dig mgg cs026 1g V1714 2 Note that Eq VI 56 is precisely the equation we determined for the 6 component of the three dimensional Schrodinger equation see Eq V 24 90 90 Using this realization in Eq VI 52 iv Now multiplying each term by sin produces d2 d sin2 6T sin6 cos6 mgg E 1g sin2 639 v This is a differential equation for 96 but we can write it in the more familiar form of i dig 2 2 i s1n6 d6 lts1n6d6gt 1 sin 6 m lgiO VI 56 means that a b C v f2 90 PM Yam As such spherical harmonics are precisely the nor malized eigenfunctions of L2 and L2 When we solved the Schrodinger equation by separation of variables in VA of the notes we were inadvertently con structing simultaneous eigenfunctions of the three com muting operators H L2 and L2 Ha Ea W E 1h2 sz mhzJ VI 58 There is a slight inconsistency here i The algebraic theory of angular momentum per mits E and hence also m to take on half integer values see Eq Vl 41 whereas the analytic method yielded eigenfunctions only for integer values see Eq V 28 V1715 v157 ii So does this mean that the half integer solutions are spurious The answer is no these half integer solutions are of profound importance as will be seen on the next section concerning spin They can however be ignored for the description of the or bital angular momentum Example V172 This example incorporates quantum mechanics in classical mechanics Assume two particles of mass M are attached to the ends of a massless rigid rod of length a The system is free to rotate in three dimensions about the center of the rod but the center point itself is xed This technique is actually used to describe rotational transitions in diatomic molecules a Show the allowed energies of this rigid rotor are hgn n 1 En for n 0 12 Hint First express the classical energy in terms of the total angular momentum Classically the energy of each mass is just going to be the kinetic energy of the mass potential energy is zero in this problem Even though the masses are revolving around a given point we will just use the standard de nition of kinetic energy here l2mv2 Then the Hamiltonian or the total energy will be 1 H 2 rm mug Now the angular momentum of a mass revolving around an axis is L Mm where 7 12 is the distance that the mass is from the axis Since there are two masses we have L 231m V1716 so L2 7 M 12 But from quantum mechanics we know the eigenvalues of L2 are hgf L2 12l2v2 and thus H l or since we usually label energies with n and E is the eigenvalue for H we get i hgnm 1 En M a2 n012 b What are the normalized eigenfunctions for this system What is the degeneracy of the nth energy level Since there is no radial component to this problem the eigenfunction is just going to be described by spherical harmonics nmga Yanx The degeneracy of the nth energy level is just the number of m values for a given n which is given in VlBil5e as D Spin and Particle Families 1 In classical mechanics a rigid body admits two kinds of angular momentum a Orbital L 739 X p associated with the motion g the center of mass Such motion is referred to as a revolution about the center of mass b Spin S Iw associated with the motion about the center of mass Such motion is referred to as a rotation about an axis V1717 c One then talks about the total angular momentum orbital L spin S angular momenta 2 By analogy we have the same description on the microscopic level for quantum mechanics a Orbital The motion of an electron about the nucleus of an atom as described by spherical harmonics with the orbital angular momentum quantum number 6 and the magnetic or azimuthal quantum number m This is sometimes referred to as the extrinsic angular momentum L b Spin Unlike the classical case this isn t the spin of the electron about an axis since the electron is a point particle even though we describe electron spin about an axis in elementary physics Here spin is nothing more than the intrinsic angular momentum S of the electron Since this is intrinsic spin angular momentum is independent of spatial coordinates r 6 3 In quantum mechanics the algebraic theory of spin is just a car bon copy of the theory of orbital angular momentum a It obeys the following fundamental commutation relations 3153 2725 3 32 ihSI 3231 ihSy VI59 b The eigenvectors 32 and 32 satisfy Sglsms 33 1h2sm8 Szlsms mshlsms VI 60 Note that the L2 and L2 eigenstates are eigenfunctions whereas the 32 and 32 eigenstates are eigenvectors As V1718 such Eq VI 60 uses the ket notation for the eigen state Note too that we could have used this ket notation for the orbital angular momentum eigenstate with the re alization that m E ng c We also can introduce ladder operators Si E SI 1 iSy where we have Liln Lil m I mHE 1 mm i 1 E m i 1 VI 61 for the orbital angular momentum ladder operators and Silsms h ss 1 m3m5 i 1 s mg i 1 VI 62 for the spin angular momentum ladder operators Exercise Prove Eq VI 61 by making use of the fact L2 and L2 are Hermitian and that Mm must be normalizable The spin angular momentum eigenvectors are not spherical har monics they are not functions of 6 and b at all As such there is no reason to exclude the half integer values of s and ms 1 3 s 0 1 a It so happens that every elementary particle has a speci c and immutable value of s which we call the spin of that m3 s s1s 13 VI 63 particular species b Particles with half integer spins are called fermions F ermions are said to have antisymmetrical wave functions c Particles with integer spins are called bosons Bosons are said to have symmetrical wave functions 1 The convolution ie joining together of antisymmetri cal wave functions go to zero as two identical particles V1719 5 approach each other As a result two fermions in the same quantum state exhibit mutual repulsion to avoid their combined wave functions going to zero This effect is known as the Pauli Exclusion Principle e Bosons have no such exclusion principle since the convo lution of symmetric wave functions do not go to zero as two identical particles approach each other Before going on further let s discuss particle physics There are two main groups of particles that make up all matter and energy a Elementary particles These are particles that make up matter They are subdivided into 3 groups i Leptons light particles include the electron 6 me 511 keV 1 keV 1000 eV 1 eV 160 X 10 19 Joules muon a mM 107 MeV and tau particle T m 1784 MeV each with a neg ative charge their respective neutrinos electron neutrino ye mike lt 30 eV muon neutrino UM myw lt 05 MeV and tau neutrino VT min lt 250 MeV each with no charge and the antiparti cles of each e called a positron Effm and 7 These particles do Lot participate in the strong interactions All leptons have spin of 1 2 hence are fermions ii Mesons are particles of intermediate mass that are made of quark antiquark pairs and include pi ons kaons and n particles All are unstable and decay Via weak or E M interactions All mesons have either 0 or integer spin hence are bosons V1720 iii Baryons heavy particles include the nucleons n neutrons 7 neutral particles and p protons 7 positive charged and the more massive hyperons re AEE and Baryons are composed of a triplet of quarks Each baryon has an antibaryon associated with it and has a spin of either 12 or 32 hence are fermions b Field particles These particles mediate the 4 natural forces and are sometimes referred to as the energy parti cles 7 all are bosons Here they are listed in order from strongest to weakest i Gluons Mediate the strong nuclear force Strength of this force is described by the color charge ii Photons Mediate the electromagnetic force Strength of this force is described by the electric charge iii Intermediate vector W amp Z bosons some times referred to as weakons Mediate the weak nuclear force Strength of this force is described by the weak charge iv Gravitons Mediate the force of gravity Strength of this force is described by the mass From the above list of elementary particles there seems to be only 2 types of basic particles leptons which do not obey the strong force and quarks which do obey the strong force Particles that participate in the strong force are also called hadrons 7 so obviously the hadrons are those particles composed of quarks re the mesons and baryons All quarks have a spin of 1 2 V1721 There are 6 flavors of leptons as describe above As such it was theorized and later observed 6 avors or colors of quarks and an additional 6 antiquarks must exist a Up u quark has a rest energy of 360 MeV 1 MeV 106 eV and a charge of e b Down d quark has a rest energy of 360 MeV and a charge of e c Charmed c quark has a rest energy of 1500 MeV and a charge of e 1 Strange s quark has a rest energy of 540 MeV and a 1 charge of 6 e Top t quark has a rest energy of 170 GeV 1 GeV 109 eV and a charge of 4 e f Bottom b quark has a rest energy of5 GeV and a charge of l e 3 Note that a proton is composed of 2 u and a d quark and a neutron composed of an u and 2 d quarks The theory on how quarks interact with each other is called quantum chromodynamics in analogy with quantum me chanics One interesting result of this theory is that quarks cannot exist in isolation they must always travel in groups of 2 to 3 quarks As described above the four forces in relativistic quantum me chanics are mediated by the exchange of integer spin particles bosons V1722 MATTER amp ENERGY FORCES CON STITUE NTS Quarks u c t dlslb Leptons e u I ve v vT The Standard Model of Particle Physics Figure Vl73 The Standard Model is the current best description of the subatomic world a Of the four forces only gravity gives rise to attractive forces between like particles same type of color charge electric charge weak charge or mass b This difference arises because the graviton is spin 2 whereas the gluon photon and weakon are spin 1 see Table VI l 10 This description of elementary and eld particles is called the Standard Model A graphical representation of this model is shown in Figure Vl 3 11 Getting back to angular momentum 5pm can take on any half integer or integer value for s As mentioned above each particle has its own speci c spin see Table VI l that remains xed In contrast the orbital angular momentum quantum number 6 can take on any integer value you please and will change from one to another when the system is perturbed V1723 Table V171 Spin quantum numbers for a sample of elementary and eld particles Common Particle Spin Spin Name Symb 01 Type 5 Family Pion 7T meson 0 boson 7T0 meson 0 boson Electron 6 lepton fermion Muon if lepton fermion Neutrino V5 lepton fermion Proton p baryon fermion Neutron n baryon fermion Gluon G eld boson Photon 7 eld 1 boson Weakon W eld 1 boson Delta A baryon fermion Graviton g eld 2 boson T 7 The superscript in the symbol corresponds to the charge of the particle UH positive 77 negative 07 neutral Symbols with no super script are neutral except for the proton which is positively charged and the weakons which can have a 7 or no electric charge E Spin 12 Particles 1 In the science of quantum mechanics particles with s 12 are the most important since these are the particles that make up ordinary matter protons neutrons and electrons as well as all quarks and leptons 2 There are just two eigenstates for these type of fermions a E the spin up state b E the spin down state 3 Using these as basis vectors the general state of a spin 12 par ticle can be expressed as a two element column matrix called a V1724 spinor X lt Z gt aX bxia VI64 with m lt 1 VI65gt representing spin up and X lt 1 VI66 for spin down The spin operators become 2 X 2 matrices which we can work out by noting their effect on X and 9 a Eq VI 60 says 3 3 32m 1h2X ngi lhgxi VI67 h h SzX X 9in X7 VI 68 b Eq VI 62 says SX7 hm 51m xi SX Sixi 0 VI 69 c Now Si SE iSy so 1 1 SI 8 and Sy 2743 3 v170 1 Using the above equations it is easy to show that h h SIM X7 Szxi 5m VI71 h h SyX 27X7 i SyXi 27X VI 72 V1725 e Thus in matrix form we can write 3 10 01 00 2 72 7 n 341201 311200 Shlt10gt VI73 while h 0 1 h 0 239 h 1 0 31210 Sy2lt Uls i 0 1 VI74 f Finally let s de ne the Pauli spin matrices 039 such that the eigenvector S h2a then 0i010i0 0i10 110 y 0 20 139 v175 Notice that 8 Sy 82 and 82 are all Hermitiom as they should be since they represent observables However 3 and S are not Hermitian evidently they are not observ able v g 5 Based on Eqs VI 65 VI 66 and VI 68 the eigenspmors of 32 are i 1 1 1 i 0 1 1 X1 7 0 eigenva ue 2 x i 1 eigenva ue 2 VI 76 a If you measure 32 on a particle in the general state X see Eq Vl 64 you get h2 with probability a2 or h2 with probability b2 b Since these are the only possibilities a2 b2 1 VI77 e the spinor must be normalized V1726 6 Bound electrons can carry out a spin ip transition When an electron in the ground state has a spin parallel to the proton s spin la both are spinning in the same direction it has a slightly higher energy than when the spins are antiparallel la the electron is spinning in the opposite direction with respect to the proton a An electron in the parallel state can spontaneously decay to the antiparallel state giving rise to an emission line at 21 cm at radio wavelengths b This transition is highly forbidden however it has a very low oscillator strength with an Einstein A value of A21 285 X 10 15 sec 1 remember the Lyman oz transition has an Einstein A value of A21 662 X 108 sec l c Such a transition is called a hyper ne transition gt transitions involving electron spin ips 1 Of course hydrogen can also absorb a 21 cm photon pro ducing an absorption line at this wavelength e Much of the interstellar medium is composed of cold hy drogen gas As such the structure of the Milky Way Galaxy has been mapped via this 21 cm line note that photons at this wavelength are unaffected by dust and as such can travel great distances through the galactic plane which is a very dusty place Example V173 Suppose a spin 12 particle is in the state i 11 What are the expectation values for SI and Sy What is the probability of V1727 getting h2 and the probability of getting h2 if you measure 32 Solution Since we are working with vectors here we use Eq IV 41 to de ne the inner product of two vectors ltXIXgt XT X where Xl 2quot As such the expectation value for spin is determined from XlSix where one substitutes x or y for the 239 in this equation Then SI 7 xlSzXlt16i 76gt ff2 hf a f 172 2 h h 7 lt66gtltawmc gt6 7076 O in h h hih 7767 67 7 ltSygt 7 XTSyXlt1Ti 16gt 432 Oiyaw 17239 2 h 39 h39 W w lti 1h2 gt6Z1gt6Z1 i h h 1 hi h 777677677 Following the formalism given in VlEl5 if you measure 32 the proba bility of obtaining h 2 is 1 21 1 1 12 i 2 7 M l 6 l 6 6 6 3 and the probability of obtaining h 2 is WHEAJ 7 6 7 6 7 339 V1728 Finally checking that Eq VI 77 is satis ed we have 1 2 Example V174 An electron is in the spin state xAltfgt a Determine the normalization constant A b Find the expectation values of SI Sy and 32 c Find the uncertainties 0505 05y and 03 d Confirm that your results are consistent with all three uncertainty prin ciples see Eq Vl 8 and its cyclic permutations only with S in place of L Solution a As always we determine the constant A by normalizing x so 1 XlX IAI2 3i3i 44l IAI2 9 16gt 25 IAIZ SO Solution b Using the de nition of the expectation value we get 1 h 0 1 3239 7 i i 7 7 39 31gt XSIX252lt 32 4 lt1 0 lt4 V1729 Elt3 4 lt 4 5LE 12 12 1h 0 z 32 7 T if ltgygtXSyX 2k392 324gtlt 0gtlt4gt h 42 h 24 12 2 Solution c For spin 12 systems it is always true that hZ 33gtltSZgtltSZgtZ SO 2 2 2 752 05 ltSzgtlt81gt Z 7 a E 512 h 122 h 49 2 i 2 27 7 2 7 2 051145 5 4 lt25 h 72500 625 576 2500h 7 03yh V1730 7 2 h 576 2 i 2 27 7 2 7 2 5143 lt3 4 lt50 h 2500 625 49 2500 12 03952 Solution d Here we want to show that 71 0305 2 E ltSkgt7 Where indices 239 j and k represent the cyclic permutations of the Cartesian coordinates As such h 771 7 h 7h 05 05y E 57 2 E E 57 right it the uncertainty limit 7 12 7 05y Us T 2 g 0 obeys the uncertainty limit 12 7 12 Us 05 2 g right it the uncertainty limit V1731 Physics 46175617 Quantum Physics Course Lecture Notes Dr Donald G Luttermoser East Tennessee State University Edition 51 Abstract These Class notes are designed for use of the instructor and students of the course Physics 46175617 Quantum Physics This edition was last modi ed for the Fall 2006 semester IV Formulism and Techniques A Linear Algebra 1 In classical mechanics vectors are typically de ned in Cartesian coordinates as A A A a 0111 oxyy azz IV 1 with the hat unit vector notation or aozxz390zyj 0ch in the ijk unit vector notation I prefer the use of the hat nota tion a Vectors are added via the component method such that A a w ax Mac my way a2 we Iv 2 b However in quantum mechanics often we will have more than 3 coordinates to worry about indeed sometimes there may be an in nite amount of coordinates c As such we will introduce a new notation the so called bra and ket notation to describe vectors lagt ket 0 bra that was rst introduced by Paul Dirac a a IV3 d The Dirac bra and ket notation has the following mean ings N WW a 04 if vectors oz and represent discrete e bound states IV4 and lta gt LOO cm d7 lVl IV 5 for continuous e free states given by functions oz and with d7 day in 1 D space and d7 dx dy dz in 3 D Cartesian space V 2 A vector space consists of a set of vectors a W W together with a set of real or complex scalars a b c which are subject to 2 operations a Vector addition The sum of any 2 vectors is another vector la W l7 IV6 i Vector addition is commutative la W W la IV7 ii Vector addition is associative 01gt gt7gt la W IV IV8 iii There exists a zero or null vector 0 with the property that la IO la a IV9 for every vector a iv For every vector a there is an associated in verse vector a such that la I oz l0 IV 10 b Scalar multiplication The product of any scalar with any vector is another vector ala ly IV 11 1V2 C i Scalar multiplication is distributive with respect to vector addition alagt W Ila al a IV12 and with respect to scalar addition a bagt ala bla IV 13 ii It is also associative abagt abagt IV 14 iii Multiplications by the null and unit vector are Ola l0 llal la IV15 Note that oz lagt A linear combination of the vectors agt W W an expression of the form aagt bl c7gt i A vector M is said to be linearly independent of the set agt W W a linear combination of them eg unit vectors 1 3 and if it cannot be written as ii A collection of vectors is said to span the space if every vector can be written as a linear combination of the members of this set iii A set of linearly independent vectors that spans the space is called a basis gt 95y de ne the Cartesian basis 1V3 l S iv The number of vectors in any basis is called the dimension of the space Here we will introduce the nite bases analogous to unit vectors 1gta 2gtaa ngt a of any given vector la a1e1 agleg anen IV 16 which is uniquely represented by the ordered 71 tuple of its components lagt H aha27 aan v It is often easier to work with components than with the abstract vectors themselves Use whatever method to which you are most comfortable ln 3 dimensions we encounter 2 kinds of vector products the dot product and the cross product The latter does not generalize in any natural way to n dimensional vector spaces but the former does and is called the inner product a The inner product of 2 vectors a and is a com pleX number which we write as oz with the following properties lt lagt OIW IV18 ala 2 0 amp ala 0 ltgt la 0 IV 19 WWW HM bltal gt Cltalvgt IV20 b A vector space with an inner product is called an inner product space c Because the inner product of any vector with itself is a non negative number Eq lV 19 its square root is real 1V4 d e f g h we call this the norm think of this as the length of the vector llall E Mala A unit vector whose norm is l is said to be normalized IV21 Two vectors whose inner product is zero are called or thogonal gt a collection of mutually orthogonal nor malized vectors dzlog 6H IV 22 is called an orthonormal set where 61 is the Kro necker delta Components of vectors can be written as a clla IV 23 The conditions given in Eqs IV 181920 give rise to the Schwarz inequality which states that ltCv gt2 S alan see its proof in Example lV l Note that the SchwarZ inequality holds only if the vectors a and W are colinear IV24 ie proportional to each other a We can de ne the complex angle between a and by the formula IV25 A linear transformation T the hat on an operator from this point forward will imply that the operator is a linear transfor mation don t confuse it with the hat of a unit vector takes 1V5 each vector in a vector space and transforms it into some other vector agt gt ogt ag with the proviso that the operator is linear TWO bl l amal Wm IV26 a We can write the linear transformation of basis vectors as ej TZ ei j l2n IV27 hence the T operator is a tensor b If agt is an arbitrary vector la a1e1gt anlengt aj jgt IV 28 then Tagt aJTle aimlea Z Z Tani leigt j1 j1 i1 i1 j1 IV29 T takes a vector with components 11 a2 an into a vec tor with components a Z Ejaj j1 c If the basis is orthonormal it follows from Eq IV 27 that Tij eilflejl a IV31 or in matrix notation T11 T12 Tln T21 T22 T2n T IV32 Tnl Tn Tnn 1V6 d e f g The sum of 2 linear transformations is 3 Tagt 5M Tloz IV33 or again in matrix notation The product of 2 linear transformations is the net effect of performing them in succession rst T the 3 ln matrix notation U ST ltgt Uik Z SijTJk IV 35 j1 this is the standard rule for matrix multiplication to nd the ikth element of the product you look at the ith row of S and the 16th column of T multiply corresponding entries and add The transpose of a matrix T is the same set of elements in T but with the rows and columns interchanged T11 T21 39 Tnl T T Tn T 12 22 IV36 Note that the transpose of a column matrix is a row ma trixl A square matrix is symmetric if it is equal to its trans pose re ection in the main diagonal upper left to lower right operation reverses the sign leaves it unchanged it is antisymmetric if this SYMMETRIC T T ANTISYMMETRIC T T IV37 1V7 h k The complex conjugate T is obtained by taking the complex conjugate of every element Th TF2 Tin ll Tquot 21 2 aquot a IV38 T31 T32 Tnln a A matrix is real if all its elements are real and imaginary if they are all imaginary REAL Tquot T lMAGlNARY Tquot T IV 39 A square matrix is Hermitian or selfadjoint as de ned by Tl E if it is equal to its Hermitian conjugate if Hermitian conjugation introduces a minus sign the ma trix is skew Hermitian or antiHermitian SKEW HERMITIAN Tl T IV40 HERMITIAN Tl T 7 With this notation the inner product of 2 vectors with respect to an orthonormal basis can be written in matrix form aw alb IV41 Matrix multiplication is not in general commutative ST y Ts commutator the difference between 2 orderings is called the 8 T E ST TS IV42 It can also be shown that one can write the following commutator relation 213 C A 1 O 4 C B IV43 Exercise Prove Eq IV 43 1V8 III II 0 The transpose of a product is the product of the transpose m reverse order SNT TS IV44 and the same goes for Hermitian conjugates STl TlSl IV45 The unit matrix is de ned by 1 0 0 1 IV46 j j 1 In other words 117 6 IV 47 The inverse of a matrix written T4 is de ned by T lT TT l 1 IV48 i A matrix has an inverse if and only if its deter minant is nonzero in fact 1 717070 detT T where C is the matrix of cofactors 1 quot IV49 ii The oofaotor of element Ti is lij times the determinant of the submatrix obtained from T by erasing the ith row by the jth column iii As an example for taking the inverse of a matrix let s assume that T is a 3x3 matrix of form T11 T12 T13 T21 T22 T23 T31 T32 T33 T IV50 1V9 Its determinant is then T11 T12 T13 detT T21 T22 T23 T31 T32 T33 T22 T23 T32 T33 T21 T22 T31 T32 T21 T23 T11 T31 T33 T1 T13 T11 T22ng T23T32 T12 T21ng T23T31 T13 T21T32 T22T31 IV 51 iv For this 3X3 matrix the matrix of cofactors is given by T22 T23 T21 T23 T21 T22 T32 T33 T31 T33 T31 T32 C T12 T13 T11T13 T11T12 T32 T33 T31 T33 T31 T32 T12 T13 T11 T13 T11 T12 T22 T23 T21 T23 T21 T22 IV52 v The transpose of this cofactor matrix is then see Eq IV 36 T22 T32 T12 T32 T12 T22 T23 T33 T13 T33 T13 T23 C T21T31 T11T31 T11T21 T23 T33 T13 T33 T13 T23 T21 T31 T11 T31 T11 T21 T22 T32 T12 T32 T12 T22 IV53 vi A matrix without an inverse is said to be singu lar IV 10 vii The inverse of a product assuming it exists is the product of the inverses m reverse order ST 1 T IS l IV54 p A matrix is unitary if its inverse is equal to its Hermitian conjugate UNITARY Ul U l IV55 q The trace of a matrix is the sum of the diagonal elements TrT E ZTZi IV 56 2391 and has the property T1quotT1T2 T1quotT2T1 A vector under a linear transformation that obeys the following equation a Ma IV58 where a is called the eigenvector of the transformation and the complex number A is called the eigenvalue Such an equa tion shows that a linear transformation creates a scaled dupli cate by a factor of A of the original vector a a Notice that any nonzero multiple of an eigenvector is still an eigenvector with the same eigenvalue b In matrix form the eigenvector equation takes the form Ta Aa IV 59 for nonzero a or T A1a 0 IV 60 here 0 is the zero matrix whose elements are all zero IV 11 C d If the matrix T A1 had an inverse we could multiply both sides of Eq IV 60 by T A1 1 and conclude that a 0 But by assumption a is not zero so the matrix T A1 must in fact be singular which means that its determinant vanishes T11 X T12 Tln T T A T n detT 1 21 l 22 l 0 Tnl Tn IV61 Expansion of the determinant yields an algebraic equation for A om Cn1 1 0M Co 0 IV62 where the coe icients C depend on the elements of T This is called the characteristic equation for the matrix its solutions determine the eigenvalues Note that it is an nth order equation so it has n complex roots i Some of these root may be duplicates so all we can say for certain is that an n x n matrix has at least one and at most 71 distinct eigenvalues ii In the cases where duplicates exist such states are said to be degenerate iii To construct the corresponding eigenvectors it is generally easiest simply to plug each A back into Eq IV 59 and solve by hand for the components of a see Examples lV 3 and lV 4 IV 12 In many physical problems involving matrices in both classical mechanics and quantum mechanics it is desirable to carry out a real orthogonal similarity transformation or a unitary transfor mation to reduce the matrix to its diagonal form e all non diagonal elements equal to zero a If eigenvectors span the space we are free to use them as a basis Tlf1gt A1If1gt Tlf2gt A2f2gt b The matrix representing T takes on a very simple form in this basis with the eigenvalues strung out along the main diagonal and all other elements zero A1 0 0 0 A 0 T 2 IV 63 0 0 An c The normalized eigenvectors are equally simple 1 0 0 0 l 0 am 0 a2 0 a 0 IV 64 0 0 l d A matrix that can be brought to diagonal form Eq lV 63 by change of basis is said to be diagonalizable e In a geometrical sense diagonaliZing a matrix is equiva lent to rotating the bases of a matrix about some point IV 13 in the space until all of the off diagonal elements go to zero If D is the diagonalized matrix of matrix M the operation that diagonalizes M is D SMS l IV65 where matrix S is called a similarity transformation Note that the inverse of the similarity matrix can be con structed by using the eigenvectors in the old basis as the columns of 84 84 am IV66 f There is great advantage in bringing a matrix to diagonal form it is much easier to work with Unfortunately not every matrix can be diagonalized the eigenvec tors have to span the space for a matrix to be diagonalizable The Hermitian conjugate of a linear transformation called a Hermitian transformation is that transformation Tl which when applied to the rst member of an inner product gives the same result as if T itself had been applied to the second vector flaw OITW IV57 for all vectors a and a Note that in the notation used in Eq IV 63 Tlon means the inner product of the vector Tla b Note that we can also write gum alTb Tlalb aw IV68 c In quantum mechanics a fundamental role is played by Hermitian transformations Tl The eigenvectors IV 14 and eigenvalues of a Hermitian transformation have 3 cru cial properties see Morrison 106 starting on page 464 for proofs to these theorems i The eigenvalues of a Hermitian transforma tion are real ii The eigenvectors of a Hermitian transfor mation belonging to distinct eigenvalues are orthogonal iii The eigenvectors of a Hermitian transfor mation span the space Example IVil Prove the Schwartz inequality Eq IV 24 Hint Define a new vector as a linear combination of oz and then use Eqs IV 18 IV 19 and IV 20 Solution Step a Let s de ne the vector W fla hence W is a linear combination of a and But what is the value of the scale factor f Its value is arbitrary here There are two ways we can determine a speci c value In both cases for convenience let s assume that a and W are real vectors and f is a real function Method 1 This method is based on the method shown in Arfken s Mathematical Methods for Physicists on page 445 Express Eq IV 19 in summation notation and take the minimum value of this equation ah zm Ev 0 Then 2 2a fatgt2 2a f 0 If iaz is constant for all i then f 8iOZ But if iaz is Lot constant IV 15 which we assume here for a more general expression then we must expand the polynomial out in the summation above and solve for f using the quadratic formula 0 2 2f0 i i fgag QfZOZi z f2 01 c bf If2 where c 33 b 2 Z ai i and a 2 01 Then the quadratic equation gives bib2 4ac f 2a i 220 i ii 42012395239 4EOZE Z2 7 220a 7 2201i ii 42012 52 4EOZiE i i 2202 i 2201i ii 42012 52 4Eaz i i 220x2 i 2Zai i 220x2 2065i i Ea M 0401 In the solution above we can write 2 01239 E g as a single summation Eai Zf since we are summing over the same index Method 2 This method is based upon the method described by Ander son s Modem Physics and Quantum Mechanics on page 217 In vector notation 0 S WW lt fal fagt lm flt agt WOW f2ltaagt lm flta gt flta gt f2ltaagt since f la amp W are real lm 2flta gt f2ltalagt IV 16 Since f is arbitrary we can determine any value for it Let s allow f to have a value when it is at a minimum or set BBflt77gt 0 Thus 0 llt l gt2fltal gtf2ltalagtl 0mmm2 mo 2fwrw awgt f M alaV which is identical to the value found from the summation approach Step b As such we can now write OW lwW lt Ola Even though we assumed f oz and to be real in order to determine a functional form for f keep in mind that it s functional form is arbitrary Step c Continuing on now with the proof of the Schwartz inequality we now will use the above functional form for W whether or not our vectors are real or complex again f is just an arbitrary function With this form for W use Eq IV 20 to show wmme ggmumm gmm Now from Eq IV18 wmvmgtme ggmuwm gmww wm IV 17 which is always real hence Doing the same thing with the agt vector gives ioz a Maa Maa ltilagt elt hgt lt ltl gt ltWI gtgt lt gt WW I gt 0gt therefore MO 0gt Finally plugging this back in to our original equation for WW gives n lt0 gt2 M WW 7 W W WWW i W and hence lt0 gt2 lt l gt 2 Ma lt0 gt2 Ma lt l gt lltal gt2 S ltaagtlt l gt QED Note that the minimum value is achieved lltal gt2 ltalagtlt l gt if ogt is proportional hence parallel to w la W where A is some scalar Example IVi2 Given the following two matrices 112 20 2 A203B010 22 22392 23932 IV 18 compute a A 132 3 A132 0 A2 Bl d A e Aquot f Al 3 THE h detB and B l Check that BB 1 1 Does A have an inverse Solution 31 Sum the respective elements of the matrix 1 1 2 2 0 2 1 1 0 AB 2 0 3 0 1 0 2 1 3 22 22 2 23 2 32 3 22 4 Solution b Multiply rows of A by columns of B 20 10132 2022 AB 4032 009 2206 42022 0 22396 204 3 132 32 432 9 6 22 62 6 22 6 Solution c A B AB BA we already have AB 202 20 2 220 2239 BA 020 000 030 2 642 20 42 194 0 0 0 2 0 3 632 32 12 3 1 32 32 0 0 0 A B 4 32 9 6 22 2 0 3 62 6 22 6 6 32 32 12 3 1 32 32 2 32 9 3 22 6 32 6 2 6 IV 19 Solution d Transpose of A ip A about the diagonal A l l 2 2 22 0 22 3 2 Solution e Complex conjugate of A A A l 2 22 l 2 0 3 22 2 Solution f Hermitian of A AlEAquot Solution g Trace of B Mw TrB Z l 2 22 1022 3 2 Solution h Determinant of B detB22 0 00 0 20 24 0 1 Solution i Inverse of B lV Bii212 20 multiply each 2 term by l in where 10 00 01 32 22 2393 202 0 2 2 2 20 C 32 2392 23 32326 0 2 2 239 20 Z 10 00 01 then 2 32 13470 30 z39 62 1 40 1 62 06z 2i0 22 BB 1 7 000 030 000 2240 21 39 12 104 OJ 1 3 0 0 1 0 0 7 g 0 3 0 0 1 0 0 0 3 0 0 1 1f detA y 0 then A has an inverse detA 10 6239 14 6239 i 4z 0 As such A does not have an inverse 6i 46i40 Example IV73 Find the eigenvalues and normalized eigenvectors of the following matrix 1 1 M 1 0 1 Can this matrix be diagonalized Solution 0detM 1 only one eigenvalue From Eq IV 59 we get 1 l 11 i 1 l1 0 l 12 i 12 i We get two equations from this eigenvector equation a1a2 a1 12 12 The second equation tells us nothing but the rst equation shows us that 12 0 We still need to gure out the value for 11 We can do this by normalizing our eigenvector a agtz 1 ltalagtilailg lall2 la2l2 lall2 or 11 1 Hence our normalized eigenvector is 33 Since these eigenvectors do not span the space as described on page lV 3 A2cii this matrix cannot be diagonalized Example IVi4 Find the eigenvalues and eigenvectors of the following matrix 2 0 2 M 2z39 239 2239 1 0 1 IV 22 Solution The characteristic equation is Z A 0 2 M 1 2239 i A 2239 1 0 l A i A 2239 2239 i A 2A 0 l A02 1 0 2 MW W l A 0 2l0 i Ml 2 A AAA22 2A 22 2z 222iAiA2 A2 A32i 2A 31i2 z 0 To nd the roots to this characteristic equation factor out a A and use the quadratic formula solution equation 0 31i z39 A21z z A1 0 1z l lz 2 4z39 A273 2 i 1z l l22 1 4z 2 i 1il 2239 2 However note that l 239 2i As such the equation above becomes 1z l 1 12 273 i 1ill z39 2 1i l z 2 71 A 2 2 i 1i1 i7 22397 A3 7 2 i 2 72 so the roots of A e the eigenvalues are 0 l and 239 Now let s call the components of the rst eigenvector agt a1 a2 13 which corresponds to eigenvalue A1 0 The eigenvector equation becomes 2 0 2 a1 a1 0 2i i 2i 12 0 12 0 1 0 1 a3 a3 0 which yield 3 equations 211 213 0 22 a1 2112 22113 0 l1 13 0 The rst equation gives 13 11 the second gives 12 0 and the third is redundant with the rst equation We can nd the values for 11 and 13 by norrnaliZing 3 2 1 Ola lail lall2 a22 132 112 lall2 2 a12 or 11 a3 Hence our eigenvector for A1 is l agtai 0 forA10 For the second eigenvector let s call it W b we have 2 0 2 b1 b1 b1 2239z39 2239 b2 1 b2 b2 l 0 1 b3 b3 b3 which yield the equations 2b1 2b3 b1 2ib1 ibg Zibg b2 b1 b3 b 7 IV 24 with the solutions b3 12b1 and b2 1 i2b1 Normalizing gives 3 1 lt l gti21bi2 l512b22b32 wlt1rgtltlgigt b12ltle1gtlb1lgilb1lg 4 2 2 2 1 2 7 7 7b 4lbil 4b1 41 7 2 77 b jllla or b1 So b2 1 and b3 giving our nal eigenvector for A2 as 2 l gtb 1 2 forA21 1 Finally the third eigenvector call it W c is 2 0 2 C1 C1 ZC1 2i i 2i C2 i C2 ZCQ l 0 1 C3 C3 ZC3 which gives the equations 2C1 2C3 iC1 223901 202 2203 202 C1 C3 ng with the solutions 03 01 0 with 02 undetermined Once again we can normalize our eigenvector to determine this undetermined 02 coef cient 3 2 1 ltvlvgt6il 012C22C32 C22 a IV 25 or Cg l which gives our third eigenvector 7c 1 forA3z39 0 B Function Spaces 1 Functions as Vectors 3 b In quantum mechanics we introduce the concept of a function space in which vectors are actually complex functions of an inner products are integrals and deriva tives appear as linear transformations The inner product of two functions and is de ned by the integral given in Eq IV 5 ltfggt merge dx Iv 69gt where the limits of this integral will depend on the domain of the functions in question i This integral may not converge gt if we want a function space with an inner product we must restrict the class of functions so as to ensure that lt f g is always well de ned ii It is clearly necessary that every admissible func tion be square integrable mmrdx lt oo otherwise the inner product of f with itser wouldn t IV70 even exist IV 26 iii It turns out that this restriction is also su icz39em if f and g are both square integrable then the integral in Eq IV 69 is necessarily nite Example IV5 Let TN be the set of all trigonometric func tions of the form x n20 an sinmrx bn cosmrxl IV 71 on the interval 1 g x g 1 Show that 1 lengt W 62mm 11 0 i1 i2 constitute an orthonormal basis for this function What is the di mension of this space Solution Rewrite the trig functions using Euler s relations N71 an inrrz finrrz bn inrrz finnz x TF0 e e gt lt3 lt6 6 7 N71 eimrz eiimrz 710 2239 2 2239 2 N71 Z en ezmrz n7Nil where cn z an bn for n 123N 1 CO be on am in for n 1 2 3 N 1 1 see lVAcii on page lV B So the set does span the space ls it orthonorrnal 1 1 iimrrz inrrz lt m ngt Eile 6 dx 1 yawn 1 0 for m y n 2 7im7n7r 1 7 fl1dx1 formn 7 6m IV 27 Yes it is orthonormal So it s also a basis e no extra func tions included since orthogonal vectors are necessarily linearly independent Looking at the cm coei cient equations above the dimensions are DN 11N 12N 11m 2 Operators as Linear Transformations a In function spaces operators such as ddx dZdxg or simply an behave as linear transformations provided that they carry functions in the space into other functions in the space and satisfy the linearity condition Eq lV 26 i For example in the polynomial space PN the derivative operator E ddx is a linear transfor mation since it takes N th order polynomials into N 1th order polynomials gt hence still in the space ii However the operator at multiplication by x is not for it takes N 1th order polynomials into N th order polynomials which is no longer in the space b In a function space the eigenvectors of an operator T are called eigenfunctions Tm A x IV73 c A Hermitian operator is one that satis es the de ning condition Eq lV 63 ltfITIggt ltTfggt IV74gt IV 28 d e for all functions and 906 in the space When dealing with operators you must always keep in mind the function space you re working in An operator may not be a legitimate linear transformation because i It carries functions out of the space ii The eigenfunctions of an operator may not reside in the space iii An operator that is Hermitian in one space may not be Hermitian in another One has to pay particular attention to transformations in in nite spaces i Remember that 02 is not a linear transformation in the space P N since multiplication by 06 increases the order of the polynomial and hence takes func tions outside the space ii However it is a linear transformation on P 00 the space of all polynomials on the interval 1 g 06 g 1 iii In fact it s a Hermitian transformation since f1 WW imam dx f1 ixfltxgtr you dx Example IV6 Show that 6422 is an eigenfunction of the op erator Q dgdxg x2 and find its eigenvalues IV 29 Solution The eigenfunction equation is Q x MW So carrying out this operation we get d2 x2gt67122 W ltx67122gt x267122 eiz22 062673522 x267122 e 22 fx So it is an eigenfunction with only one eigenvalue of Hilbert Space a We will now start to talk about wave functions in 3 dimensional space We have just been discussing the or thonormality of wave functions now we de ne the com pleteness of a function i We have shown that the generalized wave function is a linear combination of separable solutions see Eq Ill 21 In terms of the TISE we can write this condition as ii The wave function components are orthonormal to each other if they satisfy the condition V we Mr W 6m IV 76gt IV 30 iii In addition the wave function components 11 represent a complete system if it is impossible to nd an additional function b that is orthogonal to all of the zJn s in the sense of Eq IV 76 iv If this is the case the following completeness re lation is valid V 1111dV V de p42 IV77 where the an s are the expansion coe icients of the arbitrary wave function as de ned in Eq IV 75 b If completeness holds the zJn s constitute an orthonormal basis of a Hilbert space i A Hilbert space is a nite or in nite complete vec tor space on the basic eld of complex numbers ii In this space a scalar product is de ned such that it assigns a complex number to each pair of func tions and out of a set of linear functions iii This scalar product meets three requirements 1 ltwl gtw dv wdvgtltlt lwgtgt 2 lt lta 1b gt alt 1gtblt l 2gt 0r wlta 1 b 2gtdv aw 1dv b 2dV lt3 ltwlwgt WW 2 0 Note for the last requirement 0 only if 11 0 c The state vectors e wave functions of a quantum me chanical system constitute a Hilbert space hence the Hilbert space is a function space IV 31 d 4 We now recast the fundamental principles of quantum mechan ics as developed in ll lll of the notes in the more elegant language of linear algebra and function to Hilbert space Re member that the state of a particle is represented by its wave function lx t whose absolute square is the probability den sity for nding the particle at point x or 739 in 3 D at time t It follows that ll must be normalized which is possible if and only Mathematicians refer to a complete inner product space as L2 To physicists L2 is practically synonymous with Hilbert space if it is square integrable C The Generalized Statistical Interpretation 1 The state of a particle is represented by a normalized vector in the Hilbert space L2 3 b Classical dynamical quantities such as position veloc ity momentum and kinetic energy can be expressed as functions of the canonical variables x or r and p and sometimes t Qxpt To each such classical observ ables we associate a quantum mechanical operator Q ob tained from Q by the substitution 76 IV78 The expectation value of Q in the state l is Q Mm oqm dx IV79 which we now write as an inner product Q lt 1Q 1gtlt 1Q11gt IV80 Note that either notation or le1 is considered ac ceptable IV 32 c The expectation value of an observable quantity has got to be a real number so ltWQWgtlt 1Ql 1gt ltQ1 1gt IV81 for all vectors gt Q must be a Hermitz39an operator Observable quantities Qxpt are represented by Her mitian operators Qx t the expectation value of Q in the state llgt is A measurement of the observable Q on a particle in the state is certain to return the value A if and only if llgt is an eigenvector of Q with eigenvalue A a For example the TlSE Eq Ill 8 can be written in the form 19 Eu IV82 b Note however that this is nothing more than an eigen value equation for the Hamiltonian operator and the so lutions are states of determinate energy E c This third postulate can be rewritten in terms of a statis tical argument gt the generalized statistical inter pretation GSl as given in the following postulate 4 If you measure an observable Q on a particle in the state llgt you are certain to get one of the eigenvalues of The probability of getting the particular eigenvalue A is equal to the absolute square of the A component of lgt when expressed in the orthonormal basis of eigenvectors a To sustain this postulate it is essential that the eigen functions of Q span the space IV 33 b c This might not be possible however for in nite dimensional cases As such we shall take it as a restriction on the sub set of Hermitian operators that are observable that their eigenfunctions constitute a complete set though they need not fall inside L2 There are two kinds of eigenvectors which we need to treat separately The rst deals with systems whose spectra are discrete with the discrete eigenvalues separated by nite gaps i We can label their eigenvectors with an integer n Qlengt Anlengta eg bound states in an atom with n 123 IV83 ii The eigenvectors are orthonormal or rather they can always be chosen so iii The completeness relation takes the form of a sum 00 Z cnlengt7 nl with the components given by Cn ltenqjgta and the probability of getting the particular eigen value An is ICnI2 lt nqugt2 IV87 d The second deals with systems whose spectra are con tinuous eg ionization states of an atom or scattering states IV 34 i The eigenvectors are labeled by a continuous vari able k Qlekgt Adm with 00 lt k lt oo IV 88 ii Here the eigenfunctions are not normalizable but they satisfy a sort of orthonormality condition lt Zl kgt W k IV 89 or rather they can always be chosen so iii The completeness relation takes the form of an integral m LOO cm dk IV90 with the components given by Ck lm TV 91 and the probability of getting an eigenvalue in the range dk about M is ck2 dk lteklkllgt2dk IV92 e In the G81 the orthonormal eigenfunctions of the po sition operator are eggx 606 06 IV 93 and the eigenvalue 06 can take on any value between oo and 00 i The 06 componen of is ex exll 0 so xvim dx lxJ IV 94 IV 35 0 ii Thus the probability of nding the particle in the range dx about 06 is leggl2 dx lqxt2dxl IV 95 which is the original statistical interpretation of l The momentum operator in the G81 is handled in the following manner i lts orthonormal eigenfunctions are 1 epx femh i W IV 96 and the eigenvalue p can take on any value in the range oo lt p lt 00 ii The p component of is cp eplkl imh lamf dx E 131 t l 00 bmme IV97 iii Here 131 t is called the momentumspace wave function it is apart from the factors of h the F ourier transform of the position space wave function ll 06 t iv The probability of getting the momentum in the range dp and p is Pdp crgtpz2 dp IV98 IV 36 Example IV77 Problems concerning the generalized statistical inter pretation of quantum mechanics a Show that Zcn2 1 in Equation IV 85 b Show that fck2 dk 1 in Equation IV 90 c From postulate 4 ie the generalized statistical interpretation it follows that ltc2gt Anlcnr or ltc2gt Aklckrdk for discrete and continuous spectra respectively Show that this is con sistent with postulate 2 Solution a 1 WW Z Cinc emlen 1 m1 n 00 oo 6 Z cmcn mn 1 m1 n f Ian QED n1 Solution b 1 LZLchkltegekgtd dk cgckm k d dk Lck2dk QED Solution c From Eq IV 85 and postulate 2 ltQgtlt1Q11gt i i CCnltelelengt m1 n1 oo oo 96 Z Zcmcnknltemlengt m1n1 oo oo Z Cncnn6mn 1 m1n 0 2 Z nlcnl n1 IV 37 From Eq IV 90 and postulate 2 ltQgtlt1Qqfgt 11 626kltelelekgt dwk chkAkltegekgt drdk LZL cgckmaz k drdk L Akck2dk Example IV78 Confirm that epx in Eq IV 96 is the orthonormal eigenfunction of the momentum operator with eigenvalue p Solution The momentum operator is p Applying this operator on Eq IV 96 gives Pepepmh pep Hence 6 is an eigenfunction of the momentum operator with an eigen hd h 1 eimh value of p To prove that it is orthonormal we need to take the inner product of this eigenfunction 1 00 7 7 innh quh lt6p64gt 7 27 L00 6 6 dx 1 00 i 7 2qipy 7 27 L006 hdy Wlthyxh 1 oo 7 2qipy 2W L00 6 dy With the use of Plancherel s theorem see Eq Ill 142 let 606 in Eq Ill 142 so the Fourier transform of the delta function is 1 oo 1 1 Fk7 6 zkzd 772k07 x27T L00 e x x27T e x27T and the inverse Fourier transform of F 1 x27T is 66 7 ilkrd 7 zkzd f ltxgtm1wme x 2Wme an IV 38 Comparing this equation with the equation derived for lt p qgt we see that they are the same if q p x and y k Making these variable substitutions in our equation above we see that 00 1 ltepleqgt g Lwew Pgt dy 6ltq pgt As such epgt and eqgt are orthonormal hence are basis functions in the sense of Eq IV 93 IV 39 D The Uncertainty Principle 1 Proof of the Generalized Uncertainty Principle a b C For any observable A we can express the variance in the measurement of A as 0124ltA ltAgtWlt4 ltAgtgtIgt ltffgt where f221 AgtIIgt IV99 Likewise for any other observable B we have 0 ltgggt where ggt E B BgtIIgt lnvoking the Schwartz inequality Eq lV 24 we get 0340129 ltffgtltglggt 2 ltfggt2 IV100 i From the mathematics of complex variables for any complex number Mk4meWHme2ama ie Xgt 22IV 101 where 2 is the complex conjugate of 2 ii Therefore letting z y a gwm wmu wwm iii But fly lt A j ltAgt 1 f3 BM ltMmwB wm WIMB AltBgt BltAgt AMEN M21390 EMMA VONIBM ltAgtltBgtltqfqugt V13 ltBgtltAgt ltAgtltBgt ltAgtltBgt V13 ltBgtltAgt ltABgt ltAgtltBgt iv Similarly ltfggt ltgfgt ltABgt ltBflgt 43 gt where 4 3 2 AB BA IV103 is the commutator of the two operators 1 As a result Eq IV 102 becomes 03029 2 21 ED IV104 This is the uncertainty principle in its most general form i You may be asking isn t the right hand side of the equation negative The answer is m since the commutator carries its own factor of i and the two cancel out ii For example suppose the rst observable is posi tion x and the second is momentum To determine the commutator we use an arbitrary test function f A A hd hd lxaplf i xltfxgt W IV 41 8 06p ih IV 105 Accordingly 1 2 h 2 iaizl l l l or since standard deviations are by their nature positive 0350p 2 g IV 106 That proves that the original Heisenberg uncertainty prin ciple but we now see that it is just one application of a far more general theorem i There will be an uncertainty principle for any pair of observables whose corresponding operators do not commute ii We call these incompatible observables iii lncompatible observables do not have shared eigen vectors at least they cannot have a complete set of common eigenvectors Matrices representing in compatible observables cannot be simultaneously diagonalized that is they cannot both be brought to diagonal form by the same similarity transfor mation iv On the other hand compatible observables Whose operators do commute share a complete set of eigen vectors and the corresponding matrices can be si multaneously diagonalized IV 42 Example IVig Prove the famous Luttermoser uncertainty principle relating the uncertainty in position A x to the un certainty in energy B p22m V H h a a gt 7 I H 2mltvgtl For stationary states this does not tell you much why not Solution p2 1 2 7 V 7 V x2m i 2mixplix l Then from Eq IV 42 06192 mp2 p206 mp2 p061 1906 p206 0619 pim9l By making use of Eq IV 105 we get 06192 imp 19 Mp and 06V xV VxxV xV0 So 2 P i i i x V i 2m22hp i zhpm Now from Eq IV 104 030 2 gimp or W 2 lm QED For stationary states UH 0 and 0 so the Lutterrnoser uncertainty relation just says 0 2 0 for stationary states gt hence tells us nothing IV 43 2 The EnergyTime Uncertainty Principle 3 b c Compute the time derivative of the expectation value of some observable 6206 1 t d d A 81 A 862 A 81 g Q i WIQI IO BitlQl IOH IlajlqHNlQlajl Now the Schrodinger equation says Bkl A Zhait i where H p22m V is the Hamiltonian so substituting lkll for the time derivatives of the wave function in the equation above our terms become Bkl A l A A 1 A A 1911 lt leQlwgt ltHMQWgt 8 8 waifm lt57 A 9Q A mow worm lele Tommi SO d i 1 A A 1 A A BC ltQgt ltH 1 Q 1 gt lt 1 QH 1 gtlt815gt But 1 is Hermitian so thus d i 1 A A 1 A A BC ltQgt lt 1 Hqugtlt IIIQHqgtAltaitgt 1 339 A A Z A A 92 a EWIHQIM hltwlc legtltatgt 39 A A 8 1411191 gtlt gt 01quot d J A A 86 ltc2gt a E lt1Hc21gt lt87 Iv 107 IV 44

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