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Quantum Mechanics

by: Marjorie Hahn

Quantum Mechanics PHYSICS 137A

Marjorie Hahn

GPA 3.95


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This 10 page Class Notes was uploaded by Marjorie Hahn on Thursday October 22, 2015. The Class Notes belongs to PHYSICS 137A at University of California - Berkeley taught by Staff in Fall. Since its upload, it has received 14 views. For similar materials see /class/226694/physics-137a-university-of-california-berkeley in Physics 2 at University of California - Berkeley.

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Date Created: 10/22/15
Advanced Topic For Those Who May Be Interested Notes on Wave Packets their Momentum Content Fourier Series and Fourier Transforms Introduction These notes are meant to help supplement some of the qualitative discus sions we have in discussions sessions about wave packets their momentum content the uncertainty principle and Fourier series and transforms The latter two topics may be un familiar to many but I think the discussion below may help The discussion of the Fourier topics is somewhat simpli ed 7 I use a wave packet that is symmetric around z 0 and this allows me to somewhat un clutter the algebra So you should go to other sources eg Google WolframFourier series for more complete discussions if you have other needs for Fourier expansions The notes are accompanied by the Mathematica script which generated the various gures below I will post that script as those of you with access to Mathematica may want to play around eg try different wave packets from the one I have chosen 1 take a familiar Gaussian wave function centered on x0 for the examples Thus the properly normalized wave packet has one parameter the standard deviation denoted below by a that de nes the width of the probability density associated with the wave packet We are interested in the behavior in both x and Fourier p space as a is varied We take our wave function to be we 64 lt1 xEQWW Then the probability density is PW WWW 53 2 which shows that a is the standard deviation of the probability density Now we know how to calculate the expectation values ofp and p2 for our wave function ltpgt wniiw dz 0 lt3 700 2 dx The reason for this is that we have a wave centered at the origin so for every component of momentum p there is a similar component of momentum 7p the wave is a stationary one at the origin We can also calculate ltp2gt ween d we dz foo dz 4 The integral can be done by hand or in Mathematica as doing integrals is not the point of these notes please take it on faith that the integral was done properly Figure 1 The probability density PW for the Gaussian wave packet of these notes We have taken 04 04 length units Here let me emphasize something I have said in the discussions7 phrased perhaps more carefully If one wanted to ask 77 What is the typical magnitude of the momenta making up the wave packet7 the answer to that might be the root mean square ipim lt5 we see that if we 77squeeze77 the wave function in z 7 in coordinate space 7 by making 04 smaller7 we see that the typical magnitude of momenta making up the wave packet increases as 104 Of course 10gt is still zero because momentum carries a sign7 and every 10 component has a corresponding 7p component But if we envision the wave function in p space it gets broader as the coordinate space wave packet gets narrower The more sharply de ned the coordinate space shape7 the broader the momentum space shape This notion is consistent with the uncertainty principle7 and this statement can be made precise by discussing the Fourier decomposition of the wave packet done below We will discuss both Fourier series and Fourier transforms If we look at Fig 17 we notice that it has been de ned by design on a coordinate system that runs from 77T to 71 A Fourier series is a way to represent such a function or wave packet on such a nite interval7 by writing it as an appropriate sum over plane waves de ned on the same interval 1 00 00 ww an 2 an cos 7n 2 bn sin 7w 6 n1 nl These are familiar wave functions on the right They are the solutions we encountered for the square well We know z is the coordinate7 and it is easy to see that n 7 which comes in integer steps 7 represent momenta To demonstrate this7 rst recall 7r 7r sin ms sin 7w dx mlmn sin 77w sin 7w dx mlmn 7r W sinmx cos 7w dx sin ms dx cos ms dx 0 7 2 Now connecting n with momenta is a bit tricky First note einm eiirm mm 7 eiirm cos 7w f sin 7 8 2 so that for 6quot 7r 1 h d 1 7r 7 mm 7 mm p 7A e dime dx A 7171 dx 27mh 7r 1 d2 1 7r 2 mm 7h2 mm d 2752 d 2 2752 p e d e z W n x 7m 0223 ltp2gt 7 ltpgt2 0 9 while for e mm p 7re miie mm dx W7nh dx 727mb 7r 239 dx 7r p2 We imih2d7267im dx 7r 712712 dx 27Tn2h2 7r diz 7r 5 Up E P2gt lt10gt2 0 10 That is7 both the sin 7 and cos nm are states of de nite lpl they each contain one con g uration of momentum p 27mh and one of momomentum p 72mm 71 17 23 is in units of 1L for whatever length scale describes the x axis in Fig 1 The bottom line Eq 6 is a decomposition of our wave function into states of de nite Using Eqs 7 it is easy to see by placing Eq 6 successively under the integrals LY dm LY cos 71 dm and L sin 71 dm that 10 j 7 dx an r 7s cos 7w dx bn r 7 sinnz dx 11 Mathematica can do those integrals7 so we let it that frees us to think about physics How ever7 note that bn E 0 the sin 7w are not needed for our wave function because the wave function is even around zero So how good is the Fourier series expansion This is shown in Figs 2 and 37 in which just the rst four and seven nonzero terms in the Fourier series are retained The conver gence is really impressive When one sums n for 0 to 67 the answer is already essentially exact Of course7 this says something about the momentum content of The Fourier series expansion allows us to coun 77 the momentum components of the wave function Using our previous expression for p2 in which n was shown to count discrete contributions to p2 we have ltp2gt 7m 2mm lt12 where we remind ourselves that the expansion coef cients depend explicitly on the choice of our wave packet width through the parameter a You can use this expression in place of 3 2 3 Figure 2 The probability density Pz of Fig 1 blue is compared to the corresponding probability when is represent by just four terms in the Fourier series expansion brown 7 107 11 a2 and 13 Rather impressive7 eh The brown curve is underneath the blue one 1 73 72 71 1 2 3 Figure 3 The probability density Pz of Fig 1 blue is compared to the corresponding probability when is represent by just seven terms in the Fourier series expansion brown 7 a0 al 127 13 a4 a5 and 16 Really rather impressive7 eh our standard operator expression of Eq 47 which has no reference to Fourier components This is done in the accompanying Mathematica script If a suf cient number of Fourier components are added up7 the answers are identical In Fig 4 we plot the values anoz 047 as our previous gures were drawn for 0404 If we look very hard we can only six nonzero contributions7 on the scale of the graph Thus this explains in part why our Fourier series converged so well at n 6 What happens if we change a Fig 5 shows the corresponding distribution for 04 reduced by a factor of 5 to 0087 that is7 anoz 008 A wave packet 15th as wide requires many more Fourier components By counting7 one needs about 30 We see the number grows as 104 Both the typical momentum lpl and the number of needed momentum components grows as the wave packet is narrowed in coordinate space The packet widens in momentum space as it narrows in coordinate space and vice versa li l Figure 4 The Fourier series coef cients for our wave function with 0404 020 o v o o o o v Wl mw Figure 5 The Fourier series coef cients for a narrow wave function in coordinate space Oi 05 With all this wonderful success of the Fourier series why would anything else 7 like a Fourier transform 7 be needed The Fourier series is after all an old friend the basis is the basis we derived in solving the in nite square well Well there is one shortcorning If our physics were con ned in a box the Fourier series is perfect But by its construction the Fourier expansion is an expansion of a periodic function if we plot the function de ned by Eq 6 beyond the selected interval 7713711 it repeats We do this in Fig 6 plotting frorn 7337139 we nd three peaks We do not have an isolated wave packet if we think about all of our z space The Fourier transform takes care of this It can be thought of as the consequence of expanding the interval 7713711 to 700 Just as in our square well solution widening the well l l 75 5 Figure 6 The Fourier series expression for 7 is plotted on an extended interval showing that the construction produces a periodic wave function This means that a Fourier series does not reproduce an isolated wave packet when viewed over all space a Fourier transform is needed for that Nevertheless the Fourier series can be extremely useful if we are interested only in representation over a nite interval compresses the spectrum Widening it in nitely compresses it in nitely 7 to the point that n becomes a continuous not a discrete parameter Though the following notation is unusual 7 using n as the continuous parameter rather than the conventional h 7 I do so to emphasize the analogy with the Fourier series Also 7 this is important 7 I will write the Fourier transform under the assumption that the wave function I am dealing with is symmetric around 5120 Under this assumption the Fourier transform analogous to Eq 6 becomes AGO an cos in dn an E 7 cos in dx 13 Just as in Eq 6 the equation on the right decomposes the wave function into a sum 7 now a continuous sum that is an integral 7 over states of de nite n lplQirh The new an 7 the analog of the discrete an of the Fourier series 7 are given by the equation on the right In the Mathematica script this integral is done then plugged into the integral on the left to prove that 7 is then recovered We can now repeat what we did before First Figs 7 and 8 compare the discrete Fourier series coef cients an with the continuous Fourier transform coef cients an The correspon dence is perfect as expected 7 to the extent that a discrete distribution can approximate a continuous one We can also play an interesting game Just as we truncated the Fourier series to a small number of terms we can truncate the Fourier series to modest momentum in the following way 00 an cos in dn gt 7 w Anmax an cos in dn 14 0 ls this a reasonable thing to do If we truncate at the same nmax as in the Fourier se ries case what do the graphs analogous to Figs 2 and 3 look like This is shown in Figs 9 and 10 The convergence is slightly different but quite similar All of this is very satisfying Figure 7 The Fourier series coef cients can for our wave function with a04 are compared with the continuous weights an for our Fourier transform Figure 8 The Fourier series coe icients can for a narrow wave function in coordinate space 04 05 are compared with the continuous weights an for our Fourier transform Because the Fourier transform is de ned over an in nite interval it can truly be used to decompose an isolated wave function or the corresponding probability density which we are plotting into its moment components while keeping the wave function isolated This is illustrated in last gure which one should compare to the Fourier transform result of Fig 73 7 2 Figure 9 To the results of Fig 2 7 where the probability density Px of Fig 1 blue was compared to the corresponding probability when is represent by just four terms in the Fourier series expansion brown 7 107 117 12 and a3 7 we now add the results of a truncated Fourier transform green7 also stopped at 713 See text Qualitatively the discrepancies are similar in the two truncated approximations Figure 10 As in Fig 97 but with the sums or integrals extended to n 6 That is7 the truncated Fourier transform results green have been added to Fig 3 Only one curve is Visible because they lie on top of one another 95 E Figure 11 The Fourier transform result for truncated at n 6 is displayed over an extended interval The truncated Fourier transform not only reproduces the correct Gaussian7 but isolates that Gaussian over an extended region Compare with the Fourier series result of Fig 6 Physics 137A The Phase Shift Imagine a particle of mass m and energy E incident on a square potential 7 V5 if 7 a lt z lt a Va 7 0 otherwise 1 with Vb lt E Label the region z lt 7a as I7 7a lt z lt a as II and z gt a as III Since Vz is even we can decompose the solutions into even and odd parity modes which in region II are by cos I and but sin I 2 where H 2mE 7 Vb and we have set 71 1 for simplicity For now we consider only the even mode7 the argument for the odd mode is completely analogous In regions I and III7 was must match onto even functions oscillating with period 27rk where k 2mE so 1015 A coskz 7 65 3 bum A coskx 65 4 where A and 65 are constants This satis es the wave equation in all 3 regions and has the right parity The only thing left is to match the function and its rst derivative at z ia At z a we have cos m1 A coska 65 7 sin m1 7kA sinka 65 6 Cf from which we can solve for the phase shift as tanka 65 gtan m1 7 and then use Eq 5 to solve for A If we were to match at z 7a we would obtain the identical equations An argument completely analogous to the above for the odd modes gives tanka 60 5 tan m1 8 H


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