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- Falling p.11

2d fall problem the time dependent solutions are linear combinations of solutions like or in dimensionless form where the time unit for dimensionless time t is e so E t E t OK our y time dependent Schrödinger s equation has produced unnormalizable solutions like exp i k y exp i k 2 t We seek a linear combination of these solutions to make a normalizable lump of probability Classical physics suggests that this lump of probability should move at constant speed Basically the idea is to have all the simple waves constructively interfere to produce the lump and destructively interfere everywhere else Finding this desired linear combination at t 0 is easy once you remember Fourier transforms Thus if at t 0 is to be a Gaussian lump of probability say exp y 2 we can use Fourier analysis to express such a function as a superposition integral of simple waves like exp i k y y At future times each simple wave has time dependence exp i k y 2 t given by the time dependent Schrödinger s equation Thus Three details 1 According to Heisenberg s Uncertainty the size of the wavefunction in space y affects the range of momentums p y that make up the wavefunction If the particle has a large range of speeds the wavefunction must separate out i e expand Thus to get relatively constant sized wavefunctions we will want to enlarge the initial spread of the wavefunction 2 Our wavefunction should be normalized so the total probability of it being someplace on the y axis is one indeed producing such normalized solutions was the point in starting all this Fourier analysis in the first place Thus we will want to consider t 0 wavefunctions of the form N exp y 2 4

Original URL path: http://www.physics.csbsju.edu/QM/fall.11.html (2016-02-01)

Open archived version from archive - Falling p.12

k 0 3 the wavelength is about 2 Recall that the fourier transform g k shows which simple waves make up the wave packet Thus the peak of g k is at k 0 The variable determines the initial width of the lump of probability At the same time because of Heisenberg s Uncertainty it determines the range of speeds momentums energies that contribute significantly to the superposition Large spatial extent means a reduced range of wavelengths to make up the packet Thus while the peak of g k is at k 0 the width of the peak is determined by Large sized wave packets have a large and a narrow peak in g k If the spatial extent of the wavefunction is large the waves that make up the packet must have quite similar wavelengths otherwise those waves could not be in phase within the packet which is required for constructive interference Here are some pictures of a k 0 6 1 5 wavefunction at various times t 0 t 1 t 2 t 3 Notice that the central part of the wave packet moved from y 0 12 24 36 as t 0 1 2 3 that is to say that the velocity of the packet is about 12 2 k 0 Also notice that as expected the packet has expanded slightly Notice that within the lump of probability you can see the waves of the real part of In the following sequence of stills for t 0 0 1 0 2 0 3 we follow one of these peaks in Re within the wave packet Notice that this particular peak marked in red falls behind the wave packet During the 1 2 tic time period displayed the wave packet as a whole has moved forward 6 units

Original URL path: http://www.physics.csbsju.edu/QM/fall.12.html (2016-02-01)

Open archived version from archive - Falling p.13

the 1 d fall is the initial condition Before we were dealing with a particle dropped from rest Here we seek a solution that moves upwards reaches a maximum height and then falls that is we seek a solution with an upwards initial velocity The previous pages have given us a model for the initial wavefunction of a moving wave packet N exp y 2 4 2 exp i k 0 y We need to modify it only a bit to use it for the initial state of our z motion First since U for z 0 0 for z 0 The gaussian wave packet is never actually zero just very small several s away from the peak Thus if we continue to use a gaussian form we must raise its CM location by several Thus we limit the discontinuity at z 0 by making small for positive z near 0 whereas 0 for z 0 Thus we are led to an initial wavefunction f z exp z z 0 2 4 2 exp i k z0 z where z 0 1 so f 0 is small As before we must chose relatively large if we want a narrow range of velocities in the wave packet We then proceed to calculate the b n exactly as above I chose k z0 12 5 z 0 20 I cannot actually find and infinite number of b n so I restrict my sum about 150 n values whose eigenenergies are near the expected value of k z0 2 z 0 164 which occurs near n 450 Since I have not included the complete set of orthonormal wavefunctions in the sum my solution to Schrödinger s equation will not exactly equal the gaussian form Furthermore I did not normalize my above f z

Original URL path: http://www.physics.csbsju.edu/QM/fall.13.html (2016-02-01)

Open archived version from archive - Falling p.14

t 2 t 3 Here is a multiple time image showing the probability density at t 0 3 6 9 12 15 18 21 24 Notice the expansion of the wave packet with time and the classical looking trajectory If one electron were fired with this wavefunction it would hit the ground at exactly one spot If we fired electron after electron we would gradually build up the probable range of hit locations given by this wavefunction Notice that much the same statement could be made of cannons whose projectiles follow Newton s laws However we would attribute the variation in hit location to variation in the initial velocity say due to variation in the amount of propellant or uncertainty in the angle of elevation or variation in the forces active say due to varying wind conditions or atmospheric pressure In QM we state that there is irreducible uncertainty even if the initial conditions and forces are identical in each shot Thus there is not really a trajectory at all The electron may be found here one time and there the following shot All we have is a moving blob of probability density whose center may follow a path quite

Original URL path: http://www.physics.csbsju.edu/QM/fall.14.html (2016-02-01)

Open archived version from archive - Falling p.15

states in which the bouncing ball is now mostly likely to be found Calculate the weighted average and standard deviation of those energies Must we observe that the particle s energy is conserved Using those 7 states with proper time dependence plot the position probabilities for several times near when the infinite potential is removed and for a time one classical period in the two force situation and classical period after the infinite potential is removed It s easiest to plot the z 0 and z 0 separately Reproduce the falling ball solution described on p 7 with some other initial wavefunction Produce plots and tables showing the results WKB Why is it N rather than N The falling QM ball is based on a energy eigenfunction with E near 122 2 Find more significant digits for the energy and determine what n has this energy In the case of the 27 lhs force the energy of the 10th state is about 12 2842 What would WKB predict i e do WKB for two sided linear case Find the WKB approximation for the energy in the symmetric two force situation Rayleigh Ritz Confirm the reported results for f z z exp az Find the RR estimate for the groundstate energy using f z z exp az 2 Find the RR estimate for the groundstate energy using f z z exp az 3 2 Pick your own f z Plot the above four wavefunctions along with the exact Airy function solution Rayleigh Ritz There is no reason to limit the number of adjustable parameters in our trial wavefunction other than that it then becomes a more difficult problem to find the minimum Pick a trial wavefunction with with two adjustable parameters e g f z z 1 bz 2 exp az 2 and find the expectation value for H as a function of the two parameters e g a and b Make a contour plot of the result to identify an approximate best value and then use Mathematica s FindMinimum or similar program to get the best possible RR bound Plot the resulting wavefunction along with the exact wavefunction Perturbation Theory Consider a perturbing potential which for 0 z a is V a z and otherwise V 0 This results in a total potential that is constant total P E a if 0 z a and exactly equals our old potential otherwise The above total potential can be easily solved For 0 z a Schrödinger s equation solution is sin kz where k 2 E a For z a we have our usual Airy solutions Ai z E As usual we find eigenenergies by solving the equation that comes from matching the logarithmic derivatives at z a The below plot shows how the energy levels for n 1 2 change as a function of a Note that perturbation theory tracks the exact result until about a 1 5 for n 1 and a 3 5 for n 2 Note that in

Original URL path: http://www.physics.csbsju.edu/QM/fall.15.html (2016-02-01)

Open archived version from archive - SHM p.2

is to provide examples where all the dependencies are explicit so that they are available for play understanding If a problem has an analytical solution say f x where f x is some well known function consider the consequences Just about every known function requires that its argument be dimensionless I can think of only two counterexamples Thus sin x or exp x 2 cannot be proper answers if x has the units you cannot take the sine of 1 meter nor have a power that is 1 square meter Thus answers must be of the form f x L where L sets the scale of the problem i e has units that cancel those of x Thus the first point of attack on any physics problem is to determine how you could form quantities like L that allow you to form dimensionless quantities like x L In the classical mechanics problem we have givens of v 0 and x 0 From these one can form two scales for both time and length time scales 1 x 0 v 0 and length scales x 0 v 0 Thus the classical problem is actually a generally more difficult two scale problem that in this case has an easy solution because one time scale enters in only in the dimensionless ratio x 0 v 0 tan Thus we end up with simple one scale dependencies like t Similarly there is really only one length scale in the solution A which is just the square root of the sum of the squares of the above two length scales In the quantum mechanical problem we loose the givens v 0 and x 0 there is no single position or velocity and we gain a quantity with In particular Schrödinger s equation has constants 2

Original URL path: http://www.physics.csbsju.edu/QM/shm.02.html (2016-02-01)

Open archived version from archive - SHM p.3

2 This is without loss of generality wolog i e for any there is such an H x We hope however that finding H x will be easier than finding directly If we now try to write H as a polynomial we can plug the polynomial form into the differential equation The result is a two term recursion relation i e the result has just two a s so for example given a 0 we can calculate a 2 from which we can calculate a 4 etc until we re done Note that the the recursion relation connects even k to even k and odd k to odd k Thus the polynomials will have only even or odd terms It turns out that the polynomial must end if the wavefunction is to be normalizable One can show that the non terminating sum gets at least as big as exp x 2 so instead of going to zero for large x gets big like exp x 2 2 The only way the sum can end is if the numerator of the recursion relation is zero i e if E 1 2 n then a n 2 0 and hence all further a s are zero Note that if we had tried for a polynomial solution for itself rather than factoring out the exp x 2 2 first we would get an unsolvable three term recursion relation The polynomials we have found are called Hermite polynomials see for example Abramowitz Stegun 22 p 771 or Szegö Ch V Below is a plot of a Hermite polynomial H 10 x 30240 302400 x 2 403200 x 4 161280 x 6 23040 x 8 1024 x 10 from x 4 to 4 and x 2 to 2 Note that H 10 x is an

Original URL path: http://www.physics.csbsju.edu/QM/shm.03.html (2016-02-01)

Open archived version from archive - SHM p.4

a position x 8 whereas you are likely to find it at x 1 6 In general peaks where the particle is likely to be found alternate with valleys where the particle is unlikely to be found This is like fringes in a diffraction experiment bright illuminated regions alternate with dark unilluminated regions Particles are just like light quantum mechanics unifies particles and waves into a single beast for which we re invented no better name for than wave particle Note that these probability densities do not depend on time You should be wondering how it can be that something can oscillate back and forth if there is no time dependence for the particles position probabilities Let me take a second to explain this stacked probability density display We are interested in displaying where the particle is likely to be found for several different possible energies We can display as I did directly above individual probability densities on individual plots but lots of such plots are deemed a waste of space so we aim to plot several functions on one graph Imagine what would happen if we try to use the same axes and overplot the above two functions Thus different functions are moved vertically up to avoid overlapping display The convention is to move each plot up a distance corresponding to that solution s energy Since solutions rarely have the same energy we re automatically saved from overlapping plots Thus the zero line for each probability density function becomes the line of constant energy This is potentially quite confusing because the y axis is being used for two two different things with different units and zero points If we further confuse things by in addition to all of the above plotting the potential energy U x we can see

Original URL path: http://www.physics.csbsju.edu/QM/shm.04.html (2016-02-01)

Open archived version from archive