Harmonic oscillator Notes on Quantum Mechanics

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1 Hamonic oscillato Notes on Quantum Mechanics Last updated Thusday, Novembe 3, 26 3:5:3-5: Copyight 26 Dan Dill Depatment of Chemisty, Boston Univesity, Boston MA 225 à Classical hamonic motion The hamonic oscillato is one of the most impotant model systems in quantum mechanics. An hamonic oscillato is a paticle subject to a estoing foce that is popotional to the displacement of the paticle. In classical physics this means F = ma= m 2 x ÅÅÅÅÅÅÅÅÅÅÅÅ t 2 =-kx The constant k is known as the foce constant; the lage the foce constant, the lage the estoing foce fo a given displacement fom the equilibium position (hee taken to be x = ). A simple solution to this equation is that the displacement x is given by since x = sini è!!!!!!!!!! k ê m tm, m 2 x ÅÅÅÅÅÅÅÅÅÅÅÅ t 2 2 = m ÅÅÅÅÅÅÅÅÅÅÅ t siniè!!!!!!!!!! k ê m tm 2 =-mi è!!!!!!!!!! k ê m M 2 sini è!!!!!!!!!! k ê m tm =-k sini è!!!!!!!!!! k ê m tm =-kx. The quantity è!!!!!!!!!! k ê m plays the ole of an angula fequency, w=2 p n = è!!!!!!!!!! k ê m. The lage the foce constant, the highe the oscillation fequency; the lage the mass, the smalle the oscillation fequency. à Schödinge equation The study of quantum mechanical hamonic motion begins with the specification of the Schödinge equation. The linea estoing foces means the classical potential enegy is V =- F x =- H-kxL x = ÅÅÅÅÅ 2 kx2,

2 2 Hamonic oscillato and so we can wite down the Schödinge equation as i k ÅÅÅÅÅÅÅÅÅÅÅ 2 m j- Ñ2 2 ÅÅÅÅÅÅÅÅÅÅÅÅ x + ÅÅÅÅÅ 2 2 kx2y z yhxl = E yhxl. { Next, it will be helpful to tansfom this equation to dimensionless units. We could use the same length and enegy units that we have used fo the paticle in a box and fo the one-electon atom, but thee is a diffeent set of units that is moe natual to hamonic motion. Since hamonic motion has a chaacteistic angula fequency, w = è!!!!!!!!!! k ê m, it makes sense to measue enegy in tems of w. It tuns out that the choice Ñ w ê 2 woks well. (The choice Ñw would seem moe obvious, but the facto of ê 2 simplifies things somewhat.). Next, we can use the enegy unit to detemine the length unit. Specifically, let's use fo the unit of length the amount by which the oscillato must be displace fom equilibium (x = ) in ode fo the potential enegy to be equal to the enegy unit. That is, the unit of length, x, satisfies and so Ñ w ÅÅÅÅÅÅÅÅÅÅÅ 2 = ÅÅÅÅÅ 2 kx 2 = ÅÅÅÅÅ 2 m w2 2 x Ñ x = $%%%%%%%%%% ÅÅÅÅÅÅÅÅÅÅÅÅ m w This means we can expess enegy as E = Ñ w ÅÅÅÅÅÅÅÅÅÅÅ 2 e in tems of dimensionless multiples e of Ñ w ê 2, and length as Ñ x = $%%%%%%%%%% ÅÅÅÅÅÅÅÅÅÅÅÅ m w in tems of dimensionless multiples of è!!!!!!!!!!!!! Ñ ê m w. In these dimensionless units, the Schödinge equation becomes 2 ÅÅÅÅÅÅÅÅÅÅÅÅ yhl =-thl yhl, 2 in tems of the dimensionless kinetic enegy thl =e- 2. Veify that the Schödinge equation has this fom in the dimensionless units of enegy and length that we have chosen. Show that the length unit, x = è!!!!!!!!!!!!! Ñ ê m w, can be witten altenatively as è!!!!!!!!!!!! Ñ w ê k and "################# Ñ ë è!!!!!!! km. Hee is a plot of the dimensionless potential enegy.

3 Hamonic oscillato 3 25 vhl Hamonic potential enegy, in units Ñ w ê 2. Length is in units è!!!!!!!!!!!!!! Ñ ê m w. à Enegies and wavefunctions It tuns out that the quantal enegies in the hamonic potential ae e j = 2 j -, whee j is the numbe of loops in the wavefunction. Hee is the lowest enegy wavefunction the wavefunction with one loop. (This and the following example wavefunctions in this pat ae detemined by Numeov integation of the Schödinge equation.) y e= Lowest enegy hamonic oscillato wavefunction. The enegy is 2 μ - =, in units Ñ w ê 2. Displacement fom equilibium is in units è!!!!!!!!!!!!! Ñ ê m w. The vetical lines mak the classical tuning points. The vetical lines mak the classical tuning points, that is, the displacements fo which the hamonic potential equals the enegy. tun@j_d := ρê. Solve@ρ 2 == 2 j, ρd êêevaluate; tun@jd 9 è!!!!!!!!!!!!!!!!! + 2j, è!!!!!!!!!!!!!!!!! + 2j= Hee is the sixth lowest enegy wavefunction,

4 4 Hamonic oscillato y e= Sixth lowest enegy hamonic oscillato wavefunction. The enegy is 2 μ 6 - =, in units Ñ w ê 2. Displacement fom equilibium is in units è!!!!!!!!!!!!! Ñ ê m w. The vetical lines mak the classical tuning points. and hee is the 2th lowest enegy wavefunction, y e= th lowest enegy hamonic oscillato wavefunction. The enegy is 2 μ 6 - =, in units Ñ w ê 2. Displacement fom equilibium is in units è!!!!!!!!!!!!! Ñ ê m w. The vetical lines mak the classical tuning points. This wavefunction shows clealy the geneal featue of hamonic oscillato wavefunctions, that the oscillations in wavefunction have the smallest amplitude and loop length nea =, whee the kinetic enegy is lagest, and the lagest amplitude and loop length nea the classical tuning points, whee the kinetic enegy is nea zeo. Finally, hee ae the seven lowest enegy wavefunctions.

5 Hamonic oscillato 5 y Seven lowest enegy hamonic oscillato wavefunctions. The enegies ae 2 μ j - =, 3,, 3, in units Ñ w ê 2. Displacement fom equilibium is in units è!!!!!!!!!!!!!! Ñ ê m w. The vetical lines mak the classical tuning points. à Absolute units We have expessed enegy as E = Ñ w ÅÅÅÅÅÅÅÅÅÅÅ 2 e, in tems of dimensionless multiples e of Ñ w ê 2, and length as Ñ x = $%%%%%%%%%% ÅÅÅÅÅÅÅÅÅÅÅÅ m w, in tems of dimensionless multiples of è!!!!!!!!!!!!! Ñ ê m w. To get a feeling fo these units, let's see how they tanslate into actual enegies and length fo paticula molecules. The atoms in hydogen halide molecules, HF, HCl, etc., vibate appoximately hamonically about thei equilibium sepaation. The mass undegoing the hamonic motion is the educed mass of the molecule, m= m a m b ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Å m a + m b (I emembe that the poduct of the masses goes in the numeato since the atio must have units of mass.) To calculate the educed mass we need to detemine the mass of each atom, and to do this, we need to know which isotope of each atom is pesent in the molecule. Recall that isotope masses ae given in units of atomic mass, u. The atomic mass unit is defined such that the mass of exactly one gam of cabon 2 is Avogado's numbe times u. This means that the atomic mass unit is Gam ê Mole u = AvogadoConstant ê. Gam 3 Kilogam Kilogam Let's calculate the educed mass fo HCl. If we use the most stable isotope of each atom, H and 35 Cl, the esult is

6 6 Hamonic oscillato μh35cl = i j m H m Cl y z í AvogadoConstant êê. 8 k m H + m Cl { m H.78 Gam ê Mole, m Cl Gam ê Mole, Gam 3 Kilogam < Kilogam Hee ae the educed masses fo othe combinations of isotopes, togethe with that fo H 35 Cl. H 35 Cl Kilogam H 37 Cl Kilogam 2 H 35 Cl Kilogam 2 H 37 Cl Kilogam Confim that these esults ae coect. The effect a change in the lighte isotope is lage than the effect of a change in the heavie isotope. Show why this is so. The next step is to detemine the hamonic angula fequency, w =2 p n. This is done by measuing the fequency of light that causes the molecule to change its vibational wavefunction by one loop, since D E matte = Ñ w=h n. Fo H 35 Cl the measued value is n è = 299 cm -. The unit n è is the ecipocal wavelength, coesponding to the fequency, n è = ÅÅÅÅÅ l = ÅÅÅÅÅÅÅÅÅÅÅÅ c ê n = ÅÅÅÅÅ n c. This means that angula fequency is elated to wavenumbe as w=2 p n = 2 p c ÅÅÅÅÅ l = 2 p c nè. Hence, the angula fequency of hamonic motion in H 35 Cl is Second Veify this esult. Note that this value popely coesponds to the IR spectal egion. Having detemined the oscillato mass and angula fequency, we can evaluate its length unit, x = è!!!!!!!!!!!!! Ñ ê m w..729 fi To intepet this esult, ecall that we have defined the unit of length so that the when the oscillato is displaced this distance fom its equilibium point, the potential enegy equals the zeo-point enegy. That is, x is the classical tuning point of the oscillation when the oscillato wavefunction has loop. This means that when H 35 Cl is in its gound state its classically allowed egion is 2 x = 458 Þ wide. The equilibium intenuclea distance of HCl is.27 Þ, and so gound state hamonic motion expands and compesses the bond by a bit less than %.

7 Hamonic oscillato 7 Evaluate x fo H 8 B (n è = 265 cm - ) and H 27 I (n è = 23 cm - ), and analyze you esults in compaison to the value fo H 35 Cl. Hee ae the answes I get. μ ω x H 35 Cl fi H 8 B fi H 27 I fi Compaison of HCl, HB and HI Plot, on the same set of axes, the hamonic potential fo HCl, HB, and HI, Measue length in Þ. Indicate the fist fou enegy levels of each potential cuve. Do this using hoizontal lines spanning the allowed egion at each enegy on each cuve. Measue enegy in units of the zeo-point enegy of HCl. In a sepaate table give enegies (in units of the zeo-point enegy of HCl) and the ight side (x > ) classical tuning point (in Þ) fo the fist fou enegy levels of each molecule. Hee ae the expessions I get fo the potential cuve, with distance, x, in Þ and enegy in units of the zeo-point enegy of HCl. H 35 Cl x 2 H 8 B x 2 H 27 I x 2 Veify that, fo HCl, when the displacement is its distance unit, x =.729 Þ, the potential enegy is, since we ae using as enegy unit the zeo-point enegy of HCl. Hint: Evaluate k ê 2 in J m -2, divide it by the zeo point enegy, Ñ w ê 2 in J, and then convet the esult fom m -2 to Þ -2. These potential enegy expessions show that the foce constant, k, deceases going fom HCl to HI. Evaluate the foce constant fo each molecules, in J m -2 = kg s -2. Answe: 56., 4.9, Hee is the plot of the esults I get.

8 8 Hamonic oscillato EêHÑ w HCl ê2l Hx-x e LêÞ Hamonic potential enegy cuves and lowest fou hamonic enegy levels (hoizontal lines) fo H 35 Cl (n è = 299 cm - ), H 8 B (n è = 265 cm - ) and H 27 I (n è = 23 cm - ). Enegy is in units of the zeo-point enegy of H 35 Cl, μ -2 J. Hee is the tabulation of enegies and ight side tuning points fo the lowest fou levels of each molecule. H 35 Cl H 8 B H 27 I loops HjL E j êh ω HCl ê2l x tp êfi Lowest fou hamonic enegy levels and ight side classical tuning points fo H 35 Cl (n è = 299 cm - ), H 8 B (n è = 265 cm - ) and H 27 I (n è = 23 cm - ). Enegy is in units of the zeo-point enegy of H 35 Cl, μ -2 J. The esults eflect the effects of the deceasing hamonic fequency going fom HCl to HI: The foce constant deceases, and so the hamonic potential enegy cuve ises less steeply on eithe side of its minimum, with the esult that tuning points ae fathe apat and so a wide allowed egion at a given total enegy. The effect is an incease in loop length and so a loweing of enegy fo a given numbe of loops, analogous to the enegy loweing in an infinite well when the well width is inceased. Now, hee ae two final questions to conside. Show that the decease in hamonic fequency, w, and so in the foce constant, k, going fom HCl to HI cannot be due to the inceasing educed mass alone. Hint: Compae the change in hamonic fequency expected due to mass alone to the actual change in hamonic fequency. Answe: Relative fequency expected due to educed mass:,.992,.9898; actual elative fequency:,.8863, What do you suppose the decease in foce constant is due to? à Analytic wavefunctions It tuns out that the hamonic oscillato Schödinge equation can be solved analytically. The wave functions have the geneal fom

9 Hamonic oscillato 9 y j HL = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ "############################## 2 j- H j - L! è!!! H j- HL -2 ê2 p in tems of Hemite polynomials, H j- HL. Hee ae the fist few Hemite polynomials. loops Hemite polynomial 2 2ρ ρ ρ+8 ρ ρ ρ ρ 6 ρ ρ ρ 2 48 ρ ρ 6 Fist seveal Hemite polynomials H j- HL. When the polynomials ae multiplied by the facto -2 ë2 the esulting function has the numbe of loops given in the fist column. Polynomial coesponding to even numbes of loops ae even about = ; polynomials coesponding to odd numbes of loops ae odd about =. Mathematica knows about Hemite polynomials, and so it is easy to constuct a function fo hamonic oscillato wavefunctions. Hee is the Mathematica function fo the wavefunction with j loops. ψ@j_, ρ_d := "################################## 2 j Hj L! è!!!! HemiteH@j, ρd ê2 π These wavefunctions ae nomalized to ; fo example, ψ@6, ρd 2 ρ They ae also othogonal, as must be so since they ae eigenfunctions of the hamonic oscillato Hamiltonian opeaot which is hemitian; fo example ψ@6, ρd ψ@3, ρd ρ Hee is a plot of y 6 HL. y Analytic hamonic oscillato wavefunction y 6 HL. The function is nomalized to. Displacement fom equilibium is in units è!!!!!!!!!!!!!! Ñ ê m w

10 Hamonic oscillato This is the same as the function that we obtained ealie using Numeov integation, to within the accuacy of the numeical implementation. à Quantal hamonic motion To teat hamonic motion quantum mechanically, we need to constuct wavepackets. A geneal expession fo a wavepacket is YHL = N g j y j HL, j in tems of elative weights g j and the nomalization constant N = ì $%%%%%%%%%%% g j2. j Fo example, a packet composed of waves with, 2 and 3 loops, with elative weights 25%, 5% and 25% is YHL = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ 85 y HL +.5 y 2 HL + 5 y 3 HL "################################ ######## Use of the othonomality of the component waves, that is, that Ÿ y j HL y k HL =d jk, to confim that this wavepacket is nomalized, that is, that Ÿ» YHL» 2 =. Let's use Mathematica to constuct and plot hamonic oscillato wavepacket pobability densities. Fist, we can define the list of weights g j. Fo the example above, this is g = 85,.5, 5<; Next, we can constuct a list of wavefunctions. Fo the example above, this is f =ψ@#, ρd & ê@ 8, 2, 3< è!!! 2 9, 2 2 ρ π ê4 π ê4 The nomalization facto, evaluates to, ρ 2 ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ, "################################ ######## nom = è!!!!!!!!! g.g H ρ 2 L 2 è!!! = 2 π ê4 The sum of the poducts of the functions times thei weights, is g y HL + g 2 y 2 HL + g 3 y 3 HL,

11 Hamonic oscillato g.f ρ H ρ 2 L Putting eveything togethe, the nomalized wavepacket is packet = nom g.f J ρ H ρ 2 LN We can check that this packet is nomalized, as. packet 2 ρ Hee is what the pobability density (the squae of the wavepacket) looks like.»y» Pobability density» YHL» 2 of a thee component hamonic oscillato wavepacket YHL. The pobability density is nomalized to. Displacement fom equilibium is in units è!!!!!!!!!!!!! Ñ ê m w The packet is localized nea the ight classical tuning point. If we wee to add moe waves to the packet, the width of the localized egion would become smalle, in accodance with the Heisenbeg indeteminacy elation. The next thing we need to do is to make the wavepacket move. To do this we need only to inset the time dependent phase factos, -Â E j têñ = -Â Ñ w ÅÅÅÅÅÅÅÅ 2 H2 j-l têñ = -Â wh j- ÅÅÅÅ 2 L t, to the components of the packet. Fo ou example packet, the list of these phase factos is phase = 9 ω I# 2 t ω, 3 2 t ω, 5 2 M t & ê@ 8, 2, 3< 2 t ω = The list of functions times thei phase factos is f phase 2 2 t ω 9 π ê4, è!!! t ω ρ π ê4, ρ t ω 2 H ρ 2 L 2 è!!! = 2 π ê4

12 2 Hamonic oscillato The sum of the poducts of the functions times thei phase factos and thei weights, and multiplied by the nomalization facto, is ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ Ig y HL "############ j g 2 j -Â wh- ÅÅÅÅ packetwithtime = nom g.hf phasel J t ω 2 L t + g 2 y 2 HL Â wh2- ÅÅÅÅ 2 3 t ω 2 L t + g 3 y 3 HL -Â wh3- ÅÅÅÅ 2 ρ L t M, 2 5 t ω 2 H ρ 2 LN Show that this wavepacket is nomalized to, fo evey value of time, t. At t = the time-dependent wavepacket, YH, L, is identical to the time independent wavepacket, YHL, and so gives the same pobability density. Veify that this statement is coect. Hee is what the pobability density (the squae of the wavepacket) looks like at time t = 5 ê n, one quate of the way though one oscillation peiod.»y» Pobability density» YH, tl» 2 of a thee component hamonic oscillato wavepacket YH, tl at time t = 5ê n. The pobability density is nomalized to. Displacement fom equilibium is in units è!!!!!!!!!!!!! Ñ ê m w Hee is what the pobability density looks like at time t =.5 ê n, half way though one oscillation peiod.

13 Hamonic oscillato 3»Y» Pobability density» YH, tl» 2 of a thee component hamonic oscillato wavepacket YH, tl at time t =.5 ê n. The pobability density is nomalized to. Displacement fom equilibium is in units è!!!!!!!!!!!!!! Ñ ê m w Finally, hee is what the pobability density looks like at time t = 2 p ê w= ê n, afte one oscillation peiod.»y» Pobability density» YH, tl» 2 of a thee component hamonic oscillato wavepacket YH.tL at time t = ê n. The pobability density is nomalized to. Displacement fom equilibium is in units è!!!!!!!!!!!!!! Ñ ê m w The packet has etuned to its oiginal fom and location at t = ê n. à Enegy of wavepackets: expectation values The enegy of a wave packet is defined to be XH\ = - Y * H, tl H YH, tl. The notation X \ denotes "aveage" o "expectation" and such an expession is known as the aveage value of the expectation value of the physical quantity coesponding to the opeato that appeas between the backets. Fo example, the expectation value of position would be X\ = - Y * H, tl YH, tl. Since the opeato fo position is just "multiply by position," we can eaange this expession as

14 4 Hamonic oscillato X\ = - À YH, tlà 2. In this fom the aveage value of the position is seen to be just the aveage of all possible positions, weighted by the pobability that the paticle is at each position. It is this intepetation that led of the name aveage value o expectation value. If the opeato is moe complicated than "multiply by", then we cannot eaange things in this way, but we still intepet the expession in the same way. Show that the expession fo the expectation value of the squaed momentum, in dimensionless units, is Xp 2 \ =-Ÿ Y * H, tl 2 ê 2 YH, tl. Taking account of the othonomality of the hamonic oscillato wavefunctions, - y j HL y k HL = d jk, it is not too difficult to show that XH\ = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ j g j 2 k g k 2 E k, an aveage of the component enegies, each weighted by its elative contibution to the pobability density. Show that this expession is coect fo any wavepacket composed of othonomal components y k HL. Why do you suppose this expession is independent of time? Hee is a Mathematica function that takes as input a list of component wavefunctions (in tems of the numbe of loops of each component) and a list of the coesponding weights, and computes the dimensionless enegy expectation value of the packet. εavg@j_list, g_listd := g.g g2.hh2 # L & ê@ jl Fo example, the aveage enegy of a wavepacket consisting of an equal mixtue of the one-loop and two-loop wavefunctions is εavg@8, 2<, 8, <D 2 This is what we expect, since y has enegy e = and y 2 has enegy e 2 = 3. A packet composed of % y and 9% y 2 has instead the aveage enegy This is vey close to e 2, as we would expect, since the packet is most y 2. à Wavepacket machine Collecting eveything togethe, a geneal time-dependent hamonic oscillato wavepacket is

15 Hamonic oscillato 5 YH, tl = ÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅÅ "############# k g 2 k g j y j HL j -Â wh j- ÅÅÅÅ 2 L t. Show that this expession is nomalized to fo all values of t. Show that» YH, tl» 2 oscillates with peiod ê n. Hee is a Mathematica function that makes a wavepacket, fo a specified choice of component wavefunctions, y j, and weights, g j. HOPacket@j_List, g_listd := è!!!!!!!!! g.jψ@#, ρd g.g I# 2 M 2 π t & ê@ jn In this function, time is measued in dimensionless units of the oscillation peiod, 2 p ê w= ê n; this means that t = coesponds to elapsed time ê n=2 p è!!!!!!!!!! m ê k. As example of this function, hee is the two-component wavepacket composed of equal contibutions of one-loop and two-loop components. 3 πt 2 H 2 πt + è!!! 2 ρl è!!! 2 π ê4 Veify that this expession is coect, by constucting the wavepacket by hand. Now that we have a tool to constuct hamonic oscillato wavepackets, let's exploe the popeties of diffeent packets. à Gaussian wavepacket We can constuct wavepackets of essentially abitay shape by appopiate choice of weights g j. Expeimentally, this amounts to appopiate excitation of the oscillato into a coheent supeposition of wavefunctions y j HL. (Coheent means thee is a definite phase elation between the components of the packet.) One common supeposition esults in a Gaussian distibution of weights. The Gaussian distibution centeed at m and with mean squaed deviation )vaiance) s is gauss@σ_, x_, μ_d := è!!!!!!! 2 π σ Hx μl2 êh2 σ 2 L The Gaussian distibution is nomalized to ; fo example gauss@, x, D x Hee is the Gaussian distibution centeed about with mean squaed deviation (the so-called nomal density function).

16 6 Hamonic oscillato pobability x Gaussian pobability distibution with mean and vaiance. The filled cicles maked values of the distibution at 7 equally spaced values of x centeed on the mean. We can select weights g j that appoximate this Gaussian distibution by evaluating the distibution at a ange of points centeed about the mean. Fo example, hee is a set of thiteen such weights chosen to span the distibution. g = gauss@, #, D & ê@ Range@ 3, 3, D êên êê sf2 8.44,.54, 4,, 4,.54,.44< These weights ae indicated as the filled cicled on the plot of the Gaussian distibution above. Hee is a plot of a Gaussian wavepacket consisting of the fist 3 hamonic oscillato wavefunctions y j HL, thoughout one cycle of oscillation..8.6»yh,tl» t 5 Hamonic oscillato Gaussian wavepacket pobability density thoughout one cycle of oscillation. The packet is composed of the fist 3 wavefunctions y j HL.

17 Hamonic oscillato 7 Show that the dimensionless enegy expectation value of this wavepacket is XH\=3. Could you have pedicted this esult without doing a detailed calculation? Hee is a plot of a Gaussian wavepacket consisting of 3 hamonic oscillato wavefunctions y j HL centeed about j = 27, thoughout one cycle of oscillation..75»yh,tl» t Hamonic oscillato Gaussian wavepacket pobability density thoughout one cycle of oscillation. The packet is composed of 3 wavefunctions y j HL centeed at j = 27. Show that the dimensionless enegy expectation value of this wavepacket is XH\=53. Hee is a plot of a Gaussian wavepacket consisting of 3 hamonic oscillato wavefunctions y j HL centeed about j = 47, thoughout one cycle of oscillation..8.6»yh,tl» t Hamonic oscillato Gaussian wavepacket pobability density thoughout one cycle of oscillation. The packet is composed of 3 wavefunctions y j HL centeed at j = 47.

18 8 Hamonic oscillato Show that the dimensionless enegy expectation value of this wavepacket is XH\=93. These thee Gaussian hamonic oscillato wavepackets all have the same numbe of adjacent components, but the cente component is successively highe. Use the suface plots of the pobability densities to see what diffeence this coesponds to physically. à Localization of hamonic motion We have seen that adding moe components to a wavepacket localizes the pobability density to a smalle egion of space. We can illustate this by constucting Gaussian packets centeed at the same y j, but with diffeing numbes of adjacent components. Hee ae the packets consisting of 5, 3 and 2 components, fo times coesponding to the fist quate of the peiod, ê n. Hamonic oscillato Gaussian wavepacket pobability densities thoughout one quate cycle of oscillation. The packets ae each centeed about j = 3; the left column is the 5-component packet, the middle column is the 3-component packet, and the ight column is the 2-component packet. The 5-component packet is pooly localized, the 3-component packet is faily localized, and the 2-component packet is highly localized.