Curious problem using the variational method to find the ground state energy of the Harmonic oscillator.

Transcription

1 Curious proble usig the variatioal ethod to fid the groud state eergy of the Haroic oscillator. Origially appeared at: Peeter Joot Oct 3, 211 variatioharoicoscillator.tex Cotets 1 Motivatio 1 2 Recap. Variatioal ethod to fid the groud state eergy. 1 3 Recap. The variatioal ethod. 1 4 The Haroic oscillator variatioal proble. 2 5 Is our trial fuctio represetable? 4 6 Calculatig the Fourier ters 5 7 Correctig, treatig the origi this way Motivatio 2. Recap. Variatioal ethod to fid the groud state eergy. Proble 3 of 24.4 i the text [1] is a iterestig oe. It asks to use the variatioal ethod to fid the groud state eergy of a oe diesioal haroic oscillator Hailtoia. Soewhat uexpectedly, oce I take derivatives equate to zero, I fid that the variatioal paraeter beta becoes iagiary? I tried this twice o paper ad pecil, both ties gettig the sae thig. This sees like a oteworthy proble, ad oe worth reflectig o a bit. 3. Recap. The variatioal ethod. Give ay, ot ecessarily oralized wavefuctio, with a series represetatio specified usig the eergy eigevectors for the space where ψ = c ψ, (1 H ψ = E ψ, (2 1

2 ad ψ ψ = δ. (3 We ca perfor a eergy expectatio calculatio with respect to this ore geeral state ψ H ψ = = =, c ψ H c ψ c E ψ c ψ c c E ψ ψ c 2 E c 2 E = E ψ ψ This allows us to for a estiate of the groud state eergy for the syste, by usig ay state vector fored fro a superpositio of eergy eigestates, by siply calculatig E ψ H ψ ψ ψ. (4 Oe of the exaples i the text is to use this to fid a approxiatio of the groud state eergy for the Heliu ato Hailtoia H = h2 ( ( e e r 1 r 2 r 1 r 2. (5 This calculatio is perfored usig a trial fuctio that was a solutio of the iteractio free Hailtoia φ = Z3 πa 3 e Z(r 1+r 2 /a. (6 This is despite the fact that this is ot a solutio to the iteractio Hailtoia. The ed result eds up beig pretty close to the easured value (although there is a pesky error i the book that appears to require a copesatig error soewhere else. Part of the variatioal techique used i that proble, is to allow Z to vary, ad the oce the oralized expectatio is coputed, set the derivative of that equal to zero to calculate the trial wavefuctio as a paraeter of Z that has the lowest eergy eigestate for a fuctio of that for. We fid cosiderig the Haroic oscillator that this fial variatio does ot ecessarily produce eaigful results. 4. The Haroic oscillator variatioal proble. The proble asks for the use of the trial wavefuctio φ = e β x, (7 2

3 to perfor the variatioal calculatio above for the Haroic oscillator Hailtoia, which has the oe diesioal positio space represetatio H = h2 d 2 2 dx ω2 x 2. (8 We ca fid the oralizatio easily φ φ = e 2β x dx = 2 1 2β = 2 1 2β = 1 β e 2βx 2βdx e u du Usig itegratio by parts, we fid for the eergy expectatio φ H φ = = li ɛ dxe β x ( ( ɛ The first itegral we ca do h2 2 ɛ + + ɛ d 2 dx ω2 x 2 e β x dxe ( β x h2 ( = 2 dxe 2βx h2 β ω2 x 2 ( 2 dxe 2βx h2 β ω2 x 2 = h2 β 2 ɛ = h2 β 2 2 = β h2 2 + ω2 4β 3 d 2 dx ω2 x 2 e β x h2 ɛ 2 li dxe β x d 2 ɛ ɛ dx 2 e β x dxe 2βx + ω 2 dxx 2 e 2βx due u + ω2 8β 3 duu 2 e u A aive evaluatio of this itegral requires the origi to be avoided where the derivative of x becoes udefied. This also provides a ice way to evaluate this itegral because we ca double the itegral ad half the rage, eliiatig the absolute value. However, ca we assue that the reaiig itegral is zero? I thought that we could, but the ed result is curious. I tried also verified y calculatio usig Matheatica. Matheatica also required soe special had holdig to deal with the origi. Iitially I coded this by avoidig the origi i as above, but later switched to x = x 2 which Matheatica treats ore gracefully. 3

4 Without that last itegral, ivolvig our sigular x ad x ters, our groud state eergy estiatio, paraetrized by β is E[β] = β2 h ω2 4β 2. (9 Observe that if we set the derivative of this equal to zero to fid the best beta associated with this trial fuctio = E β = β h2 2 ω2 2β 3 (1 we fid that the paraeter beta that best iiizes this groud state eergy fuctio is coplex with value β 2 = ± iω 2 h. (11 It appears at first glace that we ca t iiize 9 to fid a best groud state eergy estiate associated with the trial fuctio 7. We do however, kow the exact groud state eergy hω/2 for the Haroic oscillator. Is is possible to show that for all β 2 we have hω 2 β2 h ω2 4β 2 (12? This iequality would be expected if we ca assue that the trial wavefuctio has a Fourier series represetatio utilizig the actual eergy eigefuctios for the syste. The resolutio to this questio is avoided oce we iclude the sigularity. This is explored i the last part of these otes. 5. Is our trial fuctio represetable? I thought perhaps that sice the trial wave fuctio for this proble lies outside the spa of the Hilbert space that describes the solutios to the Haroic oscillator. Aother thig of possible iterest is the trouble ear the origi for this wave fuctio, whe operated o by P 2 /2, ad this has bee (icorrectly assued to have zero cotributio above. I had iitially thought that part of the value of this variatioal ethod was that we ca use it despite ot eve kowig what the exact solutio is (ad i the case of the Heliu ato, I believe it was stated i class that a exact closed for solutio is ot eve kow. This akes e woder what restrictios ust be iposed o the trial solutios to get a eaigful aswer fro the variatioal calculatio? Suppose that the trial wavefuctio is ot represetable i the solutio space. If that is the case, we eed to adjust the treatet to accout for that. Suppose we have φ = c ψ + c ψ. (13 where ψ is ukow, ad presued ot orthogoal to ay of the eergy eigekets. We ca still calculate the or of the trial fuctio 4

5 φ φ = c ψ + c ψ c ψ + c ψ, = c 2 + c c ψ ψ + c c ψ ψ + c 2 ψ ψ = ψ ψ + c Re (c c ψ ψ. Siilarly we ca calculate the eergy expectatio for this uoralized state ad fid φ H φ = c ψ + c ψ H c ψ + c ψ, = c 2 E + c c E ψ ψ + c c E ψ ψ + c 2 ψ H ψ Our oralized eergy expectatio is therefore the cosiderably essier φ H φ φ φ = c 2 E + c c E ψ ψ + c c E ψ ψ + c 2 ψ H ψ ψ ψ + c Re (c c ψ ψ c 2 E + c c E ψ ψ + c c E ψ ψ + c 2 (14 ψ H ψ ψ ψ + c Re (c c ψ ψ With a requireet to iclude the perpedicular cross ters the or does t just cacel out, leavig us with a clea estiatio of the groud state eergy. I order to utilize this variatioal ethod, we iplicitly have a assuptio that the ψ ψ ad ψ ψ ters i the deoiator are sufficietly sall that they ca be eglected. 6. Calculatig the Fourier ters I order to see how uch a proble represetig this trial fuctio i the Haroic oscillator wavefuctio solutio space, we ca just calculate the Fourier fit. Our first few basis fuctios, with α = ω/ h are u = I geeral our wavefuctios are α π e α2 x 2 /2 α u 1 = 2 x 2 /2 π (2αxe α2 α u 2 = 8 π (4α2 x 2 2e α2 x 2 /2 u = N H (αxe α2 x 2 /2 α N = π2! d H (η = ( 1 e η2 dη e η2 5

6 Fro which we fid ψ(x = e α2 x 2 /2 (N 2 H (αx H (αxe α2 x 2 /2 ψ(xdx (15 Our wave fuctio, with β = 1 is plotted i figure (1 Figure 1: Expoetial trial fuctio with absolute expoetial die off. The zeroth order fittig usig the Gaussia expoetial is foud to be ψ (x = 2β erfc ( β 2α e α2 x 2 /2+β 2 /(2α 2 (16 With α = β = 1, this is plotted i figure (2 ad ca be see to atch fairly well Figure 2: First te orders, fittig haroic oscillator wavefuctios to this trial fuctio. The higher order ters get sall fast, but we ca see i figure (3, where a teth order fittig is depicted that it would take a uber of the to get aythig close to the sharp peak that we have i our expoetial trial fuctio. Note that the all the brakets of eve orders i with the trial fuctio are zero, which is why the teth order approxiatio is oly a su of six ters. Details for this haroic oscillator wavefuctio fittig ca be foud separately, calculated usig a Matheatica worksheet. The questio of iterest is why if we ca approxiate the trial fuctio so icely (except at the origi eve with just a first order approxiatio (polyoial ties Gaussia fuctios where the polyoials are Hakel fuctios, ad we ca get a exact value for the lowest eergy state usig the first order approxiatio of our trial fuctio, why do we get garbage fro the variatioal 6

7 Figure 3: Teth order haroic oscillator wavefuctio fittig. ethod, where eough ters are iplicitly icluded that the peak should be sharp. It ust therefore be iportat to cosider the origi, but how do we give soe eaig to the derivative of the absolute value fuctio? The key, supplied whe askig Professor Sipe i office hours for the course, is to express the absolute value fuctio i ters of Heavyside step fuctios, for which the derivative ca be idetified as the delta fuctio. 7. Correctig, treatig the origi this way. Here s how we ca express the absolute value fuctio usig the Heavyside step x = xθ(x xθ( x, (17 where the step fuctio is zero for x < ad oe for x > as plotted i figure (4. Figure 4: stepfuctio Expressed this way, with the idetificatio θ (x = δ(x, we have for the derivative of the absolute value fuctio x = x θ(x x θ( x + xθ (x xθ ( x = θ(x θ( x + xδ(x + xδ( x = θ(x θ( x + xδ(x + xδ(x 7

8 Observe that we have our expected uit derivative for x >, ad 1 derivative for x <. At the origi our θ cotributios vaish, ad we are left with x x= = 2 xδ(x x= We ve got zero ties ifiity here, so how do we give eaig to this? As with ay delta fuctioal, we ve got to apply it to a well behaved (square itegrable test fuctio f (x ad itegrate. Doig so we have dx x f (x = 2 dxxδ(x f (x = 2( f ( This equals zero for ay well behaved test fuctio f (x. Sice the delta fuctio oly picks up the cotributio at the origi, we ca therefore idetify x as zero at the origi. Usig the sae techique, we ca express our trial fuctio i ters of steps ψ = e β x = θ(xe βx + θ( xe βx. (18 This we ca ow take derivatives of, eve at the origi, ad fid ψ = θ (xe βx + θ ( xe βx βθ(xe βx + βθ( xe βx = δ(xe βx δ( xe βx βθ(xe βx + βθ( xe βx = δ(xe βx δ(xe βx βθ(xe βx + βθ( xe βx = β ( θ(xe βx + θ( xe βx Takig secod derivatives we fid ψ = β ( θ (xe βx + θ ( xe βx + βθ(xe βx + βθ( xe βx = β ( δ(xe βx δ( xe βx + βθ(xe βx + βθ( xe βx = β 2 ψ 2βδ(x Now applicatio of the Hailtoia operator o our trial fuctio gives us so Hψ = h2 ( β 2 ψ 2δ(x ω2 x 2 ψ, (19 8

9 Noralized we have ( ψ H ψ = h2 β ω2 x 2 e 2β x + h2 β δ(xe β x = β h2 2 + ω2 4β 3 + h2 β = β h2 2 + ω2 4β 3. E[β] = ψ H ψ ψ ψ = β2 h ω2 4β 2. (2 This is lookig uch ore proisig. We ll have the sig alteratio that we require to fid a positive, o-coplex, value for β whe E[β] is iiized. That is so the extreu is foud at = E β = β h2 ω2 2β 3, (21 β 4 = 2 ω 2 2 h 2. (22 Pluggig this back i we fid that our trial fuctio associated with the iiu eergy (uoralized still is ad that eergy, after substitutio, is ψ = e ωx 2 2 h, (23 E[β i ] = hω 2 (24 2 We have soethig that s 1.4 the true groud state eergy, but is at least a ball park value. However, to get this result, we have to be very careful to treat our poit of sigularity. A derivative that we d call udefied i first year calculus, is ot oly defied, but required, for this treatet to work! Refereces [1] BR Desai. Quatu echaics with basic field theory. Cabridge Uiversity Press,