Particle in a Box. and the state function is. In this case, the Hermitian operator. The b.c. restrict us to 0 x a. x A sin for 0 x a, and 0 otherwise

Transcription

1 Prticle i Box We must hve me where = 1,,3 Solvig for E, π h E = = where = 1,,3, m 8m d the stte fuctio is x A si for 0 x, d 0 otherwise x ˆ d KE V. m dx I this cse, the Hermiti opertor 0iside the box The b.c. restrict us to 0 x. ˆ p KE d ˆ p i m x

2 For the Hermiti opertor the eigefuctios re Prticle i Box (sidebr) ˆ d KE V, m dx x A si 0iside the box x Wht re the eigefuctios f(x) of pˆx? No restrictios o m f imx ˆ p f( x) i cf x f( x) e d c m x x Recll the Euler formul, si x i x e e i x i So we see tht the stte fuctio c be viewed s equl combitio of compoets with mometum d.

3 1,,3 for Prticle i Box

4 Orthogolity of the eigefuctios of Hermiti i opertor Def: Fuctios f 1 (x) d f (x) re sid to be orthogol b over the itervl (,b) if f * x f x dx. 0 1 Cosider Hermiti opertor ˆF. If FˆF i = f i i d FˆF j = f j j If f i f j, we c redily prove tht 1 d eigefuctios re orthogol, i.e. 1 * d = 0 Eigefuctios correspodig to differet eigevlues of Hermiti opertor re orthogol. The more geerl, very importt coclusios re tht The eigefuctios of Hermiti opertor either re orthogol or c be chose to be orthogol. The eigefuctios of Hermiti opertor re postulted to form complete set, spig the spce of the opertor.

5 Normliztio Problems 1, d 3 re rel fuctios tht re ormlized d orthogol Normlize the followig fuctios:. 1 + b. 1 - c d. 1/ 1/ 1 - () + (3/) 3

6 Smple clcultios for prticle i box Let test Normlizig, test orm 13.01, Possible results of mesuremet of E? Oly E 1 or E 5 or E 10 Most probble result? E 5 Averge result? 1 1 E ˆ E 3 E 0.1 E E 3E 0.1E E 9E 0.01E Life is sweet whe the stte fuctio is sum of eigefuctios, but...

7 Exmples of Sigificce of Postultes 30 Cosider fuctio x = x( x) for 0 < x <, d 0 otherwise. W 5 Is W cceptble stte fuctio? ) cotiuous d cotiuous derivtives? Yes except t 0 d b) sigle vlued? Yes ) x dx W W 5 x ) 0 c) Normlized? 30 ( 1 Yes Thus W fulfills the requiremets for stte fuctio, but it is NOT eigefuctio. Why?? Let's plot W for = w x

8 Exmples of Sigificce of Postultes So W is symmetric bout / d hs o odes. It is cceptble stte fuctio. It looks lot like 1. Wht is the expecttio vlue of eergy, E w, for w? E dx dx x ( x ) x ( x ) dx dx ˆ w 30 * * d W w w w 5 m m 0 0 x 0 Itegrtig, 5 h E W m m Is this eergy close to tht of 1? Recll for 1, π E= = m h 8m 1 This result is very close the vlue for E w foud bove. Notice, however, tht E w > E 1. Here E w = E 1 This iequlity is geerl d hs very importt cosequeces. It forms the bsis of geerl wy to systemticlly improve clcultio! (The Vritiol Priciple)

9 Exmples of Sigificce of Postultes Tht mkes sese!! Wht will we observe if we mke repeted mesuremets of E W? A distributio of eergies, sice W is ot eigefuctio of E Ech member of the distributio will be oe of the eergy eigevlues π h E = = where = 1,,3, m 8m 5 The verge of lrge umber of mesuremets is E h W m 7.9m Filly for W, expected, s W looks lot like w d w x

10 Smple Problem Clculte l the eergy differece betwee the =1 d = levels for electro (m=9.1 x kg) cofied to oedimesiol box hvig legth of 4.0 x m (this is the order of mgitude of tomic dimeter). Wht wvelegth correspods to spectrl trsitio betwee these levels? Aswer: E = 1.13 x J = 1.76 x 10 7 m

11 CocepTest #1 Cosider the eergy differece betwee the =1 d = levels for mrble of mss 1 g cofied i oe-dimesiol box of legth 0.10 m. Cosider the wvelegth tht correspods to spectrl trsitio betwee these levels. How does E betwee =1 d for the mrble compre to tht for the e i the previous problem? How does for this spectrl trsitio for the mrble compre to tht for the e i the previous problem? A. lrger, lrger D. smller, smller B. lrger, smller E. smller, lrger C. sme, sme

12 Numericl Aswer to CocepTest Clculte the eergy differece betwee the =1 d = levels for mrble of mss 1 g cofied i oe- dimesiol box of legth 0.10 m. Wht wvelegth correspods to spectrl trsitio betwee these levels? Aswer: E = 1.65 x 10 6 J = 1.0 x m Qutiztio cot be detected

13 Seprbility of Solutios If the multi coordite Hmiltoi opertor hs the specil form ( q, q, q q ) = h( ) + ( ) + ( )+ + ( ) with ech compoet 1 q h q h q h 3 q hvig the form h q T q V q i i i i i i the there exists simple, geerl solutio ( q, q,, q ) ()()() 1 1 q1 q 3 q () 3 i qi i=1 with ech k stisfyig 1D Schrödiger-like equtio h E. k k k k d the totl eergy is the sum of idividul compoets, E Ei. i 1 Thus wheever the Hmiltoi c be writte s sum, the becomes product of oe vrible terms d E becomes sum of eergies ssocited with ech degree of freedom. The difficulty of obtiig seprble solutio scles lierly with the umber of degrees of freedom, d y 1D problem is very esy! We will do (lmost) ythig to chieve seprbility!

14 Additiol Dimesios/Prticles M 1 M Two idepedet 1-D prticles i two boxes widths d b ˆ x, x V x V x M x M x Seprble Two idepedet prticles i two boxes E 1,,3 1,,3 1 totl 1 M Mb 1 x x Two qutum umbers 1 1 x, x si si 1 1 b b for 0 x d 0 y b

15 Additiol Dimesios/Prticles Two o-iterctig 1-D prticles i box of width x x V x ˆ, Seprble M M 1 1 x1 x E 1,,3 1,,3 1 totl 1 m m 1 x x 1 1 x, x si si 1 1 for 0 x d 0 x 1 Two qutum umbers