Chapter Seven The Quantum Mechanical Simple Harmonic Oscillator

Transcription

1 Capter Seven Te Quantum Mecanical Simple Harmonic Oscillator Introduction Te potential energy function for a classical, simple armonic oscillator is given by ZÐBÑ œ 5B were 5 is te spring constant. Suc a classical oscillator as an angular frequency = œ È5Î7, were 7 is te mass of te oscillator. Writing te potential energy function in terms of te angular frequency, rater tan te spring constant, gives ZÐBÑ œ B (7.1) Te time-independent Scrodinger equation can ten be written in te form or, in operator notation, ` < ÐBÑ B < ÐBÑ œ I< ÐBÑ (7.2) 7 `B s: < ÐBÑ sb < ÐBÑ œ I< ÐBÑ (7.3) 7 Te potential energy function for tis problem is different from te ones we ave considered tus far. Tis potential energy function essentially spans all space. In te diagram below, you can see tat tere are regions were te total energy is greater tan te potential (regions were te wave function would be sinusoidal) and regions were te total energy is less tan te potential (regions were proper wave functions will be exponentially decreasing). Tus, a proper solution to te simple armonic oscillator problem will be sinusoidal in te central region, but must be exponentially decreasing as B tends toward bot positive and negative infinity. Fig. 7.1 Te Simple Harmonic Oscillator Potential

2 Quantum Harmonic Oscillator 2 Analytical Metod We begin wit te time-independent Scrodinger equation for te simple armonic oscillator, expressed in terms of te angular frequency ` < ÐBÑ B < ÐBÑ œ I< ÐBÑ (7.) 7 `B In order to simplify tis equation somewat, we will let B œ α (7.5) Tis introduces an arbitrary constant α wic can ten be defined is suc a way as to create a dimensionless equation. Te Scrodinger equation becomes ` Ð Ñ 7 Ð Ñ œ I Ð Ñ 7 α ` < = α < < (7.6) Multiplying by 7α Î gives % ` 7 7 ` Ð Ñ < = α < Ð Ñ œ α I< ÐÑ (7.7) ow using our freedom to define α, we coose α to be so tat our equation simplifies to te dimensionless form α œ È Î (7.8) ` < ÐÑ < Ð Ñ œ I< Ð Ñ (7.9) ` = You can tell tis equation is indeed dimensionless, since te energy as units of =. Letting O œ IÎ= our equation reduces to ote tat ` Ð Ñ œ O Ð Ñ ` < ˆ < (7.1) For large (i.e., large displacements from te equilibrium position were p ) we expect te wave function to ave te approximate form < ÐÑ / (7.11) since we know tat te wave function must tend toward zero for bot positive and negative values of. Let's take tis approximate form of te wave function as a trial

3 Quantum Harmonic Oscillator 3 solution to our differential equation and see ow well it satisfies our differential equation. ` ` / œ / ` ` œ / ˆ % / œ % / Œ (7.12) otice tat if we set œ ß tis last equation reduces to ` / œ ˆ / (7.13) ` wic is almost exactly te form of our dimensionless oscillator equation except for te energy constant O. We expect te exact solution of our differential equation, terefore, to be of te form < ÐÑ œ 2ÐÑ/ (7.1) and ope tat pulling out te exponential term will elp us find a simple expression for 2Ð Ñ. (One can always ope) Our dimensionless diffential equation for te quantum oscillator is ` Ð Ñ œ O Ð Ñ ` < ˆ < (7.15) If we assum tat te correct solution for tis equation as te form of equation 7.1, we need te evaluate te second partial of tis function. Te first partial is given by ` ` Š 2ÐÑ/ œ 2ÐÑ / Œ 2ÐÑŠ / ` ` `2ÐÑ œ Œ 2ÐÑ / ` (7.16) Taking te partial again, we obtain 2 ` < ÐÑ ` 2ÐÑ `2ÐÑ `2ÐÑ œ / / / Š Š ` ` ` ` 2ÐÑŠ / Š / 2 ` < ÐÑ ` 2ÐÑ `2ÐÑ œ / 2Ð ` œ ш ` ` (7.17) Our dimensionless differential equation for te quantum oscillator becomes 2 ` 2Ð Ñ `2Ð Ñ / 2Ð Ñ œ O 2Ð Ñ/ œ ˆ ` ˆ (7.18) `

4 Quantum Harmonic Oscillator Tis leads to a differential equation for te function out, given by 2ÐÑß since te exponential drops ` 2 2Ð Ñ `2Ð Ñ O 2ÐÑ œ (7.19) ` ` Tis equation we now solve using te assumption tat te solution can be expressed in terms of a power series in of te form 2ÐÑ œ â œ + (7.2) Differentiating te series term by term, onceß gives `2ÐÑ ` œ + + $+ â œ + (7.21) $ Differentiating again gives 2 ` 2ÐÑ œ + $ + $ % + â ` $ % œ + (7.22) Substituting tis into te differential equation for 2ÐÑ gives + + O + œ (7.23) Collecting terms of te same power and rearranging finally gives or + + O + œ (7.2) c + + O + d œ (7.25) ow if te differential equation must be valid for all values of te variable, ten te only way tis last equation can be valid for any arbitrary coice of is for te coefficients to be indentically zero, or + + O + œ (7.26)

5 Quantum Harmonic Oscillator 5 ow tis equation relates te coefficient + to te coefficient + according to te equation O + œ + (7.27) Tis recursion formula enables us to write any coefficient if we know just two, + and +. All oter coefficents can be determined based upon tese two. We terefore write were 2ÐÑ œ 2evenÐÑ2odd ÐÑ (7.28) % 2evenÐÑ œ % â is te even function of, built upon +, and were (7.29) $ & 2oddÐÑ œ + + $ + & â (7.3) is te odd function of, built upon +. Tus, we ave te solution of te differential equation in terms of two undetermined constants, + and +, wic is wat we would expect from a second-order differential equation. Te final form of te solution, ten, is < ÐÑ œ / 2ÐÑ œ / + + Ÿ (7.31) were we use te recursion relation to determine te + 's from + and +. Before we celebrate too muc, owever, we must determine if tis wave function converges as p ; oterwise te solution we ave obtained is unacceptable. If tere is a problem wit tis function as gets larger it would obviously be due to te larger powers of, i.e., larger values of. Let's see wat te recursion formula gives us as gets large. Te recursion formula can be written O + œ + ÎOÎ œ + Î Î (7.32) or, in te limit of large lim + œ + large Ä (7.33) or lim Älarge + œ (7.3) + If we let be te value of were tis limit begins to be valid, ten we can divide te summation up into two separate pieces; te region below were te recursion formula

6 Quantum Harmonic Oscillator 6 must be used, and te region above were te approximation to te recursion formula for large can be used. Te summation over te even terms, ten, becomes œ 2evenÐÑ œ + œ + + (7.35) or % ' + œ â % ' But for large we ave + % œ + ' œ + + % â % + œ ' + % + % + % œ œ % % (7.36) (7.37) giving % + œ + + œ â % + œ + + xœ â x x x % + œ + + xœ â x x x % + œ + + xœ â x x x + œ + + x â x x x Ÿ ÐÑ + œ + + x x x x â Ÿ + œ + + x š / ÐÑ (7.38)

7 Quantum Harmonic Oscillator 7 form Te even solution to te quantum armonic oscillator problem, terefore, as te < evenðñ œ / 2evenÐÑ œ / + + x š / Ÿ < ÐÑ œ / 2 ÐÑ œ / + G / even even (7.39) wic clearly diverges. Tis same type beavior can be demonstrated for < oddðñ as well. So wat do we do? Obviously, our solution is not te infinite series we ave derived, because tis infinite series diverges. However, if te series were to terminate for some value of, ten te function would remain finite. Let's look again at te recursion formula O + œ + (7.) If we demand tat te term + œ for œ - (were + Á Ñ, we obtain -O - - œ (7.1) or - O œ (7.2) from wic we obtain an equation wic fixes te energy of te oscillator or O œ IÎ œ - = - (7.3) I- œ Œ- = were - œ ßßß$ßá (7.) Tis gives us a denumerable energy spectrum and implies tat te correct solution for our problem is not an infinite series solution for 2ÐÑ, but a polynomial solution for 2Ð Ñ, expressed as 2 ÐÑ œ â œ + (7.5) - wit different order polynomials being associated wit different energies. ow te coefficients satisfy te recursion formula -

8 Quantum Harmonic Oscillator 8 O + œ + (7.6) were O œ -, so te recursion relation depends upon - as well as. We express tis recursion formula wit tis substitution to obtain - + œ (7.7) For - œ, Ðcorresponding to O œ, wic gives I œ = ), and we see tat te recursion formular + œ + (7.8) gives zero for all even + 's oter tan +. But wat about te odd values of +? We require tat + be set equal to zeroà oterwise, we would ave an infinite series wic diverges. Likewise, + is set equal to zero for te odd polynomials. Te polynomials, ten can be expressed as: 2 ÐÑ œ + 2 ÐÑ œ + 2 ÐÑ œ ÐÑ œ + + $ $ $ $ $ % % % % 2% ÐÑ œ % ã (7.9) or, in terms only of te first coefficient in eac of te even or odd sequences (note tat % + Á + Á + Ñ 2 ÐÑ œ + 2 ÐÑ œ + 2 ÐÑ œ + (7.5) ã $ $ 2 Ð Ñ œ + $ œ $ % % 2 Ð Ñ œ + % % % œ $ Remember tat multiplying te wavefunction solution to Scrödinger's equation by some arbitrary constant does not cange te differential equation. Tis is te reason we are always at liberty to normalize te wave function. Te leading coefficient, ten, (te one outside te braces) in eac of tese polynomials will be determined by normalization of te wave function. Tis means tat we are at liberty to multiply or divide te coefficients in eac polynomial by some common factor witout canging te

9 Quantum Harmonic Oscillator 9 equation in any significant way (we just incorporate tat arbitrary cange into te arbitrary coefficient). It is customary to coose te constants in te polynomials so tat te coefficient of te igest order term of is 8. Doing tis removes te fractions in te polynomial equations. In addition, we can mulitiply by a negative so tat te sign of te igest order term is always positive. Wit tese canges, te polynomials can be written as 2 ÐÑ œ + 2 ÐÑ œ + 2 ÐÑ œ + % $ $ 2$ ÐÑ œ + ) % % 2% ÐÑ œ + ' %) ã (7.51) Te terms inside te braces turn out to be te so-called Hermite polynomials wic arise from te Taylor series expansion 8 D D D / œ L8ÐÑ (7.52) 8x 8œ and can be generated using te Rodrigues formula 8 8. L8ÐÑ œ / Œ /. (7.53) Problem 7.1 Sow tat you can obtain te terms in te braces from te Taylor series expansion. Also, prove tat te Rodrigues formula works for te first few Hermite polynomials. So, apart from te normalization constant, te solutions to te armonic oscillator problem are just te Hermite polynomials. We next need to determine te normalization constants. Te wave functions associated wit te various energy states ave te form < ÐÑ œ 2 ÐÑ/ 8 8 (7.5) Te normalization condition for te ground state wave function is * * ( < ÐBÑ< ÐBÑ.B œ Ê ( < Ð Ñ< ÐÑ. œ (7.55)

10 Quantum Harmonic Oscillator 1 were we recall tat B œ α, wit α œ ÈÎ. Substituting in for te wave function, we ave Ê ( ˆ + /. œ Ê ˆ ( 7 + / =. œ Ê ˆ + œ 1 œ 7 = (7.56) using te Gaussian integral tables. Solving for te coefficient, we ave + œ 7 = 1 ormalization of te next wave function gives Ê ( Ê ( ˆ + /. œ Ê ˆ ( 7 + % / =. œ % * < Ð Ñ< ÐÑ. œ Ê ˆ + ) œ 1 œ 7 = % (7.57) (7.58) or, solving for te coefficient, 7 = + œ œ È 1 È x 1 % % (7.59) Continuing tis process, we find tat te normalization constant can be written E œ 8 7 È 88x = 1 % (7.6) Tus, te solution to te quantum armonic oscillator problem can be written < 8ÐÑ œ L 8/ È 88x 1 % Î (7.61) or since B œ α, wit α œ ÈÎ, we write œ È ÎB so tat te equation, in terms of B can be written < 8ÐBÑ œ L8 Î B / È 88x 1 ˆ È % B Î (7.62)

11 Quantum Harmonic Oscillator 11 ote 1: ormalization of te second order wave function gives Ê ( * < Ð Ñ< ÐÑ. œ Ê ( + % /. œ Ê ˆ + ( % ' ' % /. œ 7 = Ê ˆ $ + œ ' 1 ' 1 % 1 œ 7 = ) % 1 Ê ˆ 7 = + e '% f œ 1 Ê ˆ 7 = + ) œ or, solving for te coefficient, 7 = + œ œ È) 1 È x 1 % % (7.63) wic follows te form stated above. ote 2: Two additional relationsips for te Hermite polynomials wic often prove useful in calculations are L œ L 8L (7.6) were one polynomial can be related to te sum of two previously calculated polynomials, and `L8 œ 8L ` 8 (7.65) were te derivative of a polynomial is related to te previous polynomial.