561 Fall 013 Lecture #5 page 1 Last time: Lecture #5: Begi Quatum Mechaics: Free Particle ad Particle i a 1D Box u 1 u 1-D Wave equatio = x v t * u(x,t): displacemets as fuctio of x,t * d -order: solutio is sum of liearly idepedet fuctios * geeral solutio by separatio of variables * boudary coditios give specific physical system * ormal modes octaves, odes, Fourier series, quatizatio * The pluck: superpositio of ormal modes, time-evolvig wavepacket Problem Set #: time evolutio of plucked system * More complicated for separatio of -D rectagular drum Two separatio costats Today: Begi Quatum Mechaics The 1-D Schrödiger equatio is very similar to the 1-D wave equatio It is a postulate Caot be derived, but it is motivated i Chapter 3 of McQuarrie You ca oly determie whether it fails to reproduce experimetal observatios This is oe of the weirdesses of Quatum Mechaics We are always tryig to break thigs (story about the Exploratorium i Sa Fracisco) 1 Operators: Tells us to do somethig to the fuctio o its right Examples: Af ˆ = g, operator deoted by A ˆ ( ^ hat) d f (x) = f (x) dx * take derivative d ( af (x) + bg(x)) = af (x) + bg (x ) dx liear operator ( xg * itegrate dx af (x) + bg(x)) = a dxf + b d liear operator * take square root ( af (x) + bg(x)) = [ af (x) + bg(x)] 1/ NOT liear operator We are iterested i liear operators i Quatum Mechaics (part of McQuarrie s postulate #) Eigevalue equatios revised 9/3/13 9:3 AM
561 Fall 013 Lecture #5 page Âf (x) = af (x) a is a eigevalue of the operator A A f(x) is a specific eigefuctio that belogs to the eigevalue a more explicit otatio Af ˆ (x) = a f (x) Operator A Eigefuctio Its eigevalue ˆ d A = dx ax e a d ˆB = dx sibx + cosbx b ˆ d C = x dx ax 3 Importat Operators i Quatum Mechaics (part of McQuarrie s postulate #) For every physical quatity there is a liear operator coordiate xˆ = x mometum pˆ x = i (at first glace, this seems surprisig Why?) x A kietic eergy T = p m = m x potetial eergy Vˆ(x) = V(x) eergy Ĥ = Tˆ + Vˆ = m x + V(x) (the Hamiltoia ) Note that these choices for xˆ ad pˆ are dimesioally correct, but their truthiess is based o whether they give the expected results 4 There is a very importat fudametal property that lies behid the ucertaity priciple: o-commutatio of two operators xˆpˆ pˆxˆ To fid out what this differece betwee xˆpˆ ad ˆpxˆ is, apply the commutator, [xˆ, ˆp] xˆpˆ pˆxˆ, to a arbitrary fuctio revised 9/3/13 9:3 AM
561 Fall 013 Lecture #5 page 3 df df xˆpf ˆ (x) = x( i) = ix dx dx d df pˆxf ˆ (x) = ( i) (xf ) = ( i) f + x dx dx [xˆ, pˆ ] xˆpˆ pˆxˆ = i a o-zero commutator We will evetually see that this o-commutatio is the reaso we caot sharply specify both x ad p x 5 Wavefuctios (McQuarrie s postulate #1) ψ(x): state of the system cotais everythig that ca be kow Stragely, ψ(x) itself ca ever be directly observed The cetral quatity of quatum mechaics is ot observable This should bother you! * ψ(x) is a probability amplitude similar to the amplitude of a wave (ca be positive or egative) * ψ(x) ca exhibit iterferece + revised 9/3/13 9:3 AM
561 Fall 013 Lecture #5 page 4 * probability of fidig particle betwee x, x + dx is ψ*(x)ψ(x)dx (ψ* is the complex cojugate of ψ) ˆ 6 Average value of observable A i state ψ? Expectatio value (part of McQuarrie s postulate #4) A = ψ *Aψ ˆ dx ψ *ψ dx Note that the deomiator is eeded whe the wavefuctio is ot ormalized to oe Hψ ˆ = E ψ ψ is a eigefuctio of H ˆ that belogs to the specific eergy eigevalue, E (part of McQuarrie s postulate #5) Let s look at two of the simplest quatum mechaical problems They are also very importat because they appear repeatedly 1 Free particle: V(x) = V 0 (costat potetial) d Ĥ = + V 0 m dx Ĥψ = Eψ, move V 0 to RHS d ψ = (E V 0 )ψ m dx d m(e V 0 ) ψ = ψ dx Note that if E > V 0, the o the RHS we eed ψ multiplied by a egative umber Therefore ψ must cotai complex expoetials This is the physically reasoable situatio But if E < V 0 (how is such a thig possible?), the o the RHS we eed ψ multiplied by a positive umber ψ must cotai real expoetials e + kx diverges to as x + uphysical [but useful for x fiite (tuelig)] e kx diverges to as x So, whe E > V 0, we fid ψ(x) by tryig ψ = ae +ikx + be ikx (two liearly idepedet terms) revised 9/3/13 9:3 AM
561 Fall 013 Lecture #5 page 5 d ψ ikx = k (ae + be ikx ) dx m(e V 0 ) = k Solve for E, ( k ) E k = + V 0 m * ψ = ae ikx is eigefuctio of pˆ You show that * with eigevalue k * ad p = k No quatizatio of E because k ca have ay real value ψ NON-LECTURE What is the average value of mometum for ψ = ae ikx + be ikx? p = = dxψ * pˆψ dxψ * ψ ormalizatio itegral ikx d + b * e ikx ikx dx ) + be ikx (a * e )( i ( ae ) dx ikx dx + b * e ikx ikx + be ikx ( a * e )( ae ) ikx i dx ( a * e + b * e ikx )(ik)( ae ikx be ikx ) = dx a + b + a * be ikx + ab * e ) ikx ( ikx k dx ( a b + ab * e a * be ) ikx = ikx + + a * be ikx dx ( a b + ab * e ) Itegrals from to + over oscillatory fuctios like e ±ikx are always equal to zero Why? a b p = k a + b if a = 0 if b = 0 p = k p = +k revised 9/3/13 9:3 AM
561 Fall 013 Lecture #5 page 6 a a + b is fractio of the observatios of the system i ψ which have p > 0 b is fractio of the observatios of the system a + b i ψ which have p < 0 END OF NON-LECTURE Free particle: it is possible to specify mometum sharply, but if we do that we will fid that the particle must be delocalized over all space For a free particle, ψ*(x)ψ(x)dx is delocalized over all space If we have chose oly oe value of k, ψ ψ ca be oscillatory, but it must be positive everywhere Oscillatios occur whe e ikx is added to e ikx NON-LECTURE ikx ψ = ae + be ikx ψ ψ = a + b + Re[ab*e ikx ], but if a,b are real ψ * ψ = a + b + ab coskx costat oscillatory Note that ψ ψ 0 everywhere For x where cos bx has its maximum egative value, cos kx = 1, the ψ ψ = (a b) Thus ψ ψ 0 for all x because (a b) 0 if a,b are real Sometimes it is difficult to uderstad the quatum mechaical free particle wavefuctio (because it is ot ormalized to 1 over a fiite regio of space) The particle i a box is the problem that we ca most easily uderstad completely This is where we begi to become comfortable with some of the mysteries of Quatum Mechaics * isight ito electroic absorptio spectra of cojugated molecules * derivatio of the ideal gas law i 56! * very easy itegrals Particle i a box, of legth a, with ifiitely high walls ifiite box pˆ Ĥ = + V(x) m V(x) = 0 0 x a very coveiet because dxψ *V(x)ψ = 0 V(x) = x < 0, x > a (covice yourself of this!) revised 9/3/13 9:3 AM
561 Fall 013 Lecture #5 page 7 ψ(x) must be cotiuous everywhere ψ(x) = 0 everywhere outside of box otherwise ( ) ψ *Vψ = ψ(0) = ψ(a) = 0 at edges of box Iside box, this looks like the free particle, which we have already solved Ĥψ = Eψ Schrödiger Equatio d ψ = Eψ (V(x) = 0 iside the box) m dx d ψ = m Eψ = k ψ dx me k ψ (x) = Asi kx + Bcos kx satisfies Schrödiger Equatio (it is the geeral solutio) Apply boudary coditios: ψ(0) = B = 0 therefore B = 0 ψ(a) = A si ka = 0 therefore A si ka = 0 (quatizatio!) ka = π π k = a is a iteger π ψ = Asi x a 0 a dxψ * ψ = 1 ormalize a π a A dx si x = A = 1 0 a 1/ A = a 1/ π ψ = si x is the complete set of eigefuctios for a particle i a box Now a a fid the eergies for each value of revised 9/3/13 9:3 AM
561 Fall 013 Lecture #5 page 8 d π Ĥψ = si x m dx a a 1/ π = + ψ m a h = ψ 8ma h E = = E 1 = 1,, 3 (ever forget this!) 8ma E 1 = 0 meas the box is empty what would a egative value of mea? a/3 a/3 E 3 =9E 1 two odes a/ E =4E 1 oe ode h E 1 = 8ma zero odes 0 x a 1 odes, odes are equally spaced All lobes betwee odes have the same shape revised 9/3/13 9:3 AM
561 Fall 013 Lecture #5 page 9 Summary: Some fudametal mathematical aspects of Quatum Mechaics Iitial solutios of two-simplest Quatum Mechaical problems * Free Particle * Particle i a ifiite 1-D box Next Lecture: 1 * more about the particle i 1-D box * Zero-poit eergy (this is uexpected) * ΔxΔp vs ( = 1 gives miimum ucertaity) particle i 3-D box * separatio of variables * degeeracy revised 9/3/13 9:3 AM
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