5.61 Fall, 013 Lecture #9 Page 1 Last time Lecture #9: Harmonic Oscillator: Creation and Annihilation Operators 1/ Simplified Schrödinger equation: ξ=α x, α =(kμ) 1/ n E +ξ ψ=0 (dimensionless) ξ nω reduced to Hermite differential equation by factoring out asymptotic form of ψ. The asymptotic ψ is alid as ξ. The exact ψ is ψ (x) = N H (ξ)e ξ Hermite polynomials = 0, 1,, orthonormal set of basis functions E = nω( + ½), = 0, 1,, een, een function odd, odd function = # of internal nodes what do you expect about T? V? (from classical mechanics) pictures * zero-point energy * tails in non-classical regions * nodes more closely spaced near x = 0 where classical elocity is largest * enelope (what is this? maxima of all oscillations) * semiclassical: good for pictures, insight, estimates of ˆ ψ* Opψ integrals without soling Schrödinger equation 1/ p E (x) = p classical (x) = μ(e V(x)) enelope of ψ(x) in classical region (classical mechanics) 1 k / π 1/4 ψ *ψdx, ψ(x) = 1/ for H. enelope O. E V (x) elocity spacing of nodes (quantum mechanics): # nodes between x 1 and x is x pe (x) dx (because λ(x) = h/p(x) and node spacing is λ/) h x 1 x+ (E ) # of leels below E: p E (x) dx Semi-classical quantization rule h x (E ) Action (h) integral. reised 9/0/13 :04 PM
5.61 Fall, 013 Lecture #9 Page Non-Lecture Intensities of Vibrational fundamentals and oertones from 1 μ(x) = μ 0 + μ 1 x + μ x + dx ψ* x n ψ +m selection rules m = n, n, n Today some amazing results from a,â (creation and annihilation operators) * dimensionless px, ˆ p ˆ exploit uniersal aspects of problem separate uniersal from specific a, ˆ a ˆ annihilation/creation or ladder or step-up operators * integral- and waefunction-free Quantum Mechanics * all E and ψ for Harmonic Oscillator using â,â * alues of integrals inoling all integer powers of xˆ and/or pˆ * selection rules * integrals ealuated on sight rather than by using integral tables. 1. Create dimensionless pxˆ and pˆ operators from xˆ and pˆ 1/ 1/ n mf t 1 kμ 1/4 xˆ = xpˆ, units = = f recall ξ=α 1/ x = x μω mt 1 n 1/ pˆ = [nμω p, pˆ units = [mf t 1 mt 1 ] 1/ = mf t 1 = p ] replace xˆ and pˆ by dimensionless operators pˆ 1 nμω k n H = + kxˆ = pˆ + xpˆ μ μ mω nω nω nω = pˆ + xpˆ factor this? nω = (ipˆ + xpˆ)( ipˆ + xpˆ)? does this work? No, this attempt at factorization generates a term i pˆ, pxˆ, which must be subtracted 1/ aˆ 1/ aˆ out: H = nω ââ i pˆ, pxˆ = i reised 9/0/13 :04 PM
5.61 Fall, 013 Lecture #9 Page 3 a ˆ = 1/ (xpˆ + ipˆ ) â = 1/ (xpˆ ipˆ ) x = 1/ pˆ ( â + â ) p ˆ = i 1/ ( â â) be careful about x, pˆ pˆ 0 We will find that âψ = ( ) 1/ ψ 1 annihilates one quantum 1/ â ψ = ( +1) ψ +1 creates one quantum H = nω( ââ 1/) = nω( â a ˆ +1/ ). This is astonishingly conenient. It presages a form of operator algebra that proceeds without eer looking at the form of ψ(x) and does not require direct ealuation of integrals of the form A ij = dx ψ* i Âψ j.. Now we must go back and repair our attempt to factor H for the harmonic oscillator. Instructie examples of operator algebra. * What is (ipˆ + xpˆ)( ipˆ + xpˆ)? pˆ + xpˆ + ipˆxpˆ ixpˆpˆ i pˆ,xpˆ Recall [ pˆ, xˆ] = in. (work this out by pˆxf ˆ xˆpf ˆ = [ pˆ, xˆ] f ). What is i pˆ, xpˆ? ˆ 1/ n 1/ i p, xpˆ = i[nmω] [ pˆ, xˆ] mω [ ] ( = i n 1/ in) =+1. reised 9/0/13 :04 PM
5.61 Fall, 013 Lecture #9 Page 4 So we were not quite successful in factoring H. We hae to subtract (1/)nω: 1 ˆ aa H = nω left oer This form for H is going to turn out to be ery useful. * Another trick, what about [â,â ]=? i i 1/ ), 1/ ˆ ˆ â,a = (ipˆ + xpˆ ( ipˆ + xpˆ) = p, xpˆ + xpˆ, p 1 1 = + = 1. ˆ ˆ 1 1 So we hae some nice results. H = nω â a + = nω ââ reised 9/0/13 :04 PM
5.61 Fall, 013 Lecture #9 Page 5 3. Now we will derie some amazing results almost without eer looking at a waefunction. If ψ is an eigenfunction of H with energy E, then â ψ is an eigenfunction of H belonging to eigenalue E + nω. ( ) ˆ 1 ˆ H a ψ = hω â a + a ψ Factor â out front 1 ˆ = hω â aâ + â ψ 1 = â hω ââ + ψ ââ = a, ˆ â + â â = 1+ â â 1 H (â ψ ) = â hω â ˆ a +1+ ψ H +hω and Ĥψ = E ψ, thus H (â ψ ) = â (E + hω)ψ = (E + hω)(â ψ ) Therefore â ψ is eigenfunction of H with eigenalue E + nω. So eery time we apply â to ψ, we get a new eigenfunction of H and a new eigenalue increased by nω from the preious eigenfunction. â creates one quantum of ibrational excitation. Similar result for â ψ. H (âψ ) = (E nω)(âψ ). â ψ is eigenfunction of H that belongs to eigenalue E nω. â destroys one quantum of ibrational excitation. We call â, â ladder operators or creation and annihilation operators (or step-up, step-down). reised 9/0/13 :04 PM
5.61 Fall, 013 Lecture #9 Page 6 Now, suppose I apply â to ψ many times. We know there must be a lowest energy eigenstate for the harmonic oscillator because E V(0). We hae a ladder and we know there must be a lowest rung on the ladder. If we try to step below the lowest rung we get â ψ min = 0 1/ ipˆ + xpˆ ψ min = 0 Now we bring xˆ and pˆ back. d in dx 1/ ˆ μω 1/ i(nμω) p + xˆ ψ min = 0 n 1/ 1/ n d μω + + x ψ min = 0 μω dx n 1/ 1/ dψ min μω = μω xψ min dx n n μω = xψ min. n This is a first-order, linear, ordinary differential equation. What kind of function has a first deriatie that is equal to a negatie constant times the ariable times the function itself? de cx cx = cxe dx μω c = n μω ψ min = N min e x n. The lowest ibrational leel has eigenfunction, ψ min (x), which is a simple Gaussian, centered at x = 0, and with tails extending into the classically forbidden E < V(x) regions. reised 9/0/13 :04 PM
5.61 Fall, 013 Lecture #9 Page 7 Now normalize: μω x * n dx ψ = 1 = min ψ min N dx e min gie factor of in exponent π 1/ ) 1/ ( μω n μω ψ min (x)= πn This is the lowest energy normalized waefunction. It has zero nodes. NON-LECTURE Gaussian integrals 1/4 μω x n [recall asymptotic factor of ψ(x): e ξ / ] e π 1/ dx e r x = 0 r 1 dx xe r x = 0 r π 1/ dx x e r x = 0 4r 3 n! n+1 r x dx x e = 0 r n+ n r x 1 3 5 (n 1) dx x e =π 1/ 0 n+1 n+1 r By inspection, using dimensional analysis, all of these integrals seem OK. We need to clean up a few loose ends. 1. Could there be seeral independent ladders built on linearly independent ψ min1, ψ min? Assertion: for any 1-D potential it is possible to show that the energy eigenfunctions are arranged so that the quantum numbers increase in step with the number of internal nodes. particle in box n = 1,, # nodes = 0, 1,, which translates into the general rule # nodes = n 1 harmonic oscillator = 0, 1,, reised 9/0/13 :04 PM
5.61 Fall, 013 Lecture #9 Page 8 # nodes = We hae found a ψ min that has zero nodes. It must be the lowest energy eigenstate. Call it = 0.. What is the lowest energy? We know that energy increases in steps of nω. E +n E = nnω. We get the energy of ψ min by plugging ψ min into the Schrödinger equation. BUT WE USE A TRICK: Now we also know 1 ˆ H = nω a a + 1 H ψ min = nω a â + ψ min but âψ min = 0 1 so H ψ min = nω 0 + ψ min 1 E min = nω! E min +n E min = nnω OR E 0+ E 0 = nω, thus E = nω( +1/ ) NON-LECTURE 3. We know what are c and d? â ψ = cψ +1 âψ = dψ 1 reised 9/0/13 :04 PM
5.61 Fall, 013 Lecture #9 Page 9 1 1 H = nω â â + = nω ââ H 1 H 1 ˆ = â a, + = ââ nω nω H 1 1 1 ψ = + ψ = â âψ nω for ââ we use the trick â â ψ = ψ â â is number operator, N. ˆ ââ = â a + [a, ˆ â ] = N +1 +1 Now dx ψ* ââ ψ = dx â ψ because ââ is Hermitian Prescription for operating to the left is ψ * â = (â * ψ ) * = (a ψ ) * similarly for d in âψ = d ψ 1 c +1 = ] 1/ c = [ +1 dx ψ* â âψ = Make phase choice and then erify by putting in xˆ and pˆ. dx âψ = d d = 1/ Again, erify phase choice reised 9/0/13 :04 PM
5.61 Fall, 013 Lecture #9 Page 10 â ψ = ( +1) 1/ ψ +1 âψ = ( ) 1/ ψ 1 N = â â N ψ = ψ [ â, â ] = 1, ˆ Now we are ready to exploit the a a operators. Suppose we want to look at ibrational transition intensities. μ(x) = μ 0 + μ 1 More generally, suppose we want to compute an integral inoling some integer power of xˆ (or pˆ ). â = 1/ a ˆ = 1/ N = â â ( ipˆ + xpˆ) (ipˆ + xpˆ) x pˆ = 1/ ( â + â) ˆ ( a â) p = 1/ i ˆ (number operator) ˆx + μ ˆx use â,a + 1/ 1/ μω μω n n xˆ = xpˆ = ( â + â) ] 1/ ˆ nμω 1/ pˆ = [nμω p = i( â â) reised 9/0/13 :04 PM
5.61 Fall, 013 Lecture #9 Page 11 n n n + ˆ + ˆ μω μω μω x = ( â + â)( â + â) = [â a + â â + ââ ] = [â a + â â +1] nμω nμω p = (â + a ˆ a ˆ a ˆ ââ ) = [ â + a ˆ a ˆ a ˆ 1 ] etc. p k nω = + ˆ ˆ + ˆ ˆ ˆ μ 4 4 H + x = nω ( â a â a 1)+ ( â a + â a +1) = nω( â a +1/ ) as expected. The terms in H inoling â + a ˆ exactly cancel out. Look at an ( â ) m ( â) n operator and, from m n, read off the selection rule for Δ. Integral is not zero when the selection rule is satisfied. reised 9/0/13 :04 PM
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