Physics 195aCourse NotesThe Simple Harmonic Oscillator: Creation and Destruction Operators021105 F. Porter1 IntroductionThe harmonic oscillator is an important model system pervading many areasin classical physics; it is likewise ubiquitous in quantum mechanics. The non-relativistic Scro¨dinger equation with a harmonic oscillator potential is readilysolved with standard analytic methods, whether in one or three dimensions.However, we will take a different tack here, and address the one-dimensionproblem more as an excuse to introduce the notion of “creation” and “an-nihilation” operators, or “step-up” and “step-down” operators. This is anexample of a type of operation which will repeat itself in many contexts,including the theory of angular momentum.2 Harmonic Oscillator in One DimensionConsider the Hamiltonian:H =p22m+12mω2x2. (1)This is the Hamiltonian for a particle of mass m in a harmonic oscillatorpotential with spring constant k = mω2,whereω is the “classical frequency”of the oscillator. We wish to find the eigenstates and eigenvalues of thisHamiltonian, that is, we wish to solve the Schr¨odinger equation for thissystem. We’ll use an approach due to Dirac.Define the operators:a ≡mω2x + i1√2mωp (2)a†≡mω2x − i1√2mωp. (3)Note that a†is the Hermitian adjoint of a. We can invert these equations toobtain the x and p operators:x =1√2mω(a + a†), (4)1p = −imω2(a − a†). (5)Further, defineN ≡ a†a (6)=mω2x2+12mωp2−12, (7)wherewehavemadeuseofthecommutator[x, p]=i. Thus, we may rewritethe Hamiltonian in the form:H = ωN +12 , (8)and our problem is equivalent to finding the eigenvectors and eigenvalues ofN.3 Algebraic Determination of the SpectrumLet |n denote an eigenvector of N, with eigenvalue n:N|n = n|n. (9)Then |n is also an eigenstate of H with eigenvalue (n +12)ω.SinceN isHermitian,N†=(a†a)†=(a†)(a†)†= a†a = N, (10)its eigenvalues are real. Also n ≥ 0, sincen = n| N|n =  n| a†a|n = an|an≥0. (11)Now consider the commutator:[a, a†]=mω2x + i1√2mωp,mω2x − i1√2mωp=i2{−[x, p]+[p, x]} =1. (12)Next, notice thatNa = a†aa = aa†a − a = a(N − 1), (13)Na†= a†aa†= a†a†a + a†= a†(N +1). (14)2Thus,Na|n = a(N − 1)|n =(n − 1)a|n. (15)We see that a|n is also an eigenvector of N, with eigenvalue n−1. Assuming|n is normalized, n|n =1,thenan|an = n|N|n = n, (16)or a|n =√n|n − 1, where we have also normalized the new eigenvector|n −1.We may continue this process to higher powers of a, e.g.,a2|n = a√n|n − 1 =n(n − 1)|n − 2. (17)If n is an eigenvalue of N,thensoaren −1,n−2,n−3,....Butweshowedthat all of the eigenvalues of N are ≥ 0, so this sequence cannot go on forever.In order for it to terminate, we must have n an integer, so that we reach thevalue 0 eventually, and conclude with the statesa|1 =√1|0, (18)a|0 =0. (19)Hence, the spectrum of N is {0, 1, 2, 3,...,n,?}.To investigate further, we similarly consider:Na†|n =(n +1)a†|n. (20)Hence a†|n is also an eigenvector of N with eigenvalue n +1:a†|n =√n +1|n +1. (21)The spectrum of N is thus the set of all non-negative integers. The energyspectrum is:En= ωn +12 ,n=0, 1, 2,.... (22)We notice two interesting differences between this spectrum and our clas-sical experience. First, of course, is that the energy levels are quantized.Second, the ground state energy is E0= ω/2 > 0. Zero is not an allowedenergy. This lowest energy value is referred to as “zero-point motion”. Wecannot give the particle zero energy and be consistent with the uncertaintyprinciple, since this would require p =0andx = 0, simultaneously.3We may give a “physical” intuition to the operators a and a†: The energyof the oscillator is quantized, in units of ω, starting at the ground state withenergy12ω.Thea†operator “creates” a quantum of energy when operating ona state – we call it a “creation operator” (alternatively, a “step-up operator”).Similarly, the a operator “destroys” a quantum of of energy – we call it a“destruction operator” (or “step-down”, or “annihilation operator”). Thisidea takes on greater significance when we encounter the subject of “secondquantization” and quantum field theory.4TheEigenvectorsWe have determined the eigenvalues; let us turn our attention now to theeigenvectors. Notice that we can start with the ground state |0 and generateall eigenvectors by repeated application of a†:|1 = a†|0|2 =(a†)2√2|0...|n =(a†)n√n!|0. (23)So, if we can find the ground state wave function, we have a prescription fordetermining any other state.We wish to find the ground state wave function, in the position represen-tation (for example). If | x represents the amplitude describing a particle atposition x,thenx|0 = ψ0(x) is the ground state wave function in postitionspace.1Consider:x|a|0 =0= x|mω2x +ip√2mω|0=mω2x +1√2mωddxx|0. (24)1The notion here is that the wave function of a particle at position x is (proportionalto) a δ-function in position. Hence, ψ0(x)=x|0 =δ(x − x)ψ0(x) dx.4Thus, we have the (first order!) differential equation:ddxψ0(x)=−mωxψ0(x), (25)with solution:ln ψ0(x)=−mωx22+ constant. (26)The constant is determined (up to an arbitrary phase) by normalizing, toobtain:ψ0(x)=mωπ 1/4exp−12mωx2 . (27)This is a Gaussian form. The Fourier transform of a Gaussian is also aGaussian, so the momentum space wave function is also of Gaussian form.In this case, the position-momentum uncertainty relation is an equality,∆x∆p =1/2. This is sometimes referred to as a “minimum wave packet”.Now that we have the ground state, we may follow our plan to obtain theexcited states. For example,x|1 = x|a†|0 =mω2x −1√2mωddxx|0 (28)Substituting in our above result for ψ0, we thus obtain:ψ1(x)=√2mωmωπ 1/4x exp−12mωx2 . (29)To express the general eigenvector, it is convenient to define y =√mωx,anda†=1√2(y − dy), (30)where dyis a shorthand notation meaning d/dy. The general eigenfunctionsare clearly all polynomials times e−y2/2,sinceψn(y)=(a†)n√n!ψ0(n)=1√2nn!(y − dy)n1π1/4e−y2/2. (31)Thus, we may letψn(y)=1√2nn!1π1/4Hn(y)e−y2/2, (32)5where