9. Harmonic Oscillator - MIT OpenCourseWare

9. Harmonic Oscillator9.1Harmonic Oscillator9.1.1 Classical harmonic oscillator and h.o. model9.1.2 Oscillator Hamiltonian: Position and momentum operators9.1.3 Position representation9.1.4 Heisenberg picture9.1.5 Schrödinger picture9.2 Uncertainty relationships9.3 Coherent States9.3.1 Expansion in terms of number states9.3.2 Non-Orthogonality9.3.3 Uncertainty relationships9.3.4 X-representation9.4 Phonons9.4.1 Harmonic oscillator model for a crystal9.4.2 Phonons as normal modes of the lattice vibration9.4.3 Thermal energy density and Specific Heat9.1 Harmonic OscillatorWe have considered up to this moment only systems with a finite number of energy levels; we are now going toconsider a system with an infinite number of energy levels: the quantum harmonic oscillator (h.o.).The quantum h.o. is a model that describes systems with a characteristic energy spectrum, given by a ladder ofevenly spaced energy levels. The energy difference between two consecutive levels is E. The number of levels isinfinite, but there must exist a minimum energy, since the energy must always be positive. Given this spectrum, weexpect the Hamiltonian will have the form 1H ni n ω ni ,2where each level in the ladder is identified by a number n. The name of the model is due to the analogy withcharacteristics of classical h.o., which we will review first.9.1.1 Classical harmonic oscillator and h.o. modelA classical h.o. is described by a potential energy V 21 kx2 . If the system has a finite energy E, the motion is bound2by two values x0 , such that V (x0 ) E. The equation of motion is given by m ddxx2 kx and the kinetic energy is2pof course T 12 mẋ2 2m. The energy is constant since it is a conservative system, with no dissipation. Most of thetime the particle is in the position x0 since there the velocity is zero, while at x 0 the velocity is maximum.The h.o. oscillator in QM is an important model that describes many different physical situations. We will studyin depth a particular system described by the h.o., the electromagnetic field. Another system that can be described79

by this model is solid-state crystals, where the oscillations of nuclei in the lattice can be described as a systems ofcoupled oscillators (phonons).Notice that any potential with a local minimum can be locally described by an h.o. Provided that the energy is lowenough (or x close to x0 ), any potential can in fact be expanded in series, giving: V (x) V (x0 ) b(x x0 )2 . . .2where b ddxV2 x0 .9.1.2 Oscillator Hamiltonian: Position and momentum operatorsWe can define the operators associated with position and momentum. They are two observables (p,x) with thecommutation properties: [x, p] i . With these two operators, the Hamiltonian of the quanutm h.o. is written as:p2kx2p21 mω 2 x2 ,2m22m 2pwhere we defined a parameter with units of frequency: ω k/m.We use the dimensionless variables, pP , X x mωmωH ˆ H/ω, to simplify the expression to Hˆ ω(X 2 P 2 )/2 or H ω (X 2 P 2 ).and H2ˆ (n 1 ) ni hn , if expressed in anWe have said initially that we expect the hamiltonian to have the form H2appropriate basis. This simply corresponds to diagonalizing the Hamiltonian (thus the basis { ni} is the energybasis, or the basis formed by the eigenstates of the Hamiltonian). However the diagonalization is not as intuitive asfor simple TLS (or n TLS) because we are considering a system with infinite dimensions.ˆ (n 1 ) ni hn is our guess for the diagonalized form of the Hamiltonian, which makes explicit theThe operator H2presence of energy levels, labeled by n. Correspondingly, there must be operators that act on the ladder of energylevels. For example, the fundamental operations possible are the raising or lowering of 1 quantum of energy, as wellas an operator giving the number of energy quanta: N ni n ni. The raising and lowering operators act as thefollowing: a ni n 1i and a† ni n 1i. They are also called the annihilation and creation operators, as theydestroy or create a quantum of energy.Instead of deriving rigorously these operators, we guess their form in terms of the X and P operators: ia 12 (X iP ) 12 ( mω x mωp) 11i†a 2 (X iP ) 2 ( mωx mωp),and we will check a posteriori that indeed they act as annihilation and creation operators. Notice that a, a† are nothermitian, but they are one the hermitian conj ugate of the other (a (a† )† ). Also, we define the number operatoras N a† a. The commutation properties are: a, a† 1 and [N, a] a, N, a† a† .Also we have:q x 2mω(a† a)q†p i mω 2 (a a)? Question: Prove the commutation relationships of the raising and lowering operators.11ii[X iP, X iP ] ([X, iP ] [iP, X]) [X, P ] [x, p] 12 2 So we also have aa† [a, a† ] a† a 1 a† a 1 N .[a, a† ] [N, a] [a† a, a] [a† , a]a aandNotice that from now on we will take as usual 1.80[N, a† ] [a† a, a† ] a† [a, a† ] a†

From the commutation relationships we have:a ni [a, N ] ni an ni N a ni N (a ni) (n 1)(a ni),that is, a ni is also an eigenvector of the N operator, with eigenvalue (n 1). Thus we confirm that this is thelowering operator: a ni cn n 1i. Similarly, a† ni is an eigenvector of N with eigenvalue n 1: a† ni N, a† ni N a† ni a† n ni N (a† ni) (n 1)(a ni).We thus have a ni cn n 1i and a† ni dn n 1i. What are the coefficients cn , dn ?Sincehn N ni hn a† a ni nandhn a† a ni (han )(a ni) hn 1 n 1ic2n , we must have cn n. Analogously, since aa† N 1, as seen from the commutation relationship:d2n hn 1 n 1i ha† n a† ni hn aa† ni hn (N 1) ni n 1So in the end we have :a ni n n 1i ;a† ni n 1 n 1i .All the n eigenvalues of N have to be non-negative since n hn N ni hψn1 ψn1 i 0 (this follows from theproperties of the inner product and the fact that ψn1 i a ni is just a regular state vector). However, if we applyover and over the a (lowering) operator, we could arrive at negative numbers n: we therefore require that a 0i 0to truncate this process. The action of the raising operator a† can then produce any eigenstate, starting from the 0eigenstate:(a† )n ni 0i .n!The matrix representatio n of these operator in the ni basis (with infinite-dimensional matrices) is particularly simple,since hn a n′ i δn′ ,n 1 n and hn a† n′ i δn′ ,n 1 n 1: 00 0 .01 0 . . . †a 0 02 . . . a 1 0 0 . . . 0 00 .02 0 .The Hamiltonian can be written in terms of these operators. We substitute a, a† at the place of X and P , yieldingH ω(a† a 21 ) ω(N 21 ) and the minimum energy ω/2 is called the zero point energy.9.1.3 Position representationWe have now started from a (physical) description of the h.o. Hamiltonian and made a change of basis in order toarrive at a simple diagonal form of it. Now that we know its eigenkets, we would like to go back to a more intuitivepicture of position and momentum. We thus want to express the eigenkets ni in terms of the position representation(see also section 5.5.1).RRThe position representation corresponds to expressing a state vector ψi in the position basis: ψi dx hx ψ i xi dx ψ(x) xi (where xi is the eigenstate of the position operator that is a continuous variable, hence the integral).This defines the wavefunction ψ(x) hx ψ i.The wave function description in the x representation of the quantumfound by starting with the ground h.o. can bepstate wavefunction. Since a 0i 0 we have 12 (X iP ) 0i 12 ( mωx imω) 0i 0. In the x representation,given ψ0 (x) hx 0i 1ip hx ( mωx ) 0i 0mω2 (mωx 81d)ψ0 (x) 0dx ψ0 (x) e mωx2/2

The other eigenstates are built using Hermite Polynomials Hn (x), using the formula31 ni differential equations: n 11 dψn (x) hx ni mωx ψ0 (x)mω dxn!2nwith solutions ψn (x) hx ni (a† )n n! 0i to derive 1H (x)ψ0 (x).2n n! nThe n 2 and n 3 wavefunctions are plotted in the following figure, while the second figure displays the probabilitydistribution function. Notice the different parity for even and odd number and the number of zeros of these . 12: Left: Harmonic oscillator wavefunction. Right: corresponding probability distribution function for n 2 (blue) andn 3 (Red, dotted).Classically, the probability that the oscillating particle is at a given valueqof x is simply the fraction of time that it2spends there, which is inversely proportional to its velocity v(x) x0 ω 1 xx2 at that position. For large n, the0probability distribution becomes close to the classical Fig. 13: Left: Harmonic oscillator wavefunction. Right: corresponding probability distribution function for n 40. In Red, theclassical probability.31For more details on Hermite Polynomials and their generator function, look on Cohen-Tannoudji. Online information from:Eric W. Weisstein. Hermite Polynomial. From MathWorld–A Wolfram Web Resource.82

9.1.4 Heisenberg pictureWe want now to study the time-evolution of the h.o. We first start with analyzing the evolution of the operators inthe Heisenberg picture. We haveda1 i[H, a] i[ω(a† a ), a] iωadt2 a(t) a(0)e iωtSimilarly:da†1 i[H, a† ] i[ω(a† a ), a† ] iωa† a† (t) a† (0)eiωtdt2Notice that we could have found this last relationship just by taking the hermitian conjugate of the first one.Using these results, we can also find the time evolution of the position and momentum operators:x(t) x(0) cos(ωt) p(0)sin(ωt)mωp(t) p(0) cos(ωt) mωx(0) sin(ωt)and the corresponding expectation values, e.g.hx(t)i hx(0)i cos(ωt) hp(0)isin(ωt)mω9.1.5 Schrödinger picturePAn initiaPl state can be expressed in terms of the number eigenvectors: ψi n cn ni. Then its evolution is simply: ψ(t)i n cn e inωt ni. From this expression, one can calculate e.g.Xhx(t)i cn c m hm x ni e iωt(n m) .n,mSince x only connects states that differ by n m 1, it’s easy to see that the double sum simplifies and we retrievethe expression above, found in the Heisenberg picture.9.2 Uncertainty relationshipsThe operators x and p for a quantum h.o. do not commute, so they do not share any eigenstate, nor they shareeigenstates with the Hamiltonian. In particular the diagonal elements of x and p in the ni-basis representation areboth zero, therefore the expectation values are also zero. In a series of measurements, it is possible to get a range ofvalues; we associate this dispersion of values with the root mean square value of the eigenvalues:p x hx2 i hxi2(3)p p hp2 i hpi2(4)Given the expression for x and p in terms of a and a† we can calculate x2 : hn aa a† a† a† a aa† ni2mω hn a† a aa† ni (2n 1)2mω2mωhx2 i (5)(6)and in the same way, we can calculate p2 : hp2 i m ω(2n 1). Since the expectation values are zero (hxi hpi 0),pp 2the deviations are just: x hx2 i and p hp2 i and the uncertainty relation can be expressed by: p x (2n 1)283(7)