VIEWS: 0 PAGES: 10 CATEGORY: Education POSTED ON: 1/13/2010
Section 5.8: The Harmonic Oscillator The Schrodinger wave equation (SWE) describes the time evolution of the wave function. The Hamiltonian for the harmonic oscillator describes a particle of mass m in a quadratic potential. Displacing the mass from equilibrium produces a linear restoring force. We focus on the 1-D oscillator since a 3-D oscillator can be decomposed into three 1-D oscillators. Any coupling between the three 1-D oscillators can be included in the Hamiltonian later if desired. The harmonic oscillator has important applications. Many systems have nonlinear potential functions. Expanding these nonlinear potentials in a Taylor series often produces a quadratic term as the lowest order approximation. As an example, the periodic motion of atoms about their equilibrium position can be modeled with the quadratic potential. We know this motion must be related to phonons moving through the material. We will see that the zero point motion of the atom can be described by the quantum mechanical vacuum state. Quantum optics provides a somewhat surprising application for the quadratic potential. The electromagnetic fields can be modeled by quadratic kinetic and potential terms. Of course, these do not refer to an electron in an electrostatic potential. Nor do they refer to the position or momemtum of photons. Instead, they refer to the form assumed by the fields in the Hamiltonian. The quantized form of the electromagnetic fields can be immediately written by comparison with the wavefunctions for the electron in the quadratic potential. Topic 5.8.1: Introduction to the Classical and Quantum Harmonic Oscillators For a harmonic oscillator, the quadratic potential produces a linear restoring force 1 2 kx 1 − D V = 2 1 kr 2 3 − D 2 where the equilibrium position occurs at the origin x=0, the “spring constant” must be positive k>0 and it describes the Figure 5.8.1: The curvature of the potential (i.e., magnitude of the force). quadratic potential The classical Hamiltonian has the form p2 1 2 Hc = + kx (5.8.1) 2m 2 where we consider the dynamic variables x, p to be independent of one another. Newton’s second law can be demonstrated using Hamilton’s canonical equation (refer to Section 5.2). ∂H p = − c = −kx = F ∂x The Lagrangian shows that the momentum p must be related to the velocity by p = mv = mx . 5.41 We want to compare and contrast solutions x(t) to the classical and quantum harmonic oscillators. The classical Hamiltonian (the total energy) can be rewritten using Equation 5.8.1 and p = mx 2 m d x(t) 1 + k (x ( t ) ) = E 2 (5.8.2) 2 dt 2 where E represents the total energy of the oscillator and x(t) represents the position of the electron parameterized by the time t. The solution has the form x ( t ) = A sin(ωo t ) (5.8.3a) The formula ωo = k / m relates the angular frequency of 2 oscillation ωο to the “spring constant” k. Substituting Equation 5.8.3a into Equation 5.8.2 provides 2E 2E A= = (5.8.3b) Figure 5.8.2: Motion of a harmonic k mωo 2 oscillator. The probability density ρ The amplitude A represents the points on the shows the most likely position of potential plot V(x) where the kinetic energy becomes finding the mass m is at the turning zero (see Figure 5.8.2) points where the oscillator momentumarily comes to rest. 1 2E E = kx 2 → A= 2 x =A k Classically, the particle can only be found in the region x ∈ [−A, A] and never outside that region. The probability density ρ for finding the particle at a point x appears similar to a delta function near the endpoints of the motion; this behavior occurs because the particle slows down near those points and spends more time there. Several differences exist between the classical and quantum mechanical harmonic oscillators. Figure 5.8.3 shows the quantum mechanical solution to Schrodinger’s equation with the quadratic potential. Unlike the classical particle, the quantum particle can be found in the classically forbidden region. The figure shows how the wavefunction exponentially decays in these classically forbidden regions. Classically, the particle doesn’t have enough energy to enter the forbidden region. The basis functions have the form Figure 5.8.3: The first two quantum 1 mechanical solutions to the α 2 α x 2 2 φn ( x ) = 1/ 2 n H n ( αx ) Exp − (5.8.4) harmonic oscillator. The π n! 2 2 probability density ρ for finding the particle at point x does not where α 4 = ( mωo / ) 2 . The exponential part of the resemble the classical one. solution ensures the wave function decreases in the 5.42 classically forbidden region. The Hermite polynomials H n primarily control the behavior in the classically allowed region near the center. They can be conveniently generated by differentiating an exponential according to H n (ξ ) = (− 1) exp(ξ 2 ) n exp(− ξ 2 ) n dn (5.8.5a) dξ where ξ=αx. The first three Hermite polynomials are H o (ξ ) = 1, H1 (ξ ) = 2ξ, H 2 (ξ ) = 4ξ 2 − 2 (5.8.5b) Continuing with Figure 5.8.3, perhaps most striking of all, the probability density function for the quantum particle decays to zero near the endpoints of motion and reaches its peak value (or values) near the center of the classical region [-A,A]. However, the classical probability of finding the classical particle assumes its minimum value near the origin. Here’s another difference between the classical and harmonic oscillator solutions. The classical oscillator energy can be increased by applying a driving force and increasing the oscillation amplitude E = A 2 mωo / 2 . The angular oscillation frequency 2 ωo = k / m remains constant for a fixed spring constant k. The energy of the quantum oscillator increases by also absorbing energy 1 E n = ωn = ωo n + n = 0,1, 2,... (5.8.6) 2 The integer “n” can be interpreted as either the “basis function number” or as the number of quanta stored in the motion. Contrary to the classical case, the angular frequency ωn of the quantum oscillator changes even though the value ωo remains fixed. The angular frequency does not refer to the rate at which the quantum particle bounces from side to side. We view the quantum particle as a stationary wave function. Larger numbers of quanta “n” result in larger “displacements” from equilibrium meaning the probability density has more peaks that move closer to the classically forbidden region. We find similar plots for quantized EM waves. The energy of an EM oscillator (the EM waves) can be changed by changing the angular frequency (or wavelength) or by changing the amplitude (i.e., the number of quanta in the mode). We will see that the “position x” and “momentum p” become the “in-phase” and “out-of-phase” electric fields. Therefore, the wavefunctions in the EM case describe the probability of finding a particular value of the electric field. Topic 5.8.2: The Hamiltonian for the Quantum Harmonic Oscillator The quantum mechanical Hamiltonians come from the classical ones by replacing the dynamical variables x,p with the corresponding operators x , p in ˆ ˆ H c = p 2 / 2m + kx 2 / 2 to find ˆ ∂ p2 1 2 ˆ ∂ H Ψ (t) = i Ψ (t) + kx Ψ ( t ) = i ˆ Ψ(t ) (5.8.7) ∂t 2m 2 ∂t 5.43 ∂ Operatin with the “coordinate” projection operator x produces x → x and p → ˆ ˆ i ∂x (refer to Appendix 6) to obtain the Schrodinger equation ˆ ∂Ψ (x, t) − 2 ∂2 1 2 ∂ H Ψ (x, t) = i or + kx Ψ ( x, t ) = i Ψ ( x, t ) (5.8.8) ∂t 2m ∂x 2 2 ∂t The boundary conditions for the Schrodinger wave equation for the harmonic oscillator require the wave function to approach zero as "x" goes to infinity Ψ (x → ±∞, t ) → 0 (5.8.9) There are two methods for solving the Schrodinger equation for the harmonic oscillator. The first method uses a power series solution, which becomes very algebraically involved. The solution starts by separating variables in Equation 5.8.8 and using a power series to find the solutions to the Sturm-Liouville problem (the eigenvector problem). The second method uses the linear algebra of raising and lowering operators. We present the method of raising and lowering operators (commonly referred to as the algebraic approach). We will find the stationary solutions given in Equation 5.8.4 and the energy eigenvalues in Equation 5.8.6. Topic 5.8.3: Introduction to the Operator Solution of the Harmonic-Oscillator The operator approach (i.e., algebraic approach) to solving Schrodinger's equation for the harmonic oscillator is simpler than the power series approach. In addition, it provides a great deal of insight into the mathematical structure of the quantum theory. The algebraic approach uses “raising a + and lowering a operators” (i.e., ladder ˆ ˆ operators, or sometimes called promotion and demotion operators). Later chapters demonstrate the similarity between the ladder operators and the “creation/annihilation” operators most commonly found in advanced studies of quantum theory. We will rewrite the Hamiltonian in terms of the raising and lowering operators in the form of the number operator ˆ ˆ ˆ N = a + a . The raising and lowering operators map one basis vector into another one according to a+ n = n +1 n +1 ˆ a n = n n −1 ˆ (5.8.10) as suggested by Figure 5.8.4. The lowering operator produces Figure 5.8.4: Raising and lowering operators zero when operating on the vacuum state a 0 = 0 . The number ˆ move the harmonic operator has two interpretations for the harmonic oscillator. oscillator from one state First we will show the energy eigenvectors are also eigenvectors to another. ˆ for the number operator according to N n = n n . The number operator therefore tells us the number of the eigenstate occupied by a particle. The number operator also tells us the number of energy quanta in the system as its second interpretation. We can say that a particle occupying one of the energy basis states n ∈ BV = { 0 = E 0 , 1 = E1 , …} has n quanta of energy according to 5.44 E n = ωo ( n + 1/ 2 ) . Therefore the vacuum state 0 corresponds to a particle state without any quanta of energy n=0. Interestingly, there exists energy in the vacuum state E 0 = ωo / 2 . The vacuum energy corresponds to zero point motion of atoms in a solid, for example. The atoms continue to move even though all of the extractable energy has been removed (i.e., n=0). Absolute zero can never be achieved since it is a classical concept corresponding to stationary atoms. Studies in quantum optics indicate that the electric field also experiences vacuum fluctuations; these fluctuations produce spontaneous emission from an ensemble of excited atoms. In the next few topics, we wish to find the energy eigenvectors B V = { 0 = E 0 , 1 = E1 ,…} and eigenvalues for the harmonic oscillator. We assume non-degenerate eigenvalues En, which means that for each energy En, there corresponds ˆ exactly one eigenstate φ n satisfying H φ n = E n φ n . We further assume an order for the energy levels E 0 < E1 < E 2 < . The operator approach must reproduce the results found with the power series approach. We first show how the Hamiltonian incorporates the raising-lowering operators. We briefly discuss the mathematical description of the ladder operators and demonstrate the origin of their normalization constant. We then easily solve for the energy eigenvalues and eigenvectors. Topic 5.8.4: Ladder Operators in the Hamiltonian The Hamiltonian for the harmonic oscillator is Hˆ = p + mωo x ˆ2 ˆ2 (5.8.11) 2m 2 We define the lowering a and the raising a + operators in terms ˆ ˆ of the position xˆ and momentum operators p . ˆ mωo i pˆ a= ˆ x+ ˆ (5.8.12a) 2m ωo 2m ωo mωo i p ˆ a+ = ˆ x− ˆ (5.8.12b) 2m ωo 2m ωo The raising operator in Equation 5.8.12b comes from taking the adjoint of the lowering operator in Equation 5.8.12a and using ˆ ˆ the fact that both x , p must be Hermitian since they correspond Figure 5.8.5: Physical to observables. Notice that the raising and lowering operators examples showing the are not Hermitian a ≠ a + . These two equations for the lowering effect of a raising ˆ ˆ and raising operators can be solved for the position and operators defined for an atom (top) and square momentum operators to find well (bottom) rather mωo than for the harmonic x= ˆ 2mωo ( a + a + ) p = −i 2 ( a − a + ) (5.8.13) oscillator. ˆ ˆ ˆ ˆ ˆ We need the Hamiltonian written in terms of the ladder operators. We must first determine the commutation relations. We can demonstrate that the raising operator 5.45 commutes with itself as does the lowering operator while the raising operator does not commute with the lowering operator [a, a ] = 0 = a + , a + ˆ ˆ ˆ ˆ a, a + = 1 ˆ ˆ (5.8.14) These last two relations can be proven using the commutation relations between the position and momentum operators [x, x ] = 0 = [p, p] [x, p] = i ˆ ˆ ˆ ˆ ˆ ˆ (5.8.15) We prove a, a + = 1 by first substituting Equations 5.8.12. ˆ ˆ mωo ˆ i p mωo i p ˆ a, a + = ˆ ˆ x+ ˆ , x− ˆ 2m ωo 2m ωo 2m ωo 2m ωo Distributing the terms provides 2 mωo [ p, p] + i p, x − i x, p ˆ ˆ a, a + = ˆ ˆ [ x, x ] + ˆ ˆ [ˆ ˆ] [ˆ ˆ] 2m ω 2m ωo 2 2 o Substituting the commutation relations from Equation 5.8.15, we find the desired results i i a, a = 0 + 0 + 2 ( −i ) − 2 ( i ) = 1 ˆ ˆ+ In the case of an ensemble of independent harmonic oscillators, each one has its own ˆ ˆ degrees of freedom x i , p i that obey their own commutation relations. ˆ ˆ ˆ ˆ x i , x j = 0 = pi , p j ˆ ˆ x i , p j = i δij As a result, there will be raising and lowering operators for each oscillator a i , a j = 0 = a i + , a j+ ˆ ˆ ˆ ˆ a i , a j+ = δij ˆ ˆ Using the definitions of the position and momentum operators, the Hamiltonian for the single harmonic oscillator can be rewritten by substituting relations 5.8.27. 2 2 p2 1 ˆ 1 mωo H =ˆ 2m 2 + mωo x 2 = 2 ˆ 2m −i 2 ( a − a + ) + 1 mωo2 2mω ( a + a + ) (5.8.16a) ˆ ˆ ˆ ˆ 2 o Squaring the constants provides ω ω H = − o (a − a+ ) + o (a + a+ ) 2 2 ˆ ˆ ˆ ˆ ˆ 4 4 Squaring the operators and taking care not to commute them gives us ˆ ω 4 { ˆ ˆ ˆ 2 H = o −a 2 + aa + + a + a − a + + a 2 + aa + + a + a + a + ˆ ˆˆ ˆ ˆˆ ˆ ˆ ˆ 2 } Combining the squared terms ω H = o { aa + + a + a} ˆ ˆˆ ˆ ˆ (5.8.16b) 2 We must always use commutation relation to change the order of operators. Finally, by using the commutation relation a, a + = 1 → aa + = 1 + a + a , the Hamiltonian becomes ˆ ˆ ˆˆ ˆ ˆ ω ω H = o {aa + + a + a} = o {2a + a + 1} ˆ ˆˆ ˆ ˆ ˆ ˆ 2 2 As a result, the Hamiltonian for the single harmonic oscillator can be written as 5.46 ˆ ˆ ˆ 1 H = ωo a + a + (5.8.17a) 2 ˆ ˆ ˆ+ We can define the number operator N = a a and rewrite Equation 5.8.17a as ˆ ˆ 1 H = ωo N + (5.8.17b) 2 Topic 5.8.5: Properties of the Raising and Lowering Operators Next, we demonstrate the relations a+ n = n +1 n +1 ˆ a n = n ˆ n −1 by first showing n − 1 ~ a n and n + 1 ~ a + n are eigenvectors of the number ˆ ˆ ˆ operator N corresponding to the eigenvalues n-1 and n+1, respectively. We next find the constants of proportionality. We will need two commutation relations. Using ˆˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ AB, C = A B, C + A, C B and A, B = − B, A and Equation 5.8.14, we find ˆ ˆ N, a = a + a, a = a + , a a = −a ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ N, a + = a + a, a + = a + a, a + = a + (5.8.18) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ We now show N = a + a and H = ˆ ω N + 1/ 2 have eigenvectors n − 1 ~ a n o ˆ( ) ˆ + and n + 1 ~ a n . Suppose n represents one eigenvector then ˆ ˆˆ { N a n = Na n = N, a + aN n = −a + aN n = {−a + a n} n = (n − 1) a n ˆ ˆ ˆ ˆ ˆˆ } { ˆ ˆˆ }ˆ ˆ ˆ Therefore a n must be an eigenvector of N with eigenvalue ( n − 1) . We can similarly ˆ ˆ show that N a + n = ( n + 1) a + n (see the chapter review exercises). Therefore, we ˆ ˆ ˆ + conclude a n = Cn n + 1 and a n = D n n − 1 since the eigenvalues are not ˆ ˆ degenerate where Cn and Dn denote constants of proportionality. ˆ ˆ ˆ ˆ ˆ 1 The eigenvalues of N = a + a and H = ωo N + must be real because 2 N = a a is Hermitian according to N = ( a a ) = a a = N . Further the eigenvalues n + ˆ ˆ ˆ + ˆ + + ˆ ˆ ˆ ˆ ˆ + must be greater than or equal to zero since the length of a vector must always be positive 2 ˆ n = n N n = n a + a n = a n ≥ 0 . We can also show that only integers represent the ˆ ˆ ˆ eigenvalues n. Next, we find the normalization constants Cn and Dn occurring in the relations. a + n = Cn n + 1 ˆ a n = Dn n − 1 ˆ Let’s work with the lowering operator. To find Dn, consider the string of equalities + + ˆ ˆ D* D n n − 1 n − 1 = D n n − 1 Dn n − 1 = a n a n n ˆ = n a +a n = n N n = n n n = n n n Now use the fact that all eigenvectors are normalized to one so that n −1 n −1 = 1 = n n 5.47 Therefore, the coefficient Dn must be 2 Dn = n → Dn = n where a phase factor has been ignored. Similarly, an expression for Cn can be developed + + C* C n n + 1 n + 1 = Cn n + 1 Cn n + 1 = a + n a + n n ˆ ˆ = n a a n = n a a +1 n = n N + 1 n = n n + 1 n = ( n + 1) n n + + ˆ where a commutator has been used in the fifth term. Once again using the eigenvector normalization conditions and comparing both sides of the last equation 2 Cn = n + 1 → Cn = n + 1 as expected. We therefore have the required relations. a+ n = n +1 n +1 ˆ a n = n ˆ n −1 (5.8.19) The set of eigenvectors 0 , 1 ,... can be obtained by repeatedly using the relation a + n = n + 1 n + 1 as ˆ ( a + ) 0 , …, n = ( a + ) 0 ,…. 2 n a+ ˆ a+ ˆ ˆ ˆ 1 = 0 , 2 = 1 = (5.8.20) 1 2 2 1 n! Some Commutation Relations H , a = ωo a + a, a = ωo a + [ a, a ] + ωo a + , a a = − ωo a ˆ (1) ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ (2) ˆ ˆ H , a + = ωo a + a, a + = ωo a + ˆ ˆ ˆ ˆ (3) ˆ ˆ ˆ ˆ ˆ N, a = −a N, a + = a + ˆ Topic 5.8.6: The Energy Eigenvalues The Hamiltonian for the harmonic oscillator can be written in terms of the ladder operators as given in Topic 5.8.4, Equation 5.8.17b. ˆ ˆ 1 H = ωo N + (5.8.21) 2 ˆ We already know the eigenvalues of the number operator to be N n = n n . The allowed energy values can be found as follows ˆ ˆ 1 1 H n = ωo N + n = ωo n + n 2 2 Therefore the energy values must be 5.48 1 E n = ωo n + (5.8.22) 2 Topic 5.8.7: The Energy Eigenfunctions We know the energy eigenvectors can be listed in the sequence ( a + ) 0 , …, n = ( a + ) 0 ,…. 2 n a+ ˆ a+ ˆ ˆ ˆ 1 = 0 , 2 = 1 = (5.8.23) 1 2 2 1 n! from Equation 5.8.20. However, we would like to know the functional form of these functions. There exists a simple method for finding the energy eigenfunctions for the harmonic oscillator using the ladder operators. Starting with 0=a 0 ˆ operate on both sides using the bra operator x and insert the definition for the lowering operator mωo xˆ ˆ ip mωo xˆ ˆ ip 0= x a 0 = x ˆ + 0 = x 0 + x 0 (5.8.24) 2 mωo 2 mωo 2 mωo 2 mωo Factor out the constants from the brackets and use the relations ∂ ∂ x x 0 = x x 0 = x φ0 ( x ) ˆ x p0 = ˆ x 0 = φ0 ( x ) i ∂x i ∂x where x 0 = φ 0 (x ) is the first energy eigenfunction in the set of eigenfunctions given by { φ 0 (x ), φ1 (x ),… } Equation 5.8.24 now provides mωo x ∂ 0= xa 0 = φ o (x ) + φ o (x ) 2 mωo 2 mωo ∂x which is a simple first-order differential equation dφ 0 mωo + xφ o = 0 dx We can easily find the solution mωo 2 φ 0 (x ) = φ 0 (0) exp − x 2 which represents the first energy eigenfunction. The normalization constant φ 0 (0) is found by requiring the wave function to have unit length 1 = φ 0 (x ) φ 0 (x ) which gives 1/ 4 mωo mωo 2 φ 0 (x ) = exp − x π 2 Now the other eigenfunctions can be found from φ1 (x ) using the raising operator a+ mωo x ˆ ˆ ip φ1 (x ) = x 1 = x 0 = x − 0 1 2 mωo 2 mωo 5.49 where the constants can be factored out and the coordinate representation can be substituted for the operators to get mωo x ∂ mωo x ∂ φ1 ( x ) = x 0 − x 0 = φ0 ( x ) − φ0 ( x ) 2 mωo 2 mωo ∂x 2 mωo 2 mωo ∂x Notice that we do not need to solve a differential equation to find the eigenfunctions φ1 , φ2 ,... in the basis set. Differentiating φ 0 (x ) provides 1/ 4 ∂φ o ∂ mωo mωo 2 mωo x = exp − x =− φo (x) ∂x ∂x π 2 Consequently the n=1 energy eigenfunction becomes mωo x mωo x 2 mωo φ1 (x ) = φ 0 (x ) + φ 0 (x ) = x φ 0 (x ) 2 mωo 2 mωo 2 1/ 4 2 mωo mωo mωo x 2 = x exp − 2 π 2 The n=2 energy eigenfunctions can be found repeating the procedure using a+ ˆ φ2 (x) = φ1 ( x ) 2 Notice that the above procedure only requires the relation between the ladder operators and the momentum/position operators. At this point, the energy eigenvalues can be found using the time independent Schrodinger equation. Special Integrals The raising and lowering operators can be used to show the following integrals. ∞ d n +1 n d i (1) ∫ dx φn ( x ) φm ( x ) = α δm,n +1 − δ m,n −1 since = pˆ −∞ dx 2 2 dx ∞ (2) ∫ dx φn ( x ) x φm ( x ) = δ m,n +1 n + 1 + δ m,n −1 n where Problem 5.24 was −∞ 2mωo combined with integral (1) above and with E n +1 − E n = ωo ∞ n +1 ( n +1)( n + 2 ) (3) ∫ dx φn ( x ) x 2 φm ( x ) = δm,n + δm,n ± 2 The closure relation can −∞ 2α 2 2α 2 be used to prove this last one. 5.50