Harmonic Oscillator Physics

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1 Physics 34 Lecture 9 Harmonic Oscillator Physics Lecture 9 Physics 34 Quantum Mechanics I Friday, February th, 00 For the harmonic oscillator potential in the time-independent Schrödinger equation: d ψx m dx + m ω x ψx E ψx, 9. we found a ground state ψ 0 x A e m ω x 9. with energy E 0 ω. Using the raising and lowering operators a + a m ω i p + m ω x m ω i p + m ω x, 9.3 we found we could construct additional solutions with increasing energy using a +, and we could take a state at a particular energy E and construct solutions with lower energy using a. The existence of a minimum energy state ensured that no solutions could have negative energy and was used to define ψ 0 : a ψ 0 0 H a n +ψ 0 + n ω a n + ψ The operators a + and a are Hermitian conjugates of one another for any I am leaving the hats off, from here on we understand that H represents a differential operator given by H`x, i x for a classical Hamiltonian. of 0

2 9.. NORMALIZATION Lecture 9 fx and gx vanishing at spatial infinity, the inner product: fx a ± gx dx fx gx x + m ω x gx dx ± fx gx + fx m ω x gx dx x a fx gx dx, 9.5 integration by parts or, in words, we can act on gx with a ± or act on fx with a in the inner product. 9. Normalization The state ψ n that comes from n applications of a n + is not normalized, nor does the eigenvalue form of the time-independent Schrödinger equation demand that it be. Normalization is the manifestation of our probabilistic interpretation of Ψx, t. Consider the ground state, that has an undetermined constant A. If we want Ψx, t to represent a probability density, then A e m ω x dx, 9.6 and the left-hand side is a Gaussian integral: A e m ω x π dx A m ω, 9.7 so the normalization is A m ω /4. π Just because we have normalized the ground state does not mean that ψ a + ψ 0 x is normalized. Indeed, we have to normalize each of the ψ n x separately. Fortunately, this can be done once and for all. Suppose we have a normalized set of ψ n, i.e. ψ n dx for all n 0. We know that ψ n± α ± a ± ψ n where the goal is to find the constants α ± associated with raising and lowering while keeping the wavefunctions normalized. Take the norm of the resulting raised or lowered state: ψ n± dx α ± α ± of 0 a ± ψ n x a ± ψ n x dx a a ± ψ n x ψ n x dx, 9.8

3 9.. ORTHONORMALITY Lecture 9 and the operator a a ± is related to the Hamiltonian, as we saw last time: H ω a ± a ±. Then H a a ± ψ n ω ± ψ n + n ± ψ n, 9.9 so 9.8 becomes α± a a ± ψ n x ψ n x dx α± + n ± ψ n x dx. 9.0 The integral ψ nx dx by assumption, so we have and our final relation is ψ n+ α + n + α n, 9. a + ψ n ψ n a ψ n. 9. n + n Starting from the ground state, for which we know the normalization, ψ a + ψ 0 ψ a + ψ ψ 3 3 a + ψ a + ψ 0 3 a 3 + ψ The general case is ψ n n! a n + ψ for the appropriately normalized ψ 0, m ω /4 ψ 0 x e m ω x. 9.5 π 9. Orthonormality The states described by ψ n are complete we assume and orthonormal take our usual inner product for ψ m and ψ n : ψ n ψ m ψ n x ψ m x dx a n +ψ 0 a m + ψ 0 dx. m! n! 3 of 0 9.6

4 9.. ORTHONORMALITY Lecture 9 Now, suppose m > n, then using the fact that a + and a are Hermitian conjugates, we can flip the a m + onto the other term: a n +ψ 0 a m + ψ 0 dx a m a n +ψ 0 ψ 0 dx, m! n! m! n! 9.7 and we know that a n + ψ 0 ψ n and a m ψ n ψ n m but for m > n, we have a m n ψ 0 0, the defining property of the ground state. In the case n > m, we just use the conjugate in the other direction, and make the same argument: a n +ψ 0 a m + ψ 0 dx ψ 0 a n a m + ψ 0 dx 0. m! n! m! n! 9.8 Only when m n will we get a non-zero result, and of course, ψ m dx by construction. So ψ n x ψ m x dx δ mn Hermite Polynomials The prescription for generating ψ n does not provide a particularly easy way to obtain the functional form for an arbitrary n we have to repeatedly apply the raising operator to the ground state. There is a connection between the Hermite polynomials and our procedure of lifting up the ground state. Using the Frobenius method, it is possible to solve Schrödinger s equation as a power series expansion described in Griffiths, and we won t re-live that argument yet. But it is important to understand the connection between the algebraic, operational approach and the brute force series expansion. Starting from the ground state, let s act with a + call the normalization constant A again, just to make the expressions more compact: ψ a + ψ 0 m ω x + m ω x A e m ω x A 9.0 m ω x e m ω x. 4 of 0

5 9.. ORTHONORMALITY Lecture 9 Consider the second state, ψ a + ψ m ω A e m ω x + m ω e m ω x A A x x x m ω + A 4 m ω x e m ω x. [ A x + m ω x [ A x + m ω x e m ω x m ω x ] m ω x e m ω x e ] m ω x 9. The pattern continues we always have some polynomial in x multiplying the exponential factor. That polynomial, for the n th wave function is called H n, the n th Hermite polynomial. In the dimensionless variable ξ m ω x, we can read off the first two H ξ ξ, and H ξ 4 ξ. Normalized, we have the expression with ψ n x m ω /4 π n n! H nξ e ξ 9. H 0 ξ H ξ ξ H ξ 4 ξ H 3 ξ 8 ξ 3 ξ. 9.3 This set of polynomials is well-known, and they have a number of interesting recursion and orthogonality properties many of which can be developed from the a + and a operators. One still needs a table of these in order to write down a particular ψ n, but that s better than taking n successive derivatives of ψ 0 in essence, the Hermite polynomials have accomplished that procedure for you. Once again, we can plot the first few wavefunctions see Figure 9., and as we increase in energy, we see a pattern similar to the infinite square well case note that for the harmonic oscillator, we start with n 0 as the ground state rather than. 5 of 0

6 9.3. EXPECTATION VALUES Lecture 9 n 3 n Energy n n 0 Figure 9.: The first four stationary states: ψ n x of the harmonic oscillator. 9.3 Expectation Values 9.3. Classical Case The classical motion for an oscillator that starts from rest at location x 0 is xt x 0 cosω t. 9.4 The probability that the particle is at a particular x at a particular time t is given by ρx, t δx xt, and we can perform the temporal average to get the spatial density. Our natural time scale for the averaging is a half cycle, take t 0 π ω, ρx π ω π ω 0 δx x 0 cosω t dt. 9.5 We perform the change of variables to allow access to the δ, let y x 0 cosω t so that ρx ω x0 δx y π x 0 x 0 ω sinω t dy x0 δx y π x 0 x 0 cos ω t dy x0 9.6 δx y π x 0 x 0 y dy π x 0. x 6 of 0

7 9.3. EXPECTATION VALUES Lecture 9 This has x 0 x 0 ρx dx as expected note that classically, the particle remains between x 0 and x 0. The expectation value for position is then zero, since ρx is symmetric, x ρx antisymmetric, and the limits of integration are symmetric. The variance is σ x x x x0 x 0 x π x 0 x dx x Quantum Case Referring to the definition of the a + and a operators in terms of x and p, we can invert and find x and p in terms of a + and a these are all still operators, but we are treating them algebraically. The inversion is simple m ω x m ω a + + a p i a + a, 9.8 and these facilitate the expectation value calculations. For example, we can find x for the n th stationary state: x ψ n x a + + a ψ n x dx 0, 9.9 m ω by orthogonality. Similarly, p 0. Those are not particularly surprising. The variance for position can be calculated by squaring the position operator expressed in terms of a ± σ x x x m ω m ω n + m ω using 9.. m ω n + ψ n+ x dx + ψ n x a + a + + a + a + a a + + a a ψ n x dx ψ n x a + a + a a + ψ n x dx [ ] n ψ n x dx 9.30 It is interesting to compare the quantum variance with the classical one. In the case of the above, we can write σ x in terms of the energy E n 7 of 0

8 9.4. MIXED STATES Lecture 9 ω n +, just σ x E n m ω. 9.3 For the classical variance, we had σx x 0, but this is related to the classical energy. Remember we start from rest at x 0, so the total energy which is conserved is just E m ω x 0, indicating that we can write the variance as σx E m ω. 9.3 This is interesting, but we must keep in mind a number of caveats:. the classical density is time-dependent, and we have chosen to average over the natural timescale in the system, if no such scale presented itself, we would be out of luck making these comparisons,. Our classical temporal averaging is very different in spirit than the statistical information carried in the quantum mechanical wavefunction remember that expectation values and variances refer to observations made multiple times on identically prepared systems, and most definitely not observations made over time for a single system. We will return to this point later on, for now, the comparison between quantum and classical probabilities is mainly a vehicle for motivating the notion of density as a good descriptor of physics. Finally, we should ask where n needs to be to achieve a particular classical energy. For example, if we have a mass m with spring constant k m ω and initial extension x 0, then the total energy of the oscillator is E m ω x 0. For what n is this equal to E n n + ω? If we put an m kg mass on the end of a spring with frequency ω π 5 /s and full extension at x 0 m, then we can get a sense of the classical energy range E.8 J. At what value of n would a quantum mechanical system approach this energy? Using J s, we have n We can compare the classical and quantum probability densities in the large n quantum limit, an example is shown in Figure Mixed States The stationary states of the harmonic oscillator are useful in characterizing some of the peculiarities of the statistical interpretation, but as with the infinite square well, the utility of these states is in their completeness, and our ability to construct arbitrary initial waveforms. Again, let s look at 8 of 0

9 9.4. MIXED STATES Lecture 9 ρx QM CM x Figure 9.: The quantum mechanical density for n 50 compared to the classical density black. a two state admixture, just to get a feel for the time-dependence of the expectation values. Take Ψx, t α e i 3 ω t ψ x + β e i 5 ω t ψ x 9.33 with α + β. Then the expectation value for x is x Ψx, t x Ψx, t dx [ α ψ x ψ + α β e i ω t ψ x ψ + α β e i ω t ψ x ψ + β ψ x ψ ] dx Using our relation for x in terms of raising and lowering operators: x m ω a + + a, we can see that the integral of ψ x ψ will be zero we will end up with a term like ψ ψ 0 which integrates to zero by orthogonality, and another term that goes like ψ ψ, similarly zero. The same is true for ψ x ψ, so we only need to evaluate the integral of ψ x ψ ψ x ψ dx ψ + ψ3 + ψ dx m ω 9.35 m ω using orthogonality. The integral ψ x ψ dx is also put these together x α β m ω m ω, and we can cosω t of 0

10 9.4. MIXED STATES Lecture 9 m ω A similar calculation for the operator p i a + a gives p α β m ω sinω t Notice that this expectation values satisfies Newton s second law: d p dv dt dx 9.38 for our potential V m ω x so that the derivative is linear in x, and dv dx m ω x. This correspondence is an example of Ehrenfest s theorem, relating expectation dynamics to classical dynamics. Homework Reading: Griffiths, pp we will return to the series method later on. Problem 9. a. For the stationary states of the harmonic oscillator, we know σ x from 9.3. Rewrite the momentum operator p in terms of the operators a + and a and use that to calculate σ p. b. Evaluate σ x σ p, the product of the position and momentum variances what is the minimum value this product can take? Problem 9. Griffiths.5. The probability of finding a quantum mechanically described particle in a harmonic well outside of the classically allowed region. Problem 9.3 Griffiths.3. Given the initial state, normalize the full wavefunction, and compute various expectation values for the harmonic oscillator. 0 of 0

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