Page 404. Lecture 22: Simple Harmonic Oscillator: Energy Basis Date Given: 2008/11/19 Date Revised: 2008/11/19

Transcription

1 Page 404 Lecture : Simple Harmonic Oscillator: Energy Basis Date Given: 008/11/19 Date Revised: 008/11/19

2 Coordinate Basis Section 6. The One-Dimensional Simple Harmonic Oscillator: Coordinate Basis Page 405 Properties of the SHO Solutions We note some interesting properties of the solutions: Parity The parity operator Π makes the transformation on the wavefunction x x; formally, Π x = x. Since this defines the action of Π for an orthonormal basis, it fully defines Π. We can see Π is Hermitian: x Π x = x x = δ(x + x) = x x = x Π = x Hence, it has real eigenvalues and its eigenvectors form a complete basis. The symmetry of the potential V (x) implies that the Hamiltonian and the parity operator commute, so the eigenvectors of H must also be eigenvectors of Π. Since Π is Hermitian and Π = I, the allowed eigenvalues of Π are ±1, corresponding to even and odd functions, respectively. We see this property reflected in the SHO eigenfunctions. As we explained in deriving them, each Hermite polynomial contains only even or odd powers of the argument. Hence, they are either even or odd in their argument. The Gaussian that multiplies them is even, so the solutions themselves are either even or odd in x. The same property held or the eigenfunctions of the particle in h a box, i and in fact is the reason we chose to have the box over the interval L, L rather than [0, L].

3 Coordinate Basis (cont.) Section 6. The One-Dimensional Simple Harmonic Oscillator: Coordinate Basis Page 406 Number and Positions of Zeroes The Hermite polynomial H n is a polynomial of order n, so it must have n zeros. What is interesting is that these zeroes are contained inside the classical turning points, x 0 = ±q En, the points where the kinetic energy vanishes and the k potential energy is maximized. The argument is not trivial and goes as follows. We may calculate the first derivative at any point by by d dy ψε(y Z ) = dy d Z y y d(y ) ψε(y ) = dy `(y ) ε ψ ε(y ) y Since ψ ε(y) y 0 in order for ψ ε(y) to be normalizable, it holds that d dy y ψε(y) 0 also. Thus, one end of the left side vanishes, giving Z d dy ψε(y) = dy ˆ(y ) ε ψ ε(y ) y Note that the factor in the integrand, (y ) ε, is positive for (y ) > y 0 = ε.

4 Coordinate Basis (cont.) Section 6. The One-Dimensional Simple Harmonic Oscillator: Coordinate Basis Page 407 Next, define y n, with ε = n + 1, to be the position of the last zero of ψε(y) (there is no amibiguity as to what y 0 means because we know the n = 0 mode has no zeros). For y > y n, the sign of ψ ε(y) is fixed because there are no more zeros. For specificity, suppose it is positive; we can always apply a 1 to ψ ε(y) to make this true without changing the positions of the zeros. The above equation tells us the first derivative must be negative for y max(y 0, y n) because the integrand is positive in this region. Now, suppose y n y 0. Since ψ ε(y) is positive for the regime y > y n, the d derivative at y n, ψε(yn), must be positive in order for the function to cross dy through zero from negative to positive value at y n. But we showed above that the first derivative is negative for y max(y 0, y n), which, by our assumption y n y 0, corresponds to y y n. We have a contradiction. The contradiction can be resolved by supposing instead y 0 > y n. Then, the integrand can go through zero to negative values for y > y n, making it possible for the integral to change sign and for d ψε(yn) to go from its asymptotic dy negative value to a positive value at y n. So we have y n < y 0 = ε. Putting the units back in, we have r r s x n < y 0 m ω = n + 1 «r En = = x 0 m ω k

5 Coordinate Basis (cont.) Section 6. The One-Dimensional Simple Harmonic Oscillator: Coordinate Basis Page 408 The same argument can be carried through for negative x; QED. One last note. The above implies that the wavefunctions are oscillatory only in the region inside the classical turning points, and decaying outside there. This is consistent with our generic discussion of bound states and how quantization arises in Section 5.3. There, by breaking the potential into an infinite number of infinitesimally wide piecewise constant regions, we saw that solutions ought to be oscillatory in regions where E > V (x) and decay outside there. The SHO solutions obey this rule. Position and Momentum Uncertainties Shankar goes through a straightforward demonstration q that the state that saturates the uncertainty principle, with ( X ) ( P) =, is the ground state of the SHO. This is unusual; we found for the particle in a box, that the energy of the ground state is higher than that implied by the uncertainty principle. Shankar also discusses how this implies there is zero-point energy just as for the particle in a box, the ground state has nonzero energy and that this has measurable physical consequences, such as the fact that the vibrational energy content of a crystal at absolute zero does not vanish.

6 Coordinate Basis (cont.) Section 6. The One-Dimensional Simple Harmonic Oscillator: Coordinate Basis Page 409 Classical Limit Shankar also shows by illustration for n = 11 that the higher n modes have probability disributions peaked at the turning points, corresponding to the distribution of dwell time one expects for a classical SHO, in which more time is spent near the turning points than near the origin. The lowest modes do not satisfy this, but we only expect to recover the classical dwell time in the limit of high energy, when the discretization of the allowed energies becomes unnoticeable; i.e., when E ω.

7 Coordinate Basis (cont.) Section 6. The One-Dimensional Simple Harmonic Oscillator: Coordinate Basis Page 410 The Propagator We calculate the propagator by the standard formula, Equation 4.18, and also calculate its { x }-basis matrix elements: U(t) = e i H t = X n [U(t)] xx = x U(t) x = = X e i (n+ 1 ) ω t n=0 = m ω π i sin ω t e i Ent ψ En (0) ψ En (0) = n=0 X e i (n+ 1 ) ω t x ψ En ψ En x n=0 m ω π n (n!) 1/ exp " i m ω X e i (n+ 1 ) ω t ψ En ψ En! 1/ H n(x) e m ω x H n(x ) e m ω (x ) `x + (x ) # cos ω t x x sin ω t where the sum is not obvious. Shankar derives it using path integrals, which we will unfortunately not have time to cover.

8 Energy Basis Section 6.3 The One-Dimensional Simple Harmonic Oscillator: Energy Basis Page 411 Motivation So far, we have found the SHO s energy eigenvalues and eigenstates in the { x } basis. However, because the energies go up linearly with the state index n (i.e., the energy levels are spaced equally), we are motivated to think of the eigenstate that the system is in as corresponding to a number of quanta of energy ω. Rather than thinking in terms of energy levels, we could think in terms of a number operator multiplying the energy per quantum. How do we get there? Before continuing, we remind the reader that it is generically true that the Hamiltonian can be written in the form H = X n E n ψ En ψ En X n E n E n E n X n E n n n The issue at hand is whether this offers any simplification. In general, it does not because, in general, the E n are not equally spaced. In fact, for potentials that have both bound states and free states, there are two pieces to the sum involved, one for the discretely spaced bound states and one for the continuum of free states. There is no natural idea of an energy quantum and hence no motivation to think in terms of a number operator.

9 Energy Basis (cont.) Section 6.3 The One-Dimensional Simple Harmonic Oscillator: Energy Basis Page 41 Raising and Lowering Operators Without real motivation, we define the operator Its adjoint is a = r r 1 m ω 1 X + i 1 m ω P a = r r 1 m ω 1 X i 1 m ω P (After you have seen canonical transformations in Ph106, the motivation will be clear; we will come back to this later.) One can easily check that they satisfy h a, a i = 1 H = a a + 1 «ω H b = H ω = a a + 1 «Both relations arise from [X, P] = i, which holds in any basis. We call a and a raising and lowering operators for reasons that will become apparent.

10 Energy Basis (cont.) Section 6.3 The One-Dimensional Simple Harmonic Oscillator: Energy Basis Page 413 Basis-Free Eigenstates of the Hamiltonian We want to find the eigenstates of b H. To date, we have always had to write the eigenvalue-eigenvector equation for the Hamiltonian in either the position or momentum basis; usually the former to obtain a differential equation that we can solve for the { x }-basis representation of the eigenstates E. Here, we are going to try to avoid a basis altogether. Getting used to this idea is not easy. So, we want to solve bh ε = ε ε (The use of the ε symbol is suggestive.) First, we need the following:» [a, H] b = a, a a + 1 = a a a a a a = [a, a ] + a a a a a a = a» [a, H] b = a, a a + 1 = a a a a a a = a a a a [a, a ] + a a = a

11 Energy Basis (cont.) Section 6.3 The One-Dimensional Simple Harmonic Oscillator: Energy Basis Page 414 The above lets us show that, if ε is an eigenstate of b H with eigenvalue ε, then a and a generate from ε other eigenstates of b H: That is bh a ε = [ H, b a] + a H b ε = ( a + a ε) ε = (ε 1) a ε bh a ε = [ H, b a ] + a H b ε = a + a ε ε = (ε + 1) a ε a ε = C ε ε 1 a ε = C ε+1 ε + 1 (We use the fact that there is no degeneracy of bound states in one dimension, so there is one eigenstate with eigenvalue ε.) The rationale for calling a and a raising and lowering operators now becomes clear.

12 Energy Basis (cont.) Section 6.3 The One-Dimensional Simple Harmonic Oscillator: Energy Basis Page 415 Left with the above only, we might think that there are states of arbitrarily low energy. We know from our { x }-basis solution that this is not true. We can also prove it more generally given the form of the Hamiltonian. Consider ψ H ψ for an arbitrary state ψ : ψ H ψ = 1 m ψ P ψ + 1 k ψ X ψ = 1 m P ψ + 1 k X ψ 0 Therefore, any eigenvalues of H and hence b H must be nonnegative. Notice that the proof that the kinetic term is nonnegative holds always, but the proof for the potential term is specific to the SHO potential and is not generally true. The lower limit tells us there must be a lowest energy state that satisfies a ε 0 = 0 so that we cannot obtain any lower energy states. (We should properly write 0 on the right side for the null vector, but, as we shall see, this may become confused with 0 = ε 0, the ground state ket.) We then have ε 0 a a ε 0 = 0 = ε 0 b H ε 0 = 1 and ε 0 H ε 0 = 1 ω

13 Energy Basis (cont.) Section 6.3 The One-Dimensional Simple Harmonic Oscillator: Energy Basis Page 416 Are we certain there is no parallel chain of states with different energies? Since we have started from scratch in the energy basis, we must assume we have no prior knowledge that ε must be an integer. Yes, we can eliminate the possibility of a parallel chain by calculating the energy of the lowest energy state. Suppose there were a parallel chain of states, ε, with eigenvalues ε. By the same argument as above, we are assured there is a lowest energy state in the chain, ε 0 for which a ε 0 = 0. The dimensionless energy of that state is then also ε 0 = 1 by the same argument as for ε 0. So ε 0 is degenerate with ε 0. But we have proven that one-dimensional bound states are nondegenerate, so it must be that ε 0 is proportional to ε 0 ; they are the same state. So, we have the set of eigenstates of H, connected by the raising and lowering operators.

14 Energy Basis (cont.) Section 6.3 The One-Dimensional Simple Harmonic Oscillator: Energy Basis Page 417 The Inner Product in the Energy Basis At this point, we only have made use of the following facts: X and P, and therefore H, are Hermitian operators on the linear vector space of states. Being Hermitian, we are assured that the eigenstates of H form a basis for the space. Because the potential has no maximum value, we are assured that all states are bound states and, with the addition of the fact that we are considering a one-dimensional potential, we are assured they are nondegenerate. (Though we proved nondegeneracy via the position basis, nondegeneracy is a non-basis-specific property; moreover, our use of the position basis was not specific to the SHO.) We can define raising and lowering operators a and a in terms of X and P, and we find they connect the various eigenstates.

15 Energy Basis (cont.) Section 6.3 The One-Dimensional Simple Harmonic Oscillator: Energy Basis Page 418 We have not formally defined the inner product on this space. (The inner product we have defined for our one-dimensional problems so far relies on the position basis, which we are trying to avoid.) However, we proved in Section 3.6 that, if there is a reasonable inner product for the space (recall, Hermiticity does not require the existence of an inner product, though it is much easier to prove when there is one!), then the eigenstates of a Hermitian operator are mutually orthogonal. We therefore take this as part of the definition of the inner product on the space. This leaves the normalization of the inner product unspecified, so we also assume the eigenstates are normalized. Since the C ε constants are still undetermined, we still have the necessary freedom to set the normalizations of the eigenstates individually. Let us now use that freedom to determine what the constants C ε must be in order to be consistent with the above normalization choice. We obtain a condition on C ε as follows, using a ε = C ɛ ε 1 from earlier: 1 = ε 1 ε 1 = C ε ε a a ε = C ε ε bh 1 «ε = C ε ε 1 «ε ε = C ε ε 1 «= C ε = ε 1

16 Section 6.3 The One-Dimensional Simple Harmonic Oscillator: Energy Basis Page 419 Energy Basis (cont.) At this point, life is less confusing if we define the number operator, N = a a = H b 1. It obviously has the same eigenstates as H b and H, with eigenvalues n = ε 1. It counts the quanta of energy in the oscillator. The above relation is then C n = n C n = e i φn n where φ n is arbitrary for each n. The simplest convention is φ n = 0 for all n, giving a n = n n 1 a n = n + 1 n + 1 n m = δ nm This defines the inner product for the entire space because the { n } are a basis. Any state can now be obtained from the ground state by use of raising operators: n = 1 n! a n 0 Finally, we make the important note that, given the above definition of n, there is now a notational degeneracy between the SHO ground state 0 and the null vector 0. As we did above, we are simply going to use 0 for the null vector because context will always make it clear whether 0 refers to scalar or a vector.

17 Energy Basis (cont.) Section 6.3 The One-Dimensional Simple Harmonic Oscillator: Energy Basis Page 40 Matrix Elements of Various Operators Here we list some useful matrix elements that we can derive from what we have done so far: n a m = m n m 1 = m δ n,m 1 n a m = m + 1 n m + 1 = m + 1 δ n,m+1 r r m X = a + a = n X m = + 1 δn,m+1 + m δ n,m 1 m ω m ω r m ω r m ω m P = i a a = n P m = i + 1 δn,m+1 m δ n,m 1 H = N + 1 «ω = n H m = n + 1 «ω δ n,m The marix representations are given in Shankar; they are obvious from the above.

18 Energy Basis (cont.) Section 6.3 The One-Dimensional Simple Harmonic Oscillator: Energy Basis Page 41 Shankar makes the point that working in the energy basis, which gives these simple forms for matrix elements of the fundamental X and P operators, makes matrix elements of any function of X and P (or of a and a ) easier to calculate than evaluating integrals using the { x }-basis representation of the eigenstates. We won t try to reproduce the algebra of calculating 3 X 3 here.