MIT OpenCourseWarehttp://ocw.mit.edu 18.102 Introduction to Functional Analysis Spring 2009 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.LECTURE NOTES FOR 18.102, SPRING 2009 133 Lecture 23. Tuesday, May 5: Harmonic oscillator As a second ‘serious’ application of at least the completeness part of the spectral theorem for self-adjoint compact operators, I want to discuss the Hermite basis for L2(R). Note that so far we have not found an explicit orthonormal basis on the whole real line, even though we know L2(R) to be separable, so we certainly know that such a basis exists. How to construct one explicitly and with some handy properties? One way is to simply orthonormalize – using Gramm-Schmidt – some countable set with dense span. For instance consider the basic Gaussian function 2 (23.1) exp(− x) ∈ L2(R).2 This is so rapidly decreasing at infinity that the product with any polynomial is also square integrable: 2 (23.2) x k exp(− x) ∈ L2(R) ∀ k ∈ N0 = {0, 1, 2, . . . }.2 Orthonormalizing this sequence gives an orthonormal basis, where completeness can be shown by an appropriate approximation technique. Rather than proceed directly we will discuss the invertibility of the harmonic oscillator d2 (23.3) H = − + x 2 dx2 which we want to think of as an operator – although for the moment I will leave vague the question of what it operates on. The first thing to observe is that the Gaussian is an eigenfunction of H 2d 2 2(23.4) He−x /2 = − (−xe−x /2 + x 2 e−x /2 dx − (x 2 − 1)e−x 2 /2 + x 2 e−x 2 /2 = e−x 2/2 with eigenvalue 1 – for the moment this is only in a formal sense. In this special case there is an essentially algebraic way to generate a whole sequence of eigenfunctions from the Gaussian. To do this, write d d(23.5) Hu = (− + x)( + x)u + u = (CA + 1)u,dx dx d d C = (− + x), A = ( + x)dx dx again formally as operators. Then note that (23.6) Ae−x 2/2 = 0 which again proves (23.4). The two operators in (23.5) are the ‘creation’ operator and the ‘annihilation’ operator. They almost commute in the sense that (23.7) (AC − CA)u = 2u for say any twice continuously differentiable function u. Now, set u0 = e−x 2/2 which is the ‘ground state’ and consider u1 = Cu0. From (23.7), (23.6) and (23.5), (23.8) Hu1 = (CAC + C)u0 = C2Au0 + 3Cu0 = 3u1. Thus, u1 is an eigenfunction with eigenvalue 3.� � � � � � � � � � 134 LECTURE NOTES FOR 18.102, SPRING 2009 Lemma 18. For j ∈ N0 = {0, 1, 2, . . . } the function uj = Cj u0 satisfies Huj = (2j + 1)uj . Proof. This follows by induction on j, where we know the result for j = 0 and j = 1. Then (23.9) HCuj = (CA + 1)Cuj = C(H − 1)uj + 3Cuj = (2j + 3)uj . /2Again by induction we can check that uj = (2j xj + qj (x))e−x 2where qj is a polynomial of degree at most j − 2. Indeed this is true for j = 0 and j = 1 (where q0 = q1 ≡ 0) and then (23.10) Cuj = (2j+1 xj+1 + Cqj )e−x 2 /2 . and qj+1 = Cqj is a polynomial of degree at most j − 1 – one degree higher than qj . From this it follows in fact that the finite span of the uj consists of all the products p(x)e−x 2/2 where p(x) is any polynomial. Now, all these functions are in L2(R) and we want to compute their norms. First, a standard integral computation3 shows that 2 2(23.11) (e−x /2)2 = e−x = √π R R For j > 0, integration by parts (easily justified by taking the integral over [−R, R] and then letting R → ∞) gives (23.12) (Cj u0)2 = Cj u0(x)Cj u0(x)dx = u0Aj Cj u0. R R R Now, from (23.7), we can move one factor of A through the j factors of C until it emerges and ‘kills’ u0 (23.13) ACj u0 = 2Cj−1 u+CACj−1 u0 = 2Cj−1 u0 + C2ACj−2 u0 = 2jCj−1 u0. So in fact, (23.14) (Cj u0)2 = 2j (Cj−1 u0)2 = 2j j!√π. R R A similar argument shows that (23.15) ukuj = u0AkCj u0 = 0 if k =� j. R R Thus the functions (23.16) ej = 2−j/2(j!)− 21 π− 41 Cj e−x 2/2 form an orthonormal sequence in L2(R). 3To compute the Gaussian integral, square it and write as a double integral then introduct polar coordinates Z Z Z Z 2π ( e−x 2 dx)2 = e−x 2−y 2 dxdy = ∞ e−r 2 rdrdθ = π ˆ − e−r 2 ˜∞ = π. R R2 0 0 0135 LECTURE NOTES FOR 18.102, SPRING 2009 We would like to show this is orthonormal sequence is complete. Rather than argue through approximation, we can guess that in some sense the operator (23.17) AC = ( d + x)(− d + x) = − d2 + x 2 + 1 dx dx dx2 should be invertible, so one approach ist to try to construct its ‘inverse’ and show this really is a compact, self-adjoint operator on L2(R) and that its only eigenfunc-tions are the ei in (23.16). Rather than do this I will proceed more indirectly.� � � � � � 136 LECTURE NOTES FOR 18.102, SPRING 2009 Solutions to Problem set 10 Problem P10.1 Let H be a separable, infinite dimensional Hilbert space. Show that the direct sum of two copies of H is a Hilbert space with the norm (23.18) 2 2 ) 21 H ⊕ H � (u1, u2) �−→ (�u1�H + �u2�H either by constructing an isometric isomorphism (23.19) T : H −→ H ⊕ H, 1-1 and onto, �u�H = �T u�H⊕H or otherwise. In any case, construct a map as in (23.19). Solution: Let {ei}i∈N be an orthonormal basis of H, which exists by virtue of the fact that it is an infinite-dimensional but separable Hilbert space. Define the map ∞ ∞(23.20) T : H � u −→ ( (u, e2i−1)ei, (u, e2i)ei) ∈ H ⊕ H i=1 i=1 The convergence of the Fourier Bessel series shows that this map is well-defined and linear. Injectivity similarly follows from the fact that T u = 0 in the image implies that (u, ei) = 0 for all i and hence u = 0. Surjectivity is also clear from the fact that ∞(23.21) S : H ⊕ H � (u1, u2) �−→ ((u1, ei)e2i−1 + (u2, ei)e2i) ∈ H i=1 is a 2-sided inverse and Bessel’s identity implies isometry since �S(u1, u2)�2 = �u1�2 + �u2�2 Problem P10.2 One can repeat the preceding construction any finite number of times. Show that it can be done ‘countably often’ in the sense that if H is a separable, infinite dimensional, Hilbert space then (23.22) l2(H) = {u : N −→ H; �u�l2 2(H) = �ui�H