1. 1 ۞ An eigenvalue λ and an eigenfunction f(x) of an operator Ĥ in a space S satisfy Week 6 2. Properties of self-adjoint operators where f(x) is implied to not be identically zero. Comments: 1.We shall use ‘shorthand’ terms EValue and EFunction. 2.In mathematical textbooks, EValues and EFunctions are usually defined by

2. 2 Example 1: Let (1) (2) (3) The corresponding eigenvalue problem is

3. 3 The general solution of (1) is It follows from BC (2) that A = 0, while BC (3) implies B can be omitted (EFunctions always involve an arbitrary constant, which can be omitted). Thus, Ĥ has an infinite sequence of EFunctions and EValues: where n is an integer (w.l.o.g, it can be assumed positive).

4. 4 The full set of the Fourier basis functions can be obtained as the EFunctions of Example 2:

5. 5 Lemma 1: Let Ĥ be a self-adjoint operator in a space S. Then, Proof: By definition of a self-adjoint operator, for any f(x) and g(x), Hence, if g = f, then which can hold only if is real. █

6. 6 Theorem 1: All EValues of a self-adjoint operator Ĥ are real. Proof: Consider Then consider (4) which yields Hence, real due to Lemma 1 real due to general properties of inner product

7. 7 Theorem 2: Let Ĥ be a self-adjoint operator and Proof: For real Ĥ, this theorem follows from Theorem 1 and the fact that the EFunctions satisfy a linear boundary-value problem with real coefficients. In the general case, its proof is more complicated. Then, all EFunctions of Ĥ in S can be chosen to be real. with real α 1,2 and β 1,2.

8. 8 1.A similar theorem can be proved for many other examples of S. 2.Note that, like any solution of a linear problem, an EFunction can be multiplied by an arbitrary factor, including a complex one – and, thus, can be made complex. Comments:

9. 9 Theorem 3: Let f 1,2 be EFunctions corresponding to two different EValues λ 1,2 of a self-adjoint operator Ĥ. Then Proof: Consider (5) (6) which yields (7)

10. 10 Since Ĥ is self-adjoint, we can rearrange Naw, taking into account that = *, one can re- write (7) as (8) Separating the real part of this equality, we obtain

11. 11 ۞ A generalised eigenvalue problem for an operator Ĥ in space S, with weight w(x) is Since we are considering different EValues ( λ 1 ≠ λ 2 ), (8) entails the result required. █ Comment: For real f 1,2, Theorem 3 implies that f 1 and f 2 are orthogonal.

12. 12 3. Sturm–Liouville theory ۞ The following generalised eigenvalue problem: is called a regular Sturm–Liouville (RSL) problem if (9)

13. 13 If the Wronskian of functions u(x) and v(x) equals zero, We’ll need the following lemma from the ODE theory: then u and v can only differ by a constant factor. Theorem 4: EValues of an RSL problem correspond to a single EFunction each. Proof (by contradiction): Let an EValue λ correspond to 2 EFunctions, f = u and f = v... Lemma 2: For simplicity, let p(x), q(x), and w(x) be analytic in [a, b].

14. 14 (10) (11) Consider (10) × v – (11) × u : hence, hence...

15. 15 It follows from BC (9) that (12) Let x = a in (12) and take into account (13): (13) Hence, const = 0 and (12) becomes

16. 16 Hence, according to Lemma 2, Comment: Observe that the proof of Theorem 4 fails if α 2 = 0. How can this be fixed? i.e. u(x) and v(x) represent the same EFunction. █

17. 17 Theorem 5: The EValues of an RSL problem are infinite in number. There is a smallest EValue, but not a largest one. The following theorems will be presented without proofs. Theorem 6: EFunctions of an RSL problem form a basis in the space of functions from C ∞ [a, b], satisfying BC (9). Theorem 7 (The Oscillation Theorem): Let’s number the EValues of an RSL problem starting from n = 0 in such a way that λ n+1 > λ n. Then f n (x) vanishes in the open (not including the endpoints) interval (a, b) exactly n times.