Mathematical Methods in Quantum Mechanics With Applications to Schrödinger Operators 2nd edition

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1 Errata G. Teschl, Mathematical Methods in Quantum Mechanics With Applications to Schrödinger Operators nd edition Graduate Studies in Mathematics, Vol. 157 American Mathematical Society, Providence, Rhode Island, 014 The official web page of the book: Please send comments and corrections to Updated as of September 6, 017 Errata Changes appear in yellow. Line k+ (resp., line k ) denotes the kth line from the top (resp., the bottom) of a page. My thanks go to the following individuals who have contributed to this list: Tobias Wöhrer, Simon Becker, Dennis Cutraro, Mateusz Piorkowski, Laura Kanzler, Mateus Sampaio, Laura Shou, Noema Nicolussi, Andreas Geyer-Schulz, Rene Allerstorfer. Page 16. First line: for a l p (N), b l q (N). Page 5. Proof of Theorem 0.5: and we can choose m = j u j 1. Page 36. Add the following at the end of Lemma 0.39: Moreover, if u and f both have compact support, then f k C c (R n ). Page 36. Proof of Lemma 0.41:... ϕ n Cc (R n ) with support inside some open ball X which converges... continuous functions ϕ n with support in X which converges to g... Page 54. Proof of Lemma 1.11: (ii) follows from ϕ, A ψ = A ϕ, ψ = ϕ, Aψ. 1

2 Page 60. Last sentence in the proof of Theorem 1.16: Since f ε < f zl for all z l we have f ε < f ε and we have found a required function. Page 61. Problem 1.3: Show that the span of {(t z) 1 z U} is dense in C (R). Page 66. Line after (.15): measurable function A : R d C. Page 7. Line 18+: Clearly we have αa = αa for α C\{0} and A + B=A+B provided A is closable and B is bounded (Problem.8). Page 75. Proof of Lemma.7: (A z)ψ = (A x)ψ iyψ = (A x)ψ + y ψ y ψ, (.46) Page 76. Problem.8: Suppose that if A is closable and B is bounded. Show that αa = αa for α C\{0} and A + B=A + B. Page 78. Proof of Lemma.11: D(Ã) = {ψ H A ψ H : ϕ, ψ A = ϕ, ψ, ϕ D(A)} = H A D(A ) as D(A) H A is dense and ϕ, ψ A = (A + 1)ϕ, ψ for ϕ D(A), ψ H A. Page 8. Proof of Lemma.15: Re ϕ, Aψ = 1 q(ψ + ϕ) q(ψ ϕ) q ( ψ + ϕ + ψ ϕ ) = q ( ψ + ϕ ) Page 88. Problem.18: Then so does A + B if B < A 1 1. Page 93. Paragraph after Lemma.8: A conjugate linear map C : H H is called a conjugation if it satisfies C = I and Cψ, Cϕ = ϕ, ψ. Page 135. Problem 4.11: χ Ω (A) = 1 R A (z)dz, πi Γ Page 138. A ψ = ψ, A ψ = ψ, A Aψ = Aψ, ψ D( A ) = D(A), (4.31) Page 139 : U U = P Ker(A), UU = P Ker(A ), (4.34)

3 Page 139 : Last line of Theorem 4.10: Ker(U) = Ker(A) Page 155. Problem 5.9: H rc = {ψ H lim t ψ, e ita ψ = 0} H ac, (5.7) Page 159. Theorem 6.4: ( b ) γ max a γ + b,. (6.4) 1 a Page 161. Lemma 6.8: s n (K) = min ψ 1,...,ψ n 1 Page 16. Proof of Lemma 6.9: last formula γ n = K K n = sup K(ψ ψ =1 sup Kψ, (6.9) ψ U(ψ 1,...,ψ n 1) n ϕ j, ψ ϕ j ) j=1 Page 164. Proof of Lemma 6.10: Conversely, choose ϕ i = ˆφ i Page 176. Line before equation (6.45): Furthermore, we can define C q (λ) for all z ρ(a), using Page 195. Line +: Clearly H r+1 (R n ) H r (R n ) Page 00. Discussion after Lemma 7.0: ψ(x, t) d n x = ˆψ( x t ) dn x (t) n Page 09. Last line of the proof of Theorem 8.: 0 = (z+z ) Âψ Page 11. First equation in the proof of Lemma 8.3: 1 π ϕ(x)e x k (itx) j dx = 0 j! j=0 Page. Problem 9.1: and f(d) = γ, (pf )(d) = δ. Page. Problem 9.3: Let φ L 1 loc (I) be real-valued. Page. Problem 9.4: Add the assumption that a is regular. Otherwise one can also start the integration at an arbitrary point in (a, b). Page 3. Replace the last sentence by: Moreover, the following set is a core for A D 1 = {f D(τ) x 0 I : x (a, x 0 ), V x (f) = 0, (9.1) x 1 I : x (x 1, b), W x (f) = 0}, 3

4 where we set V x (f) = W x (v, f), W x (f) = W x (w, f) if τ is l.c. at a, b and V x (f) = f(x), W x (f) = f(x) if τ is l.p. at a, b, respectively. Page 8. Theorem 9.10: Delete (which are simple). And the following claim about simplicity of eigenvalues only applies to seperated boundary conditions as in Theorem 9.6. Page 50. Second line in Section 9.7: on (a, b) = R. Page 56. Problem 9.18: Change the hint according to: (Hint: Let ϕ ε (x) = exp( ε x ) and investigate ϕ ε, Hϕ ε.) Page 61. AΦ = τφ, D(A) = {Φ L (0, π) Φ AC 1 [0, π], Φ L (0, π), Φ(0) = Φ(π), Φ (0) = Φ (π)}. (10.3) Page 68. Line 3+: Note that the L (k) j (r) are polynomials of degree j which Page 3. F (z) = Page 330. Proof of Lemma A.35: Y f(z, y) dµ(y) µ(x ) lim inf µ n (x) lim sup µ n (x) µ(x+) (A.55) Page 330. Problem A.3 can be deleted as the claim is part of Lemma A.36. Page 333. Problem A.34. This claim is clearly wrong (take a function which is constant on an interval). It should be deleted. Addendum Here is an amplification of Theorem 3.16: Theorem For every self-adjoint operator A there is an ordered spectral basis {ψ j } N j=1. Moreover, it can be chosen such that dµ ψ j = χ Ωj dµ, where µ is a maximal spectral measure and Ω j+1 Ω j. The dimension N is the spectral multiplicity of A. Proof. First of all observe that for every ϕ there is a maximal spectral vector ψ such that ϕ H ψ. To see this start with a maximal spectral vector ψ. Then dµ ϕ = f dµ ψ and we set Ω = {λ f(λ) > 0}. Then P A (Ω)ϕ = ϕ since P A (Ω)ϕ = Ω dµ ϕ = R f dµ ψ = ϕ. Now set ψ = ϕ + P (R\Ω) ψ and observe dµ ψ = dµ ϕ + χ R\Ω dµ ψ = (f + χ R\Ω )dµ ψ. Since f + χ R\Ω > 0 we see that dµ ψ is absolutely continuous with respect to dµ ψ and hence ψ is a maximal spectral vector with ϕ = P A (Ω)ψ H ψ as required. Now start with some total set { ψ j } and proceed as in Lemma 3.4 to obtain an ordered spectral basis {ψ j }. Since µ ψj+1 is absolutely continuous with respect 4

5 to µ ψj all spectral measures are absolutely continuous with respect to µ = µ ψ1, that is, dµ ψj = f j dµ. Choosing Ω j = {λ f j (λ) > 0} we can replace ψ j χ Ωj (A)f j (A) 1/ ψ j such that f j χ Ωj. Since µ ψj+1 is absolutely continuous with respect to µ ψj we can even assume Ω j+1 Ω j. Finally, we show that the spectral multiplicity of A is N. By the first part we can assume that A is multiplication by λ in N j=1 L (R, χ Ωj dµ). Let {ψ j } n j=1 be a spectral basis with n < N. We will show that there is some vector in the orthogonal complement of j H ψ j. Of course such a vector exists pointwise for every λ but it is not clear that the components can be chosen measurable. To see this we use a Gauss-type elimination: For this note that we can multiply every vector ψ j with a non-vanishing function or add multiples of the other vectors to a given one without changing j H ψ j. Hence we can first normalize the first component of every ψ j to be a characteristic function. Moreover, by adding all other vectors to ψ 1 we can assume that its first component is positive on a maximal set Ω 1. In fact, after another normalization we can assume that ψ 1,1 = and after subtracting multiples of ψ χ Ω1 1 from the remaining vectors we can assume ψ j,1 = 0 for j. If µ 1 (R\ Ω 1 ) > 0 then ϕ = (χ R\ Ω1, 0,... ) would be in the orthogonal complement and we are done. So assume χ Ω1 = 1 and continue with the other components until they satisfy ψ j,k = δ j,k for 1 j, k n. Then ϕ = ( ψ 1,n+1,..., ψ n,n+1, 1, 0,... ) is in the orthogonal complement contradicting our assumption that {ψ j } n j=1 is a spectral basis. 5