Lecture 03 - Separable Hilbert spaces (Schuller's Lectures on Quantum Theory)

Acknowledgments This set of lecture notes accompanies Frederic Schuller’s course on Quantum Theory, taught in the summer of 2015 at the Friedrich-Alexander-Universität Erlangen-Nürnberg as part of the Elite Graduate Programme. The entire course is hosted on YouTube at the following address: www.youtube.com/playlist?list=PLPH7f_7ZlzxQVx5jRjbfRGEzWY_upS5K6 These lecture notes are not endorsed by Dr. Schuller or the University. While I have tried to correct typos and errors made during the lectures (some helpfully pointed out by YouTube commenters), I have also taken the liberty to add and/or modify some of the material at various points in the notes. Any errors that may result from this are, of course, mine. If you have any comments regarding these notes, feel free to get in touch. Visit my blog for the most up to date version of these notes http://mathswithphysics.blogspot.com My gratitude goes to Dr. Schuller for lecturing this course and for making it available on YouTube.

Simon Rea

3

Separable Hilbert Spaces

3.1

Relationship between norms and inner products

A Hilbert space is a vector space (H, +, ·) equipped with a sesqui-linear inner product h·|·i which induces a norm k · kH with respect to which H is a Banach space. Note that by “being induced by h·|·i” we specifically mean that the norm is defined as k · k: V → R p f 7→ hf |f i. Recall that a sesqui-linear inner product on H is a map h·|·i : H × H → H which is conjugate symmetric, linear in the second argument and positive-definite. Note that conjugate symmetry together with linearity in the second argument imply conjugate linearity in the first argument: hzψ1 + ψ2 |ϕi = hϕ|zψ1 + ψ2 i = zhϕ|ψ1 i + hϕ|ψ2 i = zhϕ|ψ1 i + hϕ|ψ2 i = zhψ1 |ϕi + hψ2 |ϕi. Of course, since Hilbert spaces are a special case of Banach spaces, everything that we have learned about Banach spaces also applies to Hilbert paces. For instance, L(H, H), the collection of all bounded linear maps H → H, is a Banach space with respect to the operator norm. In particular, the dual of a Hilbert space H is just H∗ := L(H, C). We will see that the operator norm on H∗ is such that there exists an inner product on H which induces it, so that the dual of a Hilbert space is again a Hilbert space. First, in order to check that the norm induced by an inner product on V is indeed a norm on V , we need one of the most important inequalities in mathematics. Proposition 3.1 (Cauchy-Schawrz inequality1 ). Let h·|·i be a sesqui-linear inner product on V . Then, for any f, g ∈ V , we have |hf |gi|2 ≤ hf |f ihg|gi. Proof. If f = 0 or g = 0, then equality holds. Hence suppose that f 6= 0 and let z := 1

hf |gi ∈ C. hf |f i

Also known as the Cauchy-Bunyakovsky-Schwarz inequality in the Russian literature.

–1–

Then, by positive-definiteness of h·|·i, we have 0 ≤ hzf − g|zf − gi = |z|2 hf |f i − zhf |gi − zhg|f i + hg|gi |hf |gi|2 hf |gi hf |gi = hf |f i − hf |gi − hf |gi + hg|gi 2 hf |f i hf |f i hf |f i |hf |gi|2 |hf |gi|2 |hf |gi|2 = − − + hg|gi hf |f i hf |f i hf |f i |hf |gi|2 + hg|gi. =− hf |f i By rearranging, since hf |f i > 0, we obtain the desired inequality. p Note that, by defining kf k := hf |f i, we can write the Cauchy-Schwarz inequality as |hf |gi| ≤ kf kkgk. Proposition 3.2. The induced norm on V is a norm. Proof. Let f, g ∈ V and z ∈ C. Then p (i) kf k := hf |f i ≥ 0 (ii) kf k = 0 ⇔ kf k2 = 0 ⇔ hf |f i = 0 ⇔ f = 0 by positive-definiteness p p p p (iii) kzf k := hzf |zf i = zzhf |f i = |z|2 hf |f i = |z| hf |f i =: |z|kf k (iv) Using the fact that z + z = 2 Re z and Re z ≤ |z| for any z ∈ C and the CauchySchwarz inequality, we have kf + gk2 := hf + g|f + gi = hf |f i + hf |gi + hg|f i + hg|gi = hf |f i + hf |gi + hf |gi + hg|gi = hf |f i + 2 Rehf |gi + hg|gi ≤ hf |f i + 2|hf |gi| + hg|gi ≤ hf |f i + 2kf kkgk + hg|gi = (kf k + kgk)2 . By taking the square root of both sides, we have kf + gk ≤ kf k + kgk. Hence, we see that any inner product space (i.e. a vector space equipped with a sesquilinear inner product) is automatically a normed space under the induced norm. It is only natural to wonder whether the converse also holds, that is, whether every norm is induced by some sesqui-linear inner product. Unfortunately, the answer is negative in general. The following theorem gives a necessary and sufficient condition for a norm to be induced by a sesqui-linear inner product and, in fact, by a unique such.

–2–

Theorem 3.3 (Jordan-von Neumann). Let V be a vector space. A norm k · k on V is induced by a sesqui-linear inner product h·|·i on V if, and only if, the parallelogram identity kf + gk2 + kf − gk2 = 2kf k2 + 2kgk2 holds for all f, g ∈ V , in which case, h·|·i is determined by the polarisation identity 3

hf |gi =

1X k i kf + i4−k gk2 4 k=0

1 = (kf + gk2 − kf − gk2 + ikf − igk2 − ikf + igk2 ). 4 Proof. (⇒) If k · k is induced by h·|·i, then by direct computation kf + gk2 + kf − gk2 := hf + g|f + gi + hf − g|f − gi = hf |f i + hf |gi + hg|f i + hg|gi + hf |f i − hf |gi − hg|f i + hg|gi = 2hf |f i + 2hg|gi =: 2kf k2 + 2kgk2 , so the parallelogram identity is satisfied. We also have kf + gk2 − kf − gk2 := hf + g|f + gi − hf − g|f − gi = hf |f i + hf |gi + hg|f i + hg|gi − hf |f i + hf |gi + hg|f i − hg|gi = 2hf |gi + 2hg|f i and ikf − igk2 − ikf + igk2 := ihf − ig|f − igi − ihf + ig|f + igi = ihf |f i + hf |gi − hg|f i + ihg|gi − ihf |f i + hf |gi − hg|f i − ihg|gi = 2hf |gi − 2hg|f i. Therefore kf + gk2 − kf − gk2 + ikf − igk2 − ikf + igk2 = 4hf |gi. that is, the inner product is determined by the polarisation identity. (⇐) Suppose that k · k satisfies the parallelogram identity. Define h·|·i by 1 hf |gi := (kf + gk2 − kf − gk2 + ikf − igk2 − ikf + igk2 ). 4 We need to check that this satisfies the defining properties of a sesqui-linear inner product.

–3–

(i) For conjugate symmetry

:= = = =

1 2 2 2 2 4 kf + gk − kf − gk + ikf − igk − ikf + igk 2 2 2 2 1 4 (kf + gk − kf − gk − ikf − igk + ikf + igk ) 2 2 2 2 1 4 (kf + gk − kf − gk − ik(−i)(if + g)k + iki(−if + g)k ) 2 2 2 2 2 2 1 4 (kg + f k − kg − f k − i(| − i|) kg + if k + i(|i|) kg − if k ) 2 2 2 2 1 4 (kg + f k − kg − f k − ikg + if k + ikg − if k )




hf |gi =

=: hg|f i (ii) We will now show linearity in the second argument. This is fairly non-trivial and quite lengthy.We will focus on additivity first. We have 1 hf |g + hi := (kf + g + hk2 − kf − g − hk2 + ikf − ig − ihk2 − ikf + ig + ihk2 ). 4 Consider the real part of hf |g + hi. By successive applications of the parallelogram identity, we find Rehf |g + hi = 14 (kf + g + hk2 − kf − g − hk2 ) = 41 (kf + g + hk2 + kf + g − hk2 − kf + g − hk2 − kf − g − hk2 ) = 41 (2kf + gk2 + 2khk2 − 2kf − hk2 − 2kgk2 ) = 14 (2kf + gk2 + 2kf k2 + 2khk2 − 2kf − hk2 − 2kf k2 − 2kgk2 ) = 14 (2kf + gk2 + kf + hk2 + kf − hk2 − 2kf − hk2 − kf + gk2 − kf − gk2 ) = 41 (kf + gk2 + kf + hk2 − kf − hk2 − kf − gk2 ) = Rehf |gi + Rehf |hi. Replacing g and h with −ig and −ih respectively, we obtain Imhf |g + hi = Imhf |gi + Imhf |hi. Hence, we have hf |g + hi = Rehf |g + hi + i Imhf |g + hi = Rehf |gi + Rehf |hi + i(Imhf |gi + Imhf |hi) = Rehf |gi + i Imhf |gi + Rehf |hi + i Imhf |hi = hf |gi + hf |hi, which proves additivity. For scaling invariance, we will proceed in several steps. (a) First, note that hf |0i := 14 (kf k2 − kf k2 + ikf k2 − ikf k2 ) = 0 and hence hf |0gi = 0hf |gi holds.

–4–

(b) Suppose that hf |ngi = nhf |gi for some n ∈ N. Then, by additivity hf |(n + 1)gi = hf |ng + gi = hf |ngi + hf |gi = nhf |gi + hf |gi = (n + 1)hf |gi. Hence, by induction on n with base case (a), we have ∀ n ∈ N : hf |ngi = nhf |gi. (c) Note that by additivity (a)

hf |gi + hf | − gi = hf |g − gi = hf |0i = 0. Hence hf | − gi = −hf |gi. (d) Then, for any n ∈ N (c)

(b)

hf |−ngi = −hf |ngi = −nhf |gi and thus ∀ n ∈ Z : hf |ngi = nhf |gi. (e) Now note that for any m ∈ Z \ {0} (d)

1 1 gi = hf |m m gi = hf |gi mhf | m 1 and hence, by dividing by m, we have hf | m gi =

(f) Therefore, for any r =

n m

1 m hf |gi.

∈ Q, we have (d)

(e) n m hf |gi

n 1 hf |rgi = hf | m gi = nhf | m gi =

= rhf |gi

and hence ∀ r ∈ Q : hf |rgi = rhf |gi.

√ (g) Before we turn to R, we need to show that |hf |gi| ≤ 2kf kkgk. Note that here we cannot invoke the Cauchy-Schwarz inequality (which would provide a better estimate) since we don’t know that h·|·i is an inner product yet. First, consider the real part of hf |gi. Rehf |gi = 14 (kf + gk2 − kf − gk2 ) = 14 (2kf + gk2 − kf + gk2 − kf − gk2 ) = 41 (2kf + gk2 − 2kf k2 − 2kgk2 ) ≤ 14 (2(kf k + kgk)2 − 2kf k2 − 2kgk2 ) = 14 (2kf k2 + 4kf kkgk + 2kgk2 − 2kf k2 − 2kgk2 ) = kf kkgk.

–5–

Replacing g with −ig and noting that k − igk = | − i|kgk = kgk, we also have Imhf |gi ≤ kf kkgk. Hence, we find |hf |gi| = | Rehf |gi + i Imhf |gi| p = (Rehf |gi)2 + (Imhf |gi)2 p ≤ (kf kkgk)2 + (kf kkgk)2 √ = 2kf kkgk. (h) Let r ∈ R. Since R is the completion of Q (equivalently, Q is dense in R), there exists a sequence {rn }n∈N in Q which converges to r. Let ε > 0. Then, there exist N1 , N2 ∈ N such that ε ∀ n ≥ N1 : |rn − r| < √ 2 2kf kkgk ε ∀ n, m ≥ N2 : |rn − rm | < √ . 2 2kf kkgk Let N := max{N1 , N2 } and fix m ≥ N . Then, for all n ≥ N , we have |rn hf |gi − hf |rgi| = |rn hf |gi − rm hf |gi + rm hf |gi − hf |rgi| (f)

= |rn hf |gi − rm hf |gi + hf |rm gi − hf |rgi|

= |(rn − rm )hf |gi + hf |(rm − r)gi| ≤ |(rn − rm )hf |gi| + |hf |(rm − r)gi| (g) √ √ ≤ 2|rn − rm |kf kkgk + 2kf kk(rm − r)gk √ √ = 2|rn − rm |kf kkgk + 2|rm − r|kf kkgk √ √ ε ε < 2 √ kf kkgk + 2 √ kf kkgk 2 2kf kkgk 2 2kf kkgk = ε, that is, lim rn hf |gi = hf |rgi. n→∞

(i) Hence, for any r ∈ R, we have   (h) rhf |gi = lim rn hf |gi = lim rn hf |gi = hf |rgi n→∞

n→∞

and thus ∀ r ∈ R : rhf |gi = hf |rgi. (j) We now note that hf |igi := 14 (kf + igk2 − kf − igk2 + ikf − i2 gk2 − ikf + i2 gk2 ) =

1 4 i (−ikf

+ igk2 + kf − igk2 + ikf + gk2 − kf − gk2 )

=: ihf |gi and hence hf |igi = ihf |gi.

–6–

(k) Let z ∈ C. By additivity, we have hf |zgi = hf |(Re z + i Im z)gi = hf |(Re z)gi + hf |i(Im z)gi (j)

= hf |(Re z)gi + ihf |(Im z)gi

(i)

= Re zhf |gi + i Im zhf |gi = (Re z + i Im z)hf |gi = zhf |gi,

which shows scaling invariance in the second argument. Combining additivity and scaling invariance in the second argument yields linearity in the second argument. (iii) For positive-definiteness hf |f i := 41 (kf + f k2 − kf − f k2 + ikf − if k2 − ikf + if k2 ) = = =

2 2 2 1 4 (4kf k + i|1 − i| kf k − i|1 + 2 2 2 1 4 (4 + i|1 − i| − i|1 + i| )kf k 2 1 4 (4 + 2i − 2i)kf k 2

i|2 kf k2 )

= kf k .

Thus, hf |f i ≥ 0 and hf |f i = 0 ⇔ f = 0. Hence, h·|·i is indeed a sesqui-linear inner product. Note that, from part (iii) above, we have p hf |f i = kf k. That is, the inner product h·|·i does induce the norm from which we started, and this completes the proof. Remark 3.4 . Our proof of linearity is based on the hints given in Section 6.1, Exercise 27, from Linear Algebra (4th Edition) by Friedberg, Insel, Spence. Other proofs of the Jordan-von Neumann theorem can by found in • Kadison, Ringrose, Fundamentals of the Theory of Operator Algebras: Volume I: Elementary Theory, American Mathematical Society 1997 • Kutateladze, Fundamentals of Functional Analysis, Springer 1996. Remark 3.5 . Note that, often in the more mathematical literature, a sesqui-linear inner product is defined to be linear in the first argument rather than the second. In that case, the polarisation identity takes the form 3

hf |gi =

1X k i kf + ik gk2 4 k=0

1 = (kf + gk2 − kf − gk2 + ikf + igk2 − ikf − igk2 ). 4

–7–

Example 3.6 . Consider CC0 [0, 1] and let f (x) = x and g(x) = 1. Then kf k∞ = 1,

kgk∞ = 1,

kf + gk∞ = 2,

kf − gk∞ = 1

and hence kf + gk2∞ + kf − gk2∞ = 5 6= 4 = 2kf k2∞ + 2kgk2∞ . Thus, by the Jordan-von Neumann theorem, there is no inner product on CC0 [0, 1] which induces the supremum norm. Therefore, (CC0 [0, 1], k · k∞ ) cannot be a Hilbert space. Proposition 3.7. Let H be a Hilbert space. Then, H∗ is a Hilbert space. Proof. We already know that H∗ := L(H, C) is a Banach space. The norm on H∗ is just the usual operator norm |f (ϕ)| kf kH∗ := sup ϕ∈H kϕkH where, admittedly somewhat perversely, we have reversed our previous notation for the dual √ elements. Since the modulus is induced by the standard inner product on C, i.e. |z| = zz, it satisfies the parallelogram identity. Hence, we have     |(f1 + f2 )(ϕ)| 2 |(f1 − f2 )(ϕ)| 2 2 2 kf1 + f2 kH∗ + kf1 − f2 kH∗ := sup + sup kϕkH kϕkH ϕ∈H ϕ∈H |(f1 + f2 )(ϕ)|2 |(f1 − f2 )(ϕ)|2 + sup kϕk2H kϕk2H ϕ∈H ϕ∈H

= sup

|f1 (ϕ) + f2 (ϕ)|2 + |f1 (ϕ) − f2 (ϕ)|2 kϕk2H ϕ∈H

= sup

2|f1 (ϕ)|2 + 2|f2 (ϕ)|2 kϕk2H ϕ∈H

= sup

|f2 (ϕ)|2 |f1 (ϕ)|2 + 2 sup 2 2 ϕ∈H kϕkH ϕ∈H kϕkH

= 2 sup

=: 2kf1 k2H∗ + 2kf2 k2H∗ , where several steps are justified by the fact that the quantities involved are non-negative. Hence, by the Jordan-von Neumann theorem, the inner product on H∗ defined by the polarisation identity induces k · kH∗ . Hence, H∗ is a Hilbert space. The following useful fact is an immediate application of the Cauchy-Schwarz inequality. Proposition 3.8. Inner products on a vector space are sequentially continuous. Proof. Let h·|·i be an inner product on V . Fix ϕ ∈ V and let lim ψn = ψ. Then n→∞

|hϕ|ψn i − ϕ|ψi| = |hϕ|ψn − ψi| ≤ kϕkkψn − ψk and hence lim hϕ|ψn i = hϕ|ψi. n→∞

–8–

3.2

Hamel versus Schauder2

Choosing a basis on a vector space is normally regarded as mathematically inelegant. The reason for this is that most statements about vector spaces are much clearer and, we maintain, aesthetically pleasing when expressed without making reference to a basis. However, in addition to the fact that some statements are more easily and usefully written in terms of a basis, bases provide a convenient way to specify the elements of a vector space in terms of components. The notion of basis for a vector space that you most probably met in your linear algebra course is more properly know as Hamel basis. Definition. A Hamel basis of a vector space V is a subset B ⊆ V such that (i) any finite subset {e1 , . . . , en } ⊆ B is linearly independent, i.e. n X

λi ei = 0 ⇒ λ1 = · · · = λn = 0

i=1

(ii) the set B is a generating (or spanning) set for V . That is, for any element v ∈ V , there exist a finite subset {e1 , . . . , en } ⊆ B and λ1 , . . . , λn ∈ C such that v=

n X

λi ei .

i=1

Equivalently, by defining the linear span of a subset U ⊆ V as X  n 1 i n span U := λ ui λ , . . . , λ ∈ C, u1 , . . . , un ∈ U and n ≥ 1 , i=1

i.e. the set of all finite linear combinations of elements of U with complex coefficients, we can restate this condition simply as V = span B. Given a basis B, one can show that for each v ∈ V the λ1 , . . . , λn appearing in (ii) above are uniquely determined. They are called the components of v with respect to B. One can also show that if a vector space admits a finite Hamel basis B, then any other basis of V is also finite and, in fact, of the same cardinality as B. Definition. If a vector space V admits a finite Hamel basis, then it is said to be finitedimensional and its dimension is dim V := |B|. Otherwise, it is said to be infinitedimensional and we write dim V = ∞. Theorem 3.9. Every vector space admits a Hamel basis. For a proof of (a slightly more general version of) this theorem, we refer the interested reader to Dr Schuller’s Lectures on the Geometric Anatomy of Theoretical Physics. Note that the proof that every vector space admits a Hamel basis relies on the axiom of choice and, hence, it is non-constructive. By a corollary to Baire’s category theorem, a 2

Not a boxing competition.

–9–

Hamel basis on a Banach space is either finite or uncountably infinite. Thus, while every Banach space admits a Hamel basis, such bases on infinite-dimensional Banach spaces are difficult to construct explicitly and, hence, not terribly useful to express vectors in terms of components and perform computations. Thankfully, we can use the extra structure of a Banach space to define a more useful type of basis. Definition. Let (W, k · k) be a Banach space. A Schauder basis of W is a sequence {en }n∈N in W such that, for any f ∈ W , there exists a unique sequence {λn }n∈N in C such that f = lim

n→∞

n X

λi ei =:

i=0

∞ X

λi ei

i=0

or, by explicitly using the definition of limit in W ,

n X

i

lim f − λ ei .

n→∞

i=0

We note the following points. • Since Schauder bases require a notion of convergence, they can only be defined on a vector space equipped with a (compatible) topological structure, of which Banach spaces are a special case. • Unlike Hamel bases, Schauder bases need not exist. • Since the convergence of a series may depend on the order of its terms, Schauder bases must be considered as ordered bases. Hence, two Schauder bases that merely differ in the ordering of their elements are different bases, while permuting the elements of a Schauder basis doesn’t necessarily yield another Schauder basis. • The uniqueness requirement in the definition immediately implies that the zero vector cannot be an element of a Schauder basis. • Schauder bases satisfy a stronger linear independence property than Hamel bases, namely ∞ X λi ei = 0 ⇒ ∀ i ∈ N : λi = 0. i=0

• At the same time, they satisfy a weaker spanning condition. Rather than the linear span of the basis being equal to W , we only have that it is dense in W . Equivalently, W = span{en | n ∈ N}, where the topological closure U of a subset U ⊆ W is defined as U :=



lim un | ∀ n ∈ N : un ∈ U .

n→∞

– 10 –

Definition. A Schauder basis {en }n∈N of (W, k · k) is said to be normalised if ∀ n ∈ N : ken k = 1. Multiplying an element of a Schauder basis by a complex number gives again a Schauder basis (not the same one, of course). Since Schauder bases do not contain the zero vector, any Schauder basis {en }n∈N gives rise to a normalised Schauder basis {e en }n∈N by defining een := 3.3

en . ken k

Separable Hilbert spaces and unitary maps

Separability is a topological property. Namely, a topological space is said to be separable if it contains a dense subset which is also countable. A Banach space is said to be separable if it is separable as a topological space with the topology induced by the norm. Similarly, a Hilbert space is said to be separable if it is separable as a topological space with the topology induced by the norm induced in turn by the inner product. For infinite-dimensional Hilbert spaces, there is a much more useful characterisation of separability, which we will henceforth take as our definition. Proposition 3.10. An infinite-dimensional Hilbert space is separable if, and only if, it admits an orthonormal Schauder basis. That is, a Schauder basis {en }n∈N such that ( 1 if i = j ∀ i, j ∈ N : hei |ej i = δij := . 0 if i 6= j Whether this holds for Banach spaces or not was a famous open problem in functional analysis, problem 153 from the Scottish book. It was solved in 1972, more that three decades after it was first posed, when Swedish mathematician Enflo constructed an infinite-dimensional separable Banach space which lacks a Schauder basis. That same year, he was awarded a live goose3 for his effort. Remark 3.11 . The Kronecker symbol δij appearing above does not represent the components of the identity map on H. Instead, δij are the components of the sesqui-linear form h·|·i, which is a map H × H → C, unlike idH which is a map H → H. If not immediately understood, this remark may be safely ignored. Remark 3.12 . In finite-dimensions, since every vector space admits (by definition) a finite Hamel basis, every inner product space admits an orthonormal basis by the Gram-Schmidt orthonormalisation process. From now on, we will only consider orthonormal Schauder bases, sometimes also called Hilbert bases, and just call them bases. Lemma 3.13. Let H be a Hilbert space with basis {en }n∈N . The unique sequence in the expansion of ψ ∈ H in terms of this basis is {hen |ψi}n∈N . 3

https://en.wikipedia.org/wiki/Per_Enflo#Basis_problem_of_Banach

– 11 –

Proof. By using the continuity of the inner product, we have   X ∞ j hei |ψi = ei λ ej j=0

  n X j = ei lim λ ej n→∞ j=0

  X n j λ ej = lim ei n→∞ j=0

n X

= lim

n→∞

= lim

i=0 n X

n→∞

λj hei |ej i λj δij

i=0

= λi , which is what we wanted. While we have already used the term orthonormal, let us note that this means both orthogonal and normalised. Two vectors ϕ, ψ ∈ H are said to be orthogonal if hϕ|ψi = 0, and a subset of H is called orthogonal if its elements are pairwise orthogonal. Lemma 3.14 (Pythagoras’ theorem). Let H be a Hilbert space and let {ψ0 , . . . , ψn } ⊂ H be a finite orthogonal set. Then

n

n

X 2 X

= ψ kψi k2 . i

i=0

i=0

Proof. Using the pairwise orthogonality of {ψ0 , . . . , ψn }, we simply calculate

X

2 X X  X n n n X X

n

n

ψi := ψi ψj = hψi |ψi i + hψi |ψj i =: kψi k2 .

i=0

i=0

j=0

i=0

i6=j

i=0

Remark 3.15 . An insightful way to study a structure in mathematics is to consider maps between different instances A, B, C, . . . of that structure, and especially the structurepreserving maps. If a certain structure-preserving map A → B is invertible and its inverse is also structure-preserving, then both these maps are generically called isomorphisms and A and B are said to be isomorphic instances of that structure. Isomorphic instances of a structure are essentially the same instance of that structure, just dressed up in different ways. Typically, there are infinitely many concrete instances of any given structure. The highest form of understanding of a structure that we can hope to achieve is that of a classification of its instances up to isomorphism. That is, we would like to know how many different, non-isomorphic instances of a given structure there are. In linear algebra, the structure of interest is that of vector space over some field F. The structure-preserving maps are just the linear maps and the isomorphisms are the linear

– 12 –

bijections (whose inverses are automatically linear). Finite-dimensional vector spaces over F are completely classified by their dimension, i.e. there is essentially only one vector space over F for each n ∈ N, and Fn is everyone’s favourite. Assuming the axiom of choice, infinite-dimensional vector spaces over F are classified in the same way, namely, there is, up to linear isomorphism, only one vector space over F for each infinite cardinal. Of course, one could do better and also classify the base fields themselves. The classification of finite fields (i.e. fields with a finite number of elements) was achieved in 1893 by Moore, who proved that the order (i.e. cardinality) of a finite field is necessarily a power of some prime number, and there is only one finite field of each order, up to appropriate notion of isomorphism. The classification of infinite fields remains an open problem. A classification with far-reaching implications in physics is that of finite-dimensional, semi-simple, complex Lie algebras, which is discussed in some detail in Dr Schuller’s Lectures on the Geometric Anatomy of Theoretical Physics. The structure-preserving maps between Hilbert spaces are those that preserve both the vector space structure and the inner product. The Hilbert space isomorphisms are called unitary maps. Definition. Let H and G be Hilbert spaces. A bounded bijection U ∈ L(H, G) is called a unitary map (or unitary operator ) if ∀ ψ, ϕ ∈ H : hU ψ|U ϕiG = hψ|ϕiH . If there exists a unitary map H → G, then H and G are said to be unitarily equivalent and we write H ∼ =Hil G. There are a number of equivalent definitions of unitary maps (we will later see one involving adjoints) and, in fact, our definition is fairly redundant. Proposition 3.16. Let U : H → G be a surjective map which preserves the inner product. Then, U is a unitary map. Proof.

(i) First, let us check that U is linear. Let ψ, ϕ ∈ H and z ∈ C. Then

kU (zψ + ϕ) − zU ψ − U ϕk2G = hU (zψ + ϕ) − zU ψ − U ϕ|U (zψ + ϕ) − zU ψ − U ϕiG = hU (zψ + ϕ)|U (zψ + ϕ)iG + |z|2 hU ψ|U ψiG + hU ϕ|U ϕiG − zhU (zψ + ϕ)|U ψiG − hU (zψ + ϕ)|U ϕiG + zhU ψ|U ϕiG − zhU ψ|U (zψ + ϕ)iG − hU ϕ|U (zψ + ϕ)iG + zhU ϕ|U ψiG = hzψ + ϕ|zψ + ϕiH + |z|2 hψ|ψiH + hϕ|ϕiH − zhzψ + ϕ|ψiH − hzψ + ϕ|ϕiH + zhψ|ϕiH − zhψ|zψ + ϕiH − hϕ|zψ + ϕiH + zhϕ|ψiH = 2|z|2 hψ|ψiH + zhψ|ϕiH + zhϕ|ψiH + 2hϕ|ϕiH − |z|2 hψ|ψiH − zhϕ|ψiH − zhψ|ϕiH − hϕ|ϕiH + zhψ|ϕiH − |z|2 hψ|ψiH − zhψ|ϕiH − zhϕ|ψiH − hϕ|ϕiH + zhϕ|ψiH = 0.

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Hence kU (zψ + ϕ) − zU ψ − U ϕkG = 0, and thus U (zψ + ϕ) = zU ψ + U ϕ. (ii) For boundedness, simply note that since for any ψ ∈ H p p kU ψkG := hU ψ|U ψiG = hψ|ψiH =: kψkH , we have

kU ψkG = 1 < ∞. ψ∈H kψkH sup

Hence U is bounded and, in fact, has unit operator norm. (iii) Finally, recall that a linear map is injective if, and only if, its kernel is trivial. Suppose that ψ ∈ ker U . Then, we have hψ|ψiH = hU ψ|U ψiG = h0|0iG = 0. Hence, by positive-definiteness, ψ = 0 and thus U is injective. Since U is also surjective by assumption, it satisfies our definition of unitary map. Note that a map U : H → G is called an isometry if ∀ ψ ∈ H : kU ψkG = kψkH . Linear isometries are, of course, the structure-preserving maps between normed spaces. We have shown that every unitary map is an isometry has unit operator norm, whence the name unitary operator. Example 3.17 . Consider the set of all square-summable complex sequences   ∞ X 2 ` (N) := a : N → C |ai | < ∞ . 2

i=0

We define addition and scalar multiplication of sequences termwise, that is, for all n ∈ N and all complex numbers z ∈ C, (a + b)n := an + bn (z · a)n := zan . These are, of course, just the discrete analogues of pointwise addition and scalar multiplication of maps. The triangle inequality and homogeneity of the modulus, together with the vector space structure of C, imply that (`2 (N), +, ·) is a complex vector space. The standard inner product on `2 (N) is ha|bi`2 :=

∞ X i=0

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ai bi .

This inner product induces the norm kak`2

v u∞ p uX := ha|ai`2 = t |ai |2 , i=0

with respect to which `2 (N) is complete. Hence, (`2 (N), +, ·, h·|·i`2 ) is a Hilbert space. Consider the sequence of sequences {en }n∈N where e0 = (1, 0, 0, 0, . . .) e1 = (0, 1, 0, 0, . . .) e2 = (0, 0, 1, 0, . . .) .. . i.e. we have (en )m = δnm . Each a ∈ `2 (N) can be written uniquely as a=

∞ X

λi ei ,

i=0

where λi = hei |ai`2 = ai . The sequences en are clearly square-summable and, in fact, they are orthonormal with respect to h·|·i`2 hen |em i`2 :=

∞ X

(en )i (em )i =

i=0

∞ X

δni δmi = δnm .

i=0

Hence, the sequence {en }n∈N is an orthonormal Schauder basis of `2 (N), which is therefore an infinite-dimensional separable Hilbert space. Theorem 3.18 (Classification of separable Hilbert spaces). Every infinite-dimensional separable Hilbert space is unitarily equivalent to `2 (N). Proof. Let H be a separable Hilbert space with basis {en }n∈N . Consider the map U : H → `2 (N) ψ 7→ {hen |ψiH }n∈N . Note that, for any ψ ∈ H, the sequence {hen |ψiH }n∈N is indeed square-summable since we have

2

X

∞ 2

kψkH =

hei |ψiH ei H

i=0

2

n X

= lim hei |ψiH ei

n→∞

H i=0

n

2 X

= lim

hei |ψiH ei n→∞

i=0

– 15 –

H

n X

= lim

n→∞

= lim =

∞ X

i=0 n X

n→∞

khei |ψiH ei k2H |hei |ψiH |2 kei k2H

i=0

|hei |ψiH |2 ,

i=0

where we have used Pythagoras’ theorem and the orthonormality of the basis. Hence, since kψkH is finite, we have ∞ X |hei |ψiH |2 < ∞. i=0

By our previous proposition, in order to show that U is a unitary map, it suffices to show that it is surjective and preserves the inner product. For surjectivity, let {an }n∈N be a complex square-summable sequence. Then, by elementary analysis, we know that lim |an | = 0. This implies that, for any ε > 0, there exists N ∈ N such that n→∞

n X

∀n ≥ m ≥ N :

|ai |2 < ε.

i=m

Then, for all n, m ≥ N (without loss of generality, assume n > m), we have

X

2

X

2 m n n X X X

n

n

2 2



= a e − a e = a e |a | ke k = |ai |2 < ε. i i j j i i i i H



i=0

Pn

That is, i=0 ai ei ψ ∈ H such that

j=0

n∈N

H

H

i=m

i=m

i=m

is a Cauchy sequence in H. Hence, by completeness, there exists ψ=

∞ X

ai ei

i=0

and we have U ψ = {an }n∈N , so U is surjective. Moreover, we have ∞ X  ∞ X hψ|ϕiH = hei |ψiH ei hej |ϕiH ej = = =

i=0 ∞ X ∞ X

j=0 j=0 ∞ X ∞ X

j=0

H

hei |ψiH hej |ϕiH hei |ej iH hei |ψiH hej |ϕiH δij

j=0 j=0 ∞ X

hei |ψiH hei |ϕiH

j=0



 =: {hen |ψiH }n∈N {hen |ϕiH }n∈N `2 =: hU ψ|U ϕi`2 . Hence U preserves the inner product, and it is therefore a unitary map.

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Further readings Mathematical quantum mechanics • Ballentine, Quantum Mechanics: A Modern Development (Second edition), World Scientific 2014 • Faddeev, Yakubovskii, Lectures on Quantum Mechanics for Mathematics Students, American Mathematical Society 2009 • Folland, Quantum Field Theory: A Tourist Guide for Mathematicians, American Mathematical Society 2008 • Gieres, Mathematical surprises and Dirac’s formalism in quantum mechanics https://arxiv.org/abs/quant-ph/9907069 • Hall, Quantum Theory for Mathematicians, Springer 2013 • Mackey, Mathematical Foundations of Quantum Mechanics, Dover Publications 2004 • Moretti, Spectral Theory and Quantum Mechanics: With an Introduction to the Algebraic Formulation, Springer 2013 • Parthasarathy, Mathematical Foundations of Quantum Mechanics, Hindustan Book Agency 2005 • Strocchi, An Introduction to the Mathematical Structure of Quantum Mechanics: A Short Course for Mathematicians, World Scientific 2008 • Takhtajan, Quantum Mechanics for Mathematicians, American Mathematical Society 2008 Linear Algebra • Friedberg, Insel, Spence, Linear Algebra (4th Edition), Pearson 2002 Functional analysis • Kadison, Ringrose, Fundamentals of the Theory of Operator Algebras. Volumes I-II, American Mathematical Society 197

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Alphabetical Index P

C Cauchy-Schawrz inequality

1

H Hamel basis

parallelogram identity polarisation identity

3 3

9

S

I induced norm isometry

2 14

Schauder basis

U

J Jordan-von Neumann theorem

10

3

unitary map

13