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2.3 Function spaces

2.3.1 Continuous basis functions: Fourier Transforms

For a vector space of complex valued functions f: ℝ\mathrel{↦}ℂ one can choose basis functions3

{ e}_{k} = {ϕ}_{k}(x) = {1\over \sqrt{2π}}{e}^{ikx},\qquad −∞ < k < ∞,

and expand in these functions,

f(x) ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}dk\kern 1.66702pt {ϕ}_{ k}(x)\tilde{f}(k).
(2.7)

The expansion coefficient is nothing but the Fourier transform,

\tilde{f}(k) ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}dx\kern 1.66702pt {ϕ}_{ k}^{∗}(x)f(x).
(2.8)

In much of physics, one traditionally does not normalise the {ϕ}_{k}, but uses {ϕ}_{k}(x) = {e}^{ikx}. In that case an explicit factor 2π enters in the Fourier transforms, but not in the inverse one,

\eqalignno{ f(x) & ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}{dk\over 2π}\kern 1.66702pt {e}^{ikx}\tilde{f}(k), & & \cr \tilde{f}(k) & ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}dx\kern 1.66702pt {e}^{−ikx}f(x). & & }

In order to figure out the orthogonality relations, we substitute (2.8) into (2.7), which gives the relation

f(x) ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}dk\kern 1.66702pt {ϕ}_{ k}(x){\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}dx'\kern 1.66702pt {ϕ}_{ k}^{∗}(x')f(x'),
(2.9)

which must hold for any f(x). We now swap integrals, and find

f(x) ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}dx'\kern 1.66702pt \left [{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}dk\kern 1.66702pt {ϕ}_{ k}(x){ϕ}_{k}^{∗}(x')\right ]f(x').
(2.10)

We call the object between square brackets the “Dirac delta function” δ(x − x'), where δ(y) can be defined using the explicit definition of the functions {ϕ}_{k}, as

δ(y) = {1\over 2π}{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}dz\kern 1.66702pt {e}^{iyz}.
(2.11)

(See the appendix to this chapter, 2.4, for additional properties.)

From the definition (2.11) of the delta function, we can show that the basis states satisfy the orthonormality relation

({ϕ}_{k},{ϕ}_{k'}) ={\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}dx\kern 1.66702pt {ϕ}_{ k}^{∗}(x){ϕ}_{ k'}(x) = δ(k − k')

and the completeness relation

{\mathop{\mathop{\mathop{∫ }\nolimits }}\nolimits }_{−∞}^{∞}dk\kern 1.66702pt {ϕ}_{ k}^{∗}(x){ϕ}_{ k}(x') = δ(x − x').

2.3.2 General orthogonality and completeness in function spaces

We start with a space of functions, where the scalar product is assumed to be defined by4

(ϕ,ψ) =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits dx\kern 1.66702pt {ϕ}^{∗}(x)ψ(x),

where the space from which ϕ and ψ are chosen is such that the integral is finite for any pair. In general:

Any vector space of functions with a scalar product, where all functions have finite norm, is called a Hilbert space.

Suppose in this space we have a discrete (i.e., labelled by integers), but let us assume infinite, set of basis functions {ϕ}_{n}, chosen to be orthonormal (similar to Eq. (2.5))

({ϕ}_{m},{ϕ}_{n}) =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits dx\kern 1.66702pt {ϕ}_{m}^{∗}(x){ϕ}_{ n}(x) = {δ}_{nm},

using orthogonalisation if necessary. Then an arbitrary ψ(x) can be decomposed as

ψ(x) ={ \mathop{∑ }}_{n}{c}_{n}{ϕ}_{n}(x).
(2.12)

The coefficients {c}_{n} can be determined from the orthogonality relation, multiplying (2.12) from the left with {ψ}_{m}^{∗}, integrating with respect to x, exchanging the order of the summation and the integration on the right-hand side, we find that

({ϕ}_{m},ψ) ={ \mathop{∑ }}_{n}{c}_{n}({ϕ}_{m},{ϕ}_{n}) ={ \mathop{∑ }}_{n}{c}_{n}{δ}_{mn},

from which we conclude that

{c}_{m} = ({ϕ}_{m},ψ).
(2.13)

Substituting Eq. (2.13) into Eq. (2.12) we find

\eqalignno{ ψ(x) & ={ \mathop{∑ }}_{n}\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits dx'\kern 1.66702pt {ϕ}_{n}{(x')}^{∗}ψ(x'){ϕ}_{ n}(x) & & \cr & =\mathop{ \mathop{\mathop{∫ }\nolimits }}\nolimits dx'\kern 1.66702pt \left [{\mathop{∑ }}_{n}{ϕ}_{n}{(x')}^{∗}{ϕ}_{ n}(x)\right ]ψ(x)\quad , & & }

where we have interchanged the summation and integration (which mathematicians will tell you may be incorrect!). From this we conclude that

{\mathop{∑ }}_{n}{ϕ}_{n}{(x')}^{∗}{ϕ}_{ n}(x) = δ(x − x'),

which is the form of the completeness relation for a basis labelled by a continuous variable. If the basis is labelled by a continuous label, as for the Fourier transformation, we get the completeness relations discussed for that case, Eqs. (2.9,2.10). This leads to

({ϕ}_{k},{ϕ}_{k}) = δ(0).

In this case we do not speak of a Hilbert space, since the basis is not normalisable, the norm of the basis states is necessarily infinite.

2.3.3 Example from Quantum Mechanics

Much of what we have stated in this section can be illustrated for quantum mechanical problems. Like all linear wave problems, QM relies heavily on the principle of superposition:

For any physical system, if {ψ}_{1}(x,t) and {ψ}_{2}(x,t) are possible wave functions, then so is

ψ(x,t) = λ{ψ}_{1}(x,t) + μ{ψ}_{2}(x,t),

where λ and μ are arbitrary complex numbers.5

This implies that the space of all ψ is a linear vector space (over x, since t is a parameter that describes the time evolution).

A very important statement can now be found in the fact that:

Theorem 2.3. The eigenfunctions {ψ}_{n}(x) of any physical operator form a complete set. (Can be proven for specific cases only.)

This implies that

ψ(x,t) ={ \mathop{∑ }}_{n}{c}_{n}(t){ψ}_{n}(x).

A very clear example are the eigenstates of the harmonic oscillator Hamiltonian,

\hat{H} = −{{ℏ}^{2}\over 2m} {{d}^{2}\over d{x}^{2}} + {1\over 2}m{ω}^{2}{x}^{2}.

The solutions of the Schrödinger equation

\hat{H}ψ(x,t) = ℏi{∂}_{t}ψ(x,t)

are {ψ}_{n}(x) =\mathop{ exp}\nolimits (−{x}^{2}∕(2{b}^{2})){H}_{n}(x∕b), with {H}_{n} a (Hermite) polynomial. In this case the time-dependence is determined by the eigenenergies, and we conclude that

ψ(x,t) ={ \mathop{∑ }}_{n}{a}_{n}{e}^{−i(n+1∕2)ωt}{ψ}_{ n}(x).

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