Hilbert Spaces and Dirac Notation

1. Generalized continuous basis of $L^2(\mathbb{R})$

We have already detailed the model space $L^2(\R)$ in Section 3.3 (Theme 2, Lesson 1). In particular, we exhibited a countable Hilbert basis for it. Every physical state then decomposes as a series on the basis of Hermite functions. Unfortunately the explicit expressions of these functions are rather horrible.

Another “basis” was therefore invented, which is not really one in the sense of a Hilbert basis, but which is what one calls a generalized continuous basis. The passage to infinite dimension is delicate here, and its rigorous construction will require the tools of the next theme, devoted to the theory of linear operators. We therefore admit for the moment the formulary that follows, bearing in mind that it is a set of instructions for Dirac notation in this "basis", and not an established mathematical construction.

1.1. Generalized basis in the position representation

Let us nevertheless try to give a few ideas. One first introduces a self-adjoint operator called the “position” operator, denoted $\hat{X}$, acting on the functions of a Hilbert space $L^2(\mathbb{R})$ by multiplication:

$$\boxed{ (\hat{X}\psi)(x) \equiv x\,\psi(x), \quad \forall x \in \R. }$$

Beware of the domain of definition. This expression only makes sense if the function $x\,\psi(x)$ remains in $L^2(\R)$. Now this is not always the case: certain square-integrable functions become divergent when multiplied by $x$.¹ Thus, the operator $\hat{X}$ cannot be defined on all of $L^2(\R)$, but only on a subspace of functions where its action remains in $L^2$.

In order to have a well-defined domain, dense in $L^2$ and stable under the usual operations (differentiation, multiplication by $x$, Fourier transform), one introduces the space of Schwartz functions, denoted $\mathcal{S}(\R)$:

$$\boxed{ \mathcal{S}(\R) = \left\{ f \in C^\infty(\R) \;\middle|\; \forall\, m,n \in \mathbb{N},\; \sup_{x \in \R} |\,x^m f^{(n)}(x)\,| < \infty \right\}. }$$

In other words, these are smooth, rapidly decreasing functions, such that $f$ and all its derivatives tend to zero faster than any inverse power of $x$.

The space $\mathcal{S}(\R)$ is:

• dense in $L^2(\R)$ (every function of $L^2$ can be approximated arbitrarily well by functions of $\mathcal{S}$),
• stable under differentiation, under multiplication by $x$, and under Fourier transform.

It is therefore the natural domain for properly defining the position operator $\hat{X}$ and the momentum operator $\hat{P}$, and it is on this space that the rigorous constructions of the Dirac formalism and of the Gelfand triplet rest.

These precautions being taken, one can then write, in a for the moment purely formal way², the eigenstates of $\hat{X}$:

$$\boxed{ \hat{X}\ket{x} = x \ket{x}. }$$

These kets $\ket{x}$ constitute what one calls a generalized continuous basis, in the sense that they allow one to represent the elements of $L^2(\R)$ as continuous superpositions, although the $\ket{x}$ themselves are not vectors of the Hilbert space.

In reality, they are not kets in the proper sense, because they are not normalizable. We shall see why. This family is uncountable, and so its orthonormality cannot be written by a discrete Kronecker symbol $\delta_{ij}$, but through its « continuous version »   in the form:

$$\boxed{ \langle x | x' \rangle = \delta(x - x'), }$$

with in particular:

where we have introduced the Dirac delta $\delta(x)$, which is not an ordinary function but a distribution in the sense of the mathematical theory of distributions. It is defined by its action under the integral:

for every test function $f$ of the Schwartz space. One must understand that $\delta(x)$ individually has no meaning, it is not even a number. In reality, the $\delta$ is a linear form $\delta_{x_0}(f) = f(x_0)$ acting on test functions³. It is thus defined only by its action under the integral (2), and not by its values $\delta(x)$.

But if one absolutely insists on regarding it as a function, then one can regard it as the limit of a sequence of functions increasingly concentrated about zero, for example a limit of increasingly peaked Gaussians:

$$\delta(x) = \lim_{\epsilon \to 0^+} \frac{1}{\sqrt{\pi \epsilon}} \, e^{-x^2 / \epsilon}.$$

or again as rectangles centred at zero of width tending to zero and of height tending to infinity. In all cases, in this view, the delta appears to be in the limit a "function" that is zero everywhere except at $x = 0$, where it diverges, while keeping a total area under the curve equal to 1.

This is why, in physics, one will nevertheless write $\delta(0) = \infty$, but one must clearly understand the abuse of notation and of meaning here. In any case this shows, via eq. (1), that the ket $\ket{x}$ is not normalized. The Dirac delta also admits an integral representation, extremely useful and which is probably the most important formula of this section:

$$\boxed{ \delta(x) = \frac{1}{2\pi} \int_{-\infty}^{\infty} e^{ikx}\, dk, }$$

where here too the equality is to be understood in the sense of the theory of distributions, since this integral is manifestly not convergent and would otherwise have no meaning.

The “matrix elements” of the position operator $\hat{X}$ are $\bra{x} \hat{X} \ket{x'} = x \, \delta(x-x')$. We continue to call them that (compare with the previous section in finite dimension) even though $\hat{X}$ cannot be represented by an ordinary matrix — one could imagine a “matrix with continuous indices”, diagonal in the distributional sense, but whose value on the diagonal $x = x'$ has no proper meaning ($x\,\delta(0)$ not being a real number). It is here, no doubt, that one sees the fundamental difference between linear operators in finite dimension ( = matrices), and linear operators in infinite dimension.

In the generalized basis, the resolution of the identity is written as:

It allows one to see the decomposition of any ket $\kpsi$ of the Hilbert space on this continuous basis as an integral:

$$\boxed{\ket{\psi} = \mathbf{1} \kpsi = \int_{-\infty}^{\infty} \ket{x} \, \braket{x}{\psi} dx = \int_{-\infty}^{\infty} \psi(x) \, \ket{x} dx }$$

where we have defined the wave function

$$\boxed{\psi(x) = \braket{x}{\psi}}$$

In the same way, the bra conjugate to $\kpsi$ expands as:

$$\boxed{\bra{\psi} = \bra{\psi} \mathbf{1} = \int_{-\infty}^{\infty} \braket{\psi}{x} \, \bra{x} dx = \int_{-\infty}^{\infty} \psi^*(x) \, \bra{x} dx }$$

where we have used the Hermitian symmetry of the inner product: $\psi(x) = \braket{x}{\psi} ; \psi^*(x) = \braket{\psi}{x}$.

Note that the $x$ component of the vector $\hat{X} \psi$, which by definition equals $\braket{x}{X\psi}$, will more often be written, here too for aesthetic reasons (compare with equation (4 (Theme 2, Lesson 2))), as:

$$\braket{x}{\hat{X}\psi} \equiv \bra{x}\hat{X}\ket{\psi}$$

1.2. In the momentum representation

In the quantum mechanics of a point particle, one introduces in an analogous way the momentum operator by its action on the wave functions $\psi(x)$ belonging to $L^2(\mathbb{R})$:

$$\boxed{ (\hat{P}\psi)(x) \equiv -\,i\hbar\,\frac{d\psi}{dx}. }$$

Here too one must restrict to the Schwartz domain so that, on the one hand, the wave function is differentiable, and on the other hand, its derivative is still square-integrable.

It is a very good exercise in applying the present chapter to convince oneself that $\hat{P}$ is indeed a self-adjoint operator. To do this, let us begin by studying the differentiation operator itself, denoted:

$$\boxed{ (\hat{D}\psi)(x) \equiv \frac{d\psi}{dx}. }$$

It is a linear operator (differentiation is linear: $(f+ \alpha g)' = f' + \alpha g'$). It is obvious that this operator cannot be represented by a matrix, as in finite dimension: it acts differentially on an infinite-dimensional space of functions. One cannot therefore compute its adjoint by a transpose-conjugate operation, but one can nevertheless compute it using the definition of the adjoint, which satisfies, for two test functions of $\mathcal{S}(\R)$, $\phi$ and $\psi$:

$$\braket{\hat{D} \phi}{\psi} = \braket{\phi}{\hat{D}^\dagger \psi}$$

cf. definition 2 (Theme 2, Lesson 2). Now the left-hand side equals, on inserting the resolution of the identity:

$$\begin{aligned} \braket{\hat{D} \phi}{\psi} &= \bra{\hat{D} \phi} \times\left(\int dx \ket{x} \bra{x} \right) \times \ket{\psi} \\ &= \int dx \braket{\hat{D} \phi}{x}\braket{x}{\psi} \\ &= \int dx \braket{x}{\hat{D} \phi}^* \braket{x}{\psi} \\ &= \int dx \, \phi'^*(x) \psi(x) \\ &= - \int dx \phi^*(x) \psi'(x) + 0 \\ &= - \int dx \braket{\phi}{x} \braket{x}{\hat{D}\psi} \\ &= - \bra{\phi} \left(\int dx \ket{x} \bra{x} \right)\ket{\hat{D}\psi} \\ &= - \braket{\phi}{\hat{D}\psi} \end{aligned}$$

showing that the adjoint of the operator $\hat{D} = d/dx$ equals $-\hat{D} = -d/dx$. In the computation above, we first inserted the resolution of the identity between the bra and the ket in the first line; we distributed it in the second, we used Hermitian symmetry in the third; in the fourth line we used the definition of $\hat{D}$ and of the wave functions, in the fifth we performed an integration by parts on $\R$: the boundary terms at infinity vanish because functions of the Schwartz space decrease faster than any monomial. Finally, in the last lines, we went back from wave functions to inner products, and we factorized and then removed the resolution of the identity. Consequently, the momentum operator is indeed self-adjoint, since

$$\left(i\,\frac{d}{dx}\right)^\dagger = - i^*\,\frac{d}{dx} = i \frac{d}{dx}$$

The self-adjointness of $\hat{P}$ guarantees that its eigenvalues (the possible momenta) are real, and that its eigenfunctions form a complete basis of the state space, in the generalized sense (via Dirac distributions).

Momentum eigenstates. — In an exactly similar way, one then defines the generalized “kets” $\ket{p}$ as the eigenstates of $\hat{P}$:

$$\boxed{ \hat{P}\ket{p} = p \ket{p}, \quad p \in \mathbb{R}. }$$

In the position representation, the associated wave function is obtained by projecting onto $\bra{x}$:

$$\bra{x}\hat{P}\ket{p} = -\,i\hbar \frac{d}{dx}\braket{x}{p} = p \braket{x}{p}.$$

This is a simple differential equation whose solution is:

$$\boxed{ \braket{x}{p} = \frac{1}{\sqrt{2\pi\hbar}}\, e^{\,i\,p\,x / \hbar}. }$$

This expression shows that the transformation relating the position and momentum representations is precisely the Fourier transform.

Orthogonality and completeness. — The kets $\ket{p}$ satisfy relations analogous to those of the $\ket{x}$:

$$\boxed{ \langle p | p' \rangle = \delta(p - p'), \qquad \mathbf{1} = \int_{-\infty}^{\infty} |p\rangle \langle p| \, dp. }$$

One defines the wave function in the momentum representation as:

$$\boxed{ \tilde{\psi}(p) = \braket{p}{\psi} = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{-\,i\,p\,x / \hbar}\, \psi(x)\,dx. }$$

The transformation $\psi(x) \mapsto \tilde{\psi}(p)$ is therefore the Fourier transform, and the inverse transformation is written:

$$\boxed{ \psi(x) = \frac{1}{\sqrt{2\pi\hbar}} \int_{-\infty}^{\infty} e^{\,i\,p\,x / \hbar}\, \tilde{\psi}(p)\,dp. }$$

Fundamental commutation. — Finally, the operators $\hat{X}$ and $\hat{P}$ satisfy the canonical commutation relation:

$$\boxed{ [\hat{X}, \hat{P}] = i\hbar\,\mathbf{1}. }$$

It is this relation that founds the whole quantum mechanics of a point particle. In reality, all the formulas of this section follow from this canonical relation alone: one proves that the only possible Hilbert representation of this algebra between X and P is the one we have just written, in particular one shows that $X$ acts by multiplication on the wave functions in x, and P by differentiation, and conversely. This major result will be detailed later; it is the Stone-Von Neumann theorem.

1.3. Properties of the Dirac delta

We end this section and this chapter with a few useful formulas on the Dirac delta, always to be understood in the sense of distributions, that is, when the delta is under an integral sign:

Symmetry:

$$\delta(x) = \delta(-x).$$

Homogeneity:

for every real $a \neq 0$,

$$\delta(a x) = \frac{1}{|a|} \, \delta(x).$$

Change of variable:

if $f$ is a regular function with simple roots $\{x_i\}$ such that $f(x_i) = 0$, then

$$\delta(f(x)) = \sum_i \frac{\delta(x - x_i)}{|f'(x_i)|}.$$

Derivative of the delta:

the distributional derivative is defined by

$$\int_{-\infty}^{\infty} f(x) \, \delta'(x - x_0) \, dx = - f'(x_0).$$

2. References