Hilbert Spaces and Linear Operators

1. The idea of duality

Duality is the guiding thread of this entire lesson. Let us try to give it an intuitive interpretation. In mathematics, the notion of duality covers many situations, but in the setting of Hilbert spaces it expresses a simple and deep idea: a vector can be viewed not only as an object in its own right, but also as an operator acting on other vectors to produce numbers (that is, as a linear form). In quantum mechanics, the number thus produced is a probability amplitude.

The quantum formalism thus fully exploits these two facets, and the Dirac notation provides an ingenious system of writing that allows them to be manipulated with ease. It features kets, written $\ket{u}$, which represent vectors of the space, and bras, written $\bra{u}$, which represent linear forms acting on these vectors. Their combination $\langle u|v\rangle$ produces a scalar, called a bracket, while expressions of the form $\sum a_{ij} \ket{u_i}\bra{v_j}$ describe operators.

This notation has become the standard language of quantum physics, and full mastery of it is indispensable. A summary formula sheet of the calculation rules will be provided at the end of this lesson. In practice, it is perfectly possible to learn to manipulate these symbols without really understanding the mathematics hidden behind them. But to grasp their deep meaning, one must tackle the program we outline here: understanding how the complete identification between vectors and linear forms is constructed, and why this approach naturally leads to the definition of another essential object: the adjoint of an operator.

In the next section, we will try to establish such an isomorphism between a vector space and its algebraic dual, defined as the set of all its linear forms. We will see that this is impossible. Indeed, although a finite-dimensional vector space is in fact isomorphic to its dual, this isomorphism is not canonical (it depends on an arbitrary choice of basis). Moreover, the situation worsens in infinite dimension: such an isomorphism simply does not exist. These obstacles invalidate the automatic identification between vectors and linear forms, and justify appealing to the richer structure of Hilbert spaces.

Section 3 shows how to exploit this additional structure to overcome these difficulties. Thanks to the inner product, and by restricting attention to the continuous linear forms that make up what is called the topological dual, one reaches a remarkable result. The Riesz theorem asserts that for any Hilbert space of arbitrary dimension, there exists a canonical antilinear isometric isomorphism between the space and its topological dual.

This result rigorously grounds Dirac notation, which we will then detail in Sections 4 and 5 (kets and bras, and linear operators), before showing how the Riesz isomorphism naturally gives rise to the notion of the adjoint of an operator (Section 6). Section 7 summarizes this construction in a complete diagram. Throughout this lesson, we will use the finite-dimensional case as a concrete example to illustrate the theory, systematically working over the field $\mathbb{C}$.

2. Linear forms and the algebraic dual

Let $E$ be a vector space over $\C$. We define its algebraic dual $E^*$ as the set of linear forms on $E$, i.e. linear maps $\varphi : E \to \C.$

In finite dimension $n$, fixing a basis $(e_1,\dots,e_n)$ of $E$, one can construct a corresponding family of linear forms $(\theta_1,\dots,\theta_n)$ in $E^*$, defined by the relation:

One then shows that this family is linearly independent and generating, and therefore a basis of $E^*$ (called the dual basis). One deduces that $\dim(E^*) = \dim(E)$, which implies that $E$ and $E^*$ are isomorphic. A possible isomorphism is the map $T$ that associates with each vector $x = \sum x_i e_i$ the linear form $w_x = \sum x_i \theta_i$.

However, this isomorphism $T$ is, so to speak, artificial, in the sense that it depends on the initial choice of a basis of $E$. If one changes basis, the map $T$ also changes. Hence there is no canonical (natural) identification between a vector space and its algebraic dual.

In infinite dimension, a classical result shows that $E$ is never isomorphic to its algebraic dual, since the latter has a strictly larger dimension (cf. the Erdös-Kaplansky theorem). For example, if $E$ has countable algebraic dimension, its algebraic dual has uncountable algebraic dimension.

In both cases, then, there is no canonical identification between vectors and linear forms.

3. Topological dual and the Riesz theorem

In a Hilbert space $\H$, the norm associated with the inner product allows one to single out a particular class of linear forms: those that are continuous. One says that $\varphi : \H \to \C$ is continuous at $x_0$ if

and if it is continuous at a point, then by linearity it is continuous everywhere¹. The set of continuous linear forms is a vector subspace of the algebraic dual. It is called the topological dual and is still denoted (by abuse of notation) $\H^*$.

Remark: this norm is a special case of the operator norm that will appear later (cf. the lesson on linear operators). In itself, the above construction is valid for any normed space. What makes the Hilbert case special is the following theorem: one uses the inner product to construct a bijective identification of every continuous linear form with a vector:

We note that the map $\Phi : u \mapsto \varphi_u = \langle u, \cdot \rangle$ from $\mathcal{H}$ to $\H^*$ is injective even in a space that is merely pre-Hilbert. Indeed, if $\varphi_u = \varphi_v$, then for every $x$ in $\mathcal{H}$ we have $0 = \varphi_u(x) - \varphi_v(x) = \langle u-v , x \rangle$, and hence $u = v$ by choosing $x = u - v$ and using the positive definiteness of the inner product.

The Riesz theorem then asserts that this map is also surjective. This is not trivial, and in fact only holds for Hilbert spaces. For instance, in an incomplete pre-Hilbert space the map is not surjective, since there exist continuous linear forms that cannot be written as an inner product with a vector of the incomplete space.

The Riesz theorem thus provides the bijection $\Phi$ between $\mathcal{H}$ and $\mathcal{H}^*$ that we were seeking. We deduce the following corollary:

4. Dirac notation: bras, kets, brackets

Dirac notation is nothing other than a rewriting of the Riesz isomorphism. We first define kets, bras, and brackets. We then use them to rewrite the decomposition on a Hilbert basis seen in the previous lesson.

1. Kets. A vector $u \in \mathcal{H}$ is written in the form of a ket, denoted as follows: $$\boxed{ u \in \mathcal{H} \quad \longleftrightarrow \quad \ket{u} }$$

   The linear structure gives us the following calculation rules:

   $$\begin{aligned} \ket{u+v} &= \ket{u} + \ket{v} \\ \ket{\lambda u} &= \lambda \ket{u} \end{aligned}$$
2. Bras. To every vector $u \in \H$ we associate, via $\Phi$, a form $\varphi_u$. We write this linear form as a bra, placing the associated vector inside the bra: $$\boxed{ \varphi_{u} = \Phi(u) \in \mathcal{H}^* \quad \longleftrightarrow \quad \bra{u} }$$

   Thus $\bra{u}$ represents the vector $u$ seen as a linear form, i.e. the action "take the inner product with $u$ and return a number". We have the following calculation rules:

   $$\begin{aligned} \bra{u+v} &= \bra{u} + \bra{v} \\ \bra{\lambda u} &= \lambda^* \bra{u}, \end{aligned}$$ as a consequence of the antilinearity of $\Phi$.
3. The bracket. The two notational devices above allow one to form an inner product, which is called a bracket in English, as a product of a bra with a ket: $$\boxed{\langle u, x \rangle \equiv \braket{u}{x}}$$

   This notation in fact reflects the action of the linear form $\bra{u}$ on the vector $\ket{x}$:

   $$\bra{u}\left(\ket{x}\right) = \varphi_u(x) = \langle u, x \rangle = \braket{u}{x}.$$
4. Hilbert decomposition. Regardless of the dimension of the Hilbert space, we have seen the decomposition formula $u = \sum_{i \in I} \langle e_i, u \rangle e_i$, where the sum is finite or converges in $\mathcal{H}$ if $I$ is infinite. In Dirac notation it is written:

   $$\boxed{ \ket{u} = \sum_{i \in I} \braket{e_i}{u} \ket{e_i} = \sum_{i \in I} u_i \ket{e_i} }$$

   (1)

   where the $u_i$ are the components of the ket $u$ in the basis. Note that they are obtained by orthogonal projection onto $e_i$ in the form $u_i = \braket{e_i}{u}$, and not in the form $\braket{u}{e_i}$, which equals $u_i^*$. This is a common mistake, probably linked to the fact that in ordinary vector calculus on $\R^n$, one recovers the component $v_i$ of a vector $\vec{v}$ via $v_i = \vec{v} \cdot \vec{e}_i$, which might suggest $u_i = \braket{u}{e_i}$ in the quantum case. But this would ignore the sesquilinearity of the inner product on $\C$ and would immediately lead to a computational error.

   For bras, we have:

   $$\boxed{ \bra{u} = \sum_{i \in I} \braket{u}{e_i} \bra{e_i} = \sum_{i \in I} u_i^* \bra{e_i} }$$

   (2)

   and the squared norm reads

   $$\|u\|^2 = \braket{u}{u} = \sum_{i \in I} |u_i|^2 = \sum_{i \in I} u_i^* u_i$$

   (3)
   where the second equality is Parseval's identity.
5. The "dagger" operation. Applying the isomorphism $\Phi$ (from kets to bras) is usually denoted by the superscript symbol $\dagger$. We choose to write the inverse isomorphism $\Phi^{-1}$ from bras to kets using the same symbol. This makes the dagger operation involutive. By notational convention, we therefore have: $$\begin{aligned} \bra{u} &= \ket{u}^\dagger \quad \text{(map } \Phi)\\ \ket{u} &= \bra{u}^\dagger \quad \text{(map } \Phi^{-1})\\ \left(\ket{u}^\dagger\right)^\dagger &= \ket{u} \quad \text{(involution)} \end{aligned}$$

Let us illustrate these various points in finite dimension. Take $\H$ of dimension $n$ with a Hilbert basis $\mathcal{B} = \left(\ket{e_i}\right)_{i=1}^n$ equipped with its canonical inner product (cf. Lesson 1, the Hilbert space $\C^n$). Representing the basis kets canonically as column vectors, i.e. as $(n,1)$ matrices:

where the $1$ is in the $i$-th position, every ket $\ket{u}$ is written as a column vector

with $u_i = \braket{e_i}{u}$. The linear form $\varphi_u$ associated with the vector $u$ must satisfy, for every vector $v$:

To obtain this sum, $\bra{u}$ must therefore be represented by the $(1, n)$ row vector of the complex conjugate coordinates of $u$:

since then the inner product $\braket{u}{v}$ is obtained as the usual matrix product:

Thus, for example, in dimension 2:

We have obtained the result:

Caution: this is meaningless in infinite dimension, because the transpose is not defined. That said, in the space $\ell^2(\N)$ and in its canonical basis, everything proceeds in a relatively similar way, considering infinite column or row vectors and replacing finite sums with convergent series. This can be useful for building intuition, but strictly speaking these are neither matrices nor transpositions.

5. Operators, matrix elements, resolution of the identity

We continue our presentation of Dirac notation, this time also introducing linear maps (also called linear operators) $\hat A$ from $\mathcal{H}$ to $\mathcal{G}$, where $\mathcal{G}$ is another Hilbert space.

Writing operators. Note first that it is standard in physics to denote them with a hat. An operator $\hat A$ acting on a vector $v$ yields the vector $\hat A v$. In Dirac notation, we therefore have two equivalent ways of writing this, the second being the more commonly used:

In the same way, an inner product involving an operator can be written as:

where the second notation is more frequent for aesthetic reasons.

Matrix elements. In the formula above, the case $\bra{w} = \bra{e_i}$ and $\ket{u} = \ket{e_j}$ is particularly important, as it defines what are called the matrix elements of the operator $\hat A$:

In infinite dimension, this expression does not always make sense, since $\ket{e_i}$ must still lie in the domain of definition of the operator $\hat A$. We will return to this. In finite dimension, on the other hand, this definition is straightforward: through the representation of basis vectors as column vectors, a linear operator is in one-to-one correspondence with its representing matrix; we therefore write $\hat A \;\longleftrightarrow\; A = (A_{ij})_{1 \le i,j \le n}$ with $A_{ij} = \bra{e_i}\hat{A}\ket{e_j}$. The equation $\ket{v} = \hat A \ket{u}$ above then reads as the usual matrix product $\hat A \ket{u} = \sum_{i} \sum_j (A_{ij} u_j) \ket{e_i}$, while an inner product reads $\bra{w} \hat A \ket{u} = \transpose{\left(w^*\right)} A u = \sum_{ij} w_i^* A_{ij} u_j \in \C$.

Ket-bra operators. A bracket is a number, but a ket-bra is an operator, called an outer product (unrelated to the exterior product of differential forms). Given two vectors $\ket{u}$ and $\ket{v}$ of the same Hilbert space $\H$, one associates the operator $\ket{u}\bra{v}$ from $\H$ to $\H$ defined by:

Here too, in finite dimension, the case $\hat{E}_{ij} \equiv \ket{e_i}\bra{e_j}$ is particularly important. It is the operator whose matrix elements are all zero except for a "1" in row $i$ and column $j$; its associated matrix is therefore:

In finite dimension, every linear operator $\hat{A}$ decomposes according to its matrix elements:

where $A_{ij} = \bra{e_i}\hat{A}\ket{e_j}$. Caution: in infinite dimension this decomposition is not always possible; and when it does exist, it must be understood in the sense of strong convergence. We will return to this in subsequent lessons.

Resolution of the identity operator. The identity operator in $\mathcal{H}$, denoted $\mathbf{1}$, is defined by $\mathbf{1}\ket{x} = \ket{x}$ for every $\ket{x} \in \mathcal{H}$, trivially. In finite dimension it is, obviously, represented by the identity matrix. More generally, we have the formula

also valid in countably infinite dimension, in which case the series converges strongly. It is also called the resolution of the identity or the closure relation. This formula is extremely useful for all computations in quantum mechanics. Caution: it is false in non-separable spaces².

6. The adjoint operator

Let us return to the Riesz isomorphism and consider a continuous linear operator³ $\hat{A}$ from $\mathcal{H}$ to $\mathcal{G}$, where $\H$ and $\G$ are two Hilbert spaces, possibly with $\H = \G$.

When $\hat{A}$ acts on a ket $\ket{u} \in \mathcal{H}$, we obtain a new ket $\ket{v} = | \hat{A} u \rangle = \hat{A}\ket{u} \in \mathcal{G}$. It is natural to ask what the bra associated via the Riesz isomorphism with the vector $\ket{v}$ is, i.e. what the linear form $\bra{v} = \langle \hat{A}u |$ on $\G$ is. This will lead us naturally to consider a new operator, called the adjoint of $\hat{A}$ and likewise denoted $\hat{A}^\dagger$. This operator plays an essential role in quantum mechanics.

To determine $\bra{v} = \langle \hat{A}u |$, fix any ket $\ket{w} \in \mathcal{G}$ and consider the linear form on $\H$:

(Above, the subscript $\G$ indicates in which space the inner product is to be taken.) We will take for granted here that $\varphi$ is a continuous linear form on $\mathcal{H}$ when $\hat{A}$ is a continuous operator. By the Riesz theorem applied to $\mathcal{H}$, there then exists a unique vector $\ket{z} \in \mathcal{H}$ such that $\varphi = \varphi_z$, that is:

The above construction has thus allowed us to associate with a vector $\ket{w}$ of $\G$ a unique vector $\ket{z}$ of $\H$. This operation can formally be written as the action of an operator, called the adjoint operator and denoted $\hat{A}^\dagger$, in the form $\ket{z} = \hat{A}^\dagger \ket{w}$. Note that the operator $\hat A$ acts from $\H$ to $\G$, while its adjoint acts from $\G$ to $\H$.

Completing this construction for every $\ket{w}$, one verifies without difficulty that the operator thus defined is itself a continuous linear operator. We then obtain the following fundamental result:

The adjoint has two practical uses in formal quantum calculations.

1. The formula above shows that, so to speak, "with the positions of $u$ and $w$ fixed, the adjoint makes it possible to move the operator from the left-hand side to the right-hand side".
2. Using the Hermitian symmetry of the inner product, $\braket{\smash{\hat{A}} u}{w}_{\mathcal{G}} = \braket{w}{\smash{\hat{A}} u}_{\mathcal{G}}^*$, we also obtain:
   $$\boxed{\forall\, \ket{u} \in \mathcal{H},\ \forall\,\ket{w} \in \mathcal{G}, \quad \bra{w} \hat A \ket{u}_{\G}^* = \bra{u}\hat{A}^\dagger \ket{w}_{\mathcal{H}}}$$

   (7)

   which shows this time that "the adjoint lets one swap bra and ket up to a complex conjugate".

These two identities are equivalent. One must of course pay attention to the source and target spaces $\H$ and $\G$, and to which of the two inner products—that of $\H$ or that of $\G$—is being used. In practice, however, we will almost always have $\H = \G$, and the notation simplifies accordingly.

Let us now illustrate the above construction in the finite-dimensional case. We have the following main result:

We have not yet explicitly answered the question: what is the bra $\bra{v} = \bra{\smash{\hat A} u}$ canonically associated with the ket $\ket{v} = \ket{\smash{\hat A} u}$? Formula (8) answers this via an inner product valid for every vector $w$, which allows us to write:

where the conclusion is obtained by noting that two linear forms that coincide on every vector $w$ are necessarily equal. It is, however, crucial to clearly understand what this object is. Indeed, a common misconception is to infer from this notation that "the adjoint acts on the bras from the left". This is incorrect! The object $\bra{u} \hat{A}^\dagger$ is in fact a composition of operators, not a left action. Indeed, the source and target spaces are:

so that $\bra{u} \hat A^\dagger$ is the composition $\bra{u} \circ \hat A^\dagger \, : \, \mathcal{G} \to \mathcal{H} \to \mathbb{C}$, which is indeed a linear form on $\mathcal{G}$. The $\circ$ symbol is usually omitted, which can be confusing.

Finally, from $\bra{\smash{\hat{A}} u} = \bra{u} \hat A^\dagger$, by taking the dagger we deduce the useful formulas:

The following figure summarizes the relations between bras, kets, the operator $\hat A$ and its adjoint seen in this lesson.

We close this section with a few important properties of the adjoint operation: