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Hilbert Spaces and Dirac Notation

Hilbert space structure, topological dual, and Dirac notation in finite and infinite dimension.

Dirac Notation — Finite Dimension

Algebraic rules of quantum calculation valid in any dimension, then the matrix reference sheet in finite dimension.

Dirac notationHermitian inner productAdjointKet-bra operatorResolution of the identityFinite dimensionMatrix elementsTraceConjugate transposeReference sheet

The objects we have just defined obey a set of fundamental properties, always valid whatever the dimension of the Hilbert space, and independently of whether the operators are bounded or not. In the latter case one can still define an adjoint, but it is mathematically more subtle, and we shall return to it in the next theme, devoted to the theory of linear operators. Since the whole planet uses Dirac notation, these properties are in effect the algebraic rules of quantum calculation, and are therefore imperative to master.

This lesson gathers together all the useful formulas seen so far, even at the cost of repeating them, and adds a few more. We first give those that hold in any dimension, then the formulary specific to finite dimension, where Dirac notation reduces to an explicit matrix calculation.

1. Calculation rules in any dimension

Properties of the inner product.

Let (φ,ψ)H2(\ket{\phi}, \ket{\psi}) \in \mathcal{H}^2 be two vectors of the Hilbert space and (λ,μ)C2(\lambda, \mu) \in \mathbb{C}^2 scalars. The inner product satisfies the following properties:

  1. Hermitian symmetry:

    φ|ψ=ψ|φ.\boxed{\braket{\phi}{\psi} = \braket{\psi}{\phi}^*.}

    (1)

  2. Antilinearity on the left: the bra associated with λψ\lambda \ket{\psi} is λψ\lambda^* \bra{\psi}, that is:

    (λψ)=λψandλψ|φ=λψ|φ\boxed{(\lambda \ket{\psi})^\dagger = \lambda^* \bra{\psi} \qquad \textrm{and} \qquad \braket{\lambda \psi}{\phi} = \lambda^* \braket{\psi}{\phi} }

    (2)

  3. Linearity on the right:

    φ|λψ=λφ|ψ.\boxed{\braket{\phi}{\lambda \psi} = \lambda \braket{\phi}{\psi}.}

    (3)

  4. Linear combinations: the preceding properties allow one to expand in the usual way complicated expressions such as the following: for all α,β,γ,δC\alpha, \beta, \gamma, \delta \in \C, one has:

    αφ+βψ|γφ+δψ=αγφ|φ+αδφ|ψ+βγψ|φ+βδψ|ψ.\boxed{ \braket{\alpha \phi + \beta \psi}{\gamma \phi + \delta \psi} = \alpha^* \gamma \braket{\phi}{\phi} + \alpha^* \delta \braket{\phi}{\psi} + \beta^* \gamma \braket{\psi}{\phi} + \beta^* \delta \braket{\psi}{\psi}. }

    (4)

  5. Norm: by definition,

    ψ2=ψ|ψ0,\boxed{\|\ket{\psi}\|^2 = \braket{\psi}{\psi} \ge 0,}

    (5)

    with equality if and only if ψ=0\ket{\psi} = 0 is the null vector (sometimes, but rarely, written \ket{\varnothing}).

Properties of the adjoint.

We rewrite here the two formulas that serve as definitions:

(A^φ)=A^φ=φA^\boxed{ (\hat A\ket{\phi})^\dagger = \bra{\smash{\hat{A}} \phi} = \bra{\phi} \hat A^\dagger }

(6)

and

A^φ|ψ=φ|A^ψ=φA^ψ\boxed{ \braket{\smash{\hat{A}} \phi}{\psi} = \braket{\phi}{\smash{\hat{A}}^\dagger \psi} = \bra{\phi} \hat A^\dagger \ket{\psi} }

(7)

which also allows one to prove, via Hermitian symmetry, that:

φA^ψ=ψA^φ\boxed{ \bra{\phi} \hat A^\dagger \ket{\psi}^* = \bra{\psi} \hat A \ket{\phi} }

(8)

Properties of adjoint operators.

For all continuous linear operators A^,B^\hat{A}, \hat{B} on H\mathcal{H} and every scalar λC\lambda \in \mathbb{C}, one has:

(A^+B^)=A^+B^,(λA^)=λA^,(anti-linearity)(A^B^)=B^A^,(mind the order)(A^)=A^,(involution).\begin{aligned} (\hat{A} + \hat{B})^\dagger &= \hat{A}^\dagger + \hat{B}^\dagger, \\ (\lambda \hat{A})^\dagger &= \lambda^* \hat{A}^\dagger, \quad \quad \, \text{(anti-linearity)} \\ (\hat{A}\hat{B})^\dagger &= \hat{B}^\dagger \hat{A}^\dagger, \quad \quad \text{(mind the order)} \\ (\hat{A}^\dagger)^\dagger &= \hat{A}, \qquad \quad \, \,\, \, \text{(involution)}. \end{aligned}
Ket-bra operator.

A bracket is a number, but a ket-bra is an operator, called the outer product (unrelated to the exterior product of differential forms). To two vectors φ\ket{\phi} and ψ\ket{\psi}, one associates the operator φψ\ket{\phi}\bra{\psi} defined by:

φψ:HH,χφψ|χ.\begin{aligned} \ket{\phi}\bra{\psi} \quad : \quad &\mathcal{H} \to \mathcal{H}, \\ &\ket{\chi} \mapsto \ket{\phi}\braket{\psi}{\chi}. \end{aligned}
Decomposition of the identity operator.

The identity operator on H\Hilb is written 1\mathbf{1}, trivially defined by 1x=x,xH\mathbf{1} x = x, \forall x \in \Hilb. In a countable Hilbert basis {ei}iI\{\ket{e_i}\}_{i \in I} where II is either finite or countably infinite, the identity operator 1\mathbf{1} is written:

1=iIeiei\boxed{ \mathbf{1} = \sum_{i \in I} \ket{e_i}\bra{e_i} }

(9)

This formula is extremely useful in practice. It is also called the resolution of the identity or the closure relation. The right-hand side is a sum in finite dimension, or a convergent series in countably infinite dimension. It is false in the uncountable-dimensional case.

Hilbert decomposition.

The preceding formula allows one, for example, to recover directly the decomposition theorem seen in lesson 1, by defining ψi\psi_i as the component of the ket ψ\kpsi on ei\ket{e_i}:

ψ=1ψ=iIei|ψei=iIψiei\boxed{ \ket{\psi} = \mathbf{1} \kpsi = \sum_{i \in I} \braket{e_i}{\psi} \ket{e_i} = \sum_{i \in I} \psi_i \ket{e_i} }

(10)

This is a sum or a series depending on the dimension. The decomposition of the ket has its counterpart for bras:

ψ=ψ1=iIψiei\boxed{ \bra{\psi} = \bra{\psi} \mathbf{1} = \sum_{i \in I} \psi_i^* \bra{e_i} }

(11)

The norm is then written:

ψ2=ψ|ψ=iIψiψi=iIψi2\boxed{\|\ket{\psi}\|^2 = \braket{\psi}{\psi} = \sum_{i \in I} \psi_i^* \psi_i = \sum_{i \in I} |\psi_i|^2 }

(12)

Remark 1 (Mind the order!)
The components of the ket are ψi=ei|ψ\psi_i = \braket{e_i}{\psi}, and not ψ|ei\braket{\psi}{e_i}, which equals ψi\psi_i^*. The Hermitian inner product is not commutative, so one must pay attention to the order. In Rn\mathbb{R}^n equipped with the usual Euclidean inner product, one recovers the component viv_i of a vector v\vec{v} by vi=veiv_i = \vec{v} \cdot \vec{e}_i, which might suggest that the quantum formula would be ψi=ψ|ei\psi_i = \braket{\psi}{e_i}, but this is not the case!
A few particular operators.

The next theme presents in detail the theory of linear operators on a Hilbert space. Nevertheless we note for what follows a few very common cases. Among the operators from HH\Hilb \to \Hilb, we shall have in particular an operator:

  1. A^\hat A, Hermitian or self-adjoint if A^=A^\hat{A}^\dagger = \hat{A},
  2. U^\hat U, unitary if U^=U^1\hat{U}^\dagger = \hat{U}^{-1},
  3. P^\hat P, a projector if P^2=P^\hat{P}^2 = \hat{P},
  4. P^\hat P, an orthogonal projector: a self-adjoint projector.

Unitary operators are bijective, linear, and preserve the inner product. We have therefore already met them above: they are the isometric isomorphisms (in this instance the isometric automorphisms, since we are speaking here only of linear operators from H\Hilb into itself).

2. Formulary in finite dimension

One can further specify the rules of quantum calculation when the Hilbert space is of finite dimension nn. To begin with, all the formulas of the previous section are valid, where the sums encountered there are to be written as finite sums: ii=1n\sum_i \longrightarrow \sum_{i=1}^{n}.

Moreover, once an orthonormal basis B=(ei)i=1n\mathcal{B} = \left(\ket{e_i}\right)_{i=1}^n has been chosen, Dirac notation corresponds to an explicit matrix calculation, which we now detail (beware, what follows makes no sense at all in infinite dimension).

Kets as column vectors.

By canonically representing the kets of the basis B\mathcal{B} by (n,1)(n,1) matrices, also called column vectors:

ei=(00100)B\ket{e_i} = \begin{pmatrix} 0 \\ \vdots \\ 0 \\ 1 \\ 0 \\ \vdots \\ 0 \end{pmatrix}_\mathcal{B}

where the 11 is in ii-th position, and via the algebraic decomposition:

v=i=1nviei,\ket{v} = \sum_{i=1}^n v_i \ket{e_i},

one obtains the writing of all kets as column vectors: for vH\ket{v} \in \mathcal{H}, one writes:

v=(v1v2vn)BCn,\ket{v} = \begin{pmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{pmatrix}_\mathcal{B} \in \mathbb{C}^n,

where, unless necessary, we shall omit the mention of the basis B\mathcal{B} in which this expansion is made. Moreover, the components are obtained by orthogonal projections:

vi=ei|v\boxed{v_i = \braket{e_i}{v}}

(13)

Bras as conjugate row vectors.

The linear form φu\varphi_u associated with the vector uu must satisfy, for every vector vv:

φu(v)=u,v=i=1nuivi,\varphi_u(v) = \langle u, v \rangle = \sum_{i=1}^n u_i^* v_i,

by antilinearity of the Hermitian inner product on the left. To obtain this sum as a matrix product, u\bra{u} must be the row vector of the conjugate coordinates of uu:

u=(u1,u2,,un),\begin{aligned} \bra{u} = \begin{pmatrix} u_1^*, & u_2^*, & \cdots, & u_n^* \end{pmatrix}, \end{aligned}

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

u|v=uv=(u1,u2,,un)(v1v2vn)=i=1nuiviC.\begin{aligned} \braket{u}{v} = \bra{u} \cdot \ket{v} = \begin{pmatrix} u_1^*, & u_2^*, & \cdots, & u_n^* \end{pmatrix} \cdot \begin{pmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{pmatrix} = \sum_{i=1}^n u_i^* v_i \in \mathbb{C}. \end{aligned}

One then notes that the bra u\bra{u} is the conjugate transpose of the ket u\ket{u}. The dagger operation therefore amounts, in finite dimension, to taking the conjugate transpose:

u=u=u\boxed{\ket{u}^\dagger = \transpose{\ket{u}^*} = \bra{u}}

(14)

This is false in infinite dimension, and would not even make sense, since the transpose is not defined1.

Note 1: At a stretch, in the space 2(N)\ell^2(\N), and in its canonical basis which we detailed in Section 3.2 (Theme 2, Lesson 1), everything happens in a relatively similar way by considering infinite column or row vectors, and by replacing finite sums with convergent series. This can be useful for building intuition, but strictly speaking these are not matrices. In L2(R)L^2(\R), it no longer makes any sense at all, even though an integral analogue exists there too, see the next lesson.
Matrix representation of operators.

In an orthonormal basis {ei}i=1n\{\ket{e_i}\}_{i=1}^n in finite dimension, every linear operator A^\hat{A} decomposes as:

A^=i,j=1nAijeiej,\boxed{ \hat{A} = \sum_{i,j=1}^n A_{ij}\,\ket{e_i}\bra{e_j}, }

(15)

which leads, in finite dimension, to the natural identification between the linear operator A^\hat{A} (written with a hat), and its matrix AA (without a hat). The formula above is indeed the counterpart, in the Dirac formalism, of the matrix expansion A=ijAijEijA = \sum_{ij} A_{ij} E_{ij}, where EijE_{ij} is the elementary matrix with a 11 in position (i,j)(i,j) and zeros elsewhere. Here, EijE_{ij} is therefore the matrix representing the operator eiej\ket{e_i} \bra{e_j}. The coefficients AijA_{ij} are naturally called the matrix elements of the operator A^\hat{A} in the basis {ei}\{\ket{e_i}\}. They equal:

Aij=ei|A^ejA_{ij} = \braket{e_i}{\smash{\hat{A}} e_j}

but for aesthetic reasons one will more often write them in the form:

Aij=eiA^ej.\boxed{A_{ij} = \bra{e_i}\hat{A}\ket{e_j}.}

(16)

The identity operator has of course the identity matrix for its matrix, and its matrix elements are the δij\delta_{ij} (Kronecker symbol).

Trace of an operator.

We shall often be led in what follows to consider the trace of an operator A^\hat{A}. In Dirac notation it is computed by:

Tr(A^)=i=1neiA^ei=i=1nAii.\boxed{\mathrm{Tr}(\hat{A}) = \sum_{i=1}^n \bra{e_i}\hat{A}\ket{e_i} = \sum_{i=1}^n A_{ii}.}

(17)

This expression is independent of the choice of orthonormal basis.

Adjoint and conjugate transpose.

In finite dimension, the matrix representing the adjoint A^\hat{A}^\dagger is the conjugate transpose of the matrix representing the operator A^\hat{A}:

A=(A).\boxed{A^\dagger = \transpose{\left(A^*\right)}.}

(18)

We propose a quick proof here, because it shows how the calculation rules seen above are used in practice. The matrix elements of A^\hat{A}^\dagger are by definition:

(A^)ij=ei|A^ej.(\hat{A}^\dagger)_{ij} = \braket{e_i}{\smash{\hat{A}}^\dagger e_j}.

But, by definition of the adjoint, one has:

ei|A^ej=A^ei|ej\braket{e_i}{\smash{\hat{A}}^\dagger e_j} = \braket{\smash{\hat{A}} e_i}{e_j}

Using Hermitian symmetry, one has A^ei|ej=ej|A^ei\braket{\smash{\hat{A}} e_i}{e_j} = \braket{e_j}{\smash{\hat{A}} e_i}^*, but this is nothing other than the matrix elements AjiA_{ji}^*. Therefore:

(A)ij=Aji=(A)ij(A^\dagger)_{ij} = A_{ji}^* = \left(\transpose{A^*}\right)_{ij}

which proves the property. In finite dimension, one will therefore be able to check explicitly by a matrix calculation whether an operator is Hermitian, or unitary, etc.

3. References

No references added yet for this lesson.