Adjoints and inner products

Dual vector spaces¶

Definition¶

Let $V$ be a vector space over $\CC$. the dual vector space $V^*$ is the space of all linear maps $f:V \to \CC$.

Properties and notation¶

1. $V^*$ is a vector space.

Consider linear maps $f_{1,2}$, and $a,b \in \CC$. Then we can define a linear map $a f_1 + b f_2$ by their action on a vector $\ket{v}$.

One can show that this defines a linear map: for any $c,d\in \CC$ and $\ket{v_{1,2}} \in V$,

which follows from $f_{1,2}$ being linear maps.

2. $\text{dim}(V^*) = \text{dim}(V)$. It is instructive to show this. Consider a basis $\ket{i}$ of $V$, $i = 1,\ldots d = \text{dim}(V)$. A general vector $\ket{v}$ can be expressed as

for a unique set of coefficients $c_i \in \CC$. Now

Thus, the map $f$ is completely specified by $d$ complex numbers $f(\ket{i}) = c_i \in \CC$. Thus, if we define $f_i$ by $f_i(\ket{j}) = \delta_{ij}$, we can show that each $f_i$ is a linear map. Furthermore, $f_i$ is linearly independent of $f_{j \neq i}$ (you should convince yourself of this).

Any function $f$ can always be written as $f = \sum_{i = 1}^d c_i f_i$, so this basis is maximal, and $f_i$ form a complete basis for $V^*$. There are $d$ such independent basis functions, so $\text{dim}(V^*) = d$.

3. Notation. We can express $f \in V^*$ as a “bra vector” $\bra{f}$. We then call elements $\ket{v} \in V$ “ket vectors”. We can then write

as a “bra(c)ket”. I didn’t do this, please blame Dirac. Anyhow the notation is unfortunately standard. With this notation we can define the linear structure of $V^*$ as

4. Finally, The dual vector space for $V^*$ is $V$, or $(V^*)^* = V$: for any $\ket{v}$ the map from $V^* \to \CC$ is just $\ket{v}: f \to \brket{f}{v}$.

Example¶

If $V = \CC^3$ is represented as the space of column vectors. we can represent $V^*$ as the set of row vectors. That is, consider a basis

We can define any linear map $f$ by $f(\ket{i} = c_i$. Then if $\ket{v} = \sum_i a_i \ket{i}$,

Adjoint maps¶

Since $\dim V = \dim V^*$, we expect that there is an isomorphism (a map that is one-to-one and onto) between them. Choosing such a map leads to a choice of “inner product” on $V$ itself: a way of assigning to $\ket{v}$ a number corresponding to some notion of its length.

Definition¶

Let $V$ be a vector space over $\CC$. An adjoint map is a map $\cal{A}: V \to V^*$, which we denote by $A\ket{v} \equiv \bra{f_v}$ with the properties

1. Skew symmetry: $\brket{f_w}{v} = \brket{f_v}{w}^*$.

2. Positive semi-definiteness:

In general we write $\bra{f_v} = \bra{v}$.

Properties¶

1. Antilinearity. Using skew-symmetry you can show that for $\ket{v_{1,2}} \in V$, $a, b \in \CC$,

2. Schwarz inequality

3. Triangle inequality

Examples¶

1. $V = \CC^3$.

If

then

2. $V = M_2(\CC)$.

3. $V = L^2(\CR)$, the space of complex square-integrable functions on the real line where $\ket{\psi}$ is represented by the function $\psi(x)$. A good inner product, which defines an adjoint map, is

Additional definitions and a comment¶

1. $\brket{v}{w}$ is the inner product of $\ket{v}$, $\ket{w}$.

2. $||v||^2 = \brket{v}{v}$ is called the norm of $\ket{v}$.

3. $V$ with an adjoint map is called an inner product space.

4. An inner product space (over $\CC$) is called a Hilbert space if either:

• $\text{dim}(V) < \infty$, or

• Cauchy sequences in $V$ are complete.

To explain the last possibility, note that $\ket{v_i}$, $i = 1,\ldots,\infty$ is a Cauchy sequence if for any $\eps > 0$, there exists some integer $N$ such that

Such a sequence is complete if it converges to a vector in $V$.

4. There is no unique adjoint map.

Actions of operators¶

Given a linear operator $A$ and $\ket{v} \in V$, $A\ket{v}$ is a vector and $\bra{w} A \ket{v}$ is a complex number. We can therefore define $\bra{A w} \equiv \bra{w} A$ such that $\brket{A w}{v} = \bra{w} A \ket{v}$.

Orthonormal bases¶

Definitions¶

Let $V$ be a vector space over $\CC$.

1. $\ket{v} \in V$ is a normal vector if $||v||^2 = \brket{v}{v} = 1$.

2. $\ket{v},\ket{w} \in V$ are orthogonal if $\brket{v}{w} = 0$.

3. An orthonormal basis is a basis $\ket{i} \in V$, $i = 1,\ldots,d = \text{dim} V$ such that for $\cal{A}: \ket{i} \to \bra{i}, \brket{i}{j} = \delta_{ij}$.

Examples¶

1. We can write $\ket{v} = \sum_i v_i \ket{i}$; the antilienarity of the adjoint map means that $\bra{v} = \sum_i \bra{i} v^*_i$. This means that

Similarly, for $\ket{w} = \sum_i w_i \ket{i}$,

This works if we identify

and thus

The basis element $\ket{i}$ is a column vector with all zeros except a 1 in the $i$th row.

2. If $V = M_2(\CC)$, the space of $2\times 2$ complex matrices, a natural inner product is

where $m,n$ are $2\times 2$ matrices. This clearly defines an adjoint map from $\ket{n}$ to a linear map. An orthonormal basis is:

3. Consider the vector space of complex functions on the interval $0,L$ with Dirichlet boundary condittions. You can convince yourself that the basis

is orthonormal with respect to the inner product (17)

The Gram-Schmidt machine¶

Theorem: every finite-dimensional vector space or infinite dimensional vector space with a countable basis has an orthonormal basis.

Proof (partial): Given a basis $\ket{v_1},\ket{v_2},\ldots,\ket{v_d}$, we can construct a basis iteratively. Define

Matrix elements of operators¶

Since $\ket{i}$ is a basis, we can write the action of operators in this basis: $A\ket{j} = A_{ij}\ket{i}$. As notation, we will sometimes write

We understand this to mean

where $\ket{v} = \sum_i v_i \ket{i}$, and for dual vectors $\bra{v} = \sum_i \bra{i} v_i^*$,

Thus

A particularly important example is the identity operator $\bf{1}$ for which $\bf{1}_{ij} = \delta_{ij}$. This can be represented as above by:

for any orthonormal basis. This is called a resolution of the identity, associated to a given basis.

In this basis, an important operator on $A$ is the transpose. That is given a linear operator $A$, we can define the transpose $A^T$ via its matrix elements

In particular, we can write

Adjoints of operators¶

The vector $A\ket{v} \equiv \ket{Av} = A_{lk} v_k \ket{l}$ has a natural adjoint

which defines the Hermitian conjugate $A^{\dagger}$. We can either define it as ${\cal A}: A\ket{v} \to \bra{a} A^{\dagger}$ or via its matrix elements in an orthonormal basis,

Hermitian and unitary operators¶

1. Definition. A Hermitian operator is an operator $A = A^{\dagger}$.

Note that this does not mean the operator has real matrix elements. The following operator on $\CC^2$ is Hermitian:

2. Definition. A Unitary operator is an operator $U$ such that $U^{\dagger} = U^{-1}$.

An important property of this operator is that it is norm-preserving:

1. An example of a unitary operator acting on $\CC^2$:

As we will discuss, this implements rotations on the spin components of a spin-$\half$ system.

2. Two nontrivial Hermitian examples for $L^2(\CR)$:

• The position operator ${\hat x}: \psi(x) \to x \psi(x)$. Since

as expected for a Hermitian operator.

• the operator $\hat{p} = - i\hbar \frac{\del}{\del x}$, when acting on $\psi(x)$.

The second line follows from integration by parts, and the boundary terms vanish because $\psi$ is square integrable. In other words for every $\ket{\psi},\ket{\chi}$, $\bra{\chi} {\hat p} \ket{\psi} = \bra{\chi} {\hat p}^{\dagger} \ket{\psi}$. From this we can deduce that ${\hat p} = {\hat p}^{\dagger}$.

The same argument follows for the case of complex functions with periodic boundary conditions. For Dirichlet boundary conditions, ${\hat p}$ fails to be an operator on teh Hilbert space, as the derivative of a function with Dirichlet boundary conditions does not in general satisfy Dirichlet boundary conditions. (Similarly for Neumann boundary conditions).