The Hilbert space of quantum mechanics

Introduction

A system is an object or a collection of objects whose properties are subject to observation and measurement. The state of a given system is the minimum collection of information that completely describes the system. For a classical particle, its state is specified by giving its position and momentum at any given time. The state space of the particle is the two-dimensional phase-space of position and momentum, every point in which describes a state of the particle. On the other hand, the state of a quantum object, such as an atom, is described not in terms of position and momentum but of complex-valued functions that comprise the state space of the quantum object. The state space of a quantum object has the structure of a complex, separable Hilbert space. To describe a quantum object one needs to specify the Hilbert space that is uniquely associated with the system. Here I describe the Hilbert space that is relevant to quantum mechanics.

Vector Space

A complex vector space is a set $V=\{\psi, \phi, \varphi, \dots\}$ , together with a rule that defines the sum (denoted by $+$ ) of elements of $V$ and a rule that defines the product (denoted by $\cdot$ ) of any complex number and any element of $V$ . The sum and the product satisfy the following axioms: For every complex numbers $\alpha, \beta$ and elements $\psi,\varphi,\phi$ in $V$ ,

$\psi + \varphi = \varphi+\psi$ belongs to $V$ ,
$(\psi+\varphi)+\phi=\psi+(\varphi+\phi)$ ,
$\alpha\cdot \psi$ belongs to $V$ ,
$\alpha\cdot(\psi+\varphi)=\alpha\cdot\psi+\alpha\cdot\varphi$
$(\alpha+\beta)\cdot\psi=\alpha\cdot\psi + \beta\cdot\psi$
$(\alpha\beta)\cdot\psi=\alpha\cdot(\beta\cdot\psi)$
$1\cdot\psi=\psi$
There exists a unique element $\theta$ of $V$ such that for all $\psi$ in $V$ the equality $\psi+\theta=\psi$ holds.
To every element $\psi$ of $V$ there exists a unique element $\eta$ of $V$ such that $\psi+\eta=\theta$ . The element $\eta$ is given by $\eta=(-1)\cdot\psi$ , which we denote simply by $-\psi$ .

The ordered triplet $(V,+,\cdot)$ constitutes a complex vector space, and the elements of $V$ are called vectors. More than one pair of rules $(+,\cdot)$ can be assigned to the given set $V$ , with each pair defining a distinct vector space with $V$ . If the pair $(+,\cdot)$ is clear from the outset, the set $V$ can be simply referred to as a vector space.

Example-1. Let $\mathbb{C}^n$ denote the set of all $n$ -tuple complex numbers. A typical element $z$ of this set is given by $z=\{\alpha_1,\alpha_2,\dots,\alpha_n\}$ where the $\alpha_k$ ‘s are arbitrary complex numbers. On its own, $\mathbb{C}^n$ is just a set and not a vector space. It is turned into a vector space by defining the sum of every pair of its elements, and defining the product of its every vector with a complex number that satisfy the above axioms of a vector space. Given two arbitrary elements $z=\{\alpha_1,\alpha_2,\dots,\alpha_n\}$ and $w=\{\beta_1,\beta_2,\dots,\beta_n\}$ of $\mathbb{C}^n$ , we define their sum to be $z+w=\{\alpha_1+\beta_1,\alpha_2+\beta_2,\dots,\alpha_n+\beta_n\}$ , where $\alpha_k+\beta_k$ , for all $k=1,\dots,n$ , denote the usual addition of complex numbers. Moreover, we define the product of any element $z$ and complex number $\lambda$ to be $\lambda \cdot z = \{\lambda \alpha_1,\lambda \alpha_2,\dots,\lambda\alpha_n\}$ , where $\lambda\alpha_k$ denote the usual multiplication of complex numbers.

Using the known properties of complex numbers, it is not difficult to show that all the axioms 1 to 7 of a vector space are satisfied by these definitions. Now the $n$ -tuple of zeros, $z_0=\{0,\dots,0\}$ , belongs to
$\mathbb{C}^n$ , the numeric zero being a complex number itself; moreover, $z_0$ satisfies $z+z_0=z$ for all $z$ . Hence $z_0$ is the zero of $\mathbb{C}^n$ with respect to the given sum. From this, for every $z=\{\alpha_1,\alpha_2,\dots,\alpha_n\}$ we can find a $z'$ in $\mathbb{C}^n$ such that $z+z'=z_0$ ; it is given by $z'=\{-\alpha_1,-\alpha_2,\dots,-\alpha_n\}$ .Then $\mathbb{C}^n$ , together with the defined sum and multiplication, is a complex vector space. $\square$

Example-2. Let $C^2\!(\mathbb{R})$ denote the set of all continuous, complex valued, (Lebesque) square integrable functions $\psi(x)$ in the real line $\mathbb{R}$ ; the last property of the elements of the set means $\int_{-\infty}^{\infty} |\psi(x)|^2 dx<\infty$ . The function $O(x)=0$ for all real $x$ (the function that identically vanishes in the entire real line) belongs to this set. The set $C^2\!(\mathbb{R})$ is a vector space under the usual rule of pointwise addition of functions and the usual rule of multiplication of complex numbers with any complex valued function. That is if $\psi(x)$ and $\phi(x)$ belong to this set, then their sum is defined to be the function $\varphi(x)=\psi(x)+\phi(x)$ , where the plus sign indicate the usual addition of complex numbers.

The sum belongs to the set again. This follows from the well-known inequality, $\sqrt{\int |\psi(x)+\phi(x)|^2 dx} \leq \sqrt{\int |\psi(x)|^2 dx }+ \sqrt{\int |\phi(x)|^2 dx}$ . Since the two terms in the right hand of the side are both finite, the left hand side must be finite, too. That is $\varphi(x)=\psi(x)+\phi(x)$ is itself square integrable. Also for any complex $\lambda$ and function $\psi(x)$ in $C^2(\mathbb{R})$ , the function $\lambda \psi(x)$ is clearly in the set, too. Finally the zero vector of the set is $O(x)$ . The rest of the axioms of a vector space can be easily shown to be satisfied. $\square$

Inner Product Space

A complex inner product space is a complex vector space $V$ together with a binary operation $\left<\cdot|\cdot\right>$ taking pairs of vectors of $V$ into complex numbers such that for every complex
numbers $\alpha$ , $\beta$ and vectors $\psi$ , $\phi$ , $\varphi$ in $V$ the following axioms are satisfied:

$\left<\psi|\phi\right>=\left<\phi|\psi\right>^*$ ,
$\left<\psi|\alpha\phi + \beta\varphi\right>=\alpha\left<\psi|\phi\right>+\beta\left<\psi|\phi\right>$ ,
$0\leq \left<\psi|\psi\right>$ , with $\left<\psi|\psi\right>=0$ if and only if $\psi$ is the zero vector.

The ordered pair $(V,\left<\cdot|\cdot\right>)$ constitutes an inner product space. If from the outset the inner product is clear, the set $V$ can be referred to as an inner product space.

Example-1. For the vector space $\mathbb{C}^n$ , the following inner product can be introduced: For every pair of vectors $z=\{\alpha_1,\alpha_2,\dots,\alpha_n\}$ and $w=\{\beta_1,\beta_2,\dots,\beta_n\}$ , define

(1) $\begin{equation*} \left<z|w\right>=\sum_{k=1}^n \alpha_k^* \beta_k . \end{equation*}$

It is not difficult to show that this binary operation satisfies all the properties of an inner product. $\square$

Example-2. For the vector space $C^2\!(\mathbb{R})$ , the following inner product can be introduced: For every pair of vectors $\psi(x)$ and $\varphi(x)$ , define

(2) $\begin{equation*}\left<\psi|\varphi\right>=\int_{-\infty}^{\infty} \psi^*(x) \varphi(x) dx. \end{equation*}$

From the linear property of the integral, it can shown that this binary operation satisfies all the axioms of an inner product. $\square$ .

Then $\mathbb{C}^n$ and $C^2\!(\mathbb{R})$ , together with their respective inner products defined above, are inner product spaces.

Normed Space

A complex normed space is a complex vector space $V$ together with an operation $||\cdot||$ taking vectors of $V$ such that for every complex number $\alpha$ and vectors $\psi$ , $\phi$ in $V$ the following axioms are satisfied:

$0\leq ||\psi||$ , with $||\psi||=0$ if and only if $\psi$ is the zero vector,
$||\psi+\phi||\leq ||\psi||+||\phi||$ ,
$||\alpha||=|\alpha| \cdot ||\psi||$ .

The ordered pair $(V,||\cdot||)$ is called a complex normed space. The positive number $||\psi||$ is called the norm of the vector $\psi$ . The norm is a generalization of the concept of length in an abstract set. An inner product space can be turned into a normed space by equipping the vector space with the norm $||\psi||=\sqrt{\left<\psi|\psi\right>}$ .

Example-1. The inner product space $\mathbb{C}^n$ is a normed space under the norm induced by the inner product \eqref{c1}. The norm is given by

(3) $\begin{equation*} ||z||=\sqrt{\sum_{k=0}^k |\alpha_k|^2} . \end{equation*}$

. $\square$

Example-2. The inner product space $C_0^2(\mathbb{R})$ is a normed space under the norm induced by the inner product \eqref{c2}. The norm is given by

(4) $\begin{equation*} ||\psi|| = \sqrt{\int_{-\infty}^{\infty} |\psi(x)|^2\, dx} \end{equation*}$

. $\square$

Metric Space

A complex metric space is a complex vector space $V$ together with a binary operation $d(\cdot,\cdot)$ such that for every vectors $\psi$ , $\phi$ , $\varphi$ in $V$ the following axioms are satisfied:

$d(\phi,\psi)\geq 0$ , with $d(\psi,\phi)=0$ if and only if $\psi=\phi$ ,
$d(\psi,\phi)=d(\phi,\psi)$ ,
$d(\psi,\phi)\leq d(\psi,\varphi)+d(\varphi,\phi)$ .

The ordered set $(V,d(\cdot,\cdot))$ constitute a metric space. The metric is a generalization of the concept of distance between two points. Observe that an normed space can be made into a metric space by defining the metric as $d(\psi,\phi)=||\psi-\phi||$ . A metric defines equality of vectors, i.e. two vectors are equal if the distance between them is zero, even though they are not identical. (This point will be illustrated below.) Also a metric defines the convergence of a sequence of vectors in $V$ to a particular vector in $V$ .

Example-1. The inner product space $\mathbb{C}^n$ is a metric space under the metric induced by the norm \eqref{d1}. The metric is given by

(5) $\begin{equation*} ||z-w||=\sqrt{\sum_{k=0}^k |\alpha_k-\beta_k|^2} . \end{equation*}$

. $\square$

Example-2. The inner product space $C_0^2(\mathbb{R})$ is a metric space under the metric induced by the norm \eqref{d2}. The metric is given by

(6) $\begin{equation*} ||\psi-\varphi|| = \sqrt{\int_{-\infty}^{\infty} |\psi(x)-\varphi(x)|^2\, dx} \end{equation*}$

$\square$

Hilbert Space

A complex pre-Hilbert space $H_P$ is a complex linear vector space equipped with an inner product $\left<\cdot|\cdot\right>$ , with a norm given by $||\psi||=\sqrt{\left<\psi|\psi\right>}$ , and with a metric $d(\psi,\phi)=\sqrt{\left<\psi-\phi|\psi-\phi\right>}$ . Thus to define a pre-Hilbert space one only needs to specify the vector space involved and to define the corresponding inner product. Note that a given vector space maybe equipped with more than one inner product. In quantum mechanics the choice of inner product is not arbitrary but dictated by physical considerations.

Before we can define what is a Hilbert space, we need the concept of a Cauchy sequence. A sequence of vectors in a pre-Hilbert space $H_P$ is an indexed collection of vectors, i.e. $\{\varphi_1,\varphi_2,\varphi_3,\dots\}$ with each $\varphi_k$ belonging to $H_P$ . A sequence is called Cauchy if $\lim_{m,n\rightarrow \infty}||\varphi_m-\varphi_n||=0$ . The pre-Hilbert space $H_P$ is called complete if for every Cauchy sequence $\{\varphi_1,\varphi_2,\varphi_3,\dots\}$ in $H_P$ there exists a vector $\varphi$ in $H_P$ such that $\lim_{m\rightarrow \infty}||\varphi_m-\varphi||=0$ .

A complete pre-Hilbert space is called a Hilbert space.

A pre-Hilbert space which is not complete can be turned into a Hilbert space by adding all limits of Cauchy sequences to the pre-Hilbert space itself. The resulting Hilbert space is called the completion or closure of the pre-Hilbert space. Generally, Hilbert spaces are constructed in this way. That is one starts with a linear space, then equips that space with an inner product to make it into a pre-Hilbert space, then finally completes the pre-Hilbert space to get a Hilbert space.

Example-1. $\mathbb{C}^n$ , together with the inner product \eqref{c1}, is a pre-Hilbert space which is already a Hilbert space. $\square$

Example-2. $C_0^2(\mathbb{R})$ , together with the inner product \eqref{c2}, is a pre-Hilbert space which is not a Hilbert space. The reason is that there are Cauchy sequences in the vector space which do not converge to continuous functions. For example, the sequence of continuous functions given by

(7) $\begin{equation*} \eta_n(x)=\left\{ \begin{array}{ll} 0 &, x\leq 0\\ n x &, 0\leq x\leq \frac{1}{n} \\ 1 &, \frac{1}{n}\leq x\leq 1 \\ n+1-nx &, 1\leq x \leq 1+\frac{1}{n}\\ 0&, 1+\frac{1}{n} \leq x \end{array} \right. , \; \;\;\;\; n=1, 2, \dots \end{equation*}$

is a Cauchy sequence in $C_0^2(\mathbb{R})$ . However, the sequence converges to the function which has the value $1$ in the interval $(0,1)$ and $0$ elsewhere, which is clearly not continuous but belonging to $\mathcal{L}^2(\mathbb{R})$ . The pre-Hilbert space is then not complete. To construct a Hilbert space out of $C_0^2(\mathbb{R})$ , we adjoin to $C_0^2(\mathbb{R})$ all limits of its Cauchy sequences. The resulting Hilbert space then consists not only continuous functions but also non-continuous functions. The Hilbert space is denoted by $\mathcal{L}^2(\mathbb{R})$

A peculiarity of $\mathcal{L}^2(\mathbb{R})$ is that its vectors are not individual functions but classes of functions. Consider the function $\eta(x)\in C_0^2(\mathbb{R})$ , which, of course, belongs to $\mathcal{L}^2(\mathbb{R})$ . Now define the function $\zeta(x)=\eta(x)$ for all $x\neq x_0$ and $\zeta(x)=\infty$ for $x=x_0$ . While $\zeta(x)$ is infinite at $x_0$ , the integral $\int_{\mathbb{R}}|\zeta(x)|^2\mbox{d}x$ exists, and in fact its value is given by $\int_{\mathbb{R}}|\zeta(x)|^2 \mbox{d}x=\int_{\mathbb{R}}|\eta(x)|^2\mbox{d}x$ . That is so because the value of an integral does not change by changing the value of the integrand at isolated points. Thus $\zeta(x)$ is a vector of $\mathcal{L}^2(\mathbb{R})$ . Now $\zeta(x)-\eta(x)=0$ for all $x\neq x_0$ , so that $\int_{\mathbb{R}}|\zeta(x)-\eta(x)|^2\, \mbox{d}x=0$ or $\|\zeta-\eta\|=0$ . By the definition of equality of vectors of a metric space, the functions $\zeta(x)$ and $\eta(x)$ , while they are not equal pointwise, represent the same vector in $\mathcal{L}^2(\mathbb{R})$ . Also we can introduce the new function $\omega(x)=\eta(x)$ for all $x\neq x_1,x_2$ , with $x_1\neq x_2$ , and assign arbitrary values to $\omega(x)$ at $x_1$ and $x_2$ . Again $\eta(x),\zeta(x),\omega(x)$ are not equal pointwise, but $\|\eta-\zeta\|=0$ , $\|\eta-\omega\|$ , $\|\omega-\zeta\|=0$ so that they represent the same vector in $\mathcal{L}^2(\mathbb{R})$ . Clearly we can go on changing the values of $\eta(x)$ at isolated points to obtain an infinite number of functions that differ with $\eta(x)$ at isolated points only but with zero distance with $\eta(x)$ . We can now see that a class of functions represents a vector of $\mathcal{L}^2(\mathbb{R})$ . $\square$

Separable Hilbert Spaces

Some collection of vectors, $\{\varphi_1,\varphi_2,\varphi_3,\cdots\,\varphi_n\}$ , of a Hilbert space $\mathcal{H}$ is called linearly independent if the sum

(8) $\begin{equation*} \alpha_1\varphi_1 + \alpha_2\varphi_2 + \dots +\alpha_n \varphi_n =0 \end{equation*}$

implies that $\alpha_1=\alpha_2=\dots=\alpha_n=0$ . The largest number of linearly independent vectors in a Hilbert space is the dimension $N$ of the Hilbert space. If $N$ is finite, the Hilbert space is finite dimensional; otherwise, it is infinite dimensional. A Hilbert space $\mathcal{H}$ is called separable if there exists a set of linearly independent vectors in $\mathcal{H}$ , $\{\phi_1,\phi_2,\phi_3,\dots\}$ , such that for every vector $\psi$ in $\mathcal{H}$ there exists a set of complex numbers $\{\alpha_1,\alpha_2,\alpha_3,\dots\}$ such that

(9) $\begin{equation*} \psi=\alpha_a\phi_1 + \alpha_2\phi_2 + \alpha_3\phi_3 + \dots \end{equation*}$

The number of such independent vectors must necessarily equal the dimensionality of the the Hilbert space. When such set satisfies the condition, the set is said to span the Hilbert space. Not all Hilbert spaces are separable. The Hilbert space of quantum mechanics is separable; because of that whenever we refer to a Hilbert space from hereon we mean separable Hilbert space.

Two vectors $\phi$ and $\psi$ are called orthogonal if $\left<\phi|\psi\right>=0$ . A set of independent vectors, which we denote shorthand by $\{\phi_k\}$ , is called an orthogonal set if $<\phi_i|\phi_j>=0$ when $i\neq j$ . Moreover, the set is called an orthonormal set if $<\phi_i|\phi_j>=\delta_{ij}$ where $\delta_{ij}$ is the Kronecher delta function, $\delta_{i\neq j}=0$ and $\delta_{i=j}=1$ . When the set $\{\phi_k\}$ spans the Hilbert space, the set is called an orthonormal basis for the Hilbert space $\mathcal{H}$ . Thus when $\{\phi_k\}$ is an orthonormal basis, every vector $\psi$ in $\mathcal{H}$ can be written as the sum given by equation-\eqref{vectorsum}. The coefficients $\alpha_k$ are found using the orthonormality of the $\phi_k$ ‘s and the linearity of the inner product,

(10) $\begin{eqnarray*} \left<\phi_k|\psi\right>&=&\sum_i\alpha_i\left<\phi_k|\phi_i\right>\nonumber\\ &=&\sum_i \alpha_i\delta_{ik} \nonumber\\&=&\alpha_k. \end{eqnarray*}$

Hence the coefficients are given by $\alpha_k=\left<\phi_k|\psi\right>$ . The $\alpha_k$ ‘s are called the Fourier coefficients of $\psi$ with respect to the basis set $\{\phi_k\}$ . From this definition it is clear that a given set of orthonormal vectors form a basis if and only if $\left<\psi|\varphi_k\right>=0$ for all $k$ implies that $\varphi=0$ .

Introduction

Vector Space

Inner Product Space

Normed Space

Metric Space

Hilbert Space

Separable Hilbert Spaces

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