01. The Mathematical Formalism of Quantum Mechanics

Introduction¶

The prerequisites for Physics 221A include a full year of undergraduate quantum mechanics. In this semester we will survey that material, organize it in a more logical and coherent way than the first time you saw it, and pay special attention to fundamental principles. We will also present some new material. Physics 221B will largely consist of new material.

We begin by presenting some of the mathematical formalism of quantum mechanics. We will introduce more mathematics later in the course as needed, but for now we will concentrate on the linear algebra of spaces of wave functions, which are called Hilbert spaces. These notes gather together and summarize most of what you will need to know of this subject, so we can proceed with other matters. In the next set of notes we will turn to the physical postulates of quantum mechanics, which allow us to connect experimental results with the mathematics presented here.

Introductory courses on linear algebra are usually limited to finite-dimensional, real vector spaces. Making the vector spaces complex is a small change, but making them infinite-dimensional is a big step if one wishes to be rigorous. We will make no attempt to be rigorous in the following---to do so would require more than one course in mathematics and leave no time for the physics. Instead, we will follow the usual procedure in physics courses when encountering new mathematics, which is to proceed by example and analogy, attempting to gain an intuitive understanding for some of the main ideas without going into technical proofs. Specifically, in dealing with Hilbert spaces we will try to apply what we know about finite-dimensional vector spaces to the infinite-dimensional case, often using finite-dimensional intuition in infinite dimensions, and we will try to learn where things are different in infinite dimensions and where one must be careful. Fortunately, it is one of the consequences of the mathematical definition of a Hilbert space that many of its properties are the obvious generalizations of those that hold in finite dimensions, so much of finite-dimensional intuition does carry over. We will use what we know about spaces of wave functions (an example of a Hilbert space) to gain some intuition about where this is not so.

Hilbert Spaces¶

Let us consider a wave function $\psi(x)$ for a one-dimensional quantum mechanical problem. In practice it is common to require $\psi(x)$ to be normalized, but for the purpose of the following discussion we will require only that it be normalizable, that is, that

(which means that the integral is finite; usually when we write an integral without limits like the one above, we mean that the limits are $-\infty$ to $+\infty$, or over all of space if the integral is multidimensional). Wave functions that are not normalizable cannot represent physically realizable states, because the probability of finding a real particle somewhere in space must be unity. Nevertheless, wave functions that are not normalizable, such as the free particle energy eigenfunctions $e^{ikx}$, are definitely useful for some purposes, and we will have to say something about them later. For now, however, we will stick to normalizable wave functions.

Mathematically speaking, the space of complex functions that are normalizable (or square integrable) in the sense of (1) constitutes a complex vector space. This is because if $\psi(x)$ is square-integrable, then so is $c\psi(x)$, where $c$ is any complex number, and if $\psi_1(x)$ and $\psi_2(x)$ are square-integrable, then so is $\psi_1(x) + \psi_2(x)$. Moreover, the space is infinite-dimensional (see Mathematical Notes). This vector space is an example of a Hilbert space; in general, a Hilbert space is a complex, inner product vector space (a vector space upon which an inner or scalar product is defined) with certain additional properties that will not concern us in this course (see Mathematical Notes). Often the term “Hilbert space” is defined to be an infinite-dimensional space, but in this course we will refer to any of the vector spaces of wave functions that occur in quantum mechanics as Hilbert spaces, even when finite-dimensional.

As you know, given a normalized wave function $\psi(x)$ in configuration space, $|\psi(x)|^2$ is the probability density for making measurements of position on an ensemble of systems. The wave function $\psi(x)$ is connected physically with position measurements. (See Constrained Systems, Configuration Space, and Generalized Coordinates for the definition of “configuration space.”)

Given $\psi(x)$ in configuration space, one can compute the corresponding wave function $\phi(p)$ in momentum space, by a version of the Fourier transform:

The Fourier transform is linear and invertible (via the inverse Fourier transform), so it provides a one-to-one, linear map between the space of wave functions $\psi(x)$ and that of $\phi(p)$. Moreover, according to the Parseval identity, the norm of $\psi$ in configuration space and the norm of $\phi$ in momentum space are equal,

so if we consider the Hilbert space of normalizable wave functions $\{\psi(x)\}$ in configuration space, then the set $\{\phi(p)\}$ of corresponding momentum space wave functions also forms a Hilbert space. (The “norm” of a wave function is the square root of the quantity shown in (3); it is like the length of a vector.)

The momentum space wave function $\phi(p)$ is connected with the measurement of momentum, in particular, if $\phi(p)$ is normalized then $|\phi(p)|^2$ is the probability density for momentum measurements on an ensemble of systems. The Parseval identity expresses the equality of the total probability for measurements of either position or momentum.

Likewise, let $\{u_n(x), n=1,2,\ldots\}$ be an orthonormal set of energy eigenfunctions, $Hu_n(x) = E_nu_n(x)$ for some Hamiltonian $H$. An arbitrary wave function $\psi(x)$ can be expanded as a linear combination of the functions $u_n(x)$, that is, we can write

where the expansion coefficients $c_n$ are given by

(4) can be regarded as a linear transformation taking us from the sequence of coefficients $(c_1,c_2,\ldots)$ to the wave function $\psi(x)$, while (5) goes the other way.

The sequence of coefficients $(c_1,c_2,\ldots)$ uniquely characterizes the state of the system, and we can think of it as the “wave function in energy space.” Notice that if the sequence is normalized, $\sum_n|c_n|^2=1$, then the probability of making a measurement $E=E_n$ of energy is $|c_n|^2$. Also, the norm of the original wave function $\psi(x)$ can be expressed in terms of the expansion coefficients by

which is finite if $\psi(x)$ belongs to Hilbert space. This is another expression of conservation of probability. Mathematically speaking, the set of sequences $\{(c_1,c_2,\ldots)\}$ of complex numbers of finite norm is yet another example of a Hilbert space.

Thus, we have three Hilbert spaces, the two spaces of wave functions, $\{\psi(x)\}$ and $\{\phi(p)\}$, and the space of sequences $\{(c_1,c_2,\ldots)\}$, all assumed to be normalizable. These Hilbert spaces are isomorphic, in the sense that the vectors of each space are related to one another by invertible, norm-preserving linear transformations, so they all contain the same information. (See Mathematical Notes for some mathematical points regarding Hilbert spaces of wave functions.) Therefore none of these spaces can be taken as more fundamental than the other; any calculation that can be carried out in one space (or in one representation, as we will say), can be carried out in another.

Psychologically, there is a tendency to think of wave functions $\psi(x)$ on configuration space as fundamental, probably because we are used to thinking of fields such as electric fields defined over physical space. This is also the bias imposed on us by our first courses in quantum mechanics, as well as the historical development of the subject. Also, we live in physical space, not momentum space. But the Schr"odinger wave function is not really defined over physical space, rather it is defined over configuration space, which is the same only in the case of a single particle (see Constrained Systems, Configuration Space, and Generalized Coordinates). Furthermore, completely apart from the mathematical equivalence of the different Hilbert spaces discussed above, the physical postulates of quantum mechanics show that the different wave functions discussed above correspond to the measurement of different sets of physical observables. They also show that there is a kind of democracy among the different physical observables one can choose to measure, as long as the set of observables is compatible and complete in a sense we will discuss later.

Finally, even if we do have a single particle moving in three dimensions so the wave function $\psi(\rvec)$ can be thought of as a field over physical space, the physical measurement of $\psi(\rvec)$ is very different from the measurement of a classical field such as the electric field or the pressure of a fluid. We have thermometers that can measure the temperature at a point, we have pressure gauges, and it is possible to measure electric and magnetic fields at points of space. But there is no such thing as a “$\psi$-meter” for measuring the wave function $\psi(\rvec)$. Later we will consider how one can obtain $\psi(\rvec)$ from measurements, but the process involves several subtleties. (In particular, it requires one to understand the difference between a pure and a mixed state; only pure states have wave functions.) It is a very different business than the measurement of a classical field. See Prob. Paragraph.

In summary, it is probably best to think of the different wave functions as equivalent descriptions of the same physical reality, none of which is more fundamental than the other. This leads to the concept of the state space of a quantum mechanical system as an abstract vector space in which the state vector lies, while the various wave functions are the expansion coefficients of this state vector with respect to some basis. Different bases give different wave functions, but the state vector is the same.

This was not the point of view in the early days of quantum mechanics, when $\psi(\rvec)$ was seen as a complex-valued field on physical space, and before it was completely clear what the wave function of multiparticle systems should be. We will return to this earlier point of view when we take up the Dirac equation and field theory, but that reconsideration will not change the basic concept of the state space for the quantum mechanical system, a complex vector space in which the state vector lies.

Thus, the mathematical formalism of quantum mechanics exhibits a kind of form invariance or covariance with respect to the set of observables one chooses to represent the states of a system. This view of quantum mechanics, and the transformation theory connected with changes of representation, was worked out in the early days of the quantum theory, notably by Dirac, Born, Jordan, von Neumann and others.

Dirac Ket Notation¶

Since none of the three Hilbert spaces has any claim over the other, it is preferable to use a notation that does not prejudice the choice among them. This is what the Dirac notation does. In the Dirac notation, a state of a system is represented by a so-called ket vector, denoted by $\ket{\;}$, which stands for any of the three wave functions discussed above, $\psi(x)$, $\phi(p)$, or $\{c_n\}$, assuming that they are related by the transformations (2) and (5). It is customary to put a symbol into the ket to identify the state, to write, for example, $\ket{\psi}$ [although the choice of the letter $\psi$ seems to point to something special about the configuration space wave function $\psi(x)$, which there is not]. Some people think of a ket $\ket{\psi}$ as simply another notation for a wave function on configuration space, and I have even seen some authors write equations like

But our point of view will be that the space of kets is yet another Hilbert space, an abstract space that is isomorphic to but distinct from any of the concrete Hilbert spaces of wave functions discussed above. Thus, we will never use equations like (7). I think this approach is appropriate in view of the physical postulates of quantum mechanics, which show how one can construct a Hilbert space out of the results of physical measurements. There is no need to start with wave functions. Therefore we will now put wave functions aside, and discuss the mathematical properties of ket spaces. Later, in 04. Spatial Degrees of Freedom, we will return to wave functions, in effect deriving them from the ket formalism. In this set of notes, we use wave functions only for the purpose of illustrating some general features of Hilbert spaces, calling on your experience with them in a previous course in quantum mechanics.

Kets, Ket Spaces and Rays¶

Kets and ket spaces arise out of the physical postulates of quantum mechanics, which we will discuss in 02. The Postulates of Quantum Mechanics. For now we will just say enough to get started.

A basic postulate of quantum mechanics is that a given physical system is associated with a certain complex vector space, which we will denote by $\ES$. The physical principles by which this vector space is constructed will be discussed later, but for now we simply note that the properties of this space such as its dimensionality are determined by the physics of the system under consideration. For some systems, $\ES$ is finite-dimensional (notably spin systems), while for others it is infinite-dimensional. The vectors of the space $\ES$ are called kets, and they are denoted in the Dirac notation by $\ket{\;}$, as discussed above. Since $\ES$ is a complex vector space, the multiplication of kets by complex numbers and the addition of kets is defined. The product of a complex number $c$ and a ket $\ket{\psi}$ can be written in either order,

In addition, there exist physical principles that allow us to associate a definite (pure) state of a physical system with a ray in $\ES$. We will say more about this association later, when we will define what we mean by a pure state of a physical system, but for now we simply note that a ray in $\ES$ is a 1-dimensional vector subspace of $\ES$, that is, it consists of kets $\ket{\psi}$ related to some nonzero ket $\ket{\psi_0}$ by a complex multiplicative factor, that is,

for some complex number $c$. The kets $\ket{\psi}$ in the ray differ from one another only by normalization and overall phase, which have no physical significance; thus, we say that a (pure) physical state corresponds to the ray (not to any particular ket in the ray).

The Dual Space and Bras¶

You are probably familiar with the fact that kets correspond to wave functions $\psi(x)$, while bras correspond to complex conjugated wave functions $\psi(x)^\cc$. But if we are doing without wave functions until we can properly derive them, how do we describe bras and the complex conjugation of kets? The answer is the dual correspondence.

Also, it would appear that $\psi(x)^\cc$ is just another wave function, so it should belong to the same space of wave functions as $\psi(x)$. But it turns out that there is great advantage in regarding kets and bras as belonging to two different spaces. Kets belong to the space $\ES$, while bras belong to the so-called dual space, denoted $\ES^\cc$.

If $\ES$ is any (complex) vector space, then the dual space, denoted $\ES^\cc$, is the space of complex-valued, linear operators that act on $\ES$. In the Dirac notation, such operators are called bras, and are denoted $\bra{\;}$. (In mathematics, they are called forms, covectors or dual vectors.) As with kets, it is customary to insert some symbol to identify a bra and distinguish it from other bras, such as $\bra{\alpha}$. In mathematical language, we write

which means that $\bra{\alpha}$ maps a ket into a complex number. To be a bra, the map must be linear. If $\ket{\psi}$ is a ket, then the complex number that results by letting $\bra{\alpha}$ act on it can be written $\bra{\alpha}(\ket{\psi})$. Usually we drop the parentheses write the complex number simply as $\braket{\alpha}{\psi}$, but if we wish to emphasize the interpretation of the bra as an operator acting on kets we will keep the parentheses.

Since the map is required to be linear, we have

A bra, as a linear operator acting on kets, is not to be confused with the familiar linear operators used in quantum mechanics, such as the Hamiltonian operator, which also act on kets. The distinction is that a bra is a complex-valued operator, whereas the Hamiltonian and other such operators are ket-valued operators.

It is easy to see how to define the product of a bra with a complex scalar, and the sum of two bras; we simply set

and

Therefore the set of all bras acting on a given ket space forms a vector space in its own right, which by definition is the dual space $\ES^\cc$. It is easy to show that if $\ES$ is finite-dimensional, then $\ES^\cc$ is also, and has the same dimension. If $\ES$ is infinite-dimensional, then the dual space $\ES^\cc$ is also infinite-dimensional.

What we have said so far allows us to talk about bras, but not to convert a ket into a bra (the analog of the complex conjugation of wave functions). To do that we need to introduce the metric or scalar product on our ket space. In real, Euclidean vector spaces the metric is what allows us to measure distances, and from that, angles. This is an analogy to keep in mind while thinking about the metric in (complex) Hilbert spaces, because it provides useful geometrical interpretations.

The Scalar Product and Dual Correspondence¶

We postulate the existence of a metric or scalar product $g$ on $\ES$, which is a function

with certain properties. This notation means that $g$ is a function that takes two kets and produces a complex number; for example, we will write $g(\ket{\psi},\ket{\phi})$ for the complex number that results upon letting $g$ act on kets $\ket{\psi}$ and $\ket{\phi}$. The postulated properties of $g$ are the following. First, $g$ is linear in its second operand and antilinear in its first operand. Explicitly, this means

(An antilinear function requires us to take the complex conjugate of the coefficients when evaluating it on linear combinations of vectors.) Next, $g$ is symmetric or Hermitian, which means

Third, $g$ is positive definite, which means

for all $\ket{\psi}$, with equality holding if and only if $\ket{\psi}=0$. Note that by property (17), the left hand side of (18) is necessarily real.

Next, given the metric $g$, we can define the dual correspondence, which is a mapping that converts kets into bras. It is the map which in wave function language amounts to taking the complex conjuate of a wave function. We will denote the dual correspondence by $DC$; it is the map

If $\ket{\psi}$ is a ket, then we will denote the bra corresponding to it under the dual correspondence by $\bra{\psi}$, with the same identifying symbol. We define the bra $\bra{\psi}$ in terms of its action on an arbitrary ket $\ket{\phi}$; the definition is

The idea is that in the scalar product, we regard the first argument as fixed and the second as variable, which gives us a complex valued function of kets, which according to (15) is linear. Thus, the function is a bra. By way of notation, we will denote the bra resulting from a given ket under the dual correspondence by a dagger, so that if

then we will write

We will say that $\bra{\psi}$ is the Hermitian conjugate of $\ket{\psi}$. We note that since the scalar product is antilinear in its first operand, the dual correspondence is antilinear, that is,

If $\ES$ is finite-dimensional, then it is easy to prove that the dual correspondence is one-to-one and onto, so that every bra is in fact the Hermitian conjugate of some ket. This follows from the postulated properties of the metric. In infinite-dimensional spaces, this is true only if certain restrictions are placed on the bra space. We will ignore such technicalities, and proceed as if the dual correspondence is always one-to-one and onto (which, with the right understandings, it is for a Hilbert space). This means that every bra is the dagger of some unique ket, so the dual correspondence has an inverse. We will denote the inverse with the same dagger notation, so that

and so that

We now simplify the notation for the scalar product by dropping the parentheses on the left hand side of (20). That is, we write

With this notational change, we can rewrite properties (17) and (18) of the metric as

and

for all $\ket{\psi}$, with equality if and only if $\ket{\psi}=0$. The complex number $\braket{\psi}{\phi}$ is sometimes called the inner product or scalar product of $\ket{\psi}$ and $\ket{\phi}$. Using the notation shown in (26), we can henceforth dispense with the $g$-notation when talking about scalar products.

The Schwarz Inequality¶

We can now prove an important theorem, namely the Schwarz inequality. We are interested in this inequality for complex vector spaces, but it is also true in real vector spaces with a positive definite inner product, that is, in Euclidean spaces. There it is equivalent to the geometrical fact that the shortest distance between two points is a straight line. The Schwarz inequality is important in quantum mechanics because it is used to prove the Heisenberg uncertainty relations.

The Schwarz inequality says that

for all $\ket{\psi}$ and $\ket{\phi}$, with equality if and only if $\ket{\psi}$ and $\ket{\phi}$ are linearly dependent, that is, if they lie in the same ray. To prove the theorem, we set

where $\lambda$ is a complex number, and we use (28) to write,

This is true for all kets $\ket{\psi}$ and $\ket{\phi}$, and all complex numbers $\lambda$. Since the Schwarz inequality is obviously true if $\ket{\phi}=0$, we consider the case $\ket{\phi}\ne 0$, and we set

so that (31) becomes

which is equivalent to (29). Finally, it is easy to show that (33) becomes an equality if $\ket{\psi} = c\ket{\phi}$ for any complex number $c$, and conversely, if (33) is an equality, then $\braket{\alpha}{\alpha}=0$, which implies $\ket{\alpha}=0$, or $\ket{\psi} = -\lambda\ket{\phi}$. Therefore the equality holds in (29) if and only if $\ket{\psi}$ and $\ket{\phi}$ are linearly dependent.

Operators¶

Next we discuss operators. Usually in quantum mechanics we use only linear or antilinear operators. A linear operator $L$ and an antilinear operator $A$ are both mappings from the ket space into itself,

but they have different properties when acting on linear combinations of kets. In particular, we have

The only antilinear operator of interest in nonrelativistic quantum mechanics is the time-reversal operator, which we will discuss in 22. Time Reversal. For now we ignore antilinear operators, and concentrate exclusively on linear operators.

Linear operators themselves can be multiplied by complex numbers and added, so they form a complex vector space in their own right. Linear operators can also be multiplied with one another; the product $AB$ means, apply $B$ first, then $A$ (to some ket). The multiplication is associative,

for any linear operators $A$, $B$, $C$, but it is not commutative, that is

in general. The lack of commutativity of two linear operators $A$ and $B$ is measured by their commutator, defined by

For reference, we also define the anticommutator,

We use the same curly bracket notation for the anticommutator as for the Poisson bracket in classical mechanics (see The Poisson Bracket). It will usually be clear from context which is intended.

A linear operator $A$ may possess an inverse, $A^{-1}$; if it exists, it satisfies

where the identity operator is simply denoted by 1. In finite-dimensional spaces, if $A^{-1}$ exists, then it is both a right- and a left-inverse; but in infinite dimensional spaces, some operators may possess a right-inverse but not a left-inverse, or vice versa. When all indicated inverses exist, we have

A more careful treatment of the subject of operator inverses in infinite-dimensional spaces must deal with domains of definition, boundedness and other matters. We may wish to gloss over these in a first course in practical quantum mechanics.

Rules for Commutators¶

The commutator $[\;,\;]$ obeys the following properties, which are trivial consequences of the definition (39). In the following, capital letters are linear operators and lower case letters are complex numbers. First, the commutator is linear in both operands,

it is antisymmetric,

and it obeys the Jacobi identity,

Any bracket operation $[\;,\;]$ defined on any vector space (not just spaces of linear operators) that satisfies properties (43)--(45) qualifies that vector space as a Lie algebra.

In addition, the commutator satisfies the following properties, sometimes referred to as the derivation property or Leibnitz rule:

Calculations in quantum mechanics often require one to reduce complicated commutators into simpler ones that are known. Rules (46) are especially useful for this purpose. It is interesting that properties (43)--(46) are also valid for the Poisson bracket in classical mechanics [except that the ordering of the factors in (46), which must be respected in quantum mechanics, is immaterial in classical mechanics]. See The Poisson Bracket.

The Action of Operators on Bras¶

The operators we have introduced begin life as operators that act on kets, mapping them into other kets; but the definition of these operators is easily extended to allow them to act on bras, whereupon they produce other bras. If $\bra{\psi}$ is a bra and $A$ an operator, we denote the action of $A$ on $\bra{\psi}$ by $\bra{\psi}A$, with the bra to the left. That is, we think of $A$ as acting on kets to the right, but on bras to the left. The definition of $\bra{\psi}A$ is as follows. Since $\bra{\psi}A$ is supposed to be a new bra, it is specified by its action on an arbitrary ket, say, $\ket{\phi}$; the definition is given by writing

Since by this definition the ordering of the parentheses is immaterial, it is customary to drop them, and to write simply

Thus, we can think of $A$ as acting either to the left or the right in the “matrix element” $\matrixelement{\psi}{A}{\phi}$.

The Outer Product¶

If $\ket{\alpha}$ and $\ket{\beta}$ are kets, then we can define a linear operator denoted by $\ketbra{\alpha}{\beta}$, whose action on an arbitrary ket $\ket{\psi}$ is given by

where the right-hand side is the product of the complex number $\braket{\beta}{\psi}$ times the ket $\ket{\alpha}$. The action of $\ketbra{\alpha}{\beta}$ on bras is given by

This is not the definition of $\ketbra{\alpha}{\beta}$, which is given by (49), rather it is a simple consequence of that definition. Its proof will be left as an exercise. The operator $\ketbra{\alpha}{\beta}$ can be viewed as the tensor product of the ket $\ket{\alpha}$ with the bra $\bra{\beta}$; this kind of product is similar to the dyadic product used in ordinary vector or tensor analysis. The operator $\ketbra{\alpha}{\beta}$ is called the outer product of $\ket{\alpha}$ and $\ket{\beta}$.

Bases and Orthonormal Bases; Resolutions of the Identity¶

As in any vector space, a basis is a set of linearly independent vectors that span the entire space, and the number of such vectors is the dimension of the space. This is a familiar concept in finite dimensions. In an infinite-dimensional space, we obviously need an infinite number of basis vectors, but is the basis countably or uncountably infinite? That is, can we label the basis vectors with a discrete index, or do we need continuous indices as well? As it turns out, it is one the mathematical properties of Hilbert spaces that they always possess a countable basis, that is, a set of linearly independent, normalizable vectors that span the space and can be indexed $n=1,2,3,\ldots$, or by some other such discrete indexing scheme. This does not mean that every basis we encounter will be labeled by a discrete index, only that such bases always exist. In fact, in practice in quantum mechanics we frequently encounter “bases” with continuous indices as well, but the basis vectors of the continuum are not normalizable and do not belong to Hilbert space. We will say more about these kinds of bases later, and for now proceed with the case of discrete bases, consisting of normalizable vectors. As an example, you may think of the harmonic oscillator wave functions $u_n(x)$ in one-dimensional quantum mechanics, which are normalizable, belong to Hilbert space, and form a discrete, complete set.

Let us denote a discrete basis in a ket space by $\{\ket{n}, n=1,\ldots\}$. We say the basis is orthonormal if

We usually use orthonormal bases in quantum mechanics, but non-orthonormal bases are useful sometimes as well (if more difficult to work with). In any basis, an arbitrary ket $\ket{\psi}$ can be represented as a linear combination of the basis vectors,

and if the basis is orthonormal, the expansion coefficients are given by

Substituting this into (52), we have

or, since $\ket{\psi}$ is arbitrary,

where 1 is the identity operator. This is the resolution of the identity associated with the orthonormal basis $\{ \ket{n} \}$.

Resolutions of the identity associated with bases with continuous indices will be discussed below.

Hermitian Conjugation¶

We have already defined Hermitian conjugation (denoted by the dagger $\dagger$) of bras and kets. We now define Hermitian conjugation of operators. If $A:\ES\to\ES$ is a linear operator, then $A^\hc:\ES\to\ES$ is another linear operator, defined by its action on an arbitrary ket $\ket{\psi}$ by

We should note that $\bra{\psi}$ is an antilinear function of $\ket{\psi}$ (because of the dual correspondence), but when we do the (second) Hermitian conjugation indicated on the right hand side of (56), the result is a linear function of $\ket{\psi}$. Thus, $A^\hc$ is a linear operator.

There are a number of immediate consequences of this definition, which we list here. The proofs are left as simple exercises.

We have now defined the action of the Hermitian conjugate operation ${}^\hc$ on kets, bras, and operators. It is convenient also to define the action of the Hermitian conjugate on complex numbers as being identical with ordinary complex conjugation. With this understanding, we can state a general rule, which is illustrated in several various ways in the results above, which says that the Hermitian conjugate of a product of objects (kets, bras, operators, complex numbers) is given by reversing the order of the objects and replacing each by its Hermitian conjugate. The products involved can be either ordinary multiplication by scalars, operator products, or inner or outer (tensor) products. We only require that the original product be meaningful.

Hermitian, Anti-Hermitian and Unitary Operators¶

We say that an operator $A$ is Hermitian if it is equal to its own Hermitian conjugate,

An equivalent definition is

for all kets $\ket{\psi}$ and $\ket{\phi}$. Similarly, an operator is said to be anti-Hermitian if it satisfies

An arbitrary operator is in general neither Hermitian nor anti-Hermitian, but can always be uniquely decomposed into a sum of a Hermitian and an anti-Hermitian operator:

If $A$ is Hermitian, then the matrix element $\matrixelement{\psi}{A}{\psi}$ is necessarily real, for all choices of $\ket{\psi}$, in accordance with (57). We say that a Hermitian operator is positive definite if it satisfies

for all nonzero kets $\ket{\psi}$. Similarly, we say that $A$ is nonnegative definite if

for all kets $\ket{\psi}$.

We have not discussed eigenvalues yet, but a positive definite, Hermitian operator is one whose eigenvalues are all positive, while a nonnegative definite, Hermitian operator is one whose eigenvalues are all nonnegative.

A simple but important theorem is the following:

The (trivial) proof is left as an exercise.

An operator $U$ is unitary if

so that

The product of two unitary operators is always unitary.

Eigenvalues and Eigenkets; the Spectrum of an Operator¶

If $A$ is an operator acting on $\ES$ and there exists a nonzero ket $\ket{u}$ and complex number $a$ such that

then we say that $\ket{u}$ is an eigenket of $A$ and $a$ is a (right) eigenvalue. Similarly, if there exists a nonzero bra $\bra{v}$ and a complex number $b$ such that

then we say that $\bra{v}$ is an eigenbra of $A$ and $b$ is a (left) eigenvalue. In finite dimensions, every right eigenvalue is also a left eigenvalue, but in infinite dimensions this need not be true.

We define the spectrum of an operator as the set of its eigenvalues, seen as a subset of the complex plane. (If necessary, we can distinguish between a left and right spectrum.) In finite dimensions, this is a set of discrete points in the complex plane.

Some examples of the spectra of familiar operators in quantum mechanics are illustrated in Figure 1. In part (a) of the figure we have the spectrum of the harmonic oscillator, whose eigenvalues $E_n = (n+\frac{1}{2})\hbar\omega$ are shown as spots on the real energy axis. This is an example of a discrete spectrum. Only real energies are physically meaningful, but we show the whole complex energy plane because in general eigenvalues of operators are complex. (We will see complex energies play a role in scattering and resonance theory later in the course.) In part (b) we have the spectrum of the free particle Hamiltonian $H=p^2/2m$. It consists of all real energies $E\ge0$, indicated by the thick line on the positive real axis. This is an example of the continuous spectrum. In part (c) we have the spectrum of the hydrogen atom. It consists of a discrete set of spots at negative energy, representing the bound states, an infinite number of which accumulate as we approach $E=0$. As with the free particle, there is a continuous spectrum above $E=0$. Altogether, the hydrogen atom has a mixed spectrum (discrete and continuous). Finally, in part (d) we have the spectrum of a unitary operator on a finite-dimensional space. Its eigenvalues are phase factors that lie on the unit circle in the complex plane.

In the case of the discrete spectrum, the set of kets $\ket{u}$ that satisfy (70) for a given eigenvalue $a$ forms a vector subspace of the ket space. This subspace is at least 1-dimensional, but the dimensionality may be higher (even infinite, on infinite-dimensional vector spaces). We will call this subspace the eigenspace of the operator corresponding to the given eigenvalue. If the eigenspace is 1-dimensional, then we say the eigenvalue is nondegenerate, otherwise, that it is degenerate. We refer to the dimensionality of the eigenspace as the order of the degeneracy. This is the same as the number of linearly independent eigenvectors of a given eigenvalue. We can imagine the eigenspaces of the discrete spectrum as illustrated in Figure 2, which shows the eigenspaces for a two-fold degenerate eigenvalue $a_1$ and a nondegenerate eigenvalue $a_2$ ($\dim\ES_1=2$, $\dim\ES_2=1$).

The Case of Hermitian Operators¶

In general, there is no simple relation between the eigenkets and eigenbras of an operator, even in finite dimensions. But if an operator $A$ is Hermitian, then there are a number of simplifications. We speak for the time being of the discrete spectrum.

First, the eigenvalue $a$ is real, so the spectrum in the complex eigenvalue plane is a subset of the real axis. To prove this we take $A\ket{u} = a\ket{u}$ and multiply on the left by $\bra{u}$, obtaining

Now taking the complex conjugate of this, the left-hand side goes into itself, while the right becomes $a^\cc \braket{u}{u}$. Thus we have

Since $\ket{u}\ne 0$, this implies $a=a^\cc$.

Second, if $A\ket{u}=a\ket{u}$, then $\bra{u}A = a\bra{u}$, so that the eigenbras are the Hermitian conjugates of the eigenkets, and the left and right eigenvalues are equal.

Third, any two eigenkets corresponding to distinct eigenvalues are orthogonal. We state this final property as a theorem:

To prove this, we let $A\ket{u} =a\ket{u}$ and $A\ket{u'} = a'\ket{u'}$ for two eigenkets $\ket{u}$ and $\ket{u'}$ and two eigenvalues $a$ and $a'$. Multiplying the first equation by $\bra{u'}$ and the second by $\bra{u}$, we obtain

Now taking the Hermitian conjugate of the second equation and subtracting from the first, we find

showing that either $a=a'$ or else $\braket{u}{u'}=0$.

In view of this theorem, Figure 3 gives a more accurate picture of the eigenspaces of a Hermitian operator than does Figure 2.

Completeness¶

In looking at Figs.~Figure 2 and Figure 3 we may imagine that the space in which the eigenspaces lie is three dimensional, so the two eigenspaces (of dimension 2 and 1) “fill up” the whole space. Do the eigenspaces of an operator always “fill up” the space upon which the operator acts? An operator that does this is said to be complete. The answer is no, not even in finite dimensions. For example, the matrix

possesses only one eigenvector, namely,

so it has only a single one-dimensional eigenspace. The matrix (76) is not Hermitian.

On the other hand, in finite dimensions every Hermitian operator is complete. Thus, the sum of the dimensionalities of the eigenspaces is the dimension of the whole space. Since the eigenspaces are orthogonal, by choosing an orthonormal basis in each eigenspace we obtain an orthonormal eigenbasis for the whole space, that is, a set of orthonormal basis vectors that are also eigenvectors of the operator. This is a consequence of the standard methods taught in linear algebra courses for diagonalizing Hermitian matrices, which explicitly produce the orthonormal eigenbasis.

Notice that the orthonormal eigenbasis is not unique. First, if there is any degeneracy (any eigenspace with dimensionality $\ge2$), then obviously there are an infinite number of ways of choosing an orthonormal basis in that eigenspace. But even in a nondegenerate (one-dimensional) eigenspace, a normalized eigenvector is only determined to within a phase factor.

These nice properties of Hermitian operators are shared by a larger class of normal operators, which will be studied in a homework exercise. Occasionally in quantum mechanics we must deal with the eigenvectors of non-Hermitian operators and worry about their completeness properties. This happens, for example, in the theory of the Dirac equation.

The Direct Sum; Orthonormal Eigenbases¶

Suppose we have some vector space $\ES$, and suppose $\ES_1$ and $\ES_2$ are vector subspaces of $\ES$ that have no element in common except 0, such as the spaces we see in Figs.~Figure 2 and Figure 3. Then we say that the direct sum of $\ES_1$ and $\ES_2$, denoted $\ES_1\oplus \ES_2$, is the set of vectors that can be formed by taking linear combinations of vectors in the two subspaces. That is,

The direct sum is itself a vector subspace of $\ES$. One can also think of the direct sum as the span of the set of vectors formed when we throw together a basis in $\ES_1$ with a basis in $\ES_2$. Thus, we have

This is not the most abstract definition of the direct sum, but it is the one we will use.

We can now express the completeness property in finite dimensions in terms of the direct sum. Let $A$ be an operator acting on a finite-dimensional space $\ES$. Let the eigenvalues of $A$ be $a_1,\ldots, a_N$, and let $\ES_n$ be the $n$-th eigenspace, that is,

Then $A$ is complete if

In finite dimensions, any Hermitian $A$ is complete, and moreover it decomposes the original Hilbert space $\ES$ into a collection of orthogonal subspaces, the eigenspaces of the operator.

Let us choose an orthonormal basis, say $\ket{nr}$, in each of the subspaces $\ES_n$, where $r$ is an index labeling the orthonormal vectors in the subspace, so that $r=1,\ldots,\dim\ES_n$. This can always be done, in many ways. Then, since the various subspaces are themselves orthogonal, the collection of vectors $\ket{nr}$ for all $n$ and $r$ forms an orthonormal basis on the entire space $\ES$, so that

These kets are eigenkets of $A$,

that is, they constitute an orthonormal eigenbasis. Furthermore, we have the resolution of the identity,

corresponding to the chosen orthonormal eigenbasis of the complete Hermitian operator $A$. Also, the operator itself can be written in terms of this eigenbasis,

This follows if we let both sides of this equation act on a basis vector $\ket{n'r'}$, whereupon the answers are the same. By linear superposition, the same applies to any vector. Thus, both sides are the same operator.

Spectral Properties in Infinite Dimensions¶

Here we make some remarks about infinite-dimensional ket spaces and illustrate the kinds of new phenomena that can arise.

In infinite dimensions, there exist Hermitian operators that are not complete. To cover this case, we define a complete Hermitian operator as an observable. This terminology is related to the physical interpretation of observables in quantum mechanics, as explained in 02. The Postulates of Quantum Mechanics. In practice we seldom deal with incomplete Hermitian operators in quantum mechanics.

We now use examples from one-dimensional wave mechanics to illustrate some phenomena that can occur in infinite dimensions. The Hilbert space of wave functions in one dimension consists of normalizable functions $\psi(x)$ with the scalar product,

We will present four examples of operators and their spectral properties.

The first is the one-dimensional harmonic oscillator with Hamiltonian,

which is a Hermitian operator. The energy eigenvalues are

so that the spectrum consists of a discrete set of isolated points on the positive energy axis, as illustrated in Figure 1(a). The eigenfunctions $u_n(x)$ are normalizable (they belong to Hilbert space) and orthogonal for different values of the energy. Thus they can be normalized so that

These eigenfunctions are nondegenerate and complete, so we have a resolution of the identity,

Not all Hermitian operators on an infinite dimensional space have a discrete spectrum. As a second example, consider the momentum operator in one dimension, which is $\phat=-i\hbar\, d/dx$. (We put a hat on the operator, to distinguish it from the eigenvalue $p$.) The momentum is a Hermitian operator (see below). The eigenvalue-eigenfunction equation for this operator is

where $u_p(x)$ is the eigenfunction with eigenvalue $p$. This equation has a solution for all values, real and complex, of the parameter $p$,

where $c$ is a constant. If $p$ has a nonzero imaginary part, then $u_p(x)$ diverges exponentially as $x\to\infty$ or $x\to -\infty$, so the “eigenfunction” is rather badly behaved. It is certainly not normalizable in this case. But even if $p$ is purely real, the function $u_p(x)$ is still not normalizable, so it does not represent a physically allowable state (a vector of Hilbert space). The momentum operator $\phat$ does not have any eigenvectors that belong to Hilbert space, much less a basis.

One way to handle this case, which provides a kind of generalization of the nice situation we find in finite dimensions, is to regard every real number $p$ as an eigenvalue of $\phat$, which means that the spectrum of $p$ is the entire real axis (in the complex $p$-plane). See Figure 1(b) for an illustration of the spectrum of the free particle Hamiltonian, which is the portion $E\ge0$ of the real axis. Recall that in finite dimensions a Hermitian operator has only real eigenvalues. This is an example of the continuous spectrum. Also, the eigenfunctions that result in this case are orthonormal and complete in a certain sense, but as noted they do not belong to Hilbert space. We obtain an orthonormal basis, but it consists of vectors that lie outside Hilbert space.

We adopt a normalization of the eigenfunctions (that is, we choose the constant $c$ in (92) so that

because this implies a convenient normalization relation,

where we write the function $u_p(x)$ in ket language as $\ket{p}$ (giving only the eigenvalue). The eigenfunctions $u_p(x)$ of the continuous spectrum are orthonormal only in the $\delta$-function sense. They are also complete, in the sense that an arbitrary wave function $\psi(x)$ can be expanded as a linear combination of them, but since they are indexed by a continuous variable, the linear combination is an integral, not a sum. That is, given $\psi(x)$, we can find expansion coefficients $\phi(p)$ such that

We know this because (95) is essentially a Fourier transform, whose inverse is given by

We see that the “momentum space wave function” $\phi(p)$ is otherwise the scalar product $\braket{p}{\psi}$.

Another example of an operator with a continuous spectrum is the position operator $\hat x$ (again using hats to denote an operator). This operator acting on wave functions means, multiply by $x$. If we denote the eigenfunction of this operator with eigenvalue $x_0$ by $f_{x_0}(x)$, then the eigenvalue equation is

or,

This implies either $x=x_0$ or $f_{x_0}(x)=0$, so $f_{x_0}(x)$ is a “function” that is zero everywhere except possibly at $x_0$. We interpret this situation in terms of the Dirac delta function, by writing the solution as

Again, the spectrum is continuous and occupies the entire real axis; again, the “eigenfunctions” are of infinite norm and obey the continuum orthonormality condition,

where we write the function $f_{x_0}(x)$ in ket language as $\ket{x_0}$ (giving only the eigenvalue). In addition, we have the identity,

We see that the wave function $\psi(x_0)$ is otherwise the scalar product $\braket{x_0}{\psi}$. Substituting this back into (101) and using (100), we have

or, since $x_0$ is arbitrary,

Finally, since $\ket{\psi}$ is arbitrary, we can strip it off, leaving

the resolution of the identity in the case of the continuous eigenbasis of the position operator.

We can also get the resolution of the identity in the momentum basis. Since $\psi(x) = \braket{x}{\psi}$ for any $\psi$, we have $u_p(x) =\braket{x}{p}$. Substituting these and (96) into (95), we have

or, stripping off both $\bra{x}$ on the left and $\ket{\psi}$ on the right,

Equations (104) and (106) illustrate resolutions of the identity in the case of the continuous spectrum.

For a final example, let us consider the 1-dimensional potential illustrated in Figure 4, which has a well followed on the right by an infinitely high barrier. The purpose of the barrier is simply to make the unbound energy eigenfunctions nondegenerate, which simplifies the following discussion. In a potential such as this, we expect there to be some number of bound eigenfunctions $u_n(x)$ in the potential well, which have negative energies $E_n<0$. These eigenfunctions are normalizable, because the wave functions die off exponentially outside the well, and we can normalize them according to $\braket{n}{m}=\delta_{nm}$, as with the discrete spectra discussed above. But there will also be unbound eigenfunctions $u_E(x)$ corresponding to any energy $E\ge0$; these eigenfunctions become free particle wave functions at large negative $x$, and therefore have infinite norm. We normalize them according to $\braket{E}{E'} = \delta(E-E')$. Finally, all bound states are orthogonal to all continuum states, $\braket{n}{E}=0$. Thus, the completeness relation in this case has the form,