1.2 Canonical quantization

It was known by the end of the 19th century that classical physics, namely Newtonian mechanics and classical electromagnetism, contains serious inconsistencies. Later at the beginning of the 20th century, these were resolved by the discoveries of special and general relativities and quantum mechanics. So far, there is no single experiment which contradicts quantum theory. It is surprising, however, that there is no proof for quantum theory. What one can say is that quantum theory is not in contradiction to Nature. Accordingly, we do not prove quantum mechanics here but will be satisfied with outlining some ‘rules’ on which quantum theory is based.

1.2.1 Hilbert space,bras and kets

Let us consider a complex Hilbert space

$$\mathcal{H}=\{|\varphi ⟩,|\psi ⟩,\dots \}.\text{\hspace{1em}(1.26)}$$

An element of $\mathcal{H}$ is called a ket or a ket vector.

A linear function $\alpha :\mathcal{H}\to \mathbb{C}$ is defined by

$$\alpha \left(c_1\mid {\psi }_1⟩+c_2\mid {\psi }_2⟩\right)=c_1\alpha \left(|{\psi }_1⟩\right)+c_2\alpha \left(|{\psi }_2⟩\right)\forall c_i\in \mathbb{C},|{\psi }_i⟩\in \mathcal{H}.$$

We employ a special notation introduced by Dirac and write the linear function as $⟨\alpha |$ and the action as $⟨\alpha |\psi ⟩\in \mathbb{C}$ . The set of linear functions is itself a vector space called the dual vector space of $\mathcal{H}$ , denoted ${\mathcal{H}}^{\ast }$ . An element of $\mathcal{H}$ is called a bra or a bra vector.

Let $\{|e_1⟩,|e_2⟩,\dots \}$ be a basis of $\mathcal{H}$ . Any vector $|\psi ⟩\in \mathcal{H}$ is then expanded as $|\psi ⟩={\sum }_k{\psi }_k|e_k⟩$ , where ${\psi }_k\in \mathbb{C}$ is called the $k$ th component of $|\psi ⟩$ . Now let us introduce a basis $\{⟨{\epsilon }_1|,⟨{\epsilon }_2|,\dots \}$ in ${\mathcal{H}}^{\ast }$ . We require that this basis be a dual basis of $\{|e_k⟩\}$ , that is

$$⟨{\epsilon }_i|e_j⟩={\delta }_{ij}.\text{\hspace{1em}(1.27)}$$

Then an arbitrary linear function $⟨\alpha |$ is expanded as $⟨\alpha |={\sum }_k{\alpha }_k⟨{\epsilon }_k|,$ where ${\alpha }_k\in \mathbb{C}$ is the $k$ th component of $⟨\alpha |$ . The action of $⟨\alpha |\in {\mathcal{H}}^{\ast }$ on $|\psi ⟩\in \mathcal{H}$ is now expressed in terms of their components as

$$⟨\alpha |\psi ⟩=\sum _{ij}{\alpha }_i{\psi }_j⟨{\epsilon }_i\mid e_j⟩=\sum _{ij}{\alpha }_i{\psi }_j{\delta }_{ij}=\sum _i{\alpha }_i{\psi }_i.\text{\hspace{1em}(1.28)}$$

One may consider $|\psi ⟩$ as a column vector and $⟨\alpha |$ as a row vector so that $⟨\alpha |\psi ⟩$ is regarded as just a matrix multiplication of a row vector and a column vector, yielding a scalar.

It is possible to introduce a one-to-one correspondence between elements in $\mathcal{H}$ and ${\mathcal{H}}^{\ast }$ . Let us fix a basis $\{|e_k⟩\}$ of $\mathcal{H}$ and $\{⟨{\epsilon }_k|\}$ of ${\mathcal{H}}^{\ast }$ . Then corresponding to $|\psi ⟩={\sum }_k{\psi }_k|e_k⟩$ , there exists an element $⟨\psi |={\sum }_k{\psi }_k^{\ast }⟨{\epsilon }_k|\in {\mathcal{H}}^{\ast }$ . The reason for the complex conjugation of ${\psi }_k$ becomes clear shortly. Then it is possible to introduce an inner product between two elements of $\mathcal{H}$ . Let $|\varphi ⟩,|\psi ⟩\in \mathcal{H}$ . Their inner product is defined by

$$\left(|\varphi ⟩,|\psi ⟩\right)\equiv ⟨\varphi |\psi ⟩=\sum _k{\varphi }_k^{\ast }{\psi }_k.\text{\hspace{1em}(1.29)}$$

We customarily use the same letter to denote corresponding bras and kets. The norm of a vector $|\psi ⟩$ is naturally defined by the inner product. Let $\Vert |\psi ⟩\Vert =\sqrt{⟨\psi |\psi ⟩}$ . It is easy to show that this definition satisfies all the axioms of the norm. Note that the norm is real and non-negative thanks to the complex conjugation in the components of the bra vector.

By using the inner product between two ket vectors, it becomes possible to construct an orthonormal basis $\{|e_k⟩\}$ such that $(|e_i⟩,|e_j⟩)=⟨e_i|e_j⟩={\delta }_{ij}$ . Suppose $|\psi ⟩={\sum }_k{\psi }_k|e_k⟩$ . By multiplying $⟨e_k|$ from the left, one obtains $⟨e_k|\psi ⟩={\psi }_k$ . Then $|\psi ⟩$ is expressed as $|\psi ⟩={\sum }_k⟨e_k|\psi ⟩|e_k⟩={\sum }_k|e_k⟩⟨e_k|\psi ⟩$ . Since this is true for any $|\psi ⟩$ , we have obtained the completeness relation

$$\sum _k|e_k⟩⟨e_k|=I,\text{\hspace{1em}(1.30)}$$

$I$ being the identity operator in $\mathcal{H}$ (the unit matrix when $\mathcal{H}$ is finite dimensional).

1.2.2 Axioms of canonical quantization

Given an isolated classical dynamical system such as a harmonic oscillator, we can construct a corresponding quantum system following a set of axioms.

A1. There exists a Hilbert space $\mathcal{H}$ for a quantum system and the state of the system is required to be described by a vector $|\psi ⟩\in \mathcal{H}$ . In this sense, $|\psi ⟩$ is also called the state or a state vector. Moreover, two states $|\psi ⟩$ and $c|\psi ⟩$ ( $c\in \mathbb{C},c\ne 0$ ) describe the same state. The state can also be described as a ray representation of $\mathcal{H}$ .

A2. A physical quantity $A$ in classical mechanics is replaced by a Hermitian operator $\hat{A}$ acting on $\mathcal{H}$ . The operator $\hat{A}$ is often called an observable. The result obtained when $A$ is measured is one of the eigenvalues of $\hat{A}$ . (The Hermiticity of $\hat{A}$ has been assumed to guarantee real eigenvalues.)

A3. The Poisson bracket in classical mechanics is replaced by the commutator

$$[\hat{A},\hat{B}]\equiv \hat{A}\hat{B}-\hat{B}\hat{A}\text{\hspace{1em}(1.31)}$$

multiplied by $-\mathrm{i}/\hslash$ . The unit in which $\hslash =1$ will be employed hereafter unless otherwise stated explicitly. The fundamental commutation relations are (cf (1.22))

$$[{\hat{q}}_i,{\hat{q}}_j]=[{\hat{p}}_i,{\hat{p}}_j]=0[{\hat{q}}_i,{\hat{p}}_j]=\mathrm{i}{\delta }_{ij}.\text{\hspace{1em}(1.32)}$$

Under this replacement, Hamilton’s equations of motion become

$$\frac{\mathrm{d}{\hat{q}}_i}{\mathrm{d}t}=\frac{1}{\mathrm{i}}[{\hat{q}}_i,H]\frac{\mathrm{d}{\hat{p}}_i}{\mathrm{d}t}=\frac{1}{\mathrm{i}}[{\hat{p}}_i,H].\text{\hspace{1em}(1.33)}$$

When a classical quantity $A$ is independent of $t$ explicitly, $A$ satisfies the same equation as Hamilton’s equation. By analogy, for $\hat{A}$ which does not depend on $t$ explicitly, one has Heisenberg’s equation of motion:

$$\frac{\mathrm{d}\hat{A}}{\mathrm{d}t}=\frac{1}{\mathrm{i}}[\hat{A},\hat{H}].\text{\hspace{1em}(1.34)}$$

A4. Let $|\psi ⟩\in \mathcal{H}$ be an arbitrary state. Suppose one prepares many systems, each of which is in this state. Then, observation of $A$ in these systems at time $t$ yields random results in general. Then the expectation value of the results is given by

$$⟨A{⟩}_t=\frac{⟨\psi |\hat{A}(t)|\psi ⟩}{⟨\psi |\psi ⟩}.\text{\hspace{1em}(1.35)}$$

A5. For any physical state $|\psi ⟩\in \mathcal{H}$ , there exists an operator for which $|\psi ⟩$ is one of the eigenstates. ${}^{10}$

These five axioms are adopted as the rules of the game. A few comments are in order. Let us examine axiom A4 more carefully. Let us assume that $|\psi ⟩$ is normalized as $\Vert |\psi ⟩{\Vert }^2=⟨\psi |\psi ⟩=1$ for simplicity. Suppose $\hat{A}(t)$ has the set of discrete eigenvalues $\{a_n\}$ with the corresponding normalized eigenvectors $\{|n⟩\}$ :¹

$$\hat{A}(t)|n⟩=a_n|n⟩⟨n|n⟩=1.$$

Then the expectation value of $\hat{A}(t)$ with respect to an arbitrary state

$$|\psi ⟩=\sum _n{\psi }_n|n⟩{\psi }_n=⟨n|\psi ⟩$$

is

$$⟨\psi |\hat{A}(t)|\psi ⟩=\sum _{m,n}{\psi }_m^{\ast }{\psi }_n⟨m|\hat{A}(t)|n⟩=\sum _n{a}_n|{\psi }_n{|}^2.$$

From the fact that the result of the measurement of $A$ in state $|n⟩$ is always $a_n$ , it follows that the probability of the outcome of the measurement being $a_n$ , that is the probability of $|\psi ⟩$ being in $|n⟩$ , is

$$|{\psi }_n{|}^2=|⟨n|\psi ⟩{|}^2.$$

The number $⟨n|\psi ⟩$ represents the ‘weight’ of the state $|n⟩$ in the state $|\psi ⟩$ and is called the probability amplitude.

If $\hat{A}$ has a continuous spectrum $a$ , the state $|\psi ⟩$ is expanded as

$$|\psi ⟩=\int \mathrm{d}a\psi (a)|a⟩.$$

The completeness relation now takes the form

$$\int \mathrm{d}a|a⟩⟨a|=I.\text{\hspace{1em}(1.36)}$$

Then, from the identity $\int \mathrm{d}{a}^{\prime }|a^{\prime }⟩⟨a^{\prime }|a⟩=|a⟩$ , one must have the normalization

$$⟨a^{\prime }|a⟩=\delta \left(a^{\prime }-a\right),\text{\hspace{1em}(1.37)}$$

where $\delta (a)$ is the Dirac $\delta$ -function. The expansion coefficient $\psi (a)$ is obtained from this normalization condition as $\psi (a)=⟨a|\psi ⟩$ . If $|\psi ⟩$ is normalized as $⟨\psi |\psi ⟩=1$ , one should have

$$1=\int \mathrm{d}a\mathrm{d}{a}^{\prime }{\psi }^{\ast }(a)\psi (a^{\prime })⟨a|a^{\prime }⟩=\int \mathrm{d}a|\psi (a){|}^2.$$

It also follows from the relation

$$⟨\psi |\hat{A}|\psi ⟩=\int a|\psi (a){|}^2\mathrm{d}a$$

that the probability with which the measured value of $A$ is found in the interval $[a,a+\mathrm{da}]$ is $|\psi (a){|}^2\mathrm{da}$ . Therefore, the probability density is given by

$$\rho (a)=|⟨a|\psi ⟩{|}^2.\text{\hspace{1em}(1.38)}$$

Finally let us clarify why axiom A5 is required. Suppose that the system is in the state $|\psi ⟩$ and assume that the probability of the state to be in $|\varphi ⟩$ simultaneously is $|⟨\psi |\varphi ⟩{|}^2$ . This has already been mentioned, when $|\psi ⟩$ is an eigenstate of some observable. Axiom A5 asserts that this is true for an arbitrary state $|\psi ⟩$ .

1.2.3 Heisenberg equation, Heisenberg picture and Schrödinger picture

The formal solution to the Heisenberg equation of motion

$$\frac{\mathrm{d}\hat{A}}{\mathrm{d}t}=\frac{1}{\mathrm{i}}[\hat{A},\hat{H}]$$

is easily obtained as

$$\hat{A}(t)={\mathrm{e}}^{\mathrm{i}\hat{H}t}\hat{A}(0){\mathrm{e}}^{-\mathrm{i}\hat{H}t}.\text{\hspace{1em}(1.39)}$$

Therefore, the operators $\hat{A}(t)$ and $\hat{A}(0)$ are related by the unitary operator

$$\hat{U}(t)={\mathrm{e}}^{-\mathrm{i}\hat{H}t}\text{\hspace{1em}(1.40)}$$

and, hence, are unitary equivalent. This formalism, in which operators depend on $t$ , while states do not, is called the Heisenberg picture.

It is possible to introduce another picture which is equivalent to the Heisenberg picture. Let us write down the expectation value of $\hat{A}$ with respect to the state $|\psi ⟩$ as

$$\begin{array}{rl}⟨\hat{A}(t)⟩& =⟨\psi |e^{i\hat{H}t}\hat{A}(0)e^{-i\hat{H}t}|\psi ⟩\\ & =\left(⟨\psi |e^{i\hat{H}t}\right)\hat{A}(0)\left(e^{-i\hat{H}t}|\psi ⟩\right).\end{array}$$

If we write $|\psi (t)⟩\equiv {\mathrm{e}}^{-\mathrm{i}\hat{H}t}|\psi ⟩$ , we find that the expectation value at $t$ is also expressed as

$$⟨\hat{A}(t)⟩=⟨\psi (t)|\hat{A}(0)|\psi (t)⟩.\text{\hspace{1em}(1.41)}$$

Thus, states depend on $t$ while operators do not in this formalism. This formalism is called the Schrödinger picture.

Our next task is to find the equation of motion for $|\psi (t)⟩$ . To avoid confusion, quantities associated with the Schrödinger picture (the Heisenberg picture) are denoted with the subscript S (H), respectively. Thus, $|\psi (t){⟩}_{\mathrm{S}}={\mathrm{e}}^{-\mathrm{i}\hat{H}t}|\psi {⟩}_{\mathrm{H}}$ and ${\hat{A}}_{\mathrm{S}}={\hat{A}}_{\mathrm{H}}(0)$ . By differentiating $|\psi (t){⟩}_{\mathrm{S}}$ with respect to $t$ , one finds the Schrödinger equation:

$$\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}t}|\psi (t){⟩}_{\mathrm{S}}=\hat{H}|\psi (t){⟩}_{\mathrm{S}}.\text{\hspace{1em}(1.42)}$$

Note that the Hamiltonian $\hat{H}$ is the same for both the Schrödinger picture and the Heisenberg picture. We will drop the subscripts S and H whenever this does not cause confusion.

1.2.4 Wavefunction

Let us consider a particle moving on the real line $\mathbb{R}$ and let $\hat{x}$ be the position operator with the eigenvalue $y$ and the corresponding eigenvector $|y⟩$ ; $\hat{x}|y⟩=y|y⟩$ . The eigenvectors are normalized as $⟨x|y⟩=\delta (x-y)$ .

Similarly, let $q$ be the eigenvalue of $\hat{p}$ with the eigenvector $|q⟩$ ; $\hat{p}|q⟩=q|q⟩$ such that $⟨p|q⟩=\delta (p-q)$ .

Let $|\psi ⟩\in \mathcal{H}$ be a state. The inner product

$$\psi (x)\equiv ⟨x|\psi ⟩\text{\hspace{1em}(1.43)}$$

is the component of $|\psi ⟩$ in the basis $|x⟩$ ,

$$|\psi ⟩=\int |x⟩⟨x|dx|\psi ⟩=\int \psi (x)|x⟩dx.$$

The coefficient $\psi (x)\in \mathbb{C}$ is called the wavefunction. According to the earlier axioms of quantum mechanics outlined, it is the probability amplitude of finding the particle at $x$ in the state $|\psi ⟩$ , namely $|\psi (x){|}^2\mathrm{d}x$ is the probability of finding the particle in the interval $[x,x+\mathrm{d}x]$ . Then it is natural to impose the normalization condition

$$\int \mathrm{d}x|\psi (x){|}^2=⟨\psi |\psi ⟩=1\text{\hspace{1em}(1.44)}$$

since the probability of finding the particle anywhere on the real line is always unity.

Similarly, $\psi (p)=⟨p|\psi ⟩$ is the probability amplitude of finding the particle in the state with the momentum $p$ and the probability of finding the momentum of the particle in the interval $[p,p+\mathrm{d}p]$ is $|\psi (p){|}^2\mathrm{d}p$ .

The inner product of two states in terms of the wavefunctions is

$$\begin{array}{rl}⟨\psi |\varphi ⟩& =\int \mathrm{d}x⟨\psi |x⟩⟨x|\varphi ⟩=\int \mathrm{d}x{\psi }^{\ast }(x)\varphi (x),(1.45a)\\ & =\int \mathrm{d}p⟨\psi |p⟩⟨p|\varphi ⟩=\int \mathrm{d}p{\psi }^{\ast }(p)\varphi (p).(1.45b)\end{array}$$

An abstract ket vector is now expressed in terms of a more concrete wavefunction $\psi (x)$ or $\psi (p)$ . What about the operators? Now we write down the operators in the basis $|x⟩$ . From the defining equation $\hat{x}|x⟩=x|x⟩$ , one obtains $⟨x|\hat{x}=⟨x|x$ , which yields after multiplication by $|\psi ⟩$ from the right,

$$⟨x|\hat{x}|\psi ⟩=x⟨x|\psi ⟩=x\psi (x).\text{\hspace{1em}(1.46)}$$

This is often written as $(\hat{x}\psi )(x)=x\psi (x)$

What about the momentum operator $\hat{p}$ ? Let us consider the unitary operator

$$\hat{U}(a)={\mathrm{e}}^{-\mathrm{i}a\hat{p}}.$$

Lemma 1.1

The operator $\hat{U}(a)$ defined as before satisfies

$$\hat{U}(a)|x⟩=|x+a⟩.\text{\hspace{1em}(1.47)}$$

Proof

It follows from $[\hat{x},\hat{p}]=\mathrm{i}$ that $[\hat{x},{\hat{p}}^n]=\mathrm{i}n{\hat{p}}^{n-1}$ for $n=1,2,\dots$ Accordingly, we have

$$[\hat{x},\hat{U}(a)]=\left[\hat{x},\sum _n\frac{(-\mathrm{i}a{)}^n}{n!}{\hat{p}}^n\right]=a\hat{U}(a)$$

which can also be written as

$$\hat{x}\hat{U}(a)|x⟩=\hat{U}(a)(\hat{x}+a)|x⟩=(x+a)\hat{U}(a)|x⟩.$$

This shows that $\hat{U}(a)|x⟩\propto |x+a⟩$ . Since $\hat{U}(a)$ is unitary, it preserves the norm of a vector. Thus, $\hat{U}(a)|x⟩=|x+a⟩$ .

Let us take an infinitesimal number $\epsilon$ . Then

$$\hat{U}(\epsilon )|x⟩=|x+\epsilon ⟩\simeq (1-\mathrm{i}\epsilon \hat{p})|x⟩.$$

It follows from this that

$$\hat{p}|x⟩=\frac{|x+\epsilon ⟩-|x⟩}{-\mathrm{i}\epsilon }\stackrel{\epsilon \to 0}{\to }\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}x}|x⟩\text{\hspace{1em}(1.48)}$$

and its dual

$$⟨x|\hat{p}=\frac{⟨x+\epsilon |-⟨x|}{\mathrm{i}\epsilon }\stackrel{\epsilon \to 0}{\to }-\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}x}⟨x|.\text{\hspace{1em}(1.49)}$$

Therefore, for any state $|\psi ⟩$ , one obtains

$$⟨x|\hat{p}|\psi ⟩=-\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}x}⟨x|\psi ⟩=-\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}x}\psi (x).\text{\hspace{1em}(1.50)}$$

This is also written as $(\hat{p}\psi )(x)=-\mathrm{id}\psi (x)/\mathrm{d}x$

Similarly, if one uses a basis $|p⟩$ , one will have the momentum representation of the operators as

$$\hat{x}|p⟩=-\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}p}|p⟩\text{\hspace{1em}(1.51)}$$

$$\hat{p}|p⟩=p|p⟩\text{\hspace{1em}(1.52)}$$

$$⟨p|\hat{x}|\psi ⟩=\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}p}\psi (p)\text{\hspace{1em}(1.53)}$$

$$⟨p|\hat{p}|\psi ⟩=p\psi (p).\text{\hspace{1em}(1.54)}$$

Exercise 1.2

Prove (1.51)-(1.54).

Proposition 1.1

$$⟨x|p⟩=\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{\mathrm{i}px}\text{\hspace{1em}(1.55)}$$

$$⟨p|x⟩=\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-\mathrm{i}px}\text{\hspace{1em}(1.56)}$$

Proof

Take $|\psi ⟩=|p⟩$ in the relation

$$(\hat{p}\psi )(x)=⟨x|\hat{p}|\psi ⟩=-\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}x}\psi (x)$$

to find

$$p⟨x|p⟩=⟨x|\hat{p}|p⟩=-\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}x}⟨x|p⟩.$$

The solution is easily found to be

$$⟨x|p⟩=C{\mathrm{e}}^{\mathrm{i}px}.$$

The normalization condition requires that

$$\begin{array}{rl}\delta (x-y)& =⟨x|y⟩=⟨x|\int |p⟩⟨p|dp|y⟩\\ & =C^2\int \mathrm{d}p{\mathrm{e}}^{\mathrm{i}p(x-y)}\\ & =C^{2}2\pi \delta (x-y),\end{array}$$

where $C$ has been taken to be real. This shows that $C=1/\sqrt{2\pi }$ . The proof of (1.56) is left as an exercise.

Thus, $\psi (x)$ and $\psi (p)$ are related as

$$\psi (p)=⟨p|\psi ⟩=\int \mathrm{d}x⟨p|x⟩⟨x|\psi ⟩=\int \frac{\mathrm{d}x}{\sqrt{2\pi }}{\mathrm{e}}^{-\mathrm{i}px}\psi (x)\text{\hspace{1em}(1.57)}$$

which is nothing other than the Fourier transform of $\psi (x)$

Let us next derive the Schrödinger equation which $\psi (x)$ satisfies. By applying $⟨x|$ on (1.42) from the left, we obtain

$$⟨x|\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}t}|\psi (t)⟩=⟨x|\hat{H}|\psi (t)⟩$$

where the subscript S has been dropped. For a Hamiltonian of the type $\hat{H}={\hat{p}}^2/2m+V(\hat{x})$ , we obtain the time-dependent Schrödinger equation:

$$\begin{array}{rl}\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}t}\psi (x,t)& =⟨x\left|\frac{{\hat{p}}^2}{2m}+V(\hat{x})\right|\psi (t)⟩\\ & =-\frac{1}{2m}\frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}\psi (x,t)+V(x)\psi (x,t),\end{array}\text{\hspace{1em}(1.58)}$$

where $\psi (x,t)\equiv ⟨x|\psi (t)⟩$

Suppose a solution of this equation is written in the form $\psi (x,t)=T(t)\varphi (x)$ . By substituting this into (1.58) and dividing the result by $\psi (x,t)$ , we obtain

$$\frac{\mathrm{i}{T}^{\prime }(t)}{T(t)}=\frac{-{\varphi }^{\prime \prime }(x)/2m+V(x)\varphi (x)}{\varphi (x)}$$

where the prime denotes the derivative with respect to a relevant variable. Since the LHS is a function of $t$ only while the right-hand side (RHS) of $x$ only, they must be a constant, which we label $E$ . Accordingly, there are two equations, which should be solved simultaneously,

$$\mathrm{i}{T}^{\prime }(t)=ET(t)\text{\hspace{1em}(1.59)}$$

$$-\frac{1}{2m}\frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}\varphi (x)+V(x)\varphi (x)=E\varphi (x).\text{\hspace{1em}(1.60)}$$

The first equation is easily solved to yield

$$T(t)=\exp (-\mathrm{i}Et)\text{\hspace{1em}(1.61)}$$

while the second one is the eigenvalue problem of the Hamiltonian operator and called the time-independent Schrödinger equation, the stationary state Schrödinger equation or, simply, the Schrödinger equation. For three-dimensional space, it is written as

$$-\frac{1}{2m}{\nabla }^2\varphi (\mathbf{x})+V(\mathbf{x})\varphi (\mathbf{x})=E\varphi (\mathbf{x}).\text{\hspace{1em}(1.62)}$$

1.2.5 Harmonic oscillator

It is instructive to stop here for the moment and work out some non-trivial example. We take a one-dimensional harmonic oscillator as an example since it is not trivial, it is still solvable exactly and it is very important in the following applications.

The Hamiltonian operator is

$$\hat{H}=\frac{{\hat{p}}^2}{2m}+\frac{1}{2}m{\omega }^2{\hat{x}}^2[\hat{x},\hat{p}]=\mathrm{i}.\text{\hspace{1em}(1.63)}$$

The (time-independent) Schrödinger equation is

$$-\frac{1}{2m}\frac{{\mathrm{d}}^2}{\mathrm{d}{x}^2}\psi (x)+\frac{1}{2}m{\omega }^2{x}^2\psi (x)=E\psi (x).\text{\hspace{1em}(1.64)}$$

By rescaling the variables as $\xi =\sqrt{m\omega }x,\epsilon =E/\hslash \omega$ , one arrives at

$${\psi }^{\prime \prime }+(\epsilon -{\xi }^2)\psi =0.\text{\hspace{1em}(1.65)}$$

The normalizable solution of this ordinary differential equation (ODE) exists only when $\epsilon ={\epsilon }_n\equiv (n+\frac{1}{2})$ ( $n=0,1,2,\dots$ ) namely

$$E=E_n\equiv \left(n+\frac{1}{2}\right)\omega (n=0,1,2,\dots )\text{\hspace{1em}(1.66)}$$

and the normalized solution is written in terms of the Hermite polynomial

$$H_n(\xi )=(-1{)}^n{\mathrm{e}}^{{\xi }^2/2}\frac{{\mathrm{d}}^n{\mathrm{e}}^{-{\xi }^2/2}}{\mathrm{d}{\xi }^n}\text{\hspace{1em}(1.67)}$$

as

$$\psi (\xi )=\sqrt{\frac{m\omega }{2^{n}n!\sqrt{\pi }}}{H}_n(\xi ){\mathrm{e}}^{-{\xi }^2/2}.\text{\hspace{1em}(1.68)}$$

This eigenvalue problem can also be analysed by an algebraic method. Define the annihilation operator $\hat{a}$ and the creation operator ${\hat{a}}^{†}$ by

$$\hat{a}=\sqrt{\frac{m\omega }{2}}\hat{x}+\mathrm{i}\sqrt{\frac{1}{2m\omega }}\hat{p}\text{\hspace{1em}(1.69)}$$

$${\hat{a}}^{†}=\sqrt{\frac{m\omega }{2}}\hat{x}-\mathrm{i}\sqrt{\frac{1}{2m\omega }}\hat{p}.\text{\hspace{1em}(1.70)}$$

The number operator $\hat{N}$ is defined by

$$\hat{N}={\hat{a}}^{†}\hat{a}.\text{\hspace{1em}(1.71)}$$

Exercise 1.3

Show that

$$[\hat{a},\hat{a}]=[{\hat{a}}^{†},{\hat{a}}^{†}]=0[\hat{a},{\hat{a}}^{†}]=1\text{\hspace{1em}(1.72)}$$

and

$$[\hat{N},\hat{a}]=-\hat{a}[\hat{N},{\hat{a}}^{†}]={\hat{a}}^{†}.\text{\hspace{1em}(1.73)}$$

Show also that

$$\hat{H}=\left(\hat{N}+\frac{1}{2}\right)\omega .\text{\hspace{1em}(1.74)}$$

Let $|n⟩$ be a normalized eigenvector of $\hat{N}$

$$\hat{N}|n⟩=n|n⟩.$$

Then it follows from the commutation relations proved in exercise 1.3 that

$$\hat{N}(\hat{a}|n⟩)=(\hat{a}\hat{N}-\hat{a})|n⟩=(n-1)(\hat{a}|n⟩)$$

$$\hat{N}({\hat{a}}^{†}|n⟩)=({\hat{a}}^{†}\hat{N}+{\hat{a}}^{†})|n⟩=(n+1)({\hat{a}}^{†}|n⟩).$$

Therefore, $\hat{a}$ decreases the eigenvalue by one while ${\hat{a}}^{†}$ increases it by one, hence the name annihilation and creation. Note that the eigenvalue $n\ge 0$ since

$$n=⟨n|\hat{N}|n⟩=(⟨n|{\hat{a}}^{†})(\hat{a}|n⟩)=\Vert \hat{a}|n⟩{\Vert }^2\ge 0.$$

The equality holds if and only if $\hat{a}|n⟩=0$ . Take a fixed $n_0>0$ and apply $\hat{a}$ many times on $|n_0⟩$ . Eventually the eigenvalue of ${\hat{a}}^k|n_0⟩$ will be negative for some integer $k>n_0$ , which is a contradiction. This can be avoided only when $n_0$ is a non-negative integer. Thus, there exists a state $|0⟩$ which satisfies $\hat{a}|0⟩=0$ . The state $|0⟩$ is called the ground state. Since $\hat{N}|0⟩={\hat{a}}^{†}\hat{a}|0⟩=0$ , this state is

the eigenvector of $\hat{N}$ with the eigenvalue 0. The wavefunction ${\psi }_0(x)\equiv ⟨x|0⟩$ is obtained by solving the first-order ODE

$$⟨x|\hat{a}|0⟩=\sqrt{\frac{1}{2m\omega }}\left(\frac{\mathrm{d}}{\mathrm{d}x}{\psi }_0(x)+m\omega x{\psi }_0(x)\right)=0.\text{\hspace{1em}(1.75)}$$

The solution is easily found to be

$${\psi }_0(x)=C\exp (-m\omega x^2/2)\text{\hspace{1em}(1.76)}$$

where $C$ is the normalization constant given in (1.68). An arbitrary vector $|n⟩$ is obtained from $|0⟩$ by a repeated application of ${\hat{a}}^{†}$

Exercise 1.4

Show that

$$|n⟩=\frac{1}{\sqrt{n!}}({\hat{a}}^{†}{)}^n|0⟩\text{\hspace{1em}(1.77)}$$

satisfies $\hat{N}|n⟩=n|n⟩$ and is normalized.

Thus, the spectrum of $\hat{N}$ turns out to be $\mathrm{Spec}\hat{N}=\{0,1,2,\dots \}$ and hence the spectrum of the Hamiltonian is

$$\mathrm{Spec}\hat{H}=\left\{\frac{1}{2},\frac{3}{2},\frac{5}{2},\dots \right\}.\text{\hspace{1em}(1.78)}$$

Endnotes

Unplaced source note

In quantum mechanics, a Hilbert space often means the space of square integrable functions $L^2(M)$ on a space (manifold) $M$ . In the following, however, we need to deal with such functions as $\delta (x)$ and ${\mathrm{e}}^{\mathrm{i}kx}$ with infinite norm. An extended Hilbert space which contains such functions is called the rigged Hilbert space. The treatment of Hilbert spaces here is not mathematically rigorous but it will not cause any inconvenience.

Unplaced source note

We assume $\mathcal{H}$ is separable and there are, at most, a countably infinite number of vectors in the basis. Note that we cannot impose an orthonormal condition since we have not defined the norm of a vector.

Unplaced source note

An operator on $\mathcal{H}$ is denoted by $\hat{\cdot }$ . This symbol will be dropped later unless this may cause confusion.

Unplaced source note

This axiom is often ignored in the literature. The raison d’etre of this axiom will be clarified later.

来源与版本

正文：英文 · OCR 机器稿 · 待校对

核对状态：OCR 机器稿 · 待校对

原书 PDF

来源版本：second-edition-local

来源 PDF SHA-256：fcd24c3371d092a8c699ccf5cf0765462924b6c48412804231076e8084b1940d

OCR 来源 SHA-256：0b6324f01172a3548cd34382e82fd8bfae574602b6b2cfaa357ba96bc2c35478

OCR 产物 SHA-256：c7521cd3f241c8d234f9eb057867b55805ae38d4c06e80ff2eb9494baedf8a74

Footnotes

1. Since $\hat{A}(t)$ is Hermitian, it is always possible to choose $\{|n⟩\}$ to be orthonormal. ↩