1.1 Analytical mechanics

We introduce some elementary principles of Lagrangian and Hamiltonian formalisms that are necessary to understand quantum mechanics.

1.1.1 Newtonian mechanics

Let us consider the motion of a particle $m$ in three-dimensional space and let $x(t)$ denote the position of $m$ at time $t$¹ . Suppose this particle is moving under an external force $F(x)$ . Then $x(t)$ satisfies the second-order differential equation

$$m\frac{{\mathrm{d}}^2\mathbf{x}(t)}{\mathrm{d}{t}^2}=\mathbf{F}(\mathbf{x}(t))\text{\hspace{1em}(1.1)}$$

called Newton’s equation or the equation of motion.

If force $F(x)$ is expressed in terms of a scalar function $V(x)$ as $F(x)=-\nabla V(x)$ , the force is called a conserved force and the function $V(x)$ is called the potential energy or simply the potential. When $F$ is a conserved force, the combination

$$E=\frac{m}{2}{\left(\frac{\mathrm{d}\mathbf{x}}{\mathrm{d}t}\right)}^2+V(\mathbf{x})\text{\hspace{1em}(1.2)}$$

is conserved. In fact,

$$\frac{\mathrm{d}E}{\mathrm{d}t}=\sum _{k=x,y,z}\left[m\frac{\mathrm{d}{x}_k}{\mathrm{d}t}\frac{{\mathrm{d}}^2{x}_k}{\mathrm{d}{t}^2}+\frac{\partial V}{\partial x_k}\frac{\mathrm{d}{x}_k}{\mathrm{d}t}\right]=\sum _k\left(m\frac{{\mathrm{d}}^2{x}_k}{\mathrm{d}{t}^2}+\frac{\partial V}{\partial x_k}\right)\frac{\mathrm{d}{x}_k}{\mathrm{d}t}=0$$

where use has been made of the equation of motion. The function $E$ , which is often the sum of the kinetic energy and the potential energy, is called the energy.

Example 1.1 · One-dimensional harmonic oscillator

Let $x$ be the coordinate and suppose the force acting on $m$ is $F(x)=-kx$ , $k$ being a constant. This force is conservative. In fact, $V(x)=\frac{1}{2}k{x}^2$ yields $F(x)=-\mathrm{d}V(x)/\mathrm{d}x=-kx$ . In general, any one-dimensional force $F(x)$ which is a function of $x$ only is conserved and the potential is given by

$$V(x)=-{\int }^{x}F(\xi )\mathrm{d}\xi .$$

An example of a force that is not conserved is friction $F=-\eta \mathrm{d}x/\mathrm{d}t$ . We will be concerned only with conserved forces in the following.

1.1.2 Lagrangian formalism

Newtonian mechanics has the following difficulties;

1. This formalism is based on a vector equation (1.1) which is not very easy to handle unless an orthogonal coordinate system is employed.
2. The equation of motion is a second-order equation and the global properties of the system cannot be figured out easily.
3. The analysis of symmetries is not easy.
4. Constraints are difficult to take into account.

Furthermore, quantum mechanics cannot be derived directly from Newtonian mechanics. The Lagrangian formalism is now introduced to overcome these difficulties.

Let us consider a system whose state (the position of masses for example) is described by $N$ parameters $\{q_i\}$ ( $1\le i\le N$ ). The parameter is an element of some space $M$ . The space $M$ is called the configuration space and the $\{q_i\}$ are called the generalized coordinates. If one considers a particle on a circle, for example, the generalized coordinate $q$ is an angle $\theta$ and the configuration space $M$ is a circle. The generalized velocity is defined by ${\dot{q}}_i=\mathrm{d}{q}_i/\mathrm{d}t$ .

The Lagrangian $L(q,\dot{q})$ is a function to be defined in Hamilton’s principle later. We will restrict ourselves mostly to one-dimensional space but generalization to higher-dimensional space should be obvious. Let us consider a trajectory $q(t)$ ( $t\in [t_i,t_f]$ ) of a particle with conditions $q(t_i)=q_i$ and $q(t_f)=q_f$ . Consider a functional²

$$S[q(t),\dot{q}(t)]={\int }_{t_i}^{t_f}L(q,\dot{q})\mathrm{d}t\text{\hspace{1em}(1.3)}$$

called the action. Given a trajectory $q(t)$ and $\dot{q}(t)$ , the action $S[q,\dot{q}]$ produces a real number. Hamilton’s principle, also known as the principle of the least action, claims that the physically realized trajectory corresponds to an extremum of the action. Now the Lagrangian must be chosen so that Hamilton’s principle is fulfilled.

It turns out to be convenient to write Hamilton’s principle in a local form as a differential equation. Suppose $q(t)$ is a path realizing an extremum of $S$ . Consider a variation $\delta q(t)$ of the trajectory such that $\delta q(t_i)=\delta q(t_f)=0$ . The action changes under this variation by

$$\begin{array}{rl}\delta S& ={\int }_{t_i}^{t_f}L(q+\delta q,\dot{q}+\delta \dot{q})dt-{\int }_{t_i}^{t_f}L(q,\dot{q})dt\\ & ={\int }_{t_i}^{t_f}\left(\frac{\partial L}{\partial q}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}}\right)\delta q\mathrm{d}t\end{array}\text{\hspace{1em}(1.4)}$$

which must vanish because $q$ yields an extremum of $S$ . Since this is true for any $\delta q$ , the integrand of the last line of (1.4) must vanish. Thus, the Euler-Lagrange equation

$$\frac{\partial L}{\partial q}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}}=0\text{\hspace{1em}(1.5)}$$

has been obtained. If there are $N$ degrees of freedom, one obtains

$$\frac{\partial L}{\partial q_k}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial {\dot{q}}_k}=0(1\le k\le N).\text{\hspace{1em}(1.6)}$$

If we introduce the generalized momentum conjugate to the coordinate $q_k$ by

$$p_k=\frac{\partial L}{\partial {\dot{q}}_k}\text{\hspace{1em}(1.7)}$$

the Euler-Lagrange equation takes the form

$$\frac{\mathrm{d}{p}_k}{\mathrm{d}t}=\frac{\partial L}{\partial q_k}.\text{\hspace{1em}(1.8)}$$

By requiring this equation to reduce to Newton’s equation, one quickly finds the possible form of the Lagrangian in the ordinary mechanics of a particle. Let us put $L=\frac{1}{2}m{\dot{q}}^2-V(q)$ . By substituting this Lagrangian into the Euler-Lagrange equation, it is easily shown that it reduces to Newton’s equation of motion,

$$m{\ddot{q}}_k+\frac{\partial V}{\partial q_k}=0.\text{\hspace{1em}(1.9)}$$

Let us consider the one-dimensional harmonic oscillator for example. The Lagrangian is

$$L(x,\dot{x})=\frac{1}{2}m{\dot{x}}^2-\frac{1}{2}k{x}^2\text{\hspace{1em}(1.10)}$$

from which one finds $m\ddot{x}+kx=0$

It is convenient for later purposes to introduce the notion of a functional derivative. Let us consider the case with a single degree of freedom for simplicity. Define the functional derivative of $S$ with respect to $q$ by

$$\frac{\delta S[q,\dot{q}]}{\delta q(s)}\equiv \underset{\epsilon \to 0}{\lim }\frac{\left\{S[q(t)+\epsilon \delta (t-s),\dot{q}(t)+\epsilon \frac{\mathrm{d}}{\mathrm{d}t}\delta (t-s)]-S[q(t),\dot{q}(t)]\right\}}{\epsilon }.\text{\hspace{1em}(1.11)}$$

Since

$$\begin{array}{rl}& S\left[q(t)+\epsilon \delta (t-s),\dot{q}(t)+\epsilon \frac{\mathrm{d}}{\mathrm{d}t}\delta (t-s)\right]\\ & =\int \mathrm{d}tL\left(q(t)+\epsilon \delta (t-s),\dot{q}(t)+\epsilon \frac{\mathrm{d}}{\mathrm{d}t}\delta (t-s)\right)\\ & =\int \mathrm{d}tL(q,\dot{q})+\epsilon \int \mathrm{d}t\left(\frac{\partial L}{\partial q}\delta (t-s)+\frac{\partial L}{\partial \dot{q}}\frac{\mathrm{d}}{\mathrm{d}t}\delta (t-s)\right)+\mathcal{O}({\epsilon }^2)\\ & =S[q,\dot{q}]+\epsilon \left(\frac{\partial L}{\partial q}(s)-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial \dot{q}}(s)\right)+\mathcal{O}({\epsilon }^2),\end{array}$$

the Euler-Lagrange equation may be written as

$$\frac{\delta S}{\delta q(s)}=\frac{\partial L}{\partial q}(s)-\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\partial L}{\partial \dot{q}}\right)(s)=0.\text{\hspace{1em}(1.12)}$$

Let us next consider symmetries in the context of the Lagrangian formalism. Suppose the Lagrangian $L$ is independent of a certain coordinate $q_k$ . Such a coordinate is called cyclic. The momentum which is conjugate to a cyclic coordinate is conserved. In fact, the condition $\partial L/\partial q_k=0$ leads to

$$\frac{\mathrm{d}{p}_k}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial {\dot{q}}_k}=\frac{\partial L}{\partial q_k}=0.\text{\hspace{1em}(1.13)}$$

This argument can be mathematically elaborated as follows. Suppose the Lagrangian $L$ has a symmetry, which is continuously parametrized. This means, more precisely, that the action $S=\int \mathrm{d}tL$ is invariant under the symmetry operation on $q_k(t)$ . Let us consider an infinitesimal symmetry operation $q_k(t)\to q_k(t)+\delta q_k(t)$ on the path $q_k(t)$ . This implies that if $q_k(t)$ is a path producing an extremum of the action, then $q_k(t)\to q_k(t)+\delta q_k(t)$ also corresponds to an extremum. Since $S$ is invariant under this change, it follows that

$$\delta S={\int }_{t_i}^{t_f}\sum _k\delta q_k\left(\frac{\partial L}{\partial q_k}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial {\dot{q}}_k}\right)+\sum _k{\left[\delta q_k\frac{\partial L}{\partial {\dot{q}}_k}\right]}_{t_i}^{t_f}=0.$$

The first term in the middle expression vanishes since $q$ is a solution to the Euler-Lagrange equation. Accordingly, we obtain

$$\sum _k\delta q_k\left(t_i\right)p_k\left(t_i\right)=\sum _k\delta q_k\left(t_f\right)p_k\left(t_f\right)\text{\hspace{1em}(1.14)}$$

where use has been made of the definition $p_k=\frac{\partial L}{\partial {\dot{q}}_k}$ . Since $t_i$ and $t_f$ are arbitrary, this equation shows that the quantity ${\sum }_k\delta q_k(t)p_k(t)$ is, in fact, independent of $t$ and hence conserved.

Example 1.2

Let us consider a particle $m$ moving under a force produced by a spherically symmetric potential $V(r)$ , where $r,\theta ,\varphi$ are three-dimensional polar coordinates. The Lagrangian is given by

$$L=\frac{1}{2}m[{\dot{r}}^2+r^2({\dot{\theta }}^2+{\sin }^2\theta {\dot{\varphi }}^2)]-V(r).$$

Note that $q_k=\varphi$ is cyclic, which leads to the conservation law

$$\delta \varphi \frac{\partial L}{\partial \dot{\varphi }}\propto m{r}^2{\sin }^2\theta \dot{\varphi }=\text{c o n s t a n t}.$$

This is nothing but the angular momentum around the $z$ axis. Similar arguments can be employed to show that the angular momenta around the $x$ and $y$ axes are also conserved.

A few remarks are in order:

- Let $Q(q)$ be an arbitrary function of $q$ . Then the Lagrangians $L$ and $L+\mathrm{d}Q/\mathrm{d}t$ yield the same Euler-Lagrange equation. In fact,

$$\begin{array}{rl}& \frac{\partial }{\partial q_k}\left(L+\frac{\mathrm{d}Q}{\mathrm{d}t}\right)-\frac{\mathrm{d}}{\mathrm{d}t}\left[\frac{\partial }{\partial {\dot{q}}_k}\left(L+\frac{\mathrm{d}Q}{\mathrm{d}t}\right)\right]\\ & =\frac{\partial L}{\partial q_k}+\frac{\partial }{\partial q_k}\frac{\mathrm{d}Q}{\mathrm{d}t}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial L}{\partial {\dot{q}}_k}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial }{\partial {\dot{q}}_k}\left(\sum _j\frac{\partial Q}{\partial q_j}{\dot{q}}_j\right)\\ & =\frac{\partial }{\partial q_k}\frac{\mathrm{d}Q}{\mathrm{d}t}-\frac{\mathrm{d}}{\mathrm{d}t}\frac{\partial Q}{\partial q_k}=0.\end{array}$$

- An interesting observation is that Newtonian mechanics is realized as an extremum of the action but the action itself is defined for any trajectory. This fact plays an important role in path integral formation of quantum theory.

1.1.3 Hamiltonian formalism

The Lagrangian formalism yields a second-order ordinary differential equation (ODE). In contrast, the Hamiltonian formalism gives equations of motion which are first order in the time derivative and, hence, we may introduce flows in the

phase space defined later. What is more important, however, is that we can make the symplectic structure manifest in the Hamiltonian formalism, which will be shown in example 5.12 later.

Suppose a Lagrangian $L$ is given. Then the corresponding Hamiltonian is introduced via Legendre transformation of variables as

$$H(q,p)\equiv \sum _k{p}_k{\dot{q}}_k-L(q,\dot{q}),\text{\hspace{1em}(1.15)}$$

where $\dot{q}$ is eliminated in the left-hand side (LHS) in favour of $p$ by making use of the definition of the momentum $p_k=\partial L(q,\dot{q})/\partial {\dot{q}}_k$ . For this transformation to be defined, the Jacobian must satisfy

$$\det \left(\frac{\partial p_i}{\partial {\dot{q}}_j}\right)=\det \left(\frac{{\partial }^{2}L}{\partial {\dot{q}}_i{\dot{q}}_j}\right)\ne 0.$$

The space with coordinates $(q_k,p_k)$ is called the phase space.

Let us consider an infinitesimal change in the Hamiltonian induced by $\delta q_k$ and $\delta p_k$ ,

$$\begin{array}{rl}\delta H& =\sum _k\left[\delta p_k{\dot{q}}_k+p_k\delta {\dot{q}}_k-\frac{\partial L}{\partial q_k}\delta q_k-\frac{\partial L}{\partial {\dot{q}}_k}\delta {\dot{q}}_k\right]\\ & =\sum _k\left[\delta p_k{\dot{q}}_k-\frac{\partial L}{\partial q_k}\delta q_k\right].\end{array}$$

It follows from this relation that

$$\frac{\partial H}{\partial p_k}={\dot{q}}_k,\frac{\partial H}{\partial q_k}=-\frac{\partial L}{\partial q_k}\text{\hspace{1em}(1.16)}$$

which are nothing more than the replacements of independent variables. Hamilton’s equations of motion are obtained from these equations if the Euler-Lagrange equation is employed to replace the LHS of the second equation,

$${\dot{q}}_k=\frac{\partial H}{\partial p_k}{\dot{p}}_k=-\frac{\partial H}{\partial q_k}.\text{\hspace{1em}(1.17)}$$

Example 1.3

Let us consider a one-dimensional harmonic oscillator with the Lagrangian $L=\frac{1}{2}m{\dot{q}}^2-\frac{1}{2}m{\omega }^2{q}^2$ , where ${\omega }^2=k/m$ . The momentum conjugate to $q$ is $p=\partial L/\partial \dot{q}=m\ddot{q}$ , which can be solved for $\dot{q}$ to yield $\dot{q}=p/m$ . The Hamiltonian is

$$H(q,p)=p\dot{q}-L(q,\dot{q})=\frac{p^2}{2m}+\frac{1}{2}m{\omega }^2{q}^2.\text{\hspace{1em}(1.18)}$$

Hamilton’s equations of motion are:

$$\frac{\mathrm{d}p}{\mathrm{d}t}=-m{\omega }^{2}q\frac{\mathrm{d}q}{\mathrm{d}t}=\frac{p}{m}.\text{\hspace{1em}(1.19)}$$

Let us take two functions $A(q,p)$ and $B(q,p)$ defined on the phase space of a Hamiltonian $H$ . Then the Poisson bracket $[A,B]$ is defined by

$$[A,B]=\sum _k\left(\frac{\partial A}{\partial q_k}\frac{\partial B}{\partial p_k}-\frac{\partial A}{\partial p_k}\frac{\partial B}{\partial q_k}\right).\text{\hspace{1em}(1.20)}$$

Exercise 1.1

Show that the Poisson bracket is a Lie bracket, namely it satisfies

$$[A,c_1{B}_1+c_2{B}_2]=c_1[A,B_1]+c_2[A,B_2]\text{l i n e a r i t y}\text{\hspace{1em}(1.21a)}$$

$$[A,B]=-[B,A]\text{s k e w - s y m m e t r y}\text{\hspace{1em}(1.21b)}$$

$$[[A,B],C]+[[C,A],B]+[[B,C],A]=0\text{J a c o b i i d e n t i t y .}\text{\hspace{1em}(1.21c)}$$

The fundamental Poisson brackets are

$$[p_i,p_j]=[q_i,q_j]=0[q_i,p_j]={\delta }_{ij}.\text{\hspace{1em}(1.22)}$$

It is important to notice that the time development of a physical quantity $A(q,p)$ is expressed in terms of the Poisson bracket as

$$\begin{array}{rl}\frac{\mathrm{d}A}{\mathrm{d}t}& =\sum _k\left(\frac{\mathrm{d}A}{\mathrm{d}{q}_k}\frac{\mathrm{d}{q}_k}{\mathrm{d}t}+\frac{\mathrm{d}A}{\mathrm{d}{p}_k}\frac{\mathrm{d}{p}_k}{\mathrm{d}t}\right)\\ & =\sum _k\left(\frac{\mathrm{d}A}{\mathrm{d}{q}_k}\frac{\partial H}{\partial p_k}-\frac{\mathrm{d}A}{\mathrm{d}{p}_k}\frac{\partial H}{\partial q_k}\right)\\ & =[A,H].\end{array}\text{\hspace{1em}(1.23)}$$

If it happens that $[A,H]=0$ , the quantity $A$ is conserved, namely $\mathrm{d}A/\mathrm{d}t=0$ . The Hamilton equations of motion themselves are written as

$$\frac{\mathrm{d}{p}_k}{\mathrm{d}t}=\left[p_k,H\right]\frac{\mathrm{d}{q}_k}{\mathrm{d}t}=\left[q_k,H\right].\text{\hspace{1em}(1.24)}$$

Theorem 1.1 · Noether's theorem

Let $H(q_k,p_k)$ be a Hamiltonian which is invariant under an infinitesimal coordinate transformation $q_k\to q_k^{\prime }=q_k+\epsilon f_k(q)$ . Then

$$Q=\sum _k{p}_k{f}_k(q)\text{\hspace{1em}(1.25)}$$

is conserved.

Proof

One has $H(q_k,p_k)=H(q_k^{\prime },p_k^{\prime })$ by definition. It follows from $q_k^{\prime }=q_k+\epsilon f_k(q)$ that the Jacobian associated with the coordinate change is

$${\Lambda }_{ij}=\frac{\partial q_i^{\prime }}{\partial q_j}\simeq {\delta }_{ij}+\epsilon \frac{\partial f_i(q)}{\partial q_j}$$

up to $\mathcal{O}(\epsilon )$ . The momentum transforms under this coordinate change as

$$p_i\to \sum _j{p}_j{\Lambda }_{ji}^{-1}\simeq p_i-\epsilon \sum _j{p}_j\frac{\partial f_j}{\partial q_i}.$$

Then, it follows that

$$\begin{array}{rl}0& =H\left(q_k^{\prime },p_k^{\prime }\right)-H\left(q_k,p_k\right)\\ & =\frac{\partial H}{\partial q_k}\epsilon f(q)-\frac{\partial H}{\partial p_j}\epsilon p_i\frac{\partial f_i}{\partial q_j}\\ & =\epsilon \left[\frac{\partial H}{\partial q_k}{f}_k(q)-\frac{\partial H}{\partial p_j}{p}_i\frac{\partial f_i}{\partial q_j}\right]\\ & =\epsilon [H,Q]=\epsilon \frac{\mathrm{d}Q}{\mathrm{d}t},\end{array}$$

which shows that $Q$ is conserved.

$\square$

This theorem shows that to find a conserved quantity is equivalent to finding a transformation which leaves the Hamiltonian invariant.

A conserved quantity $Q$ is the ‘generator’ of the transformation under discussion. In fact,

$$[q_i,Q]=\sum _k\left[\frac{\partial q_i}{\partial q_k}\frac{\partial Q}{\partial p_k}-\frac{\partial q_i}{\partial p_k}\frac{\partial Q}{\partial q_k}\right]=\sum _k{\delta }_{ik}{f}_k(q)=f_i(q)$$

which shows that $\delta q_i=\epsilon f_i(q)=\epsilon [q_i,Q]$ .

A few examples are in order. Let $H=p^2/2m$ be the Hamiltonian of a free particle. Since $H$ does not depend on $q$ , it is invariant under $q\mapsto q+\epsilon \cdot 1$ , $p\mapsto p$ . Therefore, $Q=p\cdot 1=p$ is conserved. The conserved quantity $Q$ is identified with the linear momentum.

Example 1.4

Let us consider a particle $m$ moving in a two-dimensional plane with the axial symmetric potential $V(r)$ . The Lagrangian is

$$L(r,\theta )=\frac{1}{2}m({\dot{r}}^2+r^2{\dot{\varphi }}^2)-V(r).$$

The canonical conjugate momenta are:

$$p_r=m\dot{r}{p}_{\theta }=m{r}^2\dot{\theta }.$$

The Hamiltonian is

$$H=p_r\dot{r}+p_{\theta }\dot{\theta }-L=\frac{p_r^2}{2m}+\frac{p_{\theta }^2}{2m{r}^2}+V(r).$$

This Hamiltonian is clearly independent of $\theta$ and, hence, invariant under the transformation

$$\theta \mapsto \theta +\epsilon \cdot 1,p_{\theta }\mapsto p_{\theta }.$$

The corresponding conserved quantity is

$$Q=p_{\theta }\cdot 1=m{r}^2\dot{\theta }$$

that is the angular momentum.

Endnotes

Unplaced source note

A manifold, to be more precise, see chapter 5.

Unplaced source note

Of course, $L$ may depend on ${\dot{q}}_k$ . Otherwise, the coordinate $q_k$ is not our concern at all.

Unplaced source note

Since the symmetry is continuous, it is always possible to define such an infinitesimal operation. Needless to say, $\delta q(t_i)$ and $\delta q(t_f)$ do not, in general, vanish in the present case.

Unplaced source note

When the commutation relation $[A,B]$ of operators is introduced later, the Poisson bracket will be denoted as $[A,B{]}_{\mathrm{PB}}$ to avoid confusion.

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Footnotes

1. We call a particle with mass $m$ simply ‘a particle $m$ ’. ↩

2. A functional is a function of functions. A function $f(\bullet )$ produces a number $f(x)$ for a given number $x$ . Similarly, a functional $F[\bullet ]$ assigns a number $F[f]$ to a given function $f(x)$ . ↩