17 Jun 2020

# Introduction

### Preface

The following are course notes from Math 242 - Mathematics of Quantum Mechanics, taught by Dr. Lapidus at University of California, Riverside during the Winter 2019 quarter. The course did not have a dedicated textbook, though the book An Introduction to Quantum Theory by Keith Hannabuss was recommended as a supplemental textbook.

The course began by developing classical mechanics through the Hamiltonian formulation, which then provides the essential framework to formulate the Heisenberg picture of quantum mechanics.

### Work in Progress

Note that these notes are a currently a work in progress. If you would like to see photographic copies of my notes from the course, they are available via Google drive.

# Classical Mechanics

### Classical Observables

#### The Hamiltonian

Consider a particle of mass $$m$$ in a conservative force field $$F=-\nabla V$$ in ambient space $$\mathbb{R}^3$$. Let $$q$$ denote its position and $$p$$ its momentum. The Hamiltonian $$H$$ is the sum of kinetic and potential energy. For this single particle, $$H$$ is of the form:1 $H = \frac{p^2}{2m}+V(q)$ For a conservative system, where energy is conserved, the Hamiltonian formulation of classical mechanics yields the following equations of motion:2 \begin{aligned} \frac{\partial H}{\partial p} &= \frac{dq}{dt} \\ \frac{\partial H}{\partial q} &= -\frac{dp}{dt}\end{aligned} Note that the first term is exactly the velocity of the particle $$\frac{p}{m}$$, and the second the force $$F$$. We may recast these equations as a vector field equation: \begin{aligned} \frac{d}{dt}\begin{pmatrix} q\\ p \end{pmatrix} = \begin{pmatrix} \frac{\partial H}{\partial p} \\ -\frac{\partial H}{\partial q} \end{pmatrix}\end{aligned} Therefore, the Hamiltonian equations of motion describe the time derivative of a point $$(q,p)$$ in phase space. For an $$n$$ dimensional setting, phase space is simply $$\mathbb{R}^{2n}$$. More generally, when the configuration space is a manifold $$M$$ instead of $$\mathbb{R}^n$$, phase space is the contangent bundle $$T^*M$$. The Hamiltonian, then, is a map from phase space into $$\mathbb{R}$$.

We suppose the physical system under consideration has the following Hamiltonian: $H = \sum_{j=1}^{k}\frac{p_j^2}{2m_j} + \sum_{j=1}^{k} V(q_j) + \sum_{1\leq i< j\leq n}^{} V_{ij}(q_i-q_j)$ The new term $$V_{ij}$$ is the symmetric interaction potential between the indicated particles. For this system, there are $$kn$$ degrees of freedom, and therefore phase space is given by $$\mathbb{R}^{2kn}$$.

#### The Algebra of Classical Observables

The Hamiltonian is the first and foremost example of a physical observable. Other classical observables are defined in the same fashion: they are functions from phase space into $$\mathbb{R}$$. We shall hereforth assume that all classical observables (inculding the Hamiltonian) are smooth. Therefore, we define the algebra of classical observables:

If we denote by $$M$$ the phase space under consideration, then we define the classical observables to be: $\mathscr{A}_{cl}= C^\infty(M)$ Such smooth functions form an algebra.

More generally, one may define the algebra differently; for instance, one may take classical observables to be differentiable or perhaps measurable functions on phase space.

#### Time Evolution

We consider the following Cauchy problem: \left\{\begin{aligned} \left(\frac{dq}{dt},\frac{dp}{dt}\right) &= \left(\frac{\partial H}{\partial p},-\frac{\partial H}{\partial q}\right) \\ (q,p)\Big\vert_{t=0}&=(q_0,p_0) \end{aligned} \right. We shall assume the problem is well-posed and in particular that there is a unique global solution. Let us denote $$\mu=(q,p)$$ for a point in phase space $$\mathscr{M}=\mathbb{R}^{2n}$$. We denote by $$\mu(t)$$ to be the corresponding point in phase space at a later time $$t$$ for this time evolution, subject to the initial condition that $$\mu_0=\mu(0)$$.

Hamilton’s equations of motions generate a one-parameter group of automorphisms $$\left\{ G_t \right\}_{t\in\mathbb{R}}$$ on $$\mathscr{M}$$ with the operation of composition, $$G_{t+s}=G_t\circ G_s$$. These automorphism, in turn, generate a family $$\left\{ U_t \right\}_{t\in\mathbb{R}}$$ of transformations of $$\mathscr{A}_{cl}$$. Namely, for $$f\in C^\infty(\mathscr{M})$$ and $$\mu\in\mathscr{M}$$, $(U_tf)(\mu) = f(G_t\mu) =: f_t(\mu)$ Here we define $$f_t := f\circ G_t$$.

We can in fact find a very nice description of time evolution of observables with this formalism.

First, we start with $$G_t:\mathscr{M}{\rightarrow}\mathscr{M}$$, given by $$\mu{\rightarrow}G_t\mu$$, the group of transformations. Note that this family satisfies $$G_{s+t}=G_s\circ G_t$$; in general the indices behave as would an additive group. We defined $$U_t$$ to be the operator that precomposes $$f$$ with the map $$G_t$$ and denoted by $$f_t$$ the image of $$f$$ under this map. In summary: $f_t= U_t f = f\circ G_t$ Let $$\mu_0=\mu(0)\in\mathscr{M}$$, and write $$\mu(t)=G_t(\mu)$$. Let us compute $$\frac{df_t}{dt}$$ in the following fashion: \begin{aligned} \frac{\partial }{\partial s} f_{s+t}(\mu_0) &= \frac{\partial }{\partial s}\left( f_t \circ G_s \right)(\mu_0) \\ &= \frac{\partial }{\partial s} f_t(\mu(s)) \\ &= \sum_{i=1}^{2n} \left(\frac{\partial }{\partial x_i} f_t(\mu(s))\right) \frac{d}{ds}\mu(s) \\ &= \nabla f_t(\mu(s))\cdot \left(\frac{\partial H}{\partial p},-\frac{\partial H}{\partial q}\right)\\ &= \nabla f_{t+s}(\mu_0)\cdot \left(\frac{\partial H}{\partial p},-\frac{\partial H}{\partial q}\right)\end{aligned} We now evaluate the above expression at $$s=0$$. For simplicity of notation, let us now suppress dependence on $$\mu(0)$$. \begin{aligned} \frac{\partial }{\partial s} f_{s+t}\Big\vert_{s=0} &= \nabla f_t\cdot \left(\frac{\partial H}{\partial p},-\frac{\partial H}{\partial q}\right) \\ &= \sum_{i=1}^{n} \frac{\partial f_t}{\partial q_i} \frac{\partial H}{\partial p_i} - \sum_{i=1}^{n} \frac{\partial f_t}{\partial p_i} \frac{\partial H}{\partial q_i} \\ &= \frac{\partial f_t}{\partial q} \frac{\partial H}{\partial p} - \frac{\partial f_t}{\partial p} \frac{\partial H}{\partial q} \\ &=: \left\{ H,f_t \right\}\end{aligned} In the final line, we define a new very important operation: $$\left\{ \cdot,\cdot \right\}$$, the classical Poisson bracket.

Thus we have shown that the time derivative of an observable is given by the Poisson bracket with the Hamiltonian $$H$$. We state our result as a theorem:

Let Hamilton’s equations of motions be satisfied for points $$(p,q)\in\mathscr{M}$$. $\frac{d}{dt}(q,p) = \left(\frac{\partial H}{\partial p},-\frac{\partial H}{\partial q}\right)$ For each observable $$f\in\mathscr{A}_{cl}$$, let $$f_t=f\circ G_t$$ be the observable which measures $$f$$ at a later time $$t$$. Then $$f_t$$ will be the solution to the following Cauchy problem: \left\{\begin{aligned} \frac{df_t}{dt} &= \left\{ H,f_t \right\} = \frac{\partial H}{\partial p} \frac{\partial f_t}{\partial q} - \frac{\partial H}{\partial q} \frac{\partial f_t}{\partial p} \\ f_t\Big\vert_{t=0} &= f \end{aligned} \right.

#### The Classical Poisson Bracket

We begin by generically defining the classical Poisson bracket from the previous section.

We define a mapping $$\mathscr{A}_{cl}\times\mathscr{A}_{cl}{\rightarrow}\mathscr{A}_{cl}$$ as follows, letting $$f,g\in\mathscr{A}_{cl}$$: $\left\{ f,g \right\} := \frac{\partial f}{\partial q} \frac{\partial g}{\partial p} - \frac{\partial f}{\partial p} \frac{\partial g}{\partial q}$

We proceed to state and prove some properties of this bracket. Throughout this section, let $$f,g,h\in \mathscr{A}_{cl}$$ and $$\alpha\in \mathbb{R}$$.

##### Skew-Symmetry

The Poisson bracket is skew-symmetric. \begin{aligned} \{f,g\} &= \frac{\partial f}{\partial q} \frac{\partial g}{\partial p} - \frac{\partial f}{\partial p} \frac{\partial g}{\partial q} \\ &= -\left(\frac{\partial f}{\partial q}\frac{\partial g}{\partial p}-\frac{\partial f}{\partial p}\frac{\partial g}{\partial q}\right)\\ &= -\left(\frac{\partial g}{\partial p}\frac{\partial f}{\partial q}-\frac{\partial g}{\partial q}\frac{\partial f}{\partial p}\right)\\ &= -\{ g, f\}\end{aligned}

##### Bilinearity

The Poisson bracket is linear in both its inputs. \begin{aligned} \{f,\alpha g+ h\} &= \frac{\partial f}{\partial p}\frac{\partial (\alpha g +h)}{\partial q}-\frac{\partial f}{\partial q}\frac{\partial (\alpha g +h)}{\partial p} \\ &= \frac{\partial f}{\partial p}\left(\alpha\frac{\partial g}{\partial q}+\frac{\partial h}{\partial q}\right)-\frac{\partial f}{\partial q}\left(\alpha\frac{\partial g}{\partial p}+\frac{\partial h}{\partial p}\right)\\ &= \alpha\left(\frac{\partial f}{\partial p}\frac{\partial g}{\partial q}-\frac{\partial f}{\partial q}\frac{\partial g}{\partial p}\right)+\frac{\partial f}{\partial p}\frac{\partial h}{\partial q}-\frac{\partial f}{\partial q}\frac{\partial h}{\partial p}\\ &= \alpha\{f,g\}+\{f,h\}\\ \\ \{\alpha f + h,g\} &= -\{g,\alpha f+h\} \\ &= -\left(\alpha\{g,f\}+\{g,h\}\right)\\ &= \alpha\{f,g\}+\{h,g\}\end{aligned}

##### Jacobi Identity

The Poisson bracket satisfies the Jacobi identity: $\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\} =0$

For readability, we shall use subscripts to denote partial derivatives, e.g. $$\frac{\partial f}{\partial p_j}=f_{p_j}$$. Also in what follows, we will employ the Einstein summation convention; whenever an index is repeated, one interprets the quantity as a sum over $$n$$ coordinates. \begin{aligned} &\{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\} \\ &= \left\{ f,g_{p_j}h_{q_j}- g_{q_j}h_{p_j} \right\} + \left\{ g,h_{p_j}f_{q_j}- h_{q_j}f_{p_j} \right\} + \left\{ h,f_{p_j}g_{q_j}- f_{q_j}g_{p_j} \right\} \\ &= f_{p_k}(g_{p_j}h_{q_j}- g_{p_j}h_{q_j})_{q_k} - f_{q_k}(g_{p_j}h_{q_j}- g_{p_j}h_{q_j})_{p_k} \\ & + g_{p_k}(h_{p_j}f_{q_j}- h_{p_j}f_{q_j})_{q_k} - g_{q_k}(h_{p_j}f_{q_j}- h_{p_j}f_{q_j})_{p_k} \\ & + h_{p_k}(f_{p_j}g_{q_j}- f_{p_j}g_{q_j})_{q_k} - h_{q_k}(f_{p_j}g_{q_j}- f_{p_j}g_{q_j})_{p_k} \\\\ &= f_{p_k}(g_{p_j}h_{q_j})_{q_k}- f_{p_k}(g_{p_j}h_{q_j})_{q_k} - f_{q_k}(g_{p_j}h_{q_j})_{p_k}+ f_{q_k}(g_{p_j}h_{q_j})_{p_k} \\ & + g_{p_k}(h_{p_j}f_{q_j})_{q_k}- g_{p_k}(h_{p_j}f_{q_j})_{q_k} - g_{q_k}(h_{p_j}f_{q_j})_{p_k}+ g_{q_k}(h_{p_j}f_{q_j})_{p_k} \\ & + h_{p_k}(f_{p_j}g_{q_j})_{q_k}- h_{p_k}(f_{p_j}g_{q_j})_{q_k} - h_{q_k}(f_{p_j}g_{q_j})_{p_k}+ h_{q_k}(f_{p_j}g_{q_j})_{p_k} \\\\ &= f_{p_k}g_{p_j q_k} h_{q_j}+ f_{p_k}g_{p_j}h_{q_j q_k} - f_{p_k}g_{p_j q_k} h_{q_j}- f_{p_k}g_{p_j}h_{q_j q_k}\\ &- f_{q_k}g_{p_j p_k} h_{q_j}- f_{q_k}g_{p_j}h_{q_j p_k} + f_{q_k}g_{p_j p_k} h_{q_j}+ f_{q_k}g_{p_j}h_{q_j p_k} \\ & + g_{p_k}h_{p_j q_k} f_{q_j}+ g_{p_k}h_{p_j}f_{q_j q_k} - g_{p_k}h_{p_j q_k} f_{q_j}- g_{p_k}h_{p_j}f_{q_j q_k}\\ &- g_{q_k}h_{p_j p_k} f_{q_j}- g_{q_k}h_{p_j}f_{q_j p_k} + g_{q_k}h_{p_j p_k} f_{q_j}+ g_{q_k}h_{p_j}f_{q_j p_k} \\ & + h_{p_k}f_{p_j q_k} g_{q_j}+ h_{p_k}f_{p_j}g_{q_j q_k} - h_{p_k}f_{p_j q_k} g_{q_j}- h_{p_k}f_{p_j}g_{q_j q_k}\\ &- h_{q_k}f_{p_j p_k} g_{q_j}- h_{q_k}f_{p_j}g_{q_j p_k} + h_{q_k}f_{p_j p_k} g_{q_j}+ h_{q_k}f_{p_j}g_{q_j p_k} \\\\ &= f_{p_k}g_{p_j q_k} h_{q_j}+ f_{p_k}g_{p_j}h_{q_j q_k} - f_{p_k}g_{p_j}h_{q_j q_k}\\ &- f_{q_k}g_{p_j p_k} h_{q_j}- f_{q_k}g_{p_j}h_{q_j p_k} + f_{q_k}g_{p_j p_k} h_{q_j} \\ & + g_{p_k}h_{p_j q_k} f_{q_j}+ g_{p_k}h_{p_j}f_{q_j q_k} - g_{p_k}h_{p_j}f_{q_j q_k}\\ &- g_{q_k}h_{p_j p_k} f_{q_j}- g_{q_k}h_{p_j}f_{q_j p_k} + g_{q_k}h_{p_j p_k} f_{q_j} \\ & + h_{p_k}f_{p_j q_k} g_{q_j}+ h_{p_k}f_{p_j}g_{q_j q_k} - h_{p_k}f_{p_j}g_{q_j q_k}\\ &- h_{q_k}f_{p_j p_k} g_{q_j}- h_{q_k}f_{p_j}g_{q_j p_k} + h_{q_k}f_{p_j p_k} g_{q_j} \\\\ &= f_{p_k}g_{p_j q_k} h_{q_j}+ f_{p_k}g_{p_j}h_{q_j q_k} - f_{p_k}g_{p_j}h_{q_j q_k} - f_{q_k}g_{p_j}h_{q_j p_k} \\ & + g_{p_k}h_{p_j q_k} f_{q_j}+ g_{p_k}h_{p_j}f_{q_j q_k} - g_{p_k}h_{p_j}f_{q_j q_k} - g_{q_k}h_{p_j}f_{q_j p_k} \\ & + h_{p_k}f_{p_j q_k} g_{q_j}+ h_{p_k}f_{p_j}g_{q_j q_k} - h_{p_k}f_{p_j}g_{q_j q_k} - h_{q_k}f_{p_j}g_{q_j p_k} \\\\ &= f_{p_k}g_{p_j}h_{q_j q_k} - f_{p_k}g_{p_j}h_{q_j q_k} \\ & + g_{p_k}h_{p_j}f_{q_j q_k} - g_{p_k}h_{p_j}f_{q_j q_k} \\ & + h_{p_k}f_{p_j}g_{q_j q_k} - h_{p_k}f_{p_j}g_{q_j q_k} \\ &=0 \end{aligned}

We call attention to a critical observation which leads to the pairwise cancellations above. Let us isolate two terms in the fourth equality: the first term $$f_{p_k}h_{q_j}g_{p_j q_k}$$ and the antepenultimate term $$-f_{p_j}h_{q_k}g_{p_k q_j}$$. Note that we have rearranged the latter slightly, employing smoothness to commute the order of partial derivatives. Since under our Einstein convention each term is a double sum each of $$n$$ terms, we have the following identity: $\sum_{j,k=1}^{n} f_{p_k}h_{q_j}g_{p_j q_k}=\sum_{j,k=1}^{n} f_{p_j}h_{q_k}g_{p_k q_j}$ In other words, we may simply "switch the indices" $$j$$ and $$k$$ to obtain that $$f_{p_k}h_{q_j}g_{p_j q_k}= f_{p_j}h_{q_k}g_{p_k q_j}$$. Thus, we have two terms which are identical but with opposite signs. Since the lines in the expression cyclically repeat, we have in fact three pairs of terms that shall cancel. In this fashion, six terms vanish each step. Subsequent equality steps show more pairs that disappear, three pairs at a time.

##### Leibniz Rule

The Poisson bracket satisfies a Leibniz identity. \begin{aligned} \left\{ f,gh \right\} &= f_{p_j}(gh)_{q_j}- f_{q_j}(gh)_{p_j}\\ &= f_{p_j}g_{q_j}h + f_{p_j}g h_{q_j}- f_{q_j}g_{p_j}h - f_{q_j}g h_{p_j}\\ &= f_{p_j}g_{q_j}h - f_{q_j}g_{p_j}h + f_{p_j}h_{q_j}g - f_{q_j}h_{p_j}g\\ &= \left\{ f,g \right\}h + \left\{ f,h \right\}g\end{aligned}

##### Derivation of the Algebra

We note that the above properties imply that the Poisson bracket is a derivation of the algebra in the following sense. Let $$X_f:\mathscr{A}_{cl}{\rightarrow}\mathscr{A}_{cl}$$ be the vector field, i.e. a first order linear differential operator, given by the following: \begin{aligned} X_f :=& \frac{\partial f}{\partial p}\frac{\partial }{\partial q}-\frac{\partial f}{\partial q}\frac{\partial }{\partial p} \\ X_f\cdot g =& \left\{ f,g \right\}\end{aligned} Then $$X_f$$ acts on $$\mathscr{A}_{cl}$$. To say that $$X_f$$ is a derivation is to assert the following identity: $X_f(hg) = (X_f\cdot g)h+(X_f\cdot h)g$ We note that this identity is equivalent to the Leibniz rule. Indeed: \begin{aligned} \left\{ f,gh \right\} &= X_f\cdot (gh) \\ \left\{ f,g \right\}h + g\left\{ f,h \right\} &= (X_f\cdot g)h + (X_f\cdot h)g\end{aligned} Therefore, when one equation holds, so too must the other. Note that in a noncommutative setting, one must take care with the definitions and statement of the identities, but here we shall have no such issues.
Given two such derivations $$X_f, X_g$$ define their commutator in the usual fashion: $=X_f X_g-X_g X_f$ The following identity of these vector fields is equivalent to the Jacobi identity: $= X_{\left\{ f,g \right\}}$ To see this equivalence, we consider the action of $$[X_f,X_g]-X_{\left\{ f,g \right\}}$$ on $$h$$ and simplify using only the linearity and skew-symmetry properties. \begin{aligned} ([X_f,X_g] - X_{\left\{ f,g \right\}})h &= (X_f X_g -X_g X_f) h - X_{\left\{ f,g \right\}} h \\ &= X_f X_g h - X_g X_f h - X_{\left\{ f,g \right\}} h \\ &= X_f \left\{ g,h \right\} - X_g \left\{ f,h \right\} - \left\{ \left\{ f,g \right\},h \right\} \\ &= \{f,\{g,h\}\} - \{g,\{f,h\}\} + \{h,\{f,g\}\} \\ &= \{f,\{g,h\}\} + \left\{ g,-\left\{ f,h \right\} \right\} + \{h,\{f,g\}\}\\ &= \{f,\{g,h\}\}+\{g,\{h,f\}\}+\{h,\{f,g\}\}\end{aligned} Therefore, when one expression is zero, so too is the other. Note well that when given fixed $$f,g$$ this identity holds for any $$h\in\mathscr{A}_{cl}$$; in the reverse direction of the equivalence, this fact is needed to deduce that $$[X_f,X_g]=X_{\left\{ f,g \right\}}$$ as operators.

#### Algebraic Properties

Having established this Poisson bracket, we now summarize the following facts about this algebra. Firstly, $$\mathscr{A}_{cl}$$ is a real vector space and a commutative algebra with multiplication and addition. Together with the classical Poisson bracket, $$\mathscr{A}_{cl}$$ is moreover a Lie algebra, with the Poisson bracket as its Lie product. Furthermore, the algebraic product and the Lie product are related via the Leibniz rule, viz. the latter is a derivation.

Further, this algebra contains a distinguished element, the Hamiltonian $$H$$, by which the time evolution of any observable is determined via the action $$X_H$$.

1. We denote $$p^2:=p\cdot p$$.↩︎

2. These partial derivatives denote gradients, e.g. $$\frac{\partial H}{\partial p}=\left(\frac{\partial H}{\partial p_1},\frac{\partial H}{\partial p_2},\frac{\partial H}{\partial p_3}\right)$$.↩︎