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At constant volume, the First law of thermodynamics givesAt constant pressure,Differentiating $pV=N{k}_{B}T$ at fixed $p$ givesConsequently,and the Mayer relation isThe specific-heat ratio is therefore

For a fixed number $N$ of particles, the ideal gas equation of state isIts internal energy depends only on temperature, andWhen the heat capacity at constant volume is constant, a choice of the energy zero gives

Let $H$ be the one-dimensional Hamiltonian operator and letbe its discrete energy eigenvalues. Expanding a normalized trial wavefunction in an orthonormal energy basis,givesThus the Rayleigh-Ritz variational principle givesfor every nonzero admissible trial wavefunction. One chooses a parameterized family and minimizes the quotient.

When $V(x)=V(-x)$, parity commutes with $H$. In one dimension the nondegenerate ground state is even and the first excited state is odd. Restricting the trial family to odd wavefunctions makes every trial state orthogonal to the ground state, soThis is the odd-state variational principle for an even potential.

The unitary time evolution isFirst suppose that time reversal $T$ is linear and unitary. Differentiatingat $t=0$ givesand linearity permits the scalar $i$ to pass through $T$. HenceIf $H|E⟩=E|E⟩$, thenThus every energy has its negative as an energy. The Coulomb Hamiltonian has arbitrarily large positive energies in its continuum, so this symmetry would force energies arbitrarily far below zero. It would have no lowest energy eigenvalue and therefore no stable ground state. This is the unitary time reversal reverses the energy spectrum obstruction.

Now let $T$ be antiunitary. It is conjugate linear, soDifferentiating the same intertwining equation givesThereforeTime reversal now maps an energy eigenstate to another state with the same energy:The spectrum is preserved rather than reflected through zero, so the lower-bounded Coulomb spectrum and its stable ground state are compatible with time-reversal symmetry. This is why antiunitary time reversal preserves the energy spectrum.

The Born rule depends only on transition probabilitiesbetween rays. A physical transformation must preserve these probabilities. By Wigner theorem, such a ray transformation is induced by a unitary or antiunitary operator. For ordinary continuously connected transformations one chooses the unitary branch, so

The states $U(g_1)U(g_2)|\psi ⟩$ and $U(g_1{g}_2)|\psi ⟩$ represent the same ray. They may therefore differ by a state-independent quantum phase:Thus quantum transformations form a projective unitary representation.

If $g$ is a symmetry of the Hamiltonian operator $H$, it also obeysequivalently $[U(g),H]=0$. For a differentiable one-parameter subgroup, writewith a self-adjoint generator $Q$. Differentiating the symmetry equation at $s=0$ yieldsFor a time-independent $Q$, the Heisenberg picture equation isHence the generator is a conserved observable. Conversely, a self-adjoint conserved $Q$ commutes with $H$ and its unitary group generates symmetries. This is the conserved generator of a continuous quantum symmetry.

When $u$ is independent of $y$ and $z=x+iy$,PutThe zero-curvature equation becomesSincewe obtain the Isospectral Lax equationTherefore, for every positive integer $n$, the cyclic property of the matrix trace givesThese are the trace invariants of a Lax equation.

The PDE from part (b) reduces toWith $\varphi =2u$, this isTo extract a nontrivial invariant, use $n=2$. Writing $p=u_x$, direct multiplication givesThe terms involving $\lambda$ are constant, and $p={\varphi }^{\prime }/2$. Henceis a first integral. Direct differentiation verifies it:

SubstituteThe off-diagonal entries ofcancel identically. Its upper-left entry isand its lower-right entry is the negative of this. Hence compatibility is equivalent toThis is the Sinh-Gordon equation in complex light-cone coordinates.

The system prescribes both first derivatives of the same vector $\psi$, so it is overdetermined: the two mixed derivatives must agree. Differentiating givesThus every solution satisfiesFor the system to possess arbitrary nontrivial local solution data, the coefficient matrix must vanish:This is the zero-curvature condition, or the compatibility condition for an overdetermined linear system.

At $(1,y_0)$ the trajectory initially moves into $x>1$. Suppose $y$ had a first zero at time $t_{\ast }>0$. Before that time $y>0$, hence $x(t_{\ast })>1$. Since $U^{\prime }(x)<0$ for $x>1$,which is incompatible with crossing from positive values to zero. Thusfor all $t>0$.

The energy from part (b) decreases strictly becauseSince $U\ge 0$,Therefore

Finally, fix $\epsilon >0$. If the trajectory never entered $0<y<\epsilon$, positivity would force $y(t)\ge \epsilon$ for all $t>0$. Thensowhich becomes negative for large $t$, contradicting $E\ge 0$. Hence the trajectory must enter every such strip. This proves outward escape from the rational potential barrier.

SetThe stated identity givesHence, along the system,Near the origin, $E$ is positive definite because $U(x)>0$ for $x\ne 0$. The First Lyapunov theorem therefore gives stability.

Choose a sufficiently small compact invariant sublevel setOn $\{\dot{E}=0\}$ we have $y=0$. A trajectory can remain there only ifwhich occurs at $x=0,\pm 1$. The choice $c<1/4=U(\pm 1)$ excludes the two latter points, so the largest invariant subset of $\{\dot{E}=0\}\cap {\Omega }_c$ is the origin. LaSalle invariance principle now gives convergence to the origin. Thus, for every $k>0$,This is an instance of damped mechanical energy as a Lyapunov function.

Let $x_{\ast }$ be an equilibrium point of a dynamical system. A Lyapunov function on a neighbourhood of $x_{\ast }$ is a continuously differentiable function $V$ such thatand whose derivative along trajectories satisfies

The First Lyapunov theorem says that the existence of such a positive-definite $V$ proves Lyapunov stability of $x_{\ast }$. If $\dot{V}<0$ away from $x_{\ast }$, the Second Lyapunov theorem strengthens this to asymptotic stability.

LaSalle invariance principle says that if a trajectory remains in a compact positively invariant set $\Omega$ on which $\dot{V}\le 0$, then it approaches the largest invariant subset ofIn particular, if that largest invariant subset consists only of $x_{\ast }$, every trajectory in $\Omega$ approaches $x_{\ast }$.

The reflection map from part (d) bijects paths that hit $a$ and finish below $a$ with paths whose reflected endpoint lies strictly above $a$. Since the reflected walk has the same distribution,Subtracting the corresponding identity at level $a+1$ gives

For a simple symmetric random walk,when $|k|\le n$ and $n+k$ is even, and it is zero otherwise. Therefore the explicit answer iswhere a binomial coefficient is interpreted as zero when its lower argument is not an integer in $\{0,\dots ,n\}$. This is the point probability for the maximum of simple symmetric random walk.

DefineThe event $\{n\le T_a\}$ means that level $a$ has not been reached by time $n-1$, so it belongs to ${\mathcal{F}}_{n-1}$. Hence $H$ is a bounded predictable process. A direct check across the hitting time givesand thereforePart (c) shows that $\hat{X}$ is a martingale. Its increments still take values in $\{-1,1\}$, so part (b) shows that they are IID symmetric signs. Consequently $\hat{X}$ is a simple symmetric random walk. This is the path transformation behind the reflection principle for simple symmetric random walk.

Because $H_k$ is previsible, $H_k$ is ${\mathcal{F}}_{k-1}$-measurable. The finite sum ${\hat{X}}_n$ is therefore ${\mathcal{F}}_n$-measurable. Boundedness of $H$ and integrability of the martingale increments implyFinally,Thusso $\hat{X}$ is a martingale. It is the martingale transform of $X$ by $H$.