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In Feynman gauge, the Lagrangian differs by a total divergence fromIts Euler-Lagrange equation is $\square A^{\mu }=0$. Directly from the original gauge-fixed Lagrangian, the canonical momenta areequivalently ${\pi }^{\mu }=-{\partial }^0{A}^{\mu }$ after using the total-divergence form.

Using the mode expansions and oscillator commutator, the two nonzero cross terms combine towhere polarization completeness was used. The other equal-time field commutators vanish.

Substitution into the Hamiltonian and normal ordering giveEvery oscillator excitation carries positive energy $|\mathbf{p}|$, but the covariant state space has an indefinite inner product: time-like excitations have negative norm, and time-like and longitudinal polarizations are unphysical. Gupta-Bleuler quantization imposesand quotients by null states. The physical state space contains only the two transverse photon polarizations with positive norm, as required by gauge invariance.

The Euler-Lagrange equation is $m\ddot{x}=0$. The unique path with the prescribed endpoints is thereforeIt is a minimum of the Euclidean action and a stationary point of the real-time action. Its classical action isThe principle of stationary action consequently fixes the position-dependent phase of the semiclassical propagator asBecause the action is quadratic, the stationary-phase evaluation of the path integral is exact. Composition of propagators, or the Van Vleck determinant, gives $C(T)=\sqrt{m/(2\pi i\hslash T)}$, reproducing part i.

With a nonconstant potential energy, momentum no longer diagonalizes the Hamiltonian operator. The operator derivation must use the energy eigenfunctions of the full Hamiltonian, a Dyson series, or a time-sliced Trotter product formula. In the classical derivation, the straight paths are replaced by every solution of the nonlinear Euler-Lagrange equation with the specified endpoints. The semiclassical propagator becomes a sumwhere the prefactor is the Van Vleck determinant and ${\mu }_{\mathrm{cl}}$ is a Maslov index. Unlike a quadratic theory, the classical-path sum is generally only an asymptotic approximation: the exact path integral includes fluctuations of every order. On the circle, the sum must still include all winding number sectors.

Under the global U(1) gauge symmetry, the phases in $\bar{\psi }\psi$ cancel and $\alpha$ is constant, soBoth the fermion term and $F_{\mu \nu }{F}^{\mu \nu }$ are therefore invariant. To extract the Noether current, temporarily promote $\alpha$ to a function. The variation of the action isThe Euler-Lagrange equations then imply the current conservation law ${\partial }_{\mu }{j}^{\mu }=0$.

The Euler-Lagrange equation of the gauge-fixed Polyakov action is the two-dimensional wave equationIts general solution is a sum of left- and right-moving functions. Periodicity in $\sigma$ and a Fourier expansion give the closed-string mode expansionThe zero modes satisfy ${\alpha }_0^{\mu }={\tilde{\alpha }}_0^{\mu }=\sqrt{{\alpha }^{\prime }/2},p^{\mu }$. Here $x^{\mu }$ and $p^{\mu }$ are the center-of-mass position and momentum, while the nonzero string oscillators describe shape excitations.

The Euler-Lagrange equation isAfter a spatial Fourier transform and the change to conformal time, ${\partial }_t=a^{-1}{\partial }_{\tau }$, this becomesFor the exact de Sitter spacetime scale factor $a=-1/(H\tau )$, one has $a^{\prime \prime }/a=2a^2{H}^2$. Setting $u_{\mathbf{k}}=a{\phi }_{\mathbf{k}}$ therefore cancels both the friction term and the effective mass term, leavingThis is the special cancellation for a conformally coupled scalar field.

Write the scalar field energy as $E=T+V$, whereUnder the Derrick scaling ${\varphi }_{\lambda }(x)=\varphi (\lambda x)$, a change of variables givesA finite-energy solution of the Euler-Lagrange equation is stationary under this admissible variation. The Derrick virial identity is therefore

The static field equation is ${\varphi }^{\prime \prime }=U^{\prime }(\varphi )$, so integration givesThe polynomial before $C$ is nonnegative and vanishes, so requiring the minimum to be zero fixes $C=0$. Hence the vacuum manifold iswhich has three elements. A finite-energy scalar-field kink can join only adjacent vacua: a solution cannot cross the intermediate vacuum at finite $x$ because its first integral would have $\varphi ={\varphi }^{\prime }=0$ there and the Picard-Lindelof theorem would make it constant. There are therefore four oriented topological sectors,comprising two increasing kinks and their two antikinks. Symmetry under $\varphi \mapsto -\varphi$ and spatial reflection generates all four from one profile.

For the $0\to 1$ sector, completing the square gives the Bogomolny boundEquality holds for the Bogomolny equationWith $y={\varphi }^2$, this becomes the logistic differential equation $y^{\prime }=2y(1-y)$. Translation invariance supplies an arbitrary center $x_0$, and the explicit kink in a phi-six model isIt tends to $0$ and $1$ at the two spatial ends and saturates the bound. Its mass is consequently

The anchoring conditions are $\theta (0)=0$ modulo $\pi$ and $\theta (L)=\pi /2$ modulo $\pi$, since the nematic director identifies angles differing by $\pi$. Their difference can therefore be $m\pi /2$ for any odd integer $m$. The Euler-Lagrange equation ofis ${\theta }^{\prime \prime }=0$, so every stationary solution has the formIts free energy per unit length in the $y$ direction isThe smallest possible $m^2$ is one, giving exactly the two degenerate global minima $m=1$ and $m=-1$.

LetThe Euler-Lagrange equation givesSince ${\dot{x}}^{\mu }=Ld{x}^{\mu }/ds$ and $d/d\lambda =Ld/ds$, this becomesExpanding the derivative, multiplying by the inverse metric, and using the Christoffel symbolgives the geodesic equation

Let $(v_n)$ be a minimizing sequence. The coercive lower bound from part 2 makes it bounded in the harmonic-oscillator energy space and in $L^4$. After taking a subsequence, $v_n$ converges weakly in both spaces. The compact embedding of the harmonic-oscillator energy space gives strong convergence in $L^2$, while weak lower semicontinuity of the gradient, moment, and $L^4$ terms yieldsThus $v$ attains the infimum. Replacing $v$ by $|v|$ does not increase the gradient norm, so a minimizer may be chosen nonnegative. It is nonzero because the infimum is negative whereas $J(0)=0$.

Taking the first variation against a smooth compactly supported function gives the Euler-Lagrange equationwhich is the Schrödinger trapped defocusing stationary equation.

The Euler-Lagrange equation of each component of the string embedding map is the wave equation $({\partial }_{\tau }^2-{\partial }_{\sigma }^2)X^{\mu }=0$. Its left- and right-moving solutions are identified by the endpoint Neumann boundary conditions, so the resulting open-string mode expansion isIndeed, the Fourier cosine series makes ${\partial }_{\sigma }{X}^{\mu }$ vanish at $\sigma =0,\pi$. The canonical momentum density is ${\Pi }_{\mu }=(2\pi {\alpha }^{\prime }{)}^{-1}{\dot{X}}_{\mu }$. Imposing the equal-time canonical commutation relations $[X^{\mu }(\sigma ),{\Pi }_{\nu }({\sigma }^{\prime })]=i{\delta }_{\nu }^{\mu }\delta (\sigma -{\sigma }^{\prime })$ gives the covariant quantization of the bosonic stringwith ${\alpha }_0^{\mu }=\sqrt{2{\alpha }^{\prime }}{p}^{\mu }$.

Since ${\Phi }_{\theta }=-A\sin \theta$ and the average of ${\sin }^2\theta$ over one period is $1/2$, the leading averaged Lagrangian isIts Euler-Lagrange equation for the slowly varying amplitude is algebraic:For a nonzero wave this again gives the dispersion relation. It does not determine $A$ because the original wave equation is linear and homogeneous: the leading amplitude is fixed only by the next-order transport equation and by initial or boundary data.

WriteSinceits derivatives areThe Euler-Lagrange equation therefore givesUsing the electromagnetic field tensorand dividing by $d\tau /d\lambda$ givesRaising the first index yields the equivalent requested formThis is the Lorentz-force equation from the worldline action.

Under the stated gauge transformation, the charged-particle electromagnetic Lagrangian becomesThusThe action changes only by $q[f(t_2)-f(t_1)]$, whose variation vanishes for fixed endpoints. By total-time-derivative invariance of a Lagrangian, the Euler-Lagrange equation is unchanged.

In Cartesian components,andThe Euler-Lagrange equation therefore givesUsingand $[\dot{r}\times B{]}_i={\dot{r}}_j({\partial }_i{A}_j-{\partial }_j{A}_i)$, this becomes the Lorentz force

For a variation $y+\epsilon \eta$,Integration by parts givesFor fixed endpoints, $\eta (0)=\eta (x_0)=0$. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationA solution makes the first variation vanish for every admissible variation, so it is a stationary candidate; whether it is a minimum or maximum is decided by higher variations. If endpoint values are free, $\eta$ is arbitrary there, and the boundary term instead vanishes under the natural conditionsThus the same Euler-Lagrange solution is stationary for all free-endpoint variations. These are the natural boundary conditions for a free endpoint.

Forthe two equations areSet $u=y+z$ and $v=y-z$. ThenThe conditions $y(0)=z(0)=0$ giveand hence the most general solution is

Free conditions at $x_0$ are $y^{\prime }(x_0)=z^{\prime }(x_0)=0$. Adding and subtracting them givesThus $A=0$. The zero solution exists for every $x_0$, while nonzero solutions exist precisely whenFor those values they form the one-parameter familywhere $C$ is arbitrary. This is the free-endpoint normal mode of a coupled variational functional.

The Lagrangian isThe Euler-Lagrange equation for $X$ isExpanding the divergence and simplifying givesThe equation for $Y$ is

For the quadratic geodesic Lagrangianthe Euler-Lagrange equation isRaising $\mu$ and symmetrizing the coefficient of the two velocities produces exactlyThe quadratic action fixes an affine parameter; proper time is an affine parameter for a timelike geodesic.

Write ${\dot{x}}^{\mu }=d{x}^{\mu }/ds$ and $L=\sqrt{-g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }}$. Varying $T=\int Lds$ with fixed endpoints and then choosing proper time as parameter, for which $L=1$, gives the Euler-Lagrange equationExpanding the derivative, raising the free index with the inverse metric, and using the symmetry of ${\dot{x}}^{\alpha }{\dot{x}}^{\beta }$ gives the geodesic equationThese are the Christoffel symbols of the Levi-Civita connection.

With $u=t-z$, the quadratic geodesic Lagrangian isIts transverse Euler-Lagrange equations areThe longitudinal equations areafter using the difference of those same equations. In particular,This conserved quantity also follows from the Killing vector field ${\partial }_t+{\partial }_z$ and the geodesic conserved quantity from a Killing vector.