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For the harmonic oscillator equation $y^{\prime \prime }+{\omega }^{2}y=J\delta (t-t_{\ast })$, with $\omega >0$, the displacement is continuous and the velocity increases by $J$ at the impulse. If the incoming displacement is zero and $J=-y^{\prime }(t_{\ast }-)$, the outgoing displacement and velocity both vanish. The subsequent homogeneous solution is identically zero. Equivalently, the causal forced response $(J/\omega )H(t-t_{\ast })\sin (\omega (t-t_{\ast }))$ cancels the pre-existing oscillation, where $H$ is the Heaviside step function.

The step response of a damped oscillator is the time integral of its causal Green function of a damped oscillator. Put $s=t-b$ and define $K(s)={\int }_0^s{e}^{-ku}\sin (\omega u)/\omega du$ for $s\ge 0$. Then $K(0)=K^{\prime }(0)=0$ and its oscillator equation has constant right side $1$. Integrating explicitly givesThis formula applies when $\omega \ne 0$. It is a convolution of the causal impulse response with the shifted Heaviside step function, so the response is zero before the forcing starts and has the stated zero initial data.

The integral definition also handles the degenerate parameters without division ambiguities. For $\omega =0,k\ne 0$,for $\omega =k=0$, $y_3=H(s)s^2/2$. These are the continuous zero-frequency limits of the step response.

Before the impulse the zero initial data force $y_2=0$. An impulse modeled by the Dirac delta function requires continuity of $y_2$: a jump in $y_2$ would introduce an unwanted derivative of a delta in $y_2^{\prime \prime }$. Integrating the equation across $t=a$ then gives the velocity jump $y_2^{\prime }(a+)-y_2^{\prime }(a-)=1$.

The causal Green function of a damped oscillator isThe jump conditions give the causal impulse responseHere $H$ is the Heaviside step function. For $\omega =0$, interpret the quotient continuously as $t-a$, giving $H(t-a)(t-a)e^{-k(t-a)}$. Both initial values are zero because $a>0$.

With zero initial time derivative, the integral term in the D'Alembert formula vanishes. The Heaviside step function representation givesThe rectangular pulse splitting under the wave equation consists of two width-two pulses of height $1/2$, moving left and right at speed $c$. At $t=0$ they coincide and add to the initial height-one pulse. Before $ct=1$ the overlapping interval has height one; after $ct=1$ the two pulses are separated. Values exactly at the fronts depend on the chosen $H(0)$ convention. Away from the fronts the equation holds classically, and across the fronts it holds as a weak solution of the wave equation.

For $\mathrm{Re}s>0$, integration by parts recursively in the Laplace transform givesThe initial exponential boundary term vanishes, and $\mathcal{L}[1]=1/s$ starts the recursion. Here $H$ is the Heaviside step function.

Transform the displacement in time, writing $Y(x,s)={\mathcal{L}}_t[y](x,s)$. Zero initial displacement and velocity give $s^{2}Y=c^2{Y}_{xx}+g/s$. A particular solution is $g/s^3$, and the spatial homogeneous solutions are $e^{\pm sx/c}$. The prescribed far-field behavior excludes the growing exponential. The fixed boundary sets the remaining coefficient to $-g/s^3$, henceSince $\mathcal{L}[t^2/2]=1/s^3$, the Laplace transform time-shift rule givesEquivalently, $y=g{t}^2/2$ for $t\le x/c$, and $y=gxt/c-g{x}^2/(2c^2)$ for $t\ge x/c$. This solves the forced wave equation in each region and has continuous displacement and first derivatives across the characteristic $t=x/c$, so it also solves the equation weakly there. It satisfies the fixed-end and initial conditions and equals $g{t}^2/2$ sufficiently far away at each fixed time. The end's influence travels at speed $c$.

Away from the impulse, the harmonic oscillator equation has a homogeneous solution. Before $x=\pi /2$, the initial conditions select $y=\cos x$. The jump condition for an impulse requires continuity of $y$ and a unit jump in $y^{\prime }$: a jump in $y$ would produce a derivative of the Dirac delta distribution, which is absent from the forcing, while integrating through the impulse gives $[y^{\prime }]=1$.

Just before the impulse, $y=0$ and $y^{\prime }=-1$. Just afterwards, $y=0$ and $y^{\prime }=0$, so uniqueness for the harmonic oscillator equation makes the subsequent solution identically zero. Equivalently the causal response is $H(x-\pi /2)\sin (x-\pi /2)$, with $H$ the Heaviside step function, and it cancels the original oscillation.

The solution isThe impulse cancellation of a harmonic oscillator produces a continuous graph with a corner at $(\pi /2,0)$, followed by a horizontal zero line.

For $a>0$, the Heaviside step function vanishes before $a$ and is one afterwards. For $\mathrm{Re}p>0$,The value assigned at the single jump point does not affect this Laplace transform. The positive real part ensures convergence at infinity.

Use ${\eta }_{ab}=\mathrm{diag}(1,-1,-1,-1)$ and $\hslash =1$. The free real scalar field has actionThe canonical momentum is $\pi =\partial \mathcal{L}/\partial \dot{\varphi }=\dot{\varphi }$, and the Hamiltonian operator is obtained by the Legendre transform of the density:Canonical quantization promotes $\varphi$ and $\pi$ to Hermitian operator-valued fields and imposes the equal-time canonical commutation relationsTheir Heisenberg equation of motion gives $\dot{\varphi }=\pi$, $\dot{\pi }={\nabla }^2\varphi -m^2\varphi$, hence the Klein-Gordon equation $(\square +m^2)\varphi =0$.

Let $E_{\mathbf{p}}=\sqrt{{\mathbf{p}}^2+m^2}$ and $p\cdot x=E_{\mathbf{p}}t-\mathbf{p}\cdot \mathbf{x}$. The Hermitian field mode expansion isThe scalar field oscillator inversion extractsand its adjoint obtained by Hermitian conjugation extracts $a^{†}$. Substitute these expressions into the equal-time canonical commutation relation. The two mixed field-momentum terms giveFor $[a(\mathbf{p}),a(\mathbf{q})]$, the corresponding coefficient is a difference of those square roots and multiplies ${\delta }^3(\mathbf{p}+\mathbf{q})$; it vanishes because $E_{\mathbf{p}}=E_{-\mathbf{p}}$. Taking the adjoint gives $[a^{†},a^{†}]=0$. Thus these are bosonic annihilation operators and creation operators. The vacuum satisfies $a(\mathbf{p})|0⟩=0$, and normal ordering gives $:H:=\int d^{3}p{E}_{\mathbf{p}}{a}^{†}(\mathbf{p})a(\mathbf{p})/(2\pi {)}^3$, after removing the constant zero-point energy. A one-particle excitation has energy $E_{\mathbf{p}}$ and spin zero.

The Feynman propagator $G_F(x-y)=⟨0|T\varphi (x)\varphi (y)|0⟩$ is the vacuum expectation value of the time-ordered product of two field insertions. For $x^0>y^0$, the field at $y$ creates a one-particle excitation from the vacuum and the field at $x$ annihilates it; the opposite time ordering reverses the roles. It is a propagation amplitude and correlation function of vacuum fluctuations, rather than a transition probability. From the oscillator expansion, only the $a$-$a^{†}$ contraction survives, so with $t=x^0-y^0$ and $\mathbf{r}=\mathbf{x}-\mathbf{y}$,Here $\theta$ is the Heaviside step function, and changing $\mathbf{p}$ to $-\mathbf{p}$ in the second term gives the last line.

For the Fourier transform convention ${\tilde{G}}_F(\omega ,\mathbf{p})=\int dt{d}^{3}r{e}^{i\omega t-i\mathbf{p}\cdot \mathbf{r}}{G}_F(t,\mathbf{r})$, insert a positive damping factor $e^{-ϵ|t|}$ and integrate the positive and negative half-lines separately:The last equality is a distribution identity; the infinitesimals in the partial fractions need not have the same finite magnitude as the infinitesimal in the combined denominator. The Feynman i-epsilon prescription means that the positive-frequency pole is just below the real energy axis and the negative-frequency pole is just above it. Closing the contour integral below for $t>0$, and above for $t<0$, reproduces the oscillator result by the residue theorem. This prescription fixes which homogeneous solutions are added to the Green function. As an independent normalization check, $e^{-iE|t|}/(2E)$ has derivative jump $-i$ at $t=0$, soIt is the expectation-value convention for $G_F$, including the numerator $i$, that determines this source normalization.

The vacuum is a centered free Gaussian state. The Wick theorem expresses the four-field time-ordered product as the sum of all pair Wick contractions plus terms containing a normal-ordered product. The latter terms have zero vacuum expectation value. There are exactly three complete pairings, with no fermionic signs for this bosonic field. Thus the free scalar four-point function isThese formulas describe canonical quantization of a real scalar field; in particular, a real scalar field uses a single oscillator family rather than independent charged-particle and antiparticle families.

The PDF gives $V=H(t)$, the Heaviside step function; the TeX incorrectly changes this to $V=I(t)$. A unit voltage step has derivative the Dirac delta function, so the current is the causal Green function for $L{d}^2/d{t}^2+Rd/dt+1/C$.

The current must be continuous at zero: a jump would create a delta derivative through $L{I}^{\prime \prime }$, absent on the right-hand side. Integrating across zero therefore gives $L[I^{\prime }]+R[I]=1$, so $I(0^{+})=0$ and $I^{\prime }(0^{+})=1/L$. PutThese distinct negative characteristic roots give the overdamped RLC responseTogether with $I(t)=0$ for $t<0$, this solves the step response. The eventual current is zero: the capacitor charges and blocks steady current in the series RLC circuit.

Take $K>0$. For $t>0$, the diffusion fundamental solution of a linear differential operator $F$ satisfies $({\partial }_t-K{\partial }_x^2)F=0$ and has initial value $\delta (x)$ in the sense of distributions:for every smooth compactly supported test function $\psi$. With the usual spatial decay it has unit mass. Equivalently the causal extension $\Theta (t)F(x,t)$ is a retarded fundamental solution satisfying $({\partial }_t-K{\partial }_x^2)(\Theta F)=\delta (x)\delta (t)$.

The causal Green function is zero for $t<\tau$ and satisfiesFor $t>\tau$ it solves the homogeneous diffusion equation, and its initial jump is $G(x,{\tau }^{+};y,\tau )=\delta (x-y)$, again distributionally. The natural spatial decay and a standard bounded or integrable solution class specify the usual kernel, avoiding unrestricted growing solutions. The PDE acts on $x,t$; $y,\tau$ label the location and time of the unit impulse. Values of the Heaviside step function exactly at the jump do not affect these distribution identities.

For ordinary bosonic operators at distinct times, time ordering places the later operator to the left:Equivalently, multiply these two products by the corresponding Heaviside step functions. For two odd fermionic operators the second product receives a minus sign; more generally a graded interchange has sign $(-1{)}^{F_1{F}_2}$. This sign is the fermionic sign convention for time-ordered field products, not a minus sign for arbitrary bosonic operators. No equal-time convention is needed for the two distinct times in this clause.