Time Ordering, Contact Terms, and Green Functions

The Dyson expansion is written with the time-ordering symbol $\mathcal T$ because interaction-picture operators at different times do not generally commute. This page explains a deceptively small point that becomes one of the main engines of perturbative QFT: when derivatives act on time-ordered products, they also act on the step functions hidden inside $\mathcal T$. The resulting delta functions are called contact terms.

Those delta-functions are not optional bookkeeping. They are the reason a time-ordered two-point function is a Green function rather than a homogeneous solution of the free equation of motion. The free operator satisfies $(\partial_t^2+m^2)q(t)=0$. Nevertheless, applying that differential operator outside $\langle 0|\mathcal T q(t)q(t')|0\rangle$ differentiates the ordering step functions at $t=t'$. Their contact term produces the source in the Green-function equation.

The cleanest setting is a single harmonic oscillator. It contains the whole phenomenon without spatial integrals, spin indices, or diagrammatic combinatorics. Once the oscillator calculation is understood, the scalar-field version is just the same calculation repeated for each momentum mode.

Time ordering is made of step functions

This lesson preserves an oscillator-first derivation and its place in the course sequence. For the field-theory form of the same contact-term identity, see Coincident Products and Contact Terms; graded signs enter later through the fermionic Wick theorem.

Oscillator Green-function normalization

For the one-dimensional oscillator variable $q(t)$ we use

$$L_0=\frac12\dot q^2-\frac12m^2q^2, \qquad p=\dot q, \qquad [q,p]=i.$$

In this oscillator warm-up, $m$ is the oscillator frequency. Later, in the scalar-field formulas, the same letter denotes the scalar mass; the algebraic role is identical because each scalar momentum mode is an oscillator.

The symbol

$$G(t-t')=\langle0|\mathcal T q(t)q(t')|0\rangle$$

denotes the vacuum expectation value of the time-ordered product, with no extra factor of $i$ pulled out. With the scalar-field conventions used on qft.org, the analogous object is called $D_F(x-y)=\langle0|\mathcal T\phi(x)\phi(y)|0\rangle$ and has momentum-space form $i/(p^2-m^2+i\epsilon)$. Some books instead write $D_F=i\Delta_F$; always check where the factor of $i$ is placed before comparing Green-function equations.

For two bosonic operators $A(t)$ and $B(t')$, time ordering means

$$\mathcal T\{A(t)B(t')\} =\theta(t-t')A(t)B(t')+\theta(t'-t)B(t')A(t),$$

where

$$\theta(t)= \begin{cases} 1, & t>0,\\ 0, & t<0. \end{cases}$$

The value at $t=0$ is irrelevant for ordinary functions and is usually chosen by convention. The identity that matters is distributional:

$$\frac{d}{dt}\theta(t-t')=\delta(t-t').$$

This is why differentiating a time-ordered product is not the same as time ordering the differentiated product. The derivative sees both the operators and the hidden step functions.

Let us differentiate the two-operator product carefully:

$$\begin{aligned} \frac{d}{dt}\mathcal T\{A(t)B(t')\} &=\delta(t-t')A(t)B(t')+\theta(t-t')\dot A(t)B(t') \\ &\quad -\delta(t'-t)B(t')A(t)+\theta(t'-t)B(t')\dot A(t). \end{aligned}$$

Since $\delta(t'-t)=\delta(t-t')$, this becomes

$$\frac{d}{dt}\mathcal T\{A(t)B(t')\} =\mathcal T\{\dot A(t)B(t')\} +\delta(t-t')\,[A(t),B(t')].$$

The second term is the contact term. It is supported only at $t=t'$, and its coefficient is the discontinuity in the ordered product as $t$ crosses $t'$. For fields, this statement should be read distributionally: strictly local fields are operator-valued distributions, and the clean mathematical object is obtained after smearing the fields against test functions.

The time-ordered product is piecewise defined. As $t$ crosses $t'$, the operator order jumps from $B(t')A(t)$ to $A(t)B(t')$. Differentiating this jump produces the contact term $\delta(t-t')[A(t),B(t')]$.

The same idea works for more operators. For bosonic operators $O_j(t_j)$,

$$\begin{aligned} \frac{d}{dt}\mathcal T\{A(t)O_2(t_2)\cdots O_n(t_n)\} &=\mathcal T\{\dot A(t)O_2(t_2)\cdots O_n(t_n)\} \\ &\quad+\sum_{j=2}^{n}\delta(t-t_j)\, \mathcal T\{O_2(t_2)\cdots [A(t),O_j(t_j)]\cdots O_n(t_n)\}. \end{aligned}$$

The displayed formula is schematic in the best possible way: it says that every time the differentiated operator crosses another operator under time ordering, the derivative of a step function produces an equal-time commutator. The time-ordering symbol is therefore not a passive bracket; it carries distributional information. For fermionic operators, the same statement holds with graded signs and graded commutators.

The free oscillator two-point function

The free harmonic oscillator obeys

$$\ddot q(t)+m^2q(t)=0.$$

It is tempting to conclude that

$$\left(\frac{d^2}{dt^2}+m^2\right) \langle0|\mathcal T q(t)q(t')|0\rangle=0.$$

That conclusion is wrong. It forgets that the differential operator acts on the time-ordering step functions.

Define

$$G(t-t')=\langle0|\mathcal T q(t)q(t')|0\rangle.$$

The first derivative gives

$$\frac{d}{dt}G(t-t') =\langle0|\mathcal T \dot q(t)q(t')|0\rangle +\delta(t-t')\langle0|[q(t),q(t')]|0\rangle.$$

At equal time, $[q(t),q(t)]=0$, so the contact term vanishes:

$$\frac{d}{dt}G(t-t') =\langle0|\mathcal T \dot q(t)q(t')|0\rangle.$$

Differentiate once more:

$$\frac{d^2}{dt^2}G(t-t') =\langle0|\mathcal T \ddot q(t)q(t')|0\rangle +\delta(t-t')\langle0|[\dot q(t),q(t')]|0\rangle.$$

Now the commutator is not zero. Since $p=\dot q$ and $[q,p]=i$, we have

$$[\dot q(t),q(t)]=[p(t),q(t)]=-i.$$

Using the equation of motion $\ddot q=-m^2q$, we get

$$\frac{d^2}{dt^2}G(t-t') =-m^2G(t-t')-i\delta(t-t').$$

Therefore

$$\boxed{ \left(\frac{d^2}{dt^2}+m^2\right)G(t-t')=-i\delta(t-t') }.$$

This is the key result. The time-ordered two-point function is not an ordinary solution of the homogeneous oscillator equation. It is a Green function for the oscillator operator $d^2/dt^2+m^2$, with a delta-function source fixed by the canonical commutator.

Away from $t=t'$, the time-ordered oscillator correlator solves the free equation. At $t=t'$, the derivative of the time-ordering step functions produces a delta-function source. The source strength is fixed by $[\dot q,q]=-i$.

This calculation is the operator version of the inverse-kernel relation used later in the path integral. With $P=d^2/dt^2+m^2$, it is $iG$ that obeys $P(iG)=\delta$; the raw correlator $G$ obeys $PG=-i\delta$. The operator derivation also identifies the source: it comes from equal-time quantization and differentiation of the ordering step functions.

Momentum-space inversion and its ambiguity

Since $G$ only depends on the time difference, write

$$\tau=t-t', \qquad G(\tau)=\int_{-\infty}^{\infty}\frac{d\omega}{2\pi}\, e^{-i\omega\tau}\widetilde G(\omega).$$

Then

$$\left(\frac{d^2}{d\tau^2}+m^2\right)G(\tau) =\int\frac{d\omega}{2\pi}\,e^{-i\omega\tau}(m^2-\omega^2)\widetilde G(\omega).$$

The Green-function equation becomes

$$(m^2-\omega^2)\widetilde G(\omega)=-i,$$

so formally

$$\widetilde G(\omega)=\frac{i}{\omega^2-m^2}.$$

The word “formally” matters. The expression has poles at $\omega=\pm m$, so the inverse is not fully specified until we say how those poles are treated. Equivalently, the differential equation determines $G$ only up to a solution of the homogeneous equation:

$$\left(\frac{d^2}{d\tau^2}+m^2\right)G_{\rm hom}(\tau)=0.$$

In Fourier language one may add distributions supported at the poles:

$$\widetilde G(\omega)\longrightarrow \widetilde G(\omega)+c_+\delta(\omega-m)+c_-\delta(\omega+m).$$

The equation of motion alone cannot determine $c_+$ and $c_-$. The vacuum expectation value and the time-ordering prescription determine them. This is exactly the ambiguity that the $i\epsilon$ prescription resolves.

For now, keep the lesson separate from the prescription: a propagator is an inverse only after a boundary condition has been chosen. Retarded, advanced, and Feynman Green functions invert the same differential operator but impose different boundary conditions.

Determining the oscillator propagator from modes

The oscillator mode expansion is

$$q(t)=\frac{1}{\sqrt{2m}}\left(ae^{-imt}+a^\dagger e^{imt}\right), \qquad [a,a^\dagger]=1, \qquad a|0\rangle=0.$$

For $t>t'$,

$$\begin{aligned} G(t-t') &=\langle0|q(t)q(t')|0\rangle \\ &=\frac{1}{2m} \langle0|\left(ae^{-imt}+a^\dagger e^{imt}\right) \left(ae^{-imt'}+a^\dagger e^{imt'}\right)|0\rangle \\ &=\frac{1}{2m}e^{-im(t-t')}. \end{aligned}$$

Only the contraction $aa^\dagger$ contributes. For $t<t'$, time ordering reverses the operators:

$$G(t-t')=\langle0|q(t')q(t)|0\rangle =\frac{1}{2m}e^{-im(t'-t)}.$$

Thus

$$\boxed{ G(t-t')=\frac{1}{2m}e^{-im|t-t'|} }$$

or, equivalently,

$$G(t-t')=\frac{1}{2m} \left[\theta(t-t')e^{-im(t-t')} +\theta(t'-t)e^{-im(t'-t)}\right].$$

This expression makes the physical boundary condition visible. Positive-frequency modes propagate from the earlier insertion to the later insertion. The time-ordered product arranges the two possible orderings so that the vacuum state is used consistently.

The oscillator time-ordered correlator is built from two homogeneous pieces. For $t>t'$, the factor is $e^{-im(t-t')}$. For $t<t'$, time ordering gives $e^{-im(t'-t)}$. The discontinuity of the first derivative at $t=t'$ supplies the delta-function source.

One can check directly that this function satisfies the Green-function equation. For $\tau\neq0$, it is a sum of free oscillations, so

$$\left(\frac{d^2}{d\tau^2}+m^2\right)G(\tau)=0.$$

At $\tau=0$, the first derivative jumps:

$$G'(0^+)= -\frac{i}{2}, \qquad G'(0^-)= +\frac{i}{2}.$$

Therefore

$$G'(0^+)-G'(0^-)=-i,$$

which is precisely the coefficient of $\delta(\tau)$ in $G''(\tau)$. Hence

$$\left(\frac{d^2}{d\tau^2}+m^2\right)G(\tau)=-i\delta(\tau).$$

The next page rewrites this same statement as a contour prescription in the complex $\omega$ plane.

From the oscillator to a scalar field

For a free real scalar field, the equal-time canonical commutator is

$$[\phi(t,\mathbf x),\pi(t,\mathbf y)]=i\delta^{(3)}(\mathbf x-\mathbf y), \qquad \pi=\dot\phi.$$

The free equation of motion is

$$(\partial_t^2-\nabla^2+m^2)\phi(x)=0.$$

Define the time-ordered two-point function

$$D_F(x-y)=\langle0|\mathcal T\phi(x)\phi(y)|0\rangle.$$

Repeating the oscillator derivation gives

$$\boxed{ (\partial_t^2-\nabla^2+m^2)D_F(x-y) =-i\delta^{(4)}(x-y) }.$$

The spatial delta function enters from the equal-time commutator, while the time delta function enters from differentiating the step function in the time-ordered product.

In momentum space this becomes the formal inverse

$$\widetilde D_F(p)=\frac{i}{p^2-m^2},$$

again with the warning that the poles require a prescription. The numerator $i$ is not an arbitrary decoration: it is the same $i$ that appeared in the contact term from $[\pi,\phi]=-i\delta^{(3)}$. The fully specified Feynman propagator is

$$D_F(x-y)=\int\frac{d^4p}{(2\pi)^4}\, \frac{i\,e^{-ip\cdot(x-y)}}{p^2-m^2+i\epsilon},$$

where $i\epsilon$ denotes the $\epsilon\downarrow0$ Feynman boundary value, developed in the next page. At finite $\epsilon$ the inverse equation belongs to the correspondingly deformed kernel.

The site’s delta-normalized Feynman inverse is a distinct object:

$$G_F=iD_F, \qquad (\Box+m^2)G_F=\delta^{(4)}.$$

Thus the oscillator’s local name $G$ translates to the field correlator $D_F$, not to the inverse $G_F$. Scalar propagators develops this normalization and its boundary conditions. The contact derivation is also given in Schwartz 2014, § 14.7.1, pp. 273–274.

Interactions and equations inside correlators

Now include a quartic interaction for a single time-dependent variable,

$$L=\frac12\dot q^2-\frac12m^2q^2-\frac{\lambda}{4!}q^4.$$

The classical equation of motion is

$$\ddot q+m^2q+\frac{\lambda}{3!}q^3=0.$$

For the quantized anharmonic oscillator, with its Hamiltonian and a state in the required operator domains, the Heisenberg equation holds as an operator identity. Applying $d^2/dt^2+m^2$ outside an ordered two-point function additionally differentiates the step functions:

$$\left(\frac{d^2}{dt^2}+m^2\right) \langle\mathcal T q(t)q(t')\rangle +\frac{\lambda}{3!}\langle\mathcal T q^3(t)q(t')\rangle =-i\delta(t-t').$$

For a bare scalar field with interaction $\lambda\phi^4/4!$, retain a compatible regulator, the canonical field normalization, and the same vacuum boundary prescription. In the following continuum notation, the finite-regulator version uses its regulated kinetic operator and identity kernel in place of $\Box+m^2$ and $\delta^{(4)}$, and $\phi^3(x)$ means the regulated composite insertion. The corresponding identity is

$$(\Box_x+m^2) \langle\mathcal T\phi(x)\phi(x_1)\cdots\phi(x_n)\rangle +\frac{\lambda}{3!} \langle\mathcal T\phi^3(x)\phi(x_1)\cdots\phi(x_n)\rangle$$ $$= -i\sum_{j=1}^{n}\delta^{(4)}(x-x_j) \langle\mathcal T\phi(x_1)\cdots\widehat{\phi(x_j)}\cdots\phi(x_n)\rangle.$$

The hat means that the factor is omitted. The wave operator on the first line acts on the whole ordered correlator, not on an already ordered equation-of-motion operator. For a free field, these operations are explicitly different:

$$\langle\mathcal T\{((\Box+m^2)\phi)(x)\phi(y)\}\rangle=0, \qquad (\Box_x+m^2)D_F(x-y)=-i\delta^{(4)}(x-y).$$

This distinction also matters in the path integral, whose derivative insertions can be denoted by a covariant $\mathcal T^\ast$ prescription. It must not be identified with ordinary $\mathcal T$ after using an operator equation. Schwartz 2014, § 14.7.2, p. 275 explains the distinction; lesson 17 derives the corresponding regulated source identity.

This identity is also the first glimpse of perturbation theory in correlator language. The free kinetic operator acting on an $n$-point function produces a delta source plus a higher correlator. For $\lambda\phi^4$, the two-point equation contains a four-field expectation value with three fields at the same point. The hierarchy is exact at the fixed compatible regulator but not closed. A continuum renormalized identity requires a specified field normalization, a renormalized composite operator $[\phi^3]$, its possible mixing, and the corresponding local contact terms; the bare coincident formula cannot simply be copied unchanged. The regulated Schwinger–Dyson hierarchy makes this qualification explicit. Perturbation theory solves it order by order by expanding in $\lambda$ and evaluating free time-ordered products with Wick’s theorem.

Summary

Time ordering is a distributional operation. It is built from step functions, and derivatives of those step functions produce delta functions when operator times coincide. These delta functions are contact terms. They are fixed by equal-time commutation relations and cannot be dropped without destroying the Green-function equation.

For the harmonic oscillator,

$$G(t-t')=\langle0|\mathcal T q(t)q(t')|0\rangle$$

obeys

$$\left(\frac{d^2}{dt^2}+m^2\right)G(t-t')=-i\delta(t-t').$$

For the scalar field, the same logic gives

$$(\Box+m^2)D_F(x-y)=-i\delta^{(4)}(x-y).$$

Thus the raw propagator is $-i$ times the inverse of the free equation-of-motion operator: $G_F=iD_F$ is delta normalized. Its boundary conditions are fixed by the time-ordered vacuum expectation value. The differential equation alone leaves homogeneous solutions undetermined; the Feynman prescription, developed next, fixes that ambiguity.

Common pitfalls

A common mistake is to move derivatives through the time-ordering symbol as if $\mathcal T$ were an ordinary algebraic bracket. It is not. The step functions inside $\mathcal T$ are distributions, and their derivatives produce delta functions.

Another common mistake is to confuse ordering a zero operator with differentiating an ordered product. The free operator equation remains zero when inserted inside ordinary time ordering. The contact arises when the wave operator acts outside that ordering, differentiating its step functions.

A third trap is to identify $i/(\omega^2-m^2)$ as a complete propagator. It is only a formal inverse. The poles require a prescription, and different prescriptions correspond to different Green functions.

Finally, keep track of the Lorentzian factor. Here the oscillator correlator $G=\langle\mathcal T qq\rangle$ obeys $(d^2/dt^2+m^2)G=-i\delta$, so its delta-normalized inverse is $iG$. For the field, the corresponding names are $D_F$ and $G_F=iD_F$.

Exercises

Exercise 1: differentiating a time-ordered product

Let $A(t)$ and $B(t')$ be bosonic operators. Starting from the step-function definition of time ordering, prove

$$\frac{d}{dt}\mathcal T\{A(t)B(t')\} =\mathcal T\{\dot A(t)B(t')\}+\delta(t-t')[A(t),B(t')].$$

Solution

By definition,

$$\mathcal T\{A(t)B(t')\} =\theta(t-t')A(t)B(t')+\theta(t'-t)B(t')A(t).$$

Differentiate:

$$\begin{aligned} \frac{d}{dt}\mathcal T\{A(t)B(t')\} &=\delta(t-t')A(t)B(t')+\theta(t-t')\dot A(t)B(t')\\ &\quad -\delta(t'-t)B(t')A(t)+\theta(t'-t)B(t')\dot A(t). \end{aligned}$$

Since $\delta(t'-t)=\delta(t-t')$, the delta-function terms combine to

$$\delta(t-t')\left(A(t)B(t')-B(t')A(t)\right) =\delta(t-t')[A(t),B(t')].$$

The remaining terms are exactly

$$\mathcal T\{\dot A(t)B(t')\}.$$

Therefore

$$\frac{d}{dt}\mathcal T\{A(t)B(t')\} =\mathcal T\{\dot A(t)B(t')\}+\delta(t-t')[A(t),B(t')].$$

Exercise 2: oscillator propagator as a Green function

For the harmonic oscillator with

$$q(t)=\frac{1}{\sqrt{2m}}(ae^{-imt}+a^\dagger e^{imt}), \qquad [a,a^\dagger]=1,$$

compute $G(t-t')=\langle0|\mathcal Tq(t)q(t')|0\rangle$ and verify explicitly that

$$\left(\frac{d^2}{dt^2}+m^2\right)G(t-t')=-i\delta(t-t').$$

Solution

For $t>t'$,

$$G(t-t')=\langle0|q(t)q(t')|0\rangle.$$

Only the $aa^\dagger$ term survives, so

$$G(t-t')=\frac{1}{2m}e^{-im(t-t')}.$$

For $t<t'$,

$$G(t-t')=\langle0|q(t')q(t)|0\rangle =\frac{1}{2m}e^{-im(t'-t)}.$$

Thus

$$G(\tau)=\frac{1}{2m}e^{-im|\tau|}, \qquad \tau=t-t'.$$

For $\tau\neq0$, this is a free oscillator solution, so $(d^2/d\tau^2+m^2)G=0$. The first derivative is

$$G'(\tau)= \begin{cases} -\dfrac{i}{2}e^{-im\tau}, & \tau>0,\\[6pt] +\dfrac{i}{2}e^{im\tau}, & \tau<0. \end{cases}$$

Therefore

$$G'(0^+)-G'(0^-)=-i.$$

A function whose first derivative jumps by $J$ has a second derivative containing $J\delta(\tau)$. Hence

$$\left(\frac{d^2}{d\tau^2}+m^2\right)G(\tau)=-i\delta(\tau).$$

Exercise 3: scalar-field contact term

For a free scalar field with

$$[\phi(t,\mathbf x),\pi(t,\mathbf y)]=i\delta^{(3)}(\mathbf x-\mathbf y), \qquad \pi=\dot\phi,$$

show that

$$(\Box_x+m^2)\langle0|\mathcal T\phi(x)\phi(y)|0\rangle =-i\delta^{(4)}(x-y).$$

Solution

Let

$$D_F(x-y)=\langle0|\mathcal T\phi(x)\phi(y)|0\rangle.$$

The spatial derivatives do not act on the time-ordering step functions, so the only contact term comes from the two time derivatives. First,

$$\partial_{t_x}D_F(x-y) =\langle0|\mathcal T\dot\phi(x)\phi(y)|0\rangle$$

because $[\phi(t,\mathbf x),\phi(t,\mathbf y)]=0$. Differentiating again gives

$$\partial_{t_x}^2D_F(x-y) =\langle0|\mathcal T\ddot\phi(x)\phi(y)|0\rangle +\delta(t_x-t_y)\langle0|[\dot\phi(t_x,\mathbf x),\phi(t_y,\mathbf y)]|0\rangle.$$

At equal time,

$$[\dot\phi(t,\mathbf x),\phi(t,\mathbf y)] =[\pi(t,\mathbf x),\phi(t,\mathbf y)] =-i\delta^{(3)}(\mathbf x-\mathbf y).$$

Thus the contact term is

$$-i\delta(t_x-t_y)\delta^{(3)}(\mathbf x-\mathbf y) =-i\delta^{(4)}(x-y).$$

Using the free equation of motion $(\partial_t^2-\nabla^2+m^2)\phi=0$ inside the non-contact part, we obtain

$$(\Box_x+m^2)D_F(x-y)=-i\delta^{(4)}(x-y).$$

References

• Schwartz, Matthew D. Quantum Field Theory and the Standard Model. Cambridge University Press, 2014, § 14.7, pp. 273–275. Publisher DOI.

Further reading

• Sidney Coleman, Lectures of Sidney Coleman on Quantum Field Theory, Chapters 7–8 and 13, for time ordering, Dyson’s formula, Wick diagrams, and Green functions in the Heisenberg picture.
• Mark Srednicki, Quantum Field Theory, Sections 8 and 22, for the propagator as a Green function and the appearance of contact terms in Schwinger–Dyson and Ward identities.
• Steven Weinberg, The Quantum Theory of Fields, Volume I, Chapters 6–7, for the operator derivation of Feynman rules, propagators, and the canonical formalism.