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 operator equation $(\partial_t^2+m^2)q(t)=0$ holds away from coincident times, but the correlator $\langle 0|\mathcal T q(t)q(t')|0\rangle$ has a singularity at $t=t'$. That singularity is exactly what produces the source term 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 a fact that will later become almost automatic in the path integral: the inverse of the quadratic operator in the action is the propagator. The operator derivation has the advantage of showing where the source comes from. It comes from equal-time quantization.

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

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

Repeating the oscillator derivation gives

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

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

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

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

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

but the meaning of the $i\epsilon$ will be developed carefully in the next page.

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

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

Inside a time-ordered correlation function, the same equation holds away from coincident times, but contact terms remain. For the two-point function,

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

For a scalar field with interaction $\lambda\phi^4/4!$, 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. This is the first operator form of what will later be called a Schwinger–Dyson identity. The interacting field equation does not vanish inside a time-ordered correlator; it produces contact terms when the equation-of-motion insertion collides with another field.

This identity is also the first glimpse of perturbation theory in correlator language. The free inverse operator acting on an $n$-point function produces a delta source plus a higher correlator. For $\lambda\phi^4$, the equation for a two-point function contains a four-point expectation value with three fields at the same point. The hierarchy is exact but not closed. Perturbation theory closes 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

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

Thus a propagator is the inverse of the free equation-of-motion operator, with boundary conditions determined by the time-ordered vacuum expectation value. The equation of motion determines the inverse only up to homogeneous solutions; the Feynman $i\epsilon$ prescription, introduced next, fixes that remaining 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 say that a free field satisfies its equation of motion inside every correlator without qualification. The correct statement is that the free equation holds away from coincident insertions. At coincident insertions there are contact terms.

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 whether the propagator includes an explicit factor of $i$. In this page $G=\langle\mathcal T qq\rangle$ obeys $(d^2/dt^2+m^2)G=-i\delta$. If one defines $\Delta=-iG$ or $G=i\Delta$, the right-hand side changes accordingly. Many apparent sign disagreements in textbooks are only this convention difference.

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

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

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

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

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

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

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

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

References and 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.