research39 min studyLesson 745 in Physics

Close the time contour without doubling the physics

Construct forward and backward evolution from an initial density matrix, explain the four contour correlators, and use unitarity to check source normalization.

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What you’ll understand

  • Derive why a real-time expectation contains both forward and backward evolution.
  • Identify contour branches as bookkeeping rather than independent physical particles.
  • Check equal-source normalization and distinguish it from an unchanged observable.

An initial-state expectation value evolves both a ket and a bra. A closed real-time contour represents those two evolutions on forward and backward branches. This is useful in equilibrium and remains valid for an arbitrary specified initial density matrix. The doubled fields retain operator ordering and response information; they do not describe two copies of the laboratory or guarantee that the state is thermal.

Follow the density matrix through the trace

An initial normalized state ϱ0\varrho_0 evolves as Uϱ0U†U\varrho_0U^\dagger. Computing an expectation at a later time inserts an operator into the trace of that evolved state. The same structure can be drawn as evolution from the initial time to a final contour time and then back again. The final time is chosen later than all physical insertions. The initial density matrix supplies the boundary information and can include correlations; assuming only a few Gaussian initial correlators is an approximation to that input.

Introduce distinct sources on the forward and backward branches to generate the required orderings. Setting those sources equal restores unitary cancellation between the two evolutions, so the generating functional is one for every common source when the trace is normalized. This condition expresses conservation of probability. It does not say that an operator expectation cannot change under that common physical evolution. Observables arise through appropriately placed source derivatives before the equal-source identification, rather than by interpreting the value one as a complete state description.

Z[J+,J−]=Tr⁡ ⁣(UJ+ϱ0UJ−†),Z[J,J]=Tr⁡ϱ0=1Z[J_+,J_-]=\operatorname{Tr}\!\left(U_{J_+}\varrho_0U_{J_-}^\dagger\right),\qquad Z[J,J]=\operatorname{Tr}\varrho_0=1
UJ evolves along a physical time interval with the specified source-dependent Hamiltonian; its adjoint gives the backward branch. The equal-source identity assumes normalized initial trace and unitary evolution of the modeled closed system. An open system requires a consistent enlarged or effective description.↔ Scroll wide equations horizontally.

Organize the contour orderings

A bosonic contour two-point function has four branch assignments. Two insertions on the forward branch give time ordering, two on the backward branch give reverse time ordering, and insertions on different branches give greater or lesser products. The contour ordering, not a new physical interaction, determines their relationship. At separated times they satisfy an identity expressing that the two possible ordinary time orderings together cover the two unordered products. Fermionic versions need their matching statistical signs and definitions.

A useful change of variables uses the average field ϕr=(ϕ++ϕ−)/2\phi_r=(\phi_++\phi_-)/2 and the difference field ϕa=ϕ+−ϕ−\phi_a=\phi_+-\phi_-. It organizes causal response and fluctuations efficiently, but the letters are labels of a linear basis change. In particular, the difference field is not a second particle species, and setting its physical source to zero does not discard all fluctuations. The contour action contains a forward action minus a backward action, with matching and initial-state conditions required to reconstruct the trace.

G++=GT,G−−=GT‾,G−+=G>,G+−=G<,G+++G−−=G+−+G−+G^{++}=G_T,\quad G^{--}=G_{\overline T},\quad G^{-+}=G^>,\quad G^{+-}=G^<,\qquad G^{++}+G^{--}=G^{+-}+G^{-+}
Here all functions use the bosonic convention G = −i times the appropriate contour-ordered or raw product. The relation holds with consistent equal-time prescriptions; derivatives or composite operators can add contact terms that must also be treated consistently.↔ Scroll wide equations horizontally.

Keep equilibrium assumptions optional

The closed contour can start from a thermal density matrix, but it does not require one. Equilibrium permits KMS relations that reduce the number of independent functions, while general initial states need separate statistical information. An imaginary-time segment can encode a thermal initial ensemble in a contour construction; it is not evidence that every real-time problem must be periodic in time. A quench or driving protocol changes the actual Hamiltonian history and should not be represented by switching a parameter in a stationary equilibrium answer.

Approximations to contour dynamics should retain the normalization and causal identities relevant to the calculation. Violating them can create spurious responses or probability loss. Conservation-law checks are also valuable, but preserving one identity alone does not establish full accuracy: an approximation can conserve energy while missing scattering or long-time thermalization. Compare the intended observable, time interval, and scale hierarchy with the retained terms. The contour is an exact organizational framework; the practical closure or diagram truncation still needs a controlled justification.

Unequal sources produce a generating amplitude

Take ϱ0=diag⁡(3/4,1/4)\varrho_0=\operatorname{diag}(3/4,1/4) in a two-level z basis. Let U+=exp⁡(−iασz/2)U_+=\exp(-i\alpha\sigma_z/2) and U−=IU_-=I. Evaluate ZZ for α=π\alpha=\pi, then check what happens when both branches use the same U.

  1. At α=π\alpha=\pi, the forward evolution has diagonal entries e−iπ/2=−ie^{-i\pi/2}=-i and eiπ/2=ie^{i\pi/2}=i. With the backward evolution equal to identity, the trace weights those two phases by the initial populations.

  2. Thus Z=(3/4)(−i)+(1/4)(i)=−i/2Z=(3/4)(-i)+(1/4)(i)=-i/2. This complex generating quantity is not a probability and need not be one when the two sources differ.

  3. For equal branch evolutions, Z=Tr⁡(Uϱ0U†)=Tr⁡ϱ0=1Z=\operatorname{Tr}(U\varrho_0U^\dagger)=\operatorname{Tr}\varrho_0=1. Unitary cancellation holds for any rotation angle, independently of these particular populations.

The result

Unequal sources give minus i over two in this example, while equal sources give one. The normalization identity concerns the trace with matched branch histories, not a claim that every observable is constant.

Explore the model

Interactive lab

Interactive lab · quantum

Distinguish normalization from an unchanged expectation

Reset and set θ = 0 degrees and φ = 0 degrees, predicting basis probability P(0) = 1 and Bloch z = 1. Compare with a fresh Reset, θ = 180 degrees and φ = 0 degrees, predicting P(0) = 0 and z = −1. Both displayed states are normalized although their observables differ. This pure-state lab does not compute a mixed density matrix, contour branches, source derivatives, or a Schwinger–Keldysh generating functional.

Distinguish normalization from an unchanged expectationA normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Current readings: Probability |0⟩: 71.1%; Probability |1⟩: 28.9%; Bloch (x, y, z): 0.78, 0.45, 0.42. All parameters are adjustable with the labeled controls below.xyz|0⟩|1⟩Basis probabilitiesP(0)71.1%P(1)28.9%
Bloch vector / P(0)P(1)Phase latitude
Probability |0⟩
71.1%
Probability |1⟩
28.9%
Bloch (x, y, z)
0.78, 0.45, 0.42
t = 0 s

A normalized two-level pure state, not an interacting field simulation. Playback adds one radian per second to the relative phase; basis probabilities stay fixed. Parameter changes pause playback and recompute the model at the current elapsed time. Reset restores both parameters and returns time to zero. Playback starts paused. Adjust either parameter, then choose Play or Step.

Look forNormalized evolution can change observables; a normalized generating trace does not encode their values by itself.

Check your understanding

What does Z[J,J] = 1 establish for the normalized closed-system contour?

Try it yourself

Start again from ϱ0=diag⁡(3/4,1/4)\varrho_0=\operatorname{diag}(3/4,1/4), and apply the same physical spin-exchange unitary U=σxU=\sigma_x on both branches. Find the evolved density matrix, its trace, and the final expectation of σz\sigma_z. Explain why the result is consistent with equal-source normalization.

Take a moment to make a prediction or work it out on paper. Open a hint when you need one.

Give me a hint

Conjugation by σx exchanges the two diagonal populations while leaving their sum unchanged.

Compare with the solution

Conjugation gives Uϱ0U†=diag⁡(1/4,3/4)U\varrho_0U^\dagger=\operatorname{diag}(1/4,3/4). Its trace remains one, and the final z expectation is 1/4−3/4=−1/21/4-3/4=-1/2, compared with initial value +1/2+1/2. Thus the expectation changes even though the matched contour trace equals one. The example uses a specified mixed state and a finite unitary; it does not infer any thermalization, loss of coherence, or irreversible dynamics from the branch duplication.

Take it with you

  • The two real-time branches represent ket and bra evolution of one system.
  • Equal-source normalization is a unitarity check, while observables require appropriate derivatives.
  • KMS can simplify an equilibrium contour but must not be imposed on a general initial state.

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