5.2.2 Dyson Series#

Worked solutions for the homework problems in the 5.2.2 Dyson Series lecture. Each problem statement is reproduced verbatim, followed by a complete worked solution.

1. Volterra integral form. Verify directly that the Volterra integral equation \(\hat{U}_{\mathcal{I}}(t)=\hat{I}-(\mathrm{i}/\hbar)\int_0^t\mathrm{d}t'\,\hat{V}_{\mathcal{I}}(t')\hat{U}_{\mathcal{I}}(t')\) for \(\hat{U}_{\mathcal{I}}(t)\) is equivalent to the IP operator EOM \(\mathrm{i}\hbar\,\partial_t\hat{U}_{\mathcal{I}}(t)=\hat{V}_{\mathcal{I}}(t)\hat{U}_{\mathcal{I}}(t),\;\hat{U}_{\mathcal{I}}(0)=\hat{I}\) together with the initial condition \(\hat{U}_{\mathcal{I}}(0)=\hat{I}\), by differentiating both sides.

Solution.

The two statements to be shown equivalent are the Volterra equation

and the operator EOM with initial condition

We prove equivalence in both directions.

Volterra \(\Rightarrow\) EOM \(+\) initial condition. Set \(t=0\) in the Volterra equation. The integral \(\int_0^0\) has a vanishing domain, so it is the zero operator, leaving

The initial condition is therefore built into the Volterra form. Next differentiate with respect to the upper limit \(t\). The constant \(\hat{I}\) drops, and by the fundamental theorem of calculus \(\partial_t\int_0^t\mathrm{d}t'\,f(t')=f(t)\) applied to the integrand \(f(t')=\hat{V}_{\mathcal{I}}(t')\hat{U}_{\mathcal{I}}(t')\):

Multiplying through by \(\mathrm{i}\hbar\) gives \(\mathrm{i}\hbar\,\partial_t\hat{U}_{\mathcal{I}}=\hat{V}_{\mathcal{I}}\hat{U}_{\mathcal{I}}\), which is the operator EOM.

EOM \(+\) initial condition \(\Rightarrow\) Volterra. Conversely, take the EOM in the form \(\mathrm{i}\hbar\,\partial_{t'}\hat{U}_{\mathcal{I}}(t')=\hat{V}_{\mathcal{I}}(t')\hat{U}_{\mathcal{I}}(t')\) and integrate both sides over \(t'\) from \(0\) to \(t\). The left side is an exact differential:

Insert \(\hat{U}_{\mathcal{I}}(0)=\hat{I}\) and divide by \(\mathrm{i}\hbar\) (using \(1/\mathrm{i}=-\mathrm{i}\)):

This is the Volterra equation. The two forms are therefore equivalent: the Volterra equation packages the differential equation and its initial condition into a single relation. (This is exactly why it is the better starting point for iteration — there is no separate boundary condition to track order by order.)

2. Iteration to second order. Substitute \(\hat{U}_{\mathcal{I}}(t)=\hat{I}-(\mathrm{i}/\hbar)\int_0^t\mathrm{d}t'\,\hat{V}_{\mathcal{I}}(t')\hat{U}_{\mathcal{I}}(t')\) for \(\hat{U}_{\mathcal{I}}(t')\) on its own right-hand side once, and verify the \(k=2\) term of the Dyson series \(\hat{U}_{\mathcal{I}}(t)=\sum_{k=0}^{\infty}\Bigl(-\frac{\mathrm{i}}{\hbar}\Bigr)^{\!k}\int_0^t\!\mathrm{d}t_k\cdots\int_0^{t_2}\!\mathrm{d}t_1\;\hat{V}_{\mathcal{I}}(t_k)\cdots\hat{V}_{\mathcal{I}}(t_1)\). Pay attention to the ordering of the two integration variables and the signs.

Solution.

Start from the Volterra equation, writing the outer integration variable as \(t_1\):

The factor \(\hat{U}_{\mathcal{I}}(t_1)\) under the integral is itself given by the Volterra equation, evaluated at upper limit \(t_1\) and with inner variable \(t_2\):

Substitute this into the previous line:

Distribute \(\hat{V}_{\mathcal{I}}(t_1)\) across the bracket:

The first two terms are the \(k=0\) and \(k=1\) pieces of the Dyson series. The third term still contains the unknown \(\hat{U}_{\mathcal{I}}(t_2)\); truncating it at \(\hat{U}_{\mathcal{I}}(t_2)\to\hat{I}\) (its leading value) isolates the \(k=2\) term:

Compare with the general Dyson term in \(\hat{U}_{\mathcal{I}}(t)=\sum_{k=0}^{\infty}\Bigl(-\frac{\mathrm{i}}{\hbar}\Bigr)^{\!k}\int_0^t\!\mathrm{d}t_k\cdots\int_0^{t_2}\!\mathrm{d}t_1\;\hat{V}_{\mathcal{I}}(t_k)\cdots\hat{V}_{\mathcal{I}}(t_1)\), which for \(k=2\) reads

These are the same expression — only the dummy labels differ. In the iteration result \(t_2\) runs over \([0,t_1]\), so \(t_2\le t_1\); the operator product \(\hat{V}_{\mathcal{I}}(t_1)\hat{V}_{\mathcal{I}}(t_2)\) has the later time \(t_1\) on the left. Renaming \(t_1\leftrightarrow t_2\) turns the iteration result into the Dyson form verbatim: \(\hat{V}_{\mathcal{I}}(t_2)\hat{V}_{\mathcal{I}}(t_1)\) with \(t_1\le t_2\), again later-on-the-left.

Two points to note, as flagged in the problem:

• Ordering of integration variables. The inner integral’s upper limit is the outer variable, not \(t\). This nesting is what restricts the domain to the ordered simplex \(0\le t_2\le t_1\le t\) rather than the full square — and it is the simplex that produces the time ordering.

• Signs. Each substitution brings down one factor of \(-\mathrm{i}/\hbar\), so the \(k\)-th term carries \((-\mathrm{i}/\hbar)^k\). For \(k=2\) this is \((-\mathrm{i}/\hbar)^2=-1/\hbar^2\) — a real, negative prefactor; the individual minus signs do not cancel against the \(\mathrm{i}^2\) in a way that flips the result, they combine to \(-1/\hbar^2\).

3. Time-ordering identity. For an integrand symmetric in \((t_1,t_2)\), show that

Use this to rewrite the \(k=2\) Dyson term with the time-ordering operator \(\mathcal{T}\) as

and explain why the \(1/k!\) from the Taylor expansion of \(\mathcal{T}\exp\) matches the simplex factor from time ordering.

Solution.

The identity. The double integral runs over the full square \(S=[0,t]\times[0,t]\) in the \((t_1,t_2)\) plane. The diagonal \(t_1=t_2\) has zero area, so it splits \(S\) into two triangles:

Hence

In the second integral rename the dummy variables \(t_1\leftrightarrow t_2\). The domain \(S_{>}\) becomes \(S_{<}\), and the integrand becomes \(F(t_2,t_1)\). By the assumed symmetry \(F(t_2,t_1)=F(t_1,t_2)\), so the second integral equals the first:

Therefore

Applying it to the \(k=2\) Dyson term. The \(k=2\) term from Problem 2 is

The bare integrand \(\hat{V}_{\mathcal{I}}(t_2)\hat{V}_{\mathcal{I}}(t_1)\) is not symmetric — the operators need not commute. But the time-ordered product is. Define

This \(F(t_1,t_2):=\mathcal{T}[\hat{V}_{\mathcal{I}}(t_2)\hat{V}_{\mathcal{I}}(t_1)]\) is symmetric by construction: \(\mathcal{T}\) reorders factors by their time arguments and is indifferent to the order in which they are written. On the simplex \(t_1\le t_2\) the operator \(\mathcal{T}\) acts trivially, since \(\hat{V}_{\mathcal{I}}(t_2)\hat{V}_{\mathcal{I}}(t_1)\) is already time-ordered. So

Now apply the identity with \(F=\mathcal{T}[\hat{V}_{\mathcal{I}}(t_2)\hat{V}_{\mathcal{I}}(t_1)]\), in the form \(\int_{S_{<}}F=\tfrac12\int_S F\):

where \(\mathcal{T}\) has been pulled outside the integral (it acts on whatever operator product sits inside, ordering it by time argument). This is the required form, with \(2!=2\).

Why \(1/k!\) matches the simplex factor. The argument generalizes immediately. For the \(k\)-th order term, the hypercube \([0,t]^k\) decomposes into \(k!\) ordered simplices — one for each of the \(k!\) permutations of \(\{t_1,\dots,t_k\}\) (the lower-dimensional boundaries where two times coincide have zero measure). The time-ordering operator \(\mathcal{T}\) makes the integrand totally symmetric, so each of the \(k!\) simplices contributes the same integral. Hence

The Dyson series integrates over one simplex, with no explicit factorial:

The last sum is precisely the Taylor expansion of the exponential, so

The \(1/k!\) in \(\exp\) is the ratio of volumes \(\mathrm{vol(simplex)}/\mathrm{vol(hypercube)}=(t^k/k!)/t^k=1/k!\). Equivalently: the Dyson series carries no \(1/k!\) because it already restricts to a single ordered simplex; the time-ordered exponential carries \(1/k!\) because it integrates over the whole hypercube and over-counts each ordering \(k!\) times. The two factors are the same number, viewed from the two sides of the identity.

4. Bare Green’s function. From \(\hat{G}_0(t,t')=\hat{U}_0(t)\hat{U}_0^{\dagger}(t')=\sum_n \vert n\rangle\,\mathrm{e}^{-\mathrm{i}E_n(t-t')/\hbar}\,\langle n\vert\),

(a) Show \(\hat{G}_0(t,t'')\hat{G}_0(t'',t')=\hat{G}_0(t,t')\) using the spectral form.

(b) Show \(\hat{G}_0(t,t')^{\dagger}=\hat{G}_0(t',t)\), i.e. \(\hat{G}_0\) is unitary.

(c) Compute the matrix element \(\langle m\vert\hat{G}_0(t,t')\vert n\rangle\) and identify the Bohr phase.

Solution.

The bare Green’s function is, in the eigenbasis \(\{\vert n\rangle\}\) of \(\hat{H}_0\) with \(\hat{H}_0\vert n\rangle=E_n\vert n\rangle\),

with \(\langle m\vert n\rangle=\delta_{mn}\) and \(\sum_n\vert n\rangle\langle n\vert=\hat{I}\).

(a) Composition. Multiply two bare propagators and use orthonormality to collapse the inner sum:

The intermediate time \(t''\) cancels because the two exponents add: \((t-t'')+(t''-t')=t-t'\). Free propagation composes by simply adding elapsed times — the system has no memory of the intermediate instant.

(b) Unitarity. Take the Hermitian adjoint. Each \(E_n\) is real (eigenvalue of the Hermitian \(\hat{H}_0\)) and each \(\vert n\rangle\langle n\vert\) is self-adjoint, so only the scalar phase conjugates:

Rewrite the sign of the exponent as a swap of the two time arguments, \(+\,(t-t')=-(t'-t)\):

That \(\hat{G}_0\) is unitary now follows from (a). Combining the two results,

and \(\hat{G}_0(t',t')=\sum_n\vert n\rangle\,\mathrm{e}^{0}\,\langle n\vert=\sum_n\vert n\rangle\langle n\vert=\hat{I}\) by completeness. The same computation with the factors reversed gives \(\hat{G}_0(t,t')\hat{G}_0(t,t')^{\dagger}=\hat{I}\). Hence \(\hat{G}_0^{\dagger}=\hat{G}_0^{-1}\): the bare propagator is unitary, as it must be, since it is the genuine time-evolution operator of the closed unperturbed system.

(c) Matrix element and Bohr phase. Sandwich \(\hat{G}_0\) between two eigenstates and use orthonormality twice:

So the bare propagator is diagonal in the \(\hat{H}_0\) eigenbasis: it never mixes levels — which is exactly the statement that the unperturbed system has no transitions. Its only action on level \(n\) is to multiply by the phase factor

the Bohr phase of level \(n\) accumulated over the interval \(t-t'\). Each eigenstate winds at its own Bohr angular frequency \(E_n/\hbar\); the relative Bohr phase between two levels rotates at the Bohr (transition) frequency \(\omega_{mn}=(E_m-E_n)/\hbar\). It is this relative winding, re-expressed inside \(\hat{V}_{\mathcal{I}}=\hat{U}_0^{\dagger}\hat{V}\hat{U}_0\), that drives the resonance physics of the next section.

5. Schrödinger-picture Dyson series. Starting from \(\hat{U}(t)=\hat{U}_0(t)\hat{U}_{\mathcal{I}}(t)\) and the second-order term of \(\hat{U}_{\mathcal{I}}(t)=\sum_{k=0}^{\infty}\Bigl(-\frac{\mathrm{i}}{\hbar}\Bigr)^{\!k}\int_0^t\!\mathrm{d}t_k\cdots\int_0^{t_2}\!\mathrm{d}t_1\;\hat{V}_{\mathcal{I}}(t_k)\cdots\hat{V}_{\mathcal{I}}(t_1)\), derive the second-order term of \(\hat{U}(t)=\sum_{k=0}^{\infty}\Bigl(-\frac{\mathrm{i}}{\hbar}\Bigr)^{\!k}\int_0^t\!\mathrm{d}t_k\cdots\int_0^{t_2}\!\mathrm{d}t_1\;\hat{G}_0(t,t_k)\hat{V}(t_k)\hat{G}_0(t_k,t_{k-1})\cdots\hat{V}(t_1)\hat{G}_0(t_1,0)\) explicitly. Show step by step where each \(\hat{G}_0\) link comes from.

Solution.

The \(k=2\) term of the interaction-picture Dyson series (Problem 2) is

The Schrödinger-picture propagator is \(\hat{U}(t)=\hat{U}_0(t)\hat{U}_{\mathcal{I}}(t)\), so its second-order part is obtained by left-multiplying with \(\hat{U}_0(t)\):

Now expand each interaction-picture vertex with its definition \(\hat{V}_{\mathcal{I}}(t_j)=\hat{U}_0^{\dagger}(t_j)\,\hat{V}(t_j)\,\hat{U}_0(t_j)\):

Reading left to right, the free-evolution operators pair up into bare propagators wherever a \(\hat{U}_0\) meets a \(\hat{U}_0^{\dagger}\) across the gaps between vertices. Using \(\hat{G}_0(t_a,t_b)=\hat{U}_0(t_a)\hat{U}_0^{\dagger}(t_b)\):

• Left link. \(\hat{U}_0(t)\,\hat{U}_0^{\dagger}(t_2)=\hat{G}_0(t,t_2)\) — free propagation from the last vertex at \(t_2\) up to the final time \(t\).

• Middle link. \(\hat{U}_0(t_2)\,\hat{U}_0^{\dagger}(t_1)=\hat{G}_0(t_2,t_1)\) — free propagation between the two vertices.

• Right link. The trailing factor is \(\hat{U}_0(t_1)\) alone. Since \(\hat{U}_0(0)=\hat{I}\), write \(\hat{U}_0(t_1)=\hat{U}_0(t_1)\hat{U}_0^{\dagger}(0)=\hat{G}_0(t_1,0)\) — free propagation from the initial time \(0\) up to the first vertex at \(t_1\).

Substituting all three:

Therefore

which is exactly the \(k=2\) term of \(\hat{U}(t)=\sum_{k=0}^{\infty}\Bigl(-\frac{\mathrm{i}}{\hbar}\Bigr)^{\!k}\int_0^t\!\mathrm{d}t_k\cdots\int_0^{t_2}\!\mathrm{d}t_1\;\hat{G}_0(t,t_k)\hat{V}(t_k)\hat{G}_0(t_k,t_{k-1})\cdots\hat{V}(t_1)\hat{G}_0(t_1,0)\). Each \(\hat{G}_0\) link is a free-propagation segment; each \(\hat{V}\) is an instantaneous scattering. The expression reads, right to left along the ket, free \(\to\) scatter at \(t_1\) \(\to\) free \(\to\) scatter at \(t_2\) \(\to\) free — the propagate-scatter-propagate structure with the interaction-picture vertices traded for Schrödinger-picture vertices \(\hat{V}\) at the cost of inserting \(\hat{G}_0\) links between them.

6. Recursive Dyson equation. Show that \(\hat{G}(t,t_0)=\sum_{k=0}^{\infty}\Bigl(-\frac{\mathrm{i}}{\hbar}\Bigr)^{\!k}\int_{t_0}^{t}\!\mathrm{d}t_k\cdots\int_{t_0}^{t_2}\!\mathrm{d}t_1\;\hat{G}_0(t,t_k)\hat{V}(t_k)\hat{G}_0(t_k,t_{k-1})\cdots\hat{V}(t_1)\hat{G}_0(t_1,t_0)\) is equivalent to the closed recursion

Iterate this recursion once and verify the \(k=1\) and \(k=2\) terms of \(\hat{G}(t,t_0)=\sum_{k=0}^{\infty}\Bigl(-\frac{\mathrm{i}}{\hbar}\Bigr)^{\!k}\int_{t_0}^{t}\!\mathrm{d}t_k\cdots\int_{t_0}^{t_2}\!\mathrm{d}t_1\;\hat{G}_0(t,t_k)\hat{V}(t_k)\hat{G}_0(t_k,t_{k-1})\cdots\hat{V}(t_1)\hat{G}_0(t_1,t_0)\).

Solution.

Dyson series \(\Rightarrow\) recursion. The Dyson series for the dressed propagator is

Separate the \(k=0\) term, which is \(\hat{G}_0(t,t_0)\), from the rest. In every \(k\ge1\) term the latest interaction time is the outermost variable \(t_k\), integrated over \([t_0,t]\), and it is preceded (on the left) by the single link \(\hat{G}_0(t,t_k)\) and the vertex \(\hat{V}(t_k)\). Factor those out:

where the bracket collects everything to the right of \(\hat{V}(t_k)\), summed over all \(k\ge1\):

Set \(j=k-1\). The bracket becomes a sum over \(j\ge0\) of exactly the Dyson-series terms for a dressed propagator that runs from \(t_0\) to \(t_k\) — that is, it is \(\hat{G}(t_k,t_0)\). Hence

Renaming the dummy variable \(t_k\to t_1\) gives the stated recursion. (The recursion simply says: in any history, peel off the last scatter — free propagation from \(t_1\) to \(t\), one vertex at \(t_1\), and a fully dressed history from \(t_0\) to \(t_1\) before it.)

Iterating the recursion. Write the recursion compactly with a generic latest-vertex time \(s\):

The factor \(\hat{G}(s,t_0)\) on the right is, by the same recursion (now with upper limit \(s\) and a new latest-vertex time \(s'\)),

Substitute:

Now read off the orders in \(\hat{V}\):

• \(k=0\): \(\hat{G}_0(t,t_0)\) — matches the zeroth term of \(\hat{G}(t,t_0)=\sum_{k=0}^{\infty}\Bigl(-\frac{\mathrm{i}}{\hbar}\Bigr)^{\!k}\int_{t_0}^{t}\!\mathrm{d}t_k\cdots\int_{t_0}^{t_2}\!\mathrm{d}t_1\;\hat{G}_0(t,t_k)\hat{V}(t_k)\hat{G}_0(t_k,t_{k-1})\cdots\hat{V}(t_1)\hat{G}_0(t_1,t_0)\).

• \(k=1\): with \(s\to t_1\), \(-\dfrac{\mathrm{i}}{\hbar}\displaystyle\int_{t_0}^{t}\mathrm{d}t_1\;\hat{G}_0(t,t_1)\,\hat{V}(t_1)\,\hat{G}_0(t_1,t_0)\) — exactly the \(k=1\) term of \(\hat{G}(t,t_0)=\sum_{k=0}^{\infty}\Bigl(-\frac{\mathrm{i}}{\hbar}\Bigr)^{\!k}\int_{t_0}^{t}\!\mathrm{d}t_k\cdots\int_{t_0}^{t_2}\!\mathrm{d}t_1\;\hat{G}_0(t,t_k)\hat{V}(t_k)\hat{G}_0(t_k,t_{k-1})\cdots\hat{V}(t_1)\hat{G}_0(t_1,t_0)\).

• \(k=2\): the last line is already \(O(\hat{V}^2)\) before counting the \(\hat{G}(s',t_0)\) tail; to isolate the pure \(k=2\) term, replace that tail by its leading piece \(\hat{G}(s',t_0)\to\hat{G}_0(s',t_0)\). Renaming \(s\to t_2,\;s'\to t_1\) (so that \(t_0\le t_1\le t_2\le t\)):

  \[ \left(-\frac{\mathrm{i}}{\hbar}\right)^{\!2}\int_{t_0}^{t}\mathrm{d}t_2\int_{t_0}^{t_2}\mathrm{d}t_1\;\hat{G}_0(t,t_2)\,\hat{V}(t_2)\,\hat{G}_0(t_2,t_1)\,\hat{V}(t_1)\,\hat{G}_0(t_1,t_0), \]

  which is the \(k=2\) term of \(\hat{G}(t,t_0)=\sum_{k=0}^{\infty}\Bigl(-\frac{\mathrm{i}}{\hbar}\Bigr)^{\!k}\int_{t_0}^{t}\!\mathrm{d}t_k\cdots\int_{t_0}^{t_2}\!\mathrm{d}t_1\;\hat{G}_0(t,t_k)\hat{V}(t_k)\hat{G}_0(t_k,t_{k-1})\cdots\hat{V}(t_1)\hat{G}_0(t_1,t_0)\).

Each further substitution of the recursion peels off one more vertex and raises the order by one, regenerating the full Dyson series term by term. The recursion and the series are therefore equivalent.

7. Two-level Feynman diagrams. Take

with \(\hat{V}\) time-independent.

(a) Write the second-order (\(k=2\)) term of the survival amplitude \(\langle 0\vert\hat{U}(t)\vert 0\rangle\) from \(\hat{U}(t)=\sum_{k=0}^{\infty}\Bigl(-\frac{\mathrm{i}}{\hbar}\Bigr)^{\!k}\int_0^t\!\mathrm{d}t_k\cdots\int_0^{t_2}\!\mathrm{d}t_1\;\hat{G}_0(t,t_k)\hat{V}(t_k)\hat{G}_0(t_k,t_{k-1})\cdots\hat{V}(t_1)\hat{G}_0(t_1,0)\) as an explicit double time integral, using the spectral form of \(\hat{G}_0\).

(b) Identify the intermediate state that contributes and draw the corresponding Feynman diagram.

(c) Show that the first-order term \(\langle 0\vert\hat{U}^{(1)}(t)\vert 0\rangle\) vanishes, and explain the selection rule responsible. Conclude that the \(k=2\) term is the leading correction to the survival amplitude beyond the free-evolution phase \(\langle 0\vert\hat{G}_0(t,0)\vert 0\rangle\).

Solution.

The unperturbed Hamiltonian is diagonal in the computational basis: \(\hat{Z}\vert 0\rangle=+\vert 0\rangle\), \(\hat{Z}\vert 1\rangle=-\vert 1\rangle\), so

The bare propagator is diagonal,

and the perturbation is a pure spin-flip, \(\hat{V}=\hbar\Omega\,\hat{X}=\hbar\Omega\bigl(\vert 0\rangle\langle 1\vert+\vert 1\rangle\langle 0\vert\bigr)\).

(a) Second-order term of the survival amplitude. The \(k=2\) term of the Dyson series, sandwiched between \(\langle 0\vert\) and \(\vert 0\rangle\), is

Evaluate the operator string acting on \(\vert 0\rangle\) from right to left. Each bare propagator only attaches a phase; each \(\hat{V}=\hbar\Omega\hat{X}\) flips the qubit:

Two flips return the qubit to \(\vert 0\rangle\), and \(\langle 0\vert 0\rangle=1\), so the matrix element survives. Substituting the three phases gives

Collecting the prefactor \((-\mathrm{i}/\hbar)^2(\hbar\Omega)^2=-\Omega^2\) and the combined phase \(-\omega_0(t-t_2)/2+\omega_0(t_2-t_1)/2-\omega_0 t_1/2=-\omega_0 t/2+\omega_0(t_2-t_1)\),

The double integral is elementary; part (a) asks only for this explicit double-integral form, so we leave it unevaluated.

(b) Intermediate state and Feynman diagram. The operator string in part (a) routes the qubit through exactly one intermediate state: the first \(\hat{V}\) flips \(\vert 0\rangle\to\vert 1\rangle\), the bare propagator \(\hat{G}_0(t_2,t_1)\) carries it on level \(\vert 1\rangle\), and the second \(\hat{V}\) flips it back \(\vert 1\rangle\to\vert 0\rangle\). The single intermediate state is \(\vert 1\rangle\) — the only level \(\hat{X}\) connects to \(\vert 0\rangle\).

The Feynman diagram has its internal segment running on level \(\vert 1\rangle\):

   |0>          |1>          |0>
 0 ====●========●==== t
       t1        t2
   G0(t1,0)  V  G0(t2,t1)  V  G0(t,t2)

a \(\hat{G}_0\) line on \(\vert 0\rangle\) from \(0\) to the first vertex at \(t_1\), a \(\hat{G}_0\) line on the intermediate state \(\vert 1\rangle\) from \(t_1\) to \(t_2\), and a \(\hat{G}_0\) line back on \(\vert 0\rangle\) from \(t_2\) to the final time \(t\). Reading right to left, the operator product is \(\hat{G}_0(t,t_2)\,\hat{V}\,\hat{G}_0(t_2,t_1)\,\hat{V}\,\hat{G}_0(t_1,0)\) — the mirror image of the diagram, as required by the mirror rule.

(c) First-order term and the selection rule. The \(k=1\) term of the survival amplitude is

A single \(\hat{V}=\hbar\Omega\hat{X}\) flips \(\vert 0\rangle\to\vert 1\rangle\), and \(\hat{G}_0(t,t_1)\) keeps the state on level \(\vert 1\rangle\); projecting on \(\langle 0\vert\) then gives \(\langle 0\vert 1\rangle=0\). Every term in the integrand vanishes, so

This is a flip-parity selection rule. The perturbation \(\hat{V}\propto\hat{X}\) flips the qubit exactly once per vertex, so the order-\(k\) term carries \(k\) flips and returns \(\vert 0\rangle\) to \(\vert 0\rangle\) only when \(k\) is even. Odd orders instead carry the transition amplitude \(\langle 1\vert\hat{U}\vert 0\rangle\); even orders dress the survival amplitude \(\langle 0\vert\hat{U}\vert 0\rangle\).

The \(k=0\) term of the survival amplitude is pure free evolution,

a phase of unit modulus that carries no information about the perturbation. The \(k=1\) term vanishes. Hence the \(k=2\) term of part (a) is the leading correction to the survival amplitude beyond free evolution — the first nonzero term that depends on \(\hat{V}\), entering at order \(\Omega^2\). (For the transition amplitude \(\langle 1\vert\hat{U}\vert 0\rangle\) the hierarchy is mirrored: the \(k=1\) term is leading and the \(k=2\) term is identically zero.)

★ 8. Three-level virtual transition. Let \(\hat{H}_0=\Delta\vert 3\rangle\langle 3\vert\) with \(E_1=E_2=0\) and \(E_3=\Delta\). Take a perturbation that connects \(\vert 1\rangle\leftrightarrow\vert 3\rangle\leftrightarrow\vert 2\rangle\) but has no direct \(\vert 1\rangle\leftrightarrow\vert 2\rangle\) matrix element:

At \(t=0\) the system is prepared in \(\vert 1\rangle\); we compute the amplitude \(\langle 2\vert\hat{G}(t,0)\vert 1\rangle\) to find it in \(\vert 2\rangle\) at time \(t\). Work in units with \(\hbar=1\) throughout this problem, so \(\Delta\) is the level-3 angular frequency and the Dyson factor at order \(k\) is \((-\mathrm{i})^k\).

(a) Write down \(\hat{G}_0(t,t')\) explicitly using the spectral form.

(b) Argue from the Dyson series

that the leading nonzero contribution to \(\langle 2\vert\hat{G}(t,0)\vert 1\rangle\) is second order in \(\lambda_0\), and identify which Feynman diagram it corresponds to.

(c) Write the second-order amplitude explicitly as a double time integral; leave it unevaluated here (that comes in 5.2.3 once we know how to handle long-time limits).

Solution.

Spelling out the Hermitian conjugate, the perturbation is

Its only nonzero matrix elements are \(\langle 1\vert\hat{V}\vert 3\rangle=\langle 3\vert\hat{V}\vert 1\rangle=\lambda(t)\) and \(\langle 2\vert\hat{V}\vert 3\rangle=\langle 3\vert\hat{V}\vert 2\rangle=\lambda(t)\); in particular \(\langle 1\vert\hat{V}\vert 2\rangle=\langle 2\vert\hat{V}\vert 1\rangle=0\). Level \(\vert 3\rangle\) is the only doorway between \(\vert 1\rangle\) and \(\vert 2\rangle\).

(a) Bare Green’s function. With \(E_1=E_2=0\) and \(E_3=\Delta\), the spectral form \(\hat{G}_0(t,t')=\sum_n\vert n\rangle\,\mathrm{e}^{-\mathrm{i}E_n(t-t')}\,\langle n\vert\) gives

The two zero-energy states do not wind; only \(\vert 3\rangle\) carries a phase, at its Bohr frequency \(\Delta\).

(b) Leading order is \(\lambda_0^2\). Examine the Dyson series

for \(\langle 2\vert\hat{G}(t,0)\vert 1\rangle\) order by order; each vertex \(\hat{V}\) carries one power of \(\lambda_0\).

• \(k=0\): \(\langle 2\vert\hat{G}_0(t,0)\vert 1\rangle\). Since \(\hat{G}_0\) is diagonal, this is \(\propto\langle 2\vert 1\rangle=0\). Vanishes.

• \(k=1\): \(-\mathrm{i}\int_0^t\mathrm{d}t_1\,\langle 2\vert\hat{G}_0(t,t_1)\,\hat{V}(t_1)\,\hat{G}_0(t_1,0)\vert 1\rangle\). The bare propagators are diagonal, so this reduces to \(\langle 2\vert\hat{V}(t_1)\vert 1\rangle\) — which is zero, because \(\hat{V}\) has no direct \(\vert 1\rangle\leftrightarrow\vert 2\rangle\) element. Vanishes.

• \(k=2\): the integrand contains \(\langle 2\vert\hat{V}(t_2)\cdots\hat{V}(t_1)\vert 1\rangle\). Now there is a non-zero path: the first vertex takes \(\vert 1\rangle\to\vert 3\rangle\), the second takes \(\vert 3\rangle\to\vert 2\rangle\). Nonzero.

The leading nonzero contribution is therefore second order in \(\lambda_0\). Physically, \(\vert 1\rangle\) and \(\vert 2\rangle\) are not directly coupled; the system can only get from one to the other by a virtual excursion through the intermediate state \(\vert 3\rangle\). The corresponding Feynman diagram is the single second-order graph

   |1>          |3>          |2>
 0 ====●========●==== t
       t1        t2
  G0(t1,0)  V   G0(t2,t1)  V   G0(t,t2)
       |1>→|3>      |3>→|2>

a bare line on \(\vert 1\rangle\) from \(0\) to the vertex at \(t_1\), a bare line on the virtual state \(\vert 3\rangle\) from \(t_1\) to \(t_2\) (this internal line carries the only nontrivial phase, \(\mathrm{e}^{-\mathrm{i}\Delta(t_2-t_1)}\)), and a bare line on \(\vert 2\rangle\) from \(t_2\) to \(t\). It is the unique \(1\to3\to2\) second-order diagram.

(c) Second-order amplitude as a double integral. Specializing the Dyson series at \(k=2\) (with \(t_0=0\)):

Evaluate the matrix element, reading the operator string right to left:

and \(\langle 2\vert(\vert 1\rangle+\vert 2\rangle)=1\). The matrix element is therefore \(\lambda(t_2)\,\lambda(t_1)\,\mathrm{e}^{-\mathrm{i}\Delta(t_2-t_1)}\). With \((-\mathrm{i})^{2}=-1\),

Inserting the drive \(\lambda(t)=\lambda_0\cos(\omega t)\):

The amplitude is manifestly \(O(\lambda_0^2)\), as argued in (b). The double integral is left unevaluated; expanding the cosines into exponentials splits it into four nested integrals of the form \(\int_0^t\mathrm{d}t_2\int_0^{t_2}\mathrm{d}t_1\,\mathrm{e}^{\mathrm{i}\Omega_1 t_1}\mathrm{e}^{\mathrm{i}\Omega_2 t_2}\), whose resonant long-time behavior (and the resulting transition rate) is the subject of 5.2.3.