Feynman diagrams and perturbative expansion of the time evolution operator

In Module 1 (Lectures 1-?), you have learned the basic properties of photon and how to describe various states of light using photons and their statistics. In Module 2, you will learn how to describe atom-light interactions as exchanges of photons. Formally, this can be seen from the perturbative expansion of the unitary time evolution operator. Feynman diagrams help visualize the photon exchanges that correspond to various mathematical terms occuring in such perturbative expansion.

Virtual photons and virtual states

In atomic physics, you often hear the term "virtual state" that is connected to the initial state via an exchange of a "virtual photon." What does the term "virtual" mean? How do virtual states and virtual photons differ from real ones? You will see that the term "virtual" arises from energy non-conservation. But it doesn't mean that virtual states can be any random state - specifically, if we consider a virtual state connected from the ground state, it has the character of the excited state the ground state was "trying" to connect to. But its quantum probability amplitude carries an energy denominator due to energy non-conservation.

Let's first consider two familiar processes: one-photon absorption and emission. Consider the case where the photon is resonant with the atomic transition. In the figures below, the resonant photon is represented by a wiggly arrow which fully connects the ground state from/to the excited state. One certainly expects these processes to be "real": for example, when an excited atom spontaneously decays and emits a photon, one can "catch" the emitted photon with a photodetector (hypothetically covering the entire $4\pi$ solid angle) and measure its energy.

• Resonant one photon absorption.png

  (1.1) a resonant absorption process

• Resonant one photon emission.png

  (1.2) a resonant emission process

Now what if the photon is off-resonant with respect to the atomic transition? Let's denote the non-zero detuning between the photon frequency and the atomic transition frequency as $\Delta =\omega _{L}-\omega _{a}$, with $\Delta <0$ in this example.

• Non-resonant one photon absorption.png

  (1.3) an off-resonant absorption process (Q: is this allowed?)

Now the wiggly arrows representing the photons don't quite reach the excited state. Instead, they connect to a state represented by a dashed line - a virtual state. The virtual state in this case shares the character of the excited state (the state the off-resonant photons were "trying" to connect to): for example, the parity of the excited state, and so on. However, its time evolution differs from that of the excited state, due to energy non-conservation (the exact mathematical form for its time evolution will be dealt with soon but later). We call such states "virtual" because of the energy non-conservation - the atom cannot stay infinitely in the virtual state! It can only stay in the virtual state as long as Heisenberg's uncertainty principle allows: $\Delta t\Delta E\sim \hbar$ (we will be more quantitative when we carry out the perturbative expansion of the unitary time evolution operator).

Now, it should be clear that the off-resonant one-photon processes shown in Figure 1.3 is not complete. Because the atom cannot stay in the virtual state indefinitely, it must make a transition so that it can satisfy the law of energy conservation in the end - in this two-level example, it must go back to the ground state. Figure 1.4 and 1.5 show two examples how this can be achieved. Note now that we are distinguishing between straight and wiggly arrows. Straight arrows correspond to stimulated photon absorption/emission (stimulated by a laser shining on atom) and wiggly arrows correspond to spontaneous photon absorption (into the vacuum). Figure 1.4 describes an elastic scattering process in which atom absorbs a real photon from the laser, makes a transition to a virtual state, and emits a real photon, with the same frequency, into the vacuum. This is the well-known Rayleigh scattering. Figure 1.5 describes a stimulated absorption from and emission back to the external off-resonant laser. This gives rise to a light shift in the energy of the atom.

• Rayleigh scattering.png

  (1.4) a two-photon scattering process (Rayleigh scattering)

• Ac start shift co-rotating term.png

  (1.5) a two-photon light shift (the "co-rotating term")

It is also possible for the atom in the ground state to spontaneously emit a virtual photon and reabsorb it, as Figure 1.6 shows. The virtual state still has the character of the excited state, but its energy lies below that of the ground state. This may seem strange at first, but it is a valid process!

• Two photon process counterrotating term.png

  (1.6) Lamb shift

Feynman diagrams with time axis

In the previous section, you have seen how to visualize a basic two-photon interaction involving a ground state and a virtual state. In this section, you will visualize the same process but in a slightly different way, so that the the time dependence (and ordering) of the photon exchange is manifest.

Let's consider three examples of two-photon interaction. In Figure 1.6, the first two subfigures correspond to the co-rotating term and the counter-rotating term respectively. The last figure corresponds to the counter-rotating term for an atom in the excited state.

• Two photon process three examples.png

  (1.6) three examples of two-photon interaction

• Two photon process three examples time axis.png

  (1.7) the same three examples, but with time ordering manifest

In Figure 1.7, the time ordering of the photon exchanges is now manifest, and the character of the virtual state is now explicitly written out, although the energy mismatch is less clear. But you can tell from the figures if the process is highly energy non-conserving. Also, the rule for drawing atomic transitions is that each photon changes a ground state into an excited state, and vice versa. To clarify this rule, we remind you the electric dipole approximation of the atom-light interaction (made explicit for a two-level system):

${\begin{array}{rcl}{\mathcal {H}}_{\text{int}}&=&-{\vec {\hat {d}}}\cdot {\vec {\hat {E}}}\\&=&-{\frac {\hbar \Omega _{\text{Rabi}}}{2}}(\underbrace {{\hat {\sigma }}^{+}} _{|g{\rangle }{\langle }e|}+\underbrace {{\hat {\sigma }}^{-}} _{|e{\rangle }{\langle }g|})({\hat {a}}+{\hat {a}}^{\dagger })\\&=&-{\frac {\hbar \Omega _{\text{Rabi}}}{2}}(\underbrace {{\hat {\sigma }}^{+}{\hat {a}}+{\hat {\sigma }}^{-}{\hat {a}}^{\dagger }} _{\text{co-rotating terms}}+\underbrace {{\hat {\sigma }}^{+}{\hat {a}}^{\dagger }+{\hat {\sigma }}^{-}{\hat {a}}} _{\text{counter-rotating terms}})\,.\end{array}}$

Of course, there is some subtlety in this presentation, which is the gauge dependence. In the electric dipole gauge, each photon exchange changes the atomic internal state because every term in the atom-light interaction is a bilinear product of an atomic operator and a light mode operator. However, in the Coulomb gauge, the atom-light interaction is written as (in cgs unit and fixing charge $q=-e$ )

${\begin{array}{rcl}{\mathcal {H}}_{\text{int}}&=&{\frac {e}{mc}}{\vec {\hat {p}}}\cdot {\vec {\hat {A}}}+{\frac {e^{2}}{2mc^{2}}}|{\vec {\hat {A}}}|^{2}\end{array}}$

The first part of the interaction term, proportional to ${\vec {\hat {p}}}\cdot {\vec {\hat {A}}}$, is again a bilinear product of an atomic operator and a light mode operator, so it has a similar effect as the ${\hat {\sigma }}^{(+/-)}{\hat {a}}^{(\cdot /\dagger )}$ term in the electric dipole gauge Hamiltonian. However, the second term, proportional to $|{\vec {\hat {A}}}|^{2}$, only involves the light mode operator and hence does not change the atomic internal state even when the photons are exchanged. A corresponding Feynman diagram would look like this:

File:A squared vertex diagram.png

Note that although the atomic internal state does not change in this process, the photons can still transfer momentum and hence change the atomic external state (e.g. change the center-of-mass velocity).

So depending on the gauge choice of the interaction, you can have different kinds of diagrams. But the physical reality does not depend on the choice of gauge, so they must give the same answer, when one sums all the relevant diagrams for a chosen gauge. Sometimes, one gauge choice may prove more convenient than the other, depending on the problems. We will see when we calculate the scattering cross sections for Rayleigh scattering and Thomson scattering respectively.

One last note on the virtue of specifying the time dependence in the Feynman diagrams: with the time dependence explicit, we can track clearly what the intermediate state is at each time, and this will be useful when we carry out the perturbative calculation of the unitary time evolution operator. Consider two diagrams with where the time order of the two photon exchanges has been swapped:

File:Feynman diagram time slice.png

In the left diagram, for time $t_{1}<t<t_{2}$, the intermediate state is just $|e{\rangle }|0,0{\rangle }$. However, on the right diagram, in the same time interval, the intermediate state is $|e{\rangle }|1_{(\omega ,{\vec {k}},{\vec {\epsilon }})},1_{(\omega ,{\vec {k}}',{\vec {\epsilon }}')}{\rangle }$: we have to account for the state of the two photons that are present during this time interval.

Perturbative calculation of transition amplitudes

In this section, you will learn how to carry out perturbative expansion of the unitary time evolution operator ${\tilde {U}}(t_{f},t_{i})$ in the interaction picture and identify $n^{th}$ order terms with $n{\text{-photon}}$ exchange terms. The discussion closely follows Complement A_I of "Atom-Photon Interaction" by Cohen-Tannoudji et al. (page 23-37).

Let's start from the Schrodinger equation. The Schrodinger equation describes how the quantum state $|\psi (t){\rangle }$ at time $t$ will evolve in time.

$i\hbar {\frac {d}{dt}}|\psi (t){\rangle }={\hat {H}}|\psi (t){\rangle }\,,$

Solving this first-order differential equation (with initial condition at $t=t_{0}$ given by $|\psi (t_{0}){\rangle })$ is tantamount to obtaining a unitary time evolution operator ${\hat {U}}(t,t_{0})$:

$|\psi (t){\rangle }={\hat {U}}(t,t_{0})|\psi (t_{0}){\rangle }\,,$

One can cast the Schrodinger equation as a differential equation for the unitary time evolution operator:

$i\hbar {\frac {d}{dt}}{\hat {U}}(t,t_{i})={\hat {H}}{\hat {U}}(t,t_{i})\,,$

Once you have solved for the unitary time evolution operator, you can calculate the transition amplitude between state $|a(t_{i}){\rangle }$ and state $|b(t_{f}){\rangle }$ ($={\langle }b(t_{f})|{\hat {U}}(t_{f},t_{i})|a(t_{i}){\rangle }$), which is arguably one of the most important quantities in atomic physics, since you use this to compute quantities such as transition rate, etc. So far the discussion has been in the Schrodinger picture (time-independent operators - excepting external fields - and time-dependent states), but it's much more illuminating to work in the interaction picture.

If we can split the Hamiltonian ${\hat {H}}$ into an unperturbed term ${\hat {H_{0}}}$ and an interaction term ${\hat {V}}$, you can hide the free evolution phase factor $\exp(-iE_{0}t/\hbar )$ coming from the unperturbed term such that the only time dependence of the state comes from the interaction term. On the other hand, operators do evolve in time in the interaction picture, so it is a halfway between the Schrodinger picture and the Heisenberg picture. We put a tilde $\sim$ sign above the states and the operators to denote the interaction picture.

${\begin{array}{rcl}|{\tilde {\psi }}(t){\rangle }&=&\exp(i{\hat {H_{0}}}t/\hbar )|\psi (t){\rangle }\\{\hat {\tilde {V}}}(t)&=&\exp(i{\hat {H_{0}}}t/\hbar ){\hat {V}}\exp(-i{\hat {H_{0}}}t/\hbar )\\i\hbar {\frac {d}{dt}}|{\tilde {\psi }}(t){\rangle }&=&{\hat {\tilde {V}}}(t)|{\tilde {\psi }}(t){\rangle }\\|{\tilde {\psi }}(t_{f}){\rangle }&=&{\hat {\tilde {U}}}(t_{f},t_{i})|{\tilde {\psi }}(t_{i}){\rangle }\,\end{array}}$

Let's focus on the last two equations. Given the initial condition ${\hat {\tilde {U}}}(t_{i},t_{i})=1$, the solution to the differential equation for ${\tilde {U}}(t,t_{i})$

$i\hbar {\frac {d}{dt}}{\hat {\tilde {U}}}(t,t_{i})={\hat {\tilde {V}}}(t){\hat {\tilde {U}}}(t,t_{i})\,$

can be formally written as an integral equation (check!)

${\hat {\tilde {U}}}(t_{f},t_{i})=1+{\frac {1}{i\hbar }}\int _{t_{i}}^{t_{f}}dt\,{\hat {\tilde {V}}}(t){\hat {\tilde {U}}}(t,t_{i})\,$

We can iteratively feed in this integral equation definition of ${\hat {\tilde {U}}}$ unto itself, to write it as an infinite series of terms, which will converge when working in the perturbative regime:

${\begin{array}{rcl}{\hat {\tilde {U}}}(t_{f},t_{i})&=&1+\sum _{n=1}^{\infty }{\hat {\tilde {U}}}^{(n)}(t_{f},t_{i})\\{\hat {\tilde {U}}}^{(n)}(t_{f},t_{i})&=&\left({\frac {1}{i\hbar }}\right)^{n}\int _{t_{i}}^{t_{f}}{\underset {t_{f}\geq \tau _{n}\geq \cdots \geq \tau _{1}\geq t_{i}}{d\tau _{n}\cdots d\tau _{1}}}\,{\hat {\tilde {V}}}(\tau _{n})\cdots {\hat {\tilde {V}}}(\tau _{1})\,\end{array}}$

There are a total of $n$ factors of ${\hat {\tilde {V}}}$ in ${\hat {\tilde {U}}}^{(n)}$, and this represents a total of $n$ photon exchanges happening between time $t_{i}$ and $t_{f}$. Note the time ordering in the integrand.

Now we go back into the Schrodinger picture to see what this series expansion is in terms of ${\hat {V}}$. You have to think a little bit before applying straight the transformation rules between the interaction picture and the Schrodinger picture because the unitary time evolution operator $U(t,t_{i})$ already is time-dependent in the Schrodinger picture (the transformation rule for operators above concerned a typical time-independent operator ${\hat {V}}$). It helps to calculate the transition amplitude first, which is a physical quantity and hence does not depend on which picture you use to calculate.

Let's take the basis of the unperturbed eigenstates (eigenstates of ${\hat {H}}_{0}$) and take the matrix element of ${\hat {\tilde {U}}}(t_{f},t_{i})$ with respect to that basis. Denote our initial state as $|{\tilde {\phi }}_{i}(t_{i}){\rangle }$ and our final state as $|{\tilde {\phi }}_{f}(t_{f}){\rangle }$. In the interaction picture, the unperturbed eigenstates do not evolve in time, but to make the connection to the Schrodinger picture clear we keep the time dependence notation. Then the transition amplitude between these two states is

$\langle {\tilde {\phi }}_{f}(t_{f})|{\hat {\tilde {U}}}(t_{f},t_{i})|{\tilde {\phi }}_{i}(t_{i})\rangle _{\text{interaction}}=\langle \phi _{f}(t_{f})|e^{-i{\hat {H}}_{0}t_{f}/\hbar }{\hat {\tilde {U}}}(t_{f},t_{i})e^{i{\hat {H}}_{0}t_{i}/\hbar }|\phi _{i}(t_{i})\rangle =\langle \phi _{f}(t_{f})|{\hat {U}}(t_{f},t_{i})|\phi _{i}(t_{i})\rangle _{\text{Schrodinger}}\,$

This holds for all eigenstates $|\phi _{i}(t_{i}){\rangle }$ at time $t_{i}$ and all eigenstates $|\phi _{f}(t_{f})\rangle$ at time $t_{f}$, so we see that the series expansion for the unitary time evolution operator in the Schrodinger picture is

${\begin{array}{rcl}{\hat {U}}(t_{f},t_{i})&=&e^{-i{\hat {H}}_{0}t_{f}/\hbar }{\hat {\tilde {U}}}(t_{f},t_{i})e^{i{\hat {H}}_{0}t_{i}/\hbar }\\&=&e^{-i{\hat {H}}_{0}(t_{f}-t_{i})/\hbar }+\sum _{n=1}^{\infty }{\hat {U}}^{(n)}(t_{f},t_{i})\\{\hat {U}}^{(n)}(t_{f},t_{i})&=&\left({\frac {1}{i\hbar }}\right)^{n}\int _{t_{i}}^{t_{f}}{\underset {t_{f}\geq \tau _{n}\geq \cdots \geq \tau _{1}\geq t_{i}}{d\tau _{n}\cdots d\tau _{1}}}\,e^{-i{\hat {H}}_{0}t_{f}/\hbar }\left(e^{i{\hat {H}}_{0}\tau _{n}/\hbar }{\hat {V}}e^{-i{\hat {H}}_{0}\tau _{n}/\hbar }\right)\cdots \left(e^{i{\hat {H}}_{0}\tau _{1}/\hbar }{\hat {V}}e^{-i{\hat {H}}_{0}\tau _{1}/\hbar }\right)e^{i{\hat {H_{0}}}t_{i}/\hbar }\\&=&\left({\frac {1}{i\hbar }}\right)^{n}\int _{t_{i}}^{t_{f}}{\underset {t_{f}\geq \tau _{n}\geq \cdots \geq \tau _{1}\geq t_{i}}{d\tau _{n}\cdots d\tau _{1}}}\,\left(e^{-i{\hat {H}}_{0}(t_{f}-\tau _{n})/\hbar }\right){\hat {V}}\left(e^{-i{\hat {H}}_{0}(\tau _{n}-\tau _{n-1})/\hbar }\right){\hat {V}}\cdots {\hat {V}}\left(e^{-i{\hat {H_{0}}}(\tau _{1}-t_{i})/\hbar }\right)\,\end{array}}$

The meaning of the last expression is simple: between photon exchanges (occuring at time $\tau _{m}$), the quantum state propagates freely, with the propagation phase factor given by ${\hat {H}}_{0}$. If we again look at the expression for the $n^{th}$ order correction to the transition amplitude, now written explicitly in terms of matrix elements of ${\hat {V}}$,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array} \mathcal{J}_{fi}^{(n)} &=& \langle \phi_f (t_f) | \hat{U}^{(n)}(t_f, t_i) | \phi_i (t_i) \rangle \\ &=& \left( \frac{1}{i\hbar} \right)^{n} \int_{t_i}^{t_f} \underset{ t_f \geq \tau_n \geq \cdots \geq \tau_1 \geq t_i }{ d\tau_n \cdots d\tau_{1} } \,\sum_{k_n,k_{n-1},\ldots,k_{1}} e^{ i E_f t_f /\hbar }\left(e^{-i E_f (t_f - \tau_n)/\hbar}\right) V_{bk_{n}} \left(e^{-i E_{k_{n}} (\tau_n - \tau_{n-1})/\hbar}\right) V_{k_{n} k_{n-1}}\cdots V_{k_1 a} \left(e^{-i E_i (\tau_1 - t_i)/\hbar}\right)e^{-i E_i t_i / \hbar} \end{array}}

where $V_{k_{n}k_{n-1}}=\langle k_{n}|{\hat {V}}|k_{n-1}\rangle$ and so on. The summation is over all the intermediate states involved. NOTE: the intermediate states need not be discrete. What if the intermediate state involves a photon state with arbitrary wavevector? Also, do not be confused by the first and the last phase factor, $e^{iE_{f}t_{f}/\hbar }$ and $e^{-iE_{i}t_{i}/\hbar }$! They simply come from the fact that in the Schrodinger picture, the states evolve in time and hence the unpertubed eigenstates which form our basis also evolve in time.

Great, the derivation involved many small steps of algebra but was generally straightforward, and we have ascribed physical meaning to each term in the expansion for the unitary time evolution operator. To conclude, as an exercise let's describe ${\hat {U}}^{(2)}(t_{f},t_{i})$ term in a Feynman diagram.

File:Transition amplitude and feynman diagram.png

This describes an elastic scattering process in which a photon with (wavevector, polarization) equal to $({\vec {k}},{\vec {\epsilon }})$ is absorbed and later emitted into a different mode $({\vec {k'}},{\vec {\epsilon '}})$, at the same frequency. The quantum state $|g,1,0{\rangle }$ describes the atom in the ground state, and one photon occupying the mode $(\omega _{L},{\vec {k}},{\vec {\epsilon }})$, and so on.

Energy conservation, S- and T-matrix

We started the main topic with the discussion of virtual photons and virtual states - what they mean, and how they gained the term "virtual" from the fact that they are not energy-conserving. Invoking Heisenberg's uncertainty principle, we said earlier that virtual photons/states cannot live longer than $\Delta t\sim \hbar /\Delta E$. Now that we have worked through the perturbative series expansion of the unitary time evolution operator, we can be more quantitative with the previous statement.

Let's consider the first order transition amplitude

${\mathcal {J}}_{fi}^{(1)}={\frac {1}{i\hbar }}\int _{t_{i}}^{t_{f}}d\tau _{1}e^{iE_{f}t_{f}/\hbar }\left(e^{-iE_{f}(t_{f}-\tau _{1})/\hbar }\right)V_{fk}\left(e^{-iE_{i}(\tau _{1}-t_{i})/\hbar }\right)e^{-iE_{i}t_{i}/\hbar }={\frac {1}{i\hbar }}\int _{t_{i}}^{t_{f}}d\tau _{1}\,V_{fi}\,e^{i(E_{f}-E_{i})\tau _{1}/\hbar }\,$

Select the origin of time such that $t_{f}=-t_{i}=T/2$ and carry out the time integral. We obtain

${\mathcal {J}}_{fi}^{(1)}=-2\pi iV_{fi}\left({\frac {1}{\pi }}{\frac {\sin(E_{f}-E_{i})T/2\hbar }{(E_{f}-E_{i})}}\right)=-2\pi iV_{fi}\delta ^{(T)}(E_{f}-E_{i})\,$

As $T\to \infty$, which is most often a reasonable limit as macroscopic time (in which measurements are done) is much longer than microscopic time, the function inside the parentheses approaches the Dirac delta function $\delta (E_{f}-E_{i})$. The width of this function, measured by the distance between the first two zeros, is $4\pi \hbar /T$. Basically, this means energy conservation is satisfied up to an uncertainty of $\sim \hbar /T$, which is what the uncertainty principle says.

Now let's consider the second order transition amplitude, for which we have to start summing over intermediate states.

Failed to parse (unknown function "\begin{array}"): {\displaystyle \begin{array} \mathcal{J}_{fi}^{(2)} &=& \left(\frac{1}{i\hbar}\right)^2 \int_{t_i}^{t_f} \underset{T/2 \geq \tau_2 \geq \tau_1 \geq -T/2}{d\tau_2 d\tau_1}\,\sum_{k} e^{i E_f t_f /\hbar}\left(e^{-i E_f (t_f - \tau_2)/\hbar}\right)V_{fk} \left(e^{-i E_k (\tau_2 - \tau_1)/\hbar}\right)V_{ki} \left(e^{-iE_i(\tau_1 - t_i)/\hbar}\right)e^{-i E_i t_i / \hbar}\\ &=& \left(\frac{1}{i\hbar}\right)^2 \int_{t_i}^{t_f} \underset{T/2 \geq \tau_2 \geq \tau_1 \geq -T/2}{d\tau_2 d\tau_1} \sum_{k} V_{fk}V_{ki} e^{i(E_f-E_k)\tau_2/\hbar} e^{i(E_k - E_i)\tau_1 / \hbar} \end{array}}

A common method seen in quantum field theory to go around the time ordering restriction is to lift the restriction by multiplying the integrand with the Heaviside function, in this case $\theta (\tau _{2}-\tau _{1})$. Then to carry out the integral, one substitues an integral representation of the Heaviside function using Cauchy's integral formula for residues (see Wikipedia article on Heaviside step function). It is a relatively straightforward calculation, and it is not terribly interesting, so we will not reproduce it here. In the end, we obtain

${\mathcal {J}}_{fi}^{(2)}=-2\pi i\left[\lim _{\eta \to 0^{+}}\sum _{k}{\frac {V_{fk}V_{ki}}{E_{i}-E_{k}+i\eta }}\right]\delta ^{(T)}(E_{i}-E_{f})$

The complex term in the denominator is a remnant of the time ordering restriction. We would like to point out that the expression resembles closely the second order term from standard perturbation theory results.

References

API Ch 2