PHY2403H Quantum Field Theory. Lecture 23: QED and QCD interaction Lagrangian, Feynman propagator and rules for Fermions, hadron pair production, scattering cross section, quark pair production. Taught by Prof. Erich Poppitz

Here is a link to [a PDF with my notes for the final QFT I lecture.] That lecture followed [1] section 5.1 fairly closely (filling in some details, leaving out some others.)

This lecture

• Introduced an interaction Lagrangian with QED and QCD interaction terms
\begin{equation*}
\LL_{\text{QED}}
=
– \inv{4} F_{\mu\nu} F^{\mu\nu}
+
\overline{\Psi}_e \lr{ i \gamma^\mu \partial_\mu – m } \Psi_e

e \overline{\Psi}_e \gamma_\mu \Psi_e A^\mu
+
\overline{\Psi}_\mu \lr{ i \gamma^\mu \partial_\mu – m } \Psi_\mu

e \overline{\Psi}_\mu \gamma_\mu \Psi_\mu A^\mu,
\end{equation*}
as well as the quark interaction Lagrangian
\begin{equation*}
\LL_{\text{quarks}} = \sum_q \overline{\Psi}_q \lr{ i \gamma^\mu – m_q } \Psi_q + e Q_q \overline{\Psi}_q \gamma^\nu \Psi_q A_\nu.
\end{equation*}
• The Feynman propagator for Fermions was calculated
\begin{equation*}
\expectation{ T( \Psi_\alpha(x) \Psi_\beta(x) }_0
=
\lr{ \gamma^\mu_{\alpha\beta} \partial_\mu^{(x)} + m } D_F(x – y)
=
\int \frac{d^4 p}{(2 \pi)^4 } \frac{ i ( \gamma^\mu_{\alpha\beta} p_\mu + m ) }{p^2 – m^2 + i \epsilon} e^{-i p \cdot (x – y)}.
\end{equation*}
• We determined the Feynman rules for Fermion diagram nodes and edges.
The Feynman propagator for Fermions is
\begin{equation*}
\frac{ i \lr{ \gamma^\mu p_\mu + m } }{p^2 – m^2 + i \epsilon},
\end{equation*}
whereas the photon propagator is
\begin{equation*}
\expectation{ A_\mu A_\nu } = -i \frac{g_{\mu\nu}}{q^2 + i \epsilon}.
\end{equation*}
• Muon pair production

We then studied muon pair production in detail, and determined the form of the scattering matrix element
\begin{equation*}
i M
=
i \frac{e^2}{q^2}
\overline{v}^{s’}(p’) \gamma^\rho u^s(p)
\overline{u}^r(k) \gamma_\rho v^{r’}(k’),
\end{equation*}
where the $$(2 \pi)^4 \delta^4(…)$$ term hasn’t been made explicit, and detemined that the average of its square over all input and output polarization (spin) states was
\begin{equation*}
\inv{4} \sum_{ss’, rr’} \Abs{M}^2
=
\frac{e^4}{4 q^4}
\textrm{tr}{ \lr{
\lr{ \gamma^\alpha {k’}_\alpha – m_\mu }
\gamma_\nu
\lr{ \gamma^\beta {k}_\beta + m_\mu }
\gamma_\mu
}}
\times
\textrm{tr}{ \lr{
\lr{ \gamma^\kappa {p}_\kappa + m_e }
\gamma^\nu
\lr{ \gamma^\rho {p’}_\rho – m_e }
\gamma^\mu
}}.
\end{equation*}.
In the CM frame (neglecting the electron mass, which is small relative to the muon mass), this reduced to
\begin{equation*}
\inv{4} \sum_{\text{spins}} \Abs{M}^2
=
\frac{8 e^4}{q^4}
\lr{
p \cdot k’ p’ \cdot k
+ p \cdot k p’ \cdot k’
+ p \cdot p’ m_\mu^2
}.
\end{equation*}

• We computed the differential cross section
\begin{equation*}
{\frac{d\sigma}{d\Omega}}_{\text{CM}}
=
\frac{\alpha^2}{4 E_{\text{CM}}^2 }
\sqrt{ 1 – \frac{m_\mu^2}{E^2} }
\lr{
1 + \frac{m_\mu^2}{E^2}
+ \lr{ 1 – \frac{m_\mu^2}{E^2} } \cos^2\theta
},
\end{equation*}
and the total cross section
\begin{equation*}
\sigma_{\text{total}}
=
\frac{4 \pi \alpha^2}{3 E_{\text{CM}}^2 }
\sqrt{ 1 – \frac{m_\mu^2}{E^2} }
\lr{
1 + \inv{2} \frac{m_\mu^2}{E^2}
},
\end{equation*}
and compared that to the cross section that we was determined with the dimensional analysis handwaving at the start of the course.
• We finished off with a quick discussion of quark pair production, and how some of the calculations we performed for muon pair production can be used to measure and validate the intermediate quark states that were theorized as carriers of the strong force.

References

[1] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.

PHY2403H Quantum Field Theory. Lecture 16: Differential cross section, scattering, pair production, transition amplitude, decay rate, S-matrix, connected and amputated diagrams, vacuum fluctuation, symmetry coefficient. Taught by Prof. Erich Poppitz

Here are my [lecture notes from last Wednesday’s class], which are posted out of sequence and only in PDF format this time.

PHY2403H Quantum Field Theory. Lecture 17: Scattering, decay, cross sections in a scalar theory. Taught by Prof. Erich Poppitz

November 12, 2018 phy2403 No comments , , ,

DISCLAIMER: Very rough notes from class (today VERY VERY rough).

These are notes for the UofT course PHY2403H, Quantum Field Theory, taught by Prof. Erich Poppitz, fall 2018.

Review: S-matrix

We defined an $$S-$$matrix
\label{eqn:qftLecture17:20}
\bra{f} S \ket{i} = S_{fi} = \lr{ 2 \pi }^4 \delta^{(4)} \lr{ \sum \lr{p_i – \sum_{p_f} } } i M_{fi},

where
\label{eqn:qftLecture17:40}
i M_{fi} = \sum \text{ all connected amputated Feynman diagrams }.

The matrix element $$\bra{f} S \ket{i}$$ is the amplitude of the transition from the initial to the final state. In general this can get very complicated, as the number of terms grows factorially with the order.

Scattering in a scalar theory

Suppose that we have a scalar theory with a light field $$\Phi, M$$ and a heavy field $$\varphi, m$$, where $$m > 2 M$$. Perhaps we have an interaction with a $$z^2$$ symmetry so that the interaction potential is quadratic in $$\Phi$$
\label{eqn:qftLecture17:60}
V_{\text{int}} = \mu \varphi \Phi \Phi.

We may have $$\Phi \Phi \rightarrow \Phi \Phi$$ scattering.

We will denote diagrams using a double line for $$\phi$$ and a single line for $$\Phi$$, as sketched in

fig. 1. Particle line convention.

There are three possible diagrams:

fig 2. Possible diagrams.

The first we will call the s-channel, which has amplitude

\label{eqn:qftLecture17:80}
A(\text{s-channel}) \sim \frac{i}{p^2 – m^2 + i \epsilon} =
\frac{i}{s – m^2 + i \epsilon}

\label{eqn:qftLecture17:100}
(p_1 + p_2)^2 = s

In the centre of mass frame
\label{eqn:qftLecture17:120}
\Bp_1 = – \Bp_2,

so
\label{eqn:qftLecture17:140}
s = \lr{ p_1^0 + p_2^0 }^2 = E_{\text{cm}}^2.

To the next order we have a diagram like fig. 3.

fig. 3. Higher order.

and can have additional virtual particles created, with diagrams like fig. 4.

fig. 4. More virtual particles.

We will see (QFT II) that this leads to an addition imaginary $$i \Gamma$$ term in the propagator
\label{eqn:qftLecture17:160}
\frac{i}{s – m^2 + i \epsilon}
\rightarrow
\frac{i}{s – m^2 – i m \Gamma + i \epsilon}.

If we choose to zoom into the such a figure, as sketched in fig. 5, we find that it contains the interaction of interest for our diagram, so we can (looking forward to currently unknown material) know that our diagram also has such an imaginary $$i \Gamma$$ term in its propagator.

fig. 5. Zooming into the diagram for a higher order virtual particle creation event.

Assuming such a term, the squared amplitude becomes
\label{eqn:qftLecture17:180}
\evalbar{\sigma}{s \text{near} m^2}
\sim
\Abs{A_s}^2 \sim \inv{(s – m)^2 + m^2 \Gamma^2}

This is called a resonance (name?), and is sketched in fig. 6.

fig. 6. Resonance.

Where are the poles of the modified propagator?

\label{eqn:qftLecture17:220}
\frac{i}{s – m^2 – i m \Gamma + i \epsilon}
=
\frac{i}{p_0^2 – \Bp^2 – m^2 – i m \Gamma + i \epsilon}

The pole is found, neglecting $$i \epsilon$$, is found at
\label{eqn:qftLecture17:200}
\begin{aligned}
p_0
&= \sqrt{ \omega_\Bp^2 + i m \Gamma } \\
&= \omega_\Bp \sqrt{ 1 + \frac{i m \Gamma }{\omega_\Bp^2} } \\
&\approx \omega_\Bp + \frac{i m \Gamma }{2 \omega_\Bp}
\end{aligned}

Decay rates.

We have an initial state
\label{eqn:qftLecture17:240}
\ket{i} = \ket{k},

and final state
\label{eqn:qftLecture17:260}
\ket{f} = \ket{p_1^f, p_2^f \cdots p_n^f}.

We defined decay rate as the ratio of the number of initial particles to the number of final particles.

The probability is
\label{eqn:qftLecture17:280}
\rho \sim \Abs{\bra{f} S \ket{i}}^2
=
(2 \pi)^4 \delta^{(4)}( p_{\text{in}} – \sum p_f )
(2 \pi)^4 \delta^{(4)}( p_{\text{in}} – \sum p_f )
\times \Abs{ M_{fi} }^2

Saying that $$\delta(x) f(x) = \delta(x) f(0)$$ we can set the argument of one of the delta functions to zero, which gives us a vacuum volume element factor
\label{eqn:qftLecture17:300}
(2 \pi)^4
\delta^{(4)}( p_{\text{in}} – \sum p_f ) =
(2 \pi)^4
\delta^{(4)}( 0 )
= V_3 T,

so
\label{eqn:qftLecture17:320}
\frac{\text{probability for $$i \rightarrow f$$}}{\text{unit time}}
\sim
(2 \pi)^4 \delta^{(4)}( p_{\text{in}} – \sum p_f )
V_3
\times \Abs{ M_{fi} }^2

\label{eqn:qftLecture17:340}
\braket{\Bk}{\Bk} = 2 \omega_\Bk V_3

coming from

\label{eqn:qftLecture17:360}
\braket{k}{p} = (2 \pi)^3 2 \omega_\Bp \delta^{(3)}(\Bp – \Bk)

so
\label{eqn:qftLecture17:380}
\braket{k}{k} = 2 \omega_\Bp V_3

\label{eqn:qftLecture17:400}
\frac{\text{probability for $$i \rightarrow f$$}}{\text{unit time}}
\sim
\frac{
(2 \pi)^4 \delta^{(4)}( p_{\text{in}} – \sum p_f )
\Abs{ M_{fi} }^2 V_3
}
{
2 \omega_\Bk V_3
2 \omega_{\Bp_1}
\cdots
2 \omega_{\Bp_n} V_3^n
}

If we multiply the number of final states with $$p_i^f \in (p_i^f, p_i^f + dp_i^f)$$ for a particle in a box
\label{eqn:qftLecture17:420}
p_x = \frac{ 2 \pi n_x}{L}

\label{eqn:qftLecture17:440}
\Delta p_x = \frac{ 2 \pi }{L} \Delta n_x

\label{eqn:qftLecture17:460}
\Delta n_x
=
\frac{L}{2 \pi} \Delta p_x

and

\label{eqn:qftLecture17:480}
\Delta n_x
\Delta n_y
\Delta n_z
= \frac{V_3}{(2 \pi)^3 }
\Delta p_x
\Delta p_y
\Delta p_z

\label{eqn:qftLecture17:500}
\begin{aligned}
\Gamma
&=
\frac{\text{number of events $$i \rightarrow f$$}}{\text{unit time}} \\
&=
\prod_{f} \frac{ d^3 p}{(2 \pi)^3 2 \omega_{\Bp^f} }
\frac{ (2 \pi)^4 \delta^{(4)}( k – \sum_f p^f ) \Abs{M_{fi}}^2 }
{
2 \omega_{\Bk}
}
\end{aligned}

Note that everything here is Lorentz invariant except for the denominator of the second term ( $$2 \omega_{\Bk}$$). This is a well known result (the decay rate changes in different frames).

Cross section.

For $$2 \rightarrow \text{many}$$ transitions

\label{eqn:qftLecture17:520}
\frac{\text{probability $$i \rightarrow f$$}}{\text{unit time}}
\times \lr{
\text{ number of final states with $$p_f \in (p_f, p_f + dp_f)$$
}
}
=
\frac{ (2 \pi)^4 \delta^{(4)}( \sum p_i – \sum_f p^f ) \Abs{M_{fi}}^2 {V_3} }
{
2 \omega_{\Bk_1} V_3
2 \omega_{\Bk_2} {V_3 }
}
\prod_{f} \frac{ d^3 p}{(2 \pi)^3 2 \omega_{\Bp^f} }

We need to divide by the flux.

In the CM frame, as sketched in fig. 7, the current is
\label{eqn:qftLecture17:540}
\Bj = n \Bv_1 – n \Bv_2,

so if the density is
\label{eqn:qftLecture17:560}
n = \inv{V_3},

(one particle in $$V_3$$), then
\label{eqn:qftLecture17:580}
\Bj = \frac{\Bv_1 – \Bv_2}{V_3}.

fig. 7. Centre of mass frame.

This is where [1] stops,
\label{eqn:qftLecture17:640}
\sigma
=
\frac{ (2 \pi)^4 \delta^{(4)}( \sum p_i – \sum_f p^f ) \Abs{M_{fi}}^2 {V_3} }
{
2 \omega_{\Bk_1}
2 \omega_{\Bk_2}
\Abs{\Bv_1 – \Bv_2}
}
\prod_{f} \frac{ d^3 p}{(2 \pi)^3 2 \omega_{\Bp^f} }

There is, however, a nice Lorentz invariant generalization
\label{eqn:qftLecture17:600}
j = \inv{ V_3 \omega_{k_A} \omega_{k_B}} \sqrt{ (k_A – k_B)^2 – m_A^2 m_B^2 }

(Claim: DIY)
\label{eqn:qftLecture17:620}
\begin{aligned}
\evalbar{j}{CM}
&=
\inv{V_3}
\lr{
\frac{\Abs{\Bk}}{\omega_{k_A}}
+
\frac{\Abs{\Bk}}{\omega_{k_B}}
} \\
&=
\inv{V_3} \lr{ \Abs{\Bv_A} + \Abs{\Bv_B} } \\
&=
\inv{V_3} \Abs{\Bv_1 – \Bv_2 }.
\end{aligned}

\label{eqn:qftLecture17:660}
\sigma
=
\frac{ (2 \pi)^4 \delta^{(4)}( \sum p_i – \sum_f p^f ) \Abs{M_{fi}}^2 {V_3} }
{
4 \sqrt{ (k_A – k_B)^2 – m_A^2 m_B^2 }
}
\prod_{f} \frac{ d^3 p}{(2 \pi)^3 2 \omega_{\Bp^f} }.

References

[1] Michael E Peskin and Daniel V Schroeder. An introduction to Quantum Field Theory. Westview, 1995.

PHY2403H Quantum Field Theory. Lecture 15b: Wick’s theorem, vacuum expectation, Feynman diagrams, phi^4 interaction, tree level diagrams, scattering, cross section, differential cross section. Taught by Prof. Erich Poppitz

[Here are my notes for the second half of last wednesday’s lecture]. Because the simplewick latex package is required to format these meaningfully, and that’s not available in wordpress latex, I’m not going to attempt to make a web viewable version in addition to the pdf.

Analysis of incorrect problem solution attempt (non-relativistic QFT massive scalar field with quadratic interaction)

November 1, 2018 phy2403 No comments , ,

An incorrect start at a field theory problem.

The following is a problem from the second PHY2403 problem set, and the start of a solution attempt I made. I’m not posting this because it shows how to do the problem, but because it was a useful problem to show that I didn’t understand a lot of the problem statement.
Understanding where I went wrong is actually pretty useful in this case.

Problem: Playing with the non-relativistic limit

Consider a real scalar relativistic field theory of mass m with $$\lambda \phi^4$$ interaction. Let there be $$N$$ particles of momenta labeled by $$p_1,\cdots, p_N$$, whose energies are such that they are insufficient to create any new particles. Nevertheless, the particles can scatter and exchange momenta. In what follows you will study this N-particle nonrelativistic limit in some detail.

1. Write down the Hamiltonian of the field theory, including the interaction term, restricted to the N-particle sector of Hilbert space. (Use the creation and annihilation operator representation, i.e. write the result as sums of products of creation and annihilation operators of particles of various momenta.)
2. Does the resulting Hamiltonian preserve particle number? Is there an associated symmetry? What is the operator that generates it?
3. Consider now the interaction term in your reduced (to the N-particle sector of Hilbert space) Hamiltonian. How does a typical interaction term (for given configurations of momenta) act on an N-particle state? What kinds of scattering processes does it describe?
4. What do you think is the potential, in x-space, that allows the various particles to scatter and exchange momentum? How would you describe the resulting nonrelativistic quantum system to friends who never took QFT but are well-versed in quantum mechanics?

The Lagrangian density of a massive scalar field with a $$\lambda \phi^4$$ interaction has the form
\label{eqn:Nparticle:20}
\LL = \inv{2} \partial_\mu \phi \partial^\mu \phi – \inv{2} m^2 \phi^2 – \lambda \phi^4.

The corresponding Hamiltonian is
\label{eqn:Nparticle:40}
H = \inv{2} \int d^3x \lr{ \pi^2 + \frac{m^2}{2} (\spacegrad \phi)^2 + m^2 \phi^2 } + \lambda \int d^3 x \phi^4.

In terms of creation and annihilation operators, we know the form of the non-interaction portion of the Hamiltonian, which in normal order is
\label{eqn:Nparticle:60}
H_0 = \int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp a_\Bp^\dagger a_\Bp,

but the interaction contribution is much messier
\label{eqn:Nparticle:80}
\begin{aligned}
H_{\text{int}}
&=
\lambda \int d^3 x \frac{ d^3 p d^3 k d^3 q d^3 s}{4 (2 \pi)^{12} \sqrt{
\omega_\Bp \omega_\Bk \omega_\Bq \omega_\Bs
} }
\lr{ a_\Bp e^{-i p \cdot x} + a_\Bp e^{i p \cdot x} }
\lr{ a_\Bk e^{-i k \cdot x} + a_\Bk e^{i k \cdot x} }
\lr{ a_\Bq e^{-i q \cdot x} + a_\Bq e^{i q \cdot x} }
\lr{ a_\Bs e^{-i s \cdot x} + a_\Bs e^{i s \cdot x} } \\
&=
\lambda \int d^3 x \frac{ d^3 p d^3 k d^3 q d^3 s}{4 (2 \pi)^{12} \sqrt{
\omega_\Bp \omega_\Bk \omega_\Bq \omega_\Bs
} }
\lr{ a_\Bp e^{-i \omega_\Bp t + i \Bp \cdot \Bx} + a_\Bp e^{i \omega_\Bp t – i \Bp \cdot \Bx} }
\lr{ a_\Bk e^{-i \omega_\Bk t + i \Bk \cdot \Bx} + a_\Bk e^{i \omega_\Bk t – i \Bk \cdot \Bx} } \\
&\quad \lr{ a_\Bq e^{-i \omega_\Bq t + i \Bq \cdot \Bx} + a_\Bq e^{i \omega_\Bq t – i \Bq \cdot \Bx} }
\lr{ a_\Bs e^{-i \omega_\Bs t + i \Bs \cdot \Bx} + a_\Bs e^{i \omega_\Bs t – i \Bs \cdot \Bx} } \\
&=
\lambda \int \frac{ d^3 p d^3 k d^3 q d^3 s}{4 (2 \pi)^{9} \sqrt{
\omega_\Bp \omega_\Bk \omega_\Bq \omega_\Bs
} }
\Big(
a_\Bp a_\Bk a_\Bq a_\Bs e^{-i (\omega_\Bp + \omega_\Bk + \omega_\Bq + \omega_\Bs)t} \delta^3( \Bp + \Bk + \Bq + \Bs ) \\
a_\Bp a_\Bk a_\Bq a_\Bs^\dagger e^{-i (\omega_\Bp + \omega_\Bk + \omega_\Bq – \omega_\Bs)t} \delta^3( \Bp + \Bk + \Bq – \Bs ) \\
a_\Bp^\dagger a_\Bk^\dagger a_\Bq^\dagger a_\Bs^\dagger e^{-i (-\omega_\Bp – \omega_\Bk – \omega_\Bq – \omega_\Bs)t} \delta^3( -\Bp – \Bk – \Bq – \Bs )
\Big) \\
&=
\lambda \int \frac{ d^3 p d^3 k d^3 q }{4 (2 \pi)^{9}
}
\Big(
\inv{\sqrt{
\omega_\Bp \omega_\Bk \omega_\Bq \omega_{-\Bp – \Bk – \Bq}
}}
a_\Bp a_\Bk a_\Bq a_{-\Bp -\Bk – \Bq} e^{-i (\omega_\Bp + \omega_\Bk + \omega_\Bq + \omega_{-\Bp -\Bk -\Bq})t} \\
\inv{\sqrt{
\omega_\Bp \omega_\Bk \omega_\Bq \omega_{\Bp + \Bk + \Bq}
}}
a_\Bp a_\Bk a_\Bq a_{\Bp + \Bk + \Bq}^\dagger e^{-i (\omega_\Bp + \omega_\Bk + \omega_\Bq – \omega_{\Bp + \Bk + \Bq})t} \\
\cdots \\
\inv{\sqrt{
\omega_\Bp \omega_\Bk \omega_\Bq \omega_{-\Bp – \Bk – \Bq}
}}
a_\Bp^\dagger a_\Bk^\dagger a_\Bq^\dagger a_{-\Bp -\Bk -\Bq}^\dagger e^{-i (-\omega_\Bp – \omega_\Bk – \omega_\Bq – \omega_{-\Bp -\Bk -\Bq})t}
\Big)
\end{aligned}

Assuming we can normal order these terms as in $$H_0$$, we can rewrite the interaction as
\label{eqn:Nparticle:100}
\begin{aligned}
H_{\text{int}}
&=
\lambda \int \frac{ d^3 p d^3 k d^3 q }{4 (2 \pi)^{9} }
\Big(
\binom{4}{0}
\inv{\sqrt{
\omega_\Bp \omega_\Bk \omega_\Bq \omega_{-\Bp – \Bk – \Bq}
}}
a_\Bp a_\Bk a_\Bq a_{-\Bp -\Bk – \Bq} e^{-i (\omega_\Bp + \omega_\Bk + \omega_\Bq + \omega_{-\Bp -\Bk -\Bq})t} \\
\binom{4}{1}
\inv{\sqrt{
\omega_\Bp \omega_\Bk \omega_\Bq \omega_{\Bp – \Bk – \Bq}
}}
a_\Bp^\dagger a_\Bk a_\Bq a_{\Bp – \Bk – \Bq} e^{-i (-\omega_\Bp + \omega_\Bk + \omega_\Bq + \omega_{\Bp – \Bk – \Bq})t} \\
\binom{4}{2}
\inv{\sqrt{
\omega_\Bp \omega_\Bk \omega_\Bq \omega_{\Bp + \Bk – \Bq}
}}
a_\Bp^\dagger a_\Bk^\dagger a_\Bq a_{\Bp + \Bk – \Bq} e^{-i (-\omega_\Bp – \omega_\Bk + \omega_\Bq + \omega_{\Bp + \Bk – \Bq})t} \\
\binom{4}{3}
\inv{\sqrt{
\omega_\Bp \omega_\Bk \omega_\Bq \omega_{\Bp + \Bk _ \Bq}
}}
a_\Bp^\dagger a_\Bk^\dagger a_\Bq^\dagger a_{\Bp + \Bk + \Bq} e^{-i (-\omega_\Bp – \omega_\Bk – \omega_\Bq + \omega_{\Bp + \Bk + \Bq})t} \\
\binom{4}{4}
\inv{\sqrt{
\omega_\Bp \omega_\Bk \omega_\Bq \omega_{-\Bp – \Bk – \Bq}
}}
a_\Bp^\dagger a_\Bk^\dagger a_\Bq^\dagger a_{-\Bp -\Bk -\Bq}^\dagger e^{-i (-\omega_\Bp – \omega_\Bk – \omega_\Bq – \omega_{-\Bp -\Bk -\Bq})t}
\Big)
\end{aligned}

If we restrict the allowed momenta to the discrete set $$\Bp \in \setlr{ \Bp_1, \Bp_2, \cdots \Bp_N}$$, the total Hamiltonian including the interaction term
takes the form
\label{eqn:Nparticle:120}
\begin{aligned}
\text{$$:H:$$} &=
\sum_{i = 1}^N \omega_{\Bp_i} a_{\Bp_i}^\dagger a_{\Bp_i}
+
\frac{
\lambda
}{4 }
\sum_{j,m,n = 1}^N
\Big(
\binom{4}{0}
\inv{\sqrt{
\omega_{\Bp_j} \omega_{\Bp_m} \omega_{\Bp_n} \omega_{-{\Bp_j} – {\Bp_m} – {\Bp_n}}
}}
a_{\Bp_j} a_{\Bp_m} a_{\Bp_n} a_{-\Bp -\Bk – \Bq} e^{-i (\omega_{\Bp_j} + \omega_{\Bp_m} + \omega_{\Bp_n} + \omega_{-{\Bp_j} -{\Bp_m} -{\Bp_n}})t} \\
\binom{4}{1}
\inv{\sqrt{
\omega_{\Bp_j} \omega_{\Bp_m} \omega_{\Bp_n} \omega_{{\Bp_j} – {\Bp_m} – {\Bp_n}}
}}
a_{\Bp_j}^\dagger a_{\Bp_m} a_{\Bp_n} a_{{\Bp_j} – {\Bp_m} – {\Bp_n}} e^{-i (-\omega_{\Bp_j} + \omega_{\Bp_m} + \omega_{\Bp_n} + \omega_{{\Bp_j} – {\Bp_m} – {\Bp_n}})t} \\
\binom{4}{2}
\inv{\sqrt{
\omega_{\Bp_j} \omega_{\Bp_m} \omega_{\Bp_n} \omega_{{\Bp_j} + {\Bp_m} – {\Bp_n}}
}}
a_{\Bp_j}^\dagger a_{\Bp_m}^\dagger a_{\Bp_n} a_{{\Bp_j} + {\Bp_m} – {\Bp_n}} e^{-i (-\omega_{\Bp_j} – \omega_{\Bp_m} + \omega_{\Bp_n} + \omega_{{\Bp_j} + {\Bp_m} – {\Bp_n}})t} \\
\binom{4}{3}
\inv{\sqrt{
\omega_{\Bp_j} \omega_{\Bp_m} \omega_{\Bp_n} \omega_{{\Bp_j} + {\Bp_m} – {\Bp_n}}
}}
a_{\Bp_j}^\dagger a_{\Bp_m}^\dagger a_{\Bp_n}^\dagger a_{{\Bp_j} + {\Bp_m} + {\Bp_n}} e^{-i (-\omega_{\Bp_j} – \omega_{\Bp_m} – \omega_{\Bp_n} + \omega_{{\Bp_j} + {\Bp_m} + {\Bp_n}})t} \\
\binom{4}{4}
\inv{\sqrt{
\omega_{\Bp_j} \omega_{\Bp_m} \omega_{\Bp_n} \omega_{-{\Bp_j} – {\Bp_m} – {\Bp_n}}
}}
a_{\Bp_j}^\dagger a_{\Bp_m}^\dagger a_{\Bp_n}^\dagger a_{-{\Bp_j} -{\Bp_m} -{\Bp_n}}^\dagger e^{-i (-\omega_{\Bp_j} – \omega_{\Bp_m} – \omega_{\Bp_n} – \omega_{-{\Bp_j} -{\Bp_m} -{\Bp_n}})t}
\Big)
\end{aligned}

When we did the same sort of calculation for $$(\spacegrad \phi)^2 + m^2 \phi^2$$ all the time dependent terms cancelled nicely, but that isn’t obviously the case here.
However, we haven’t used the non-relativistic (low energy) constraint. That constraint can be expressed as $$\Bp^2 \ll m^2$$, in which case $$\omega_\Bp = \sqrt{ \Bp^2 + m^2 } \sim m$$, the mass of each of the particles. Incorporating that into our N-particle Hamiltonian, we have
\label{eqn:Nparticle:140}
\begin{aligned}
\text{$$:H:$$} &=
\sum_{i = 1}^N \omega_{\Bp_i} a_{\Bp_i}^\dagger a_{\Bp_i}
+
\frac{
\lambda
}{4 m^2 }
\sum_{j,m,n = 1}^N
\Big(
\binom{4}{0}
a_{\Bp_j} a_{\Bp_m} a_{\Bp_n} a_{-\Bp -\Bk – \Bq} e^{- 4 i m t} \\
\binom{4}{1}
a_{\Bp_j}^\dagger a_{\Bp_m} a_{\Bp_n} a_{{\Bp_j} – {\Bp_m} – {\Bp_n}} e^{-3 i m t} \\
\binom{4}{2}
a_{\Bp_j}^\dagger a_{\Bp_m}^\dagger a_{\Bp_n} a_{{\Bp_j} + {\Bp_m} – {\Bp_n}} \\
\binom{4}{3}
a_{\Bp_j}^\dagger a_{\Bp_m}^\dagger a_{\Bp_n}^\dagger a_{{\Bp_j} + {\Bp_m} + {\Bp_n}} e^{ 3 i m t } \\
\binom{4}{4}
a_{\Bp_j}^\dagger a_{\Bp_m}^\dagger a_{\Bp_n}^\dagger a_{-{\Bp_j} -{\Bp_m} -{\Bp_n}}^\dagger e^{4 i m t}
\Big).
\end{aligned}

The only annoying aspect to this Hamiltonian is the $$a_{{\Bp_j} + {\Bp_m} – {\Bp_n}}$$ operator in the interaction term, which is not clear to me how to interpret. That seems to imply that it is possible to create particles with linear combinations of momentum that may not be in the original set of $$N$$ particle momenta. I think that this can be further fudged by invoking the non-relativistic constraint again, and decreeing that each of the uniquely indexed creation and annihilation operators are distinguishable only by index, so we can write the N-particle non-relativistic sector Hamiltonian as
\label{eqn:Nparticle:170}
\text{$$:H:$$} =
\sum_{i = 1}^N \omega_{\Bp_i}
a_{i}^\dagger a_{i}
+
\frac{
3 \lambda
}{2 m^2 }
\sum_{r,s,t,u = 1}^N
a_{r}^\dagger a_{s}^\dagger a_{t} a_{u}.

Commentary.

While there are a few things that are not wrong above, those correct parts are liberally mixed with a few fundamental errors.

Even before getting to the fundamential errors, there are a few minor issues too. For example, I don’t think that there is anything strictly wrong with the expansion of \ref{eqn:Nparticle:100} for example, although it appears that I confused myself by actually evaluating the delta function instead of just identifying it, something like:
\label{eqn:Nparticle:190}
\begin{aligned}
H_{\text{int}}
&=
\frac{\lambda}{4} \int \frac{ d^3 p d^3 k d^3 q d^3 s}{(2 \pi)^{12} \sqrt{ \omega_\Bp \omega_\Bk \omega_\Bq \omega_\Bs } }
\Big(
\binom{4}{0}
\delta^3( \Bp + \Bk + \Bq + \Bs )
a_\Bp a_\Bk a_\Bq a_\Bs e^{-i (\omega_\Bp + \omega_\Bk + \omega_\Bq + \omega_\Bs) t} \\
\binom{4}{1}
\delta^3( -\Bp + \Bk + \Bq + \Bs )
a_\Bp^\dagger a_\Bk a_\Bq a_\Bs e^{-i (-\omega_\Bp + \omega_\Bk + \omega_\Bq + \omega_\Bs)t} \\
\binom{4}{2}
\delta^3( -\Bp – \Bk + \Bq + \Bs )
a_\Bp^\dagger a_\Bk^\dagger a_\Bq a_\Bs e^{-i (-\omega_\Bp – \omega_\Bk + \omega_\Bq + \omega_\Bs)t} \\
\binom{4}{3}
\delta^3( -\Bp – \Bk – \Bq + \Bs )
a_\Bp^\dagger a_\Bk^\dagger a_\Bq^\dagger a_\Bs e^{-i (-\omega_\Bp – \omega_\Bk – \omega_\Bq + \omega_\Bs)t} \\
\binom{4}{4}
\delta^3( -\Bp – \Bk – \Bq – \Bs )
a_\Bp^\dagger a_\Bk^\dagger a_\Bq^\dagger a_\Bs^\dagger e^{-i (-\omega_\Bp – \omega_\Bk – \omega_\Bq – \omega_\Bs)t}
\Big)
\end{aligned}

So much of QFT is expressed in terms of delta functions, yet I tend to view them as something to be eliminated instead of retained. One lesson is that delta functions in QFT should be embraced, and had I done so in this case, I would have had a better chance of understanding the scattering part of the question that came later. One role of the delta function actually appears to be critical to the scattering question, as it actually encodes the constraints that lead to conservation of momentum in a “scattering process” related to the interaction term of the Hamiltonian.

I didn’t understand what was meant by a scattering process, but that was related to another more fundamental misunderstanding. That screw up was in my interpretation of what was meant by the “N particle sector of the Hilbert space.” In my reading of that phrase I tossed out “Hilbert space” as irrelevant. Part of that was somewhat reactionary, as it seems to me that Hilbert space is tossed around in so much QM in a way that makes it seem more rigorous, but we never touch on the scary mathematics that rigorously defines what a Hilbert space is. It seems to me that the phrase “Hilbert space” is often used as a pretentious way of saying “complex inner product space”. It has a more precise meaning to Mathematicians, but I don’t think that most physicists understand that meaning (and most certainly most students of QM don’t).

So, long story short, I interpreted “N particle sector of the Hilbert space” as “N particle sector” and though that meant I had to somehow discretize the system, introducing N discrete momenta into the mix. Here’s an example, using the non-interaction Hamiltonian, of exactly how I did that:
\label{eqn:Nparticle:210}
\int \frac{d^3 p}{(2 \pi)^3} \omega_\Bp a_\Bp^\dagger a_\Bp \rightarrow \sum_{i = 1}^N \omega_\Bp a_\Bp^\dagger a_\Bp.

This is wrong! I actually clued into part of the trouble with this, but didn’t know what the root cause was. I fudged around that my dropping the problematic side effects of having evaluated the delta function, which lead me to have terms like $$a_{\Bp + \Bq – \Bk}$$ in the expansion of the interaction Hamiltonian. I couldn’t see how creation and annihilation operators that are associated with arbitrary linear combinations of the other momenta could maintain an $$N$$ particle system. Those linear combinations could easily lie outside of the set of the original $$N$$ particle system that I constructed in my discretization.

In office hours, Professor Poppitz had me work a small problem to illustrate the error in my ways, specifically applying the Hamiltonian to a two momentum state
\label{eqn:Nparticle:230}
\begin{aligned}
H \ket{\Bp_1 \Bp_2}
&=
\lr{
\int \frac{d^3}{(2 \pi)^3}
a_\Bp^\dagger
a_\Bp
}
a_{\Bp_1}^\dagger
a_{\Bp_2}^\dagger
\ket{0} \\
&=
\int \frac{d^3}{(2 \pi)^3} \omega_\Bp
a_\Bp^\dagger
\lr{
a_{\Bp_1}^\dagger a_\Bp
+ (2 \pi) \delta^3(\Bp -\Bp_1)
}
a_{\Bp_2}^\dagger
\ket{0} \\
&=
\int \frac{d^3}{(2 \pi)^3} \omega_\Bp
a_\Bp^\dagger
a_{\Bp_1}^\dagger a_\Bp
a_{\Bp_2}^\dagger
\ket{0}
+
\omega_{\Bp_1}
a_{\Bp_1}^\dagger
a_{\Bp_2}^\dagger
\ket{0} \\
&=
\int \frac{d^3}{(2 \pi)^3} \omega_\Bp
a_\Bp^\dagger
a_{\Bp_1}^\dagger
\lr{
a_{\Bp_2}^\dagger
a_\Bp
+ (2 \pi)^3 \delta^3(\Bp – \Bp_2)
}
\ket{0}
+
\omega_{\Bp_1}\ket{\Bp_1 \Bp_2} \\
&=
\lr{ \omega_{\Bp_1} + \omega_{\Bp_2} }
\ket{\Bp_1 \Bp_2}.
\end{aligned}

Observe that the Hamiltonian operates on a two momentum state, returning that state scaled by the energy associated with the sum of the momenta. Given an two particle subspace of all the possible two momentum states, perhaps with a basis like $$\setlr{\ket{\Bp_1 \Bp_2}, \ket{\Bp_3 \Bp_4}, \cdots }$$, the Hamiltonian provides a mapping from that set onto itself, as it scales the states in question, but does not convert two particle states into combinations of one and three momentum states (say). This is not the case with the interaction Hamiltonian, as an operation like $$\int a^\dagger a^\dagger a^\dagger a$$ doesn’t preserve a two momentum state. Example
\label{eqn:Nparticle:250}
\begin{aligned}
\int
&\frac{d^3 p}{(2\pi)^3\sqrt{ 2\omega_\Bp}}
\frac{d^3 q}{(2\pi)^3\sqrt{ 2\omega_\Bq}}
\frac{d^3 r}{(2\pi)^3\sqrt{ 2\omega_\Br}}
\frac{d^3 s}{(2\pi)^3\sqrt{ 2\omega_\Bs}}
\delta^3(\Bp + \Bq + \Br -\Bs)
a_\Bp^\dagger
a_\Bq^\dagger
a_\Br^\dagger
a_\Bs
\ket{ \Bp_1 \Bp_2 } \\
&=
\int
\frac{d^3 p}{(2\pi)^3\sqrt{ 2\omega_\Bp}}
\frac{d^3 q}{(2\pi)^3\sqrt{ 2\omega_\Bq}}
\frac{d^3 r}{(2\pi)^3\sqrt{ 2\omega_\Br}}
\frac{d^3 s}{(2\pi)^3\sqrt{ 2\omega_\Bs}}
\delta^3(\Bp + \Bq + \Br -\Bs)
a_\Bp^\dagger
a_\Bq^\dagger
a_\Br^\dagger
a_\Bs
a_{\Bp_1}^\dagger
a_{\Bp_2}^\dagger
\ket{0} \\
&=
\int
\frac{d^3 p}{(2\pi)^3\sqrt{ 2\omega_\Bp}}
\frac{d^3 q}{(2\pi)^3\sqrt{ 2\omega_\Bq}}
\frac{d^3 r}{(2\pi)^3\sqrt{ 2\omega_\Br}}
\frac{d^3 s}{(2\pi)^3\sqrt{ 2\omega_\Bs}}
\delta^3(\Bp + \Bq + \Br -\Bs)
a_\Bp^\dagger
a_\Bq^\dagger
a_\Br^\dagger
\lr{
a_{\Bp_1}^\dagger
a_\Bs
+ (2 \pi)^3 \delta^3(\Bp_1 -\Bs)
}
a_{\Bp_2}^\dagger
\ket{0} \\
&=
\int
\frac{d^3 p}{(2\pi)^3\sqrt{ 2\omega_\Bp}}
\frac{d^3 q}{(2\pi)^3\sqrt{ 2\omega_\Bq}}
\frac{d^3 r}{(2\pi)^3\sqrt{ 2\omega_\Br}}
\frac{d^3 s}{(2\pi)^3\sqrt{ 2\omega_\Bs}}
\delta^3(\Bp + \Bq + \Br -\Bs)
a_\Bp^\dagger
a_\Bq^\dagger
a_\Br^\dagger
a_{\Bp_1}^\dagger
\lr{
a_{\Bp_2}^\dagger
a_\Bs
+ (2\pi)^3 \delta^3(\Bp_2 – \Bs)
}
\ket{0} \\
\int
\frac{d^3 p}{(2\pi)^3\sqrt{ 2\omega_\Bp}}
\frac{d^3 q}{(2\pi)^3\sqrt{ 2\omega_\Bq}}
\frac{d^3 r}{(2\pi)^3\sqrt{ 2\omega_\Br}}
\inv{\sqrt{ 2\omega_{\Bp_1}}}
\delta^3(\Bp + \Bq + \Br -\Bp_1)
a_\Bp^\dagger
a_\Bq^\dagger
a_\Br^\dagger
a_{\Bp_2}^\dagger
\ket{0} \\
&=
\inv{\sqrt{ 2\omega_{\Bp_2}}}
\lr{
\int
\frac{d^3 p}{(2\pi)^3\sqrt{ 2\omega_\Bp}}
\frac{d^3 q}{(2\pi)^3\sqrt{ 2\omega_\Bq}}
\frac{d^3 r}{(2\pi)^3\sqrt{ 2\omega_\Br}}
\delta^3(\Bp + \Bq + \Br -{\Bp_2})
}
\ket{\Bp \Bq \Br \Bp_1} \\
\inv{\sqrt{ 2\omega_{\Bp_1}}}
\lr{
\int
\frac{d^3 p}{(2\pi)^3\sqrt{ 2\omega_\Bp}}
\frac{d^3 q}{(2\pi)^3\sqrt{ 2\omega_\Bq}}
\frac{d^3 r}{(2\pi)^3\sqrt{ 2\omega_\Br}}
\delta^3(\Bp + \Bq + \Br -\Bp_1)
}
\ket{ \Bp \Bq \Br \Bp_2 }.
\end{aligned}

Such a term from the interaction Hamiltonian maps a two momentum state to a four momentum state. Notice how the continuous representation is critical to this evaluation, as well as that the action of the non-interaction Hamiltonian on a two (or N) momentum state. Attempting any sort of discretization leaves you with an operator that cannot be evaluated.

Observe that only the $$(a^\dagger)^2 a^2$$ terms in the interaction will map $$(1,2,3,\cdots,N)$$-momentum states to $$(1,2,3,\cdots,N)$$-momentum states, so the language that I didn’t understand “N particle sector of the Hilbert” space was really an encoded instruction to retain only those interaction terms. I did exactly that because intuition told me to do so (and didn’t justify why I did so), but I had the wrong reasons for making that selection. I knew there was something wrong, but didn’t know what it was. What I should have done was go back to basics and root out all the aspects of the problem statement that I did not understand, and ask about those. If I had done so (in a timely fashion, and not at the last minute when I attempted this problem), then I wouldn’t have gone down a dead end path that lead to more confusion. I didn’t even get to the interesting part of this problem, which was to show the correspondence between the QFT picture and the classical QM picture. I’ll still attempt to do so, despite having lost my window to get credit for that work.