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```{code-cell}
:tags: [hide-cell]

import mmf_setup;mmf_setup.nbinit()
import logging;logging.getLogger('matplotlib').setLevel(logging.CRITICAL)
%matplotlib inline
import numpy as np, matplotlib.pyplot as plt
```

(sec:QFTNotes)=
# Notes

* {cite:p}`Donoghue:2022` and Zee use the same metric convention.
* {cite:p}`Donoghue:2022` does not clearly/quickly present the tachyonic modes of a
  photon... Show this as a motivation for Gauge invariance.
* Physics backgrounds will vary.  Need to pair up.

* Zee's baby problem is great and we should do it, but we should include the fact that
  it is an asymptotic series.
  


(sec:Prerequisites)=
# Prerequisites

## Notations

:::{margin}
In some of the mathematical literature, one might see $f_{x} = \partial_{x}f$.  We will
not use this notation here.  When a subscript is used for a derivative, we will include
a preceding comma $f_{,x}$, following Einstein.
:::
Economy of notation is important in order to simplify calculations.  In physics, we need
to take many derivatives, so we use various shorthand notation.  You might see the
following:
\begin{gather*}
  \pdiff{f(\vect{x}, t)}{x_i} = 
  \partial_{x_i}f = 
  f_{,x_i} = 
  \left[\pdiff{f(\vect{x}, t)}{\vect{x}}\right]_i
  = [\partial_{\vect{x}} f]_{i}
  = [\vect{\nabla} f]_{i},\\
  \pdiff{f(\vect{x}, t)}{t} = 
  \partial_{t}f = 
  f_{,t} = \dot{f}.
\end{gather*}
Sometimes, when it is clear, we might use $f' = \partial_{x}f$ for the spatial
derivative in 1D.

The Laplacian is sometimes written
\begin{gather*}
  ◻︎ = ∆ = \nabla^2 = \vect{\nabla}\cdot\vect{\nabla}.
\end{gather*}

## Relativity

We take the Minkowski metric to be $g_{\mu\nu} = \diag(1, -1, -1, -1)$.  This is the
convention of {cite:p}`Donoghue:2022` and {cite:p}`Zee:2010`.  Note that
{cite:p}`t-Hooft:2016` uses the opposite convention $g_{\mu\nu} = \diag(-1, 1, 1, 1)$.


## Principle of Extremal Action

In classical mechanics, Newton's law
\begin{gather*}
  m\ddot{x}(t) = F(x, t) = -V_{,x}(x, t)
\end{gather*}
for conservative forces (those that can be expressed as the gradient of a potential
function $\vect{F} = -\vect{\nabla}V(\vect{x}, t)$ as shown) can be derived from a
variational principle in terms of the **action functional** $S[x]$:
\begin{gather*}
  S[x] = \int_{t_i}^{t_f}\d{t}\;L(x, \dot{x}, t), \qquad
  L(x, \dot{x}, t) = \frac{m\dot{x}^2}{2} - V(x, t).
\end{gather*}

## Do It!

Show that Newton's law follows from the condition of extremal action
\begin{gather*}
  \delta S[x] = 0 \implies 
  \underbrace{\diff{}{t}\pdiff{L[x, \dot{x}, t]}{\dot{x}}}_{m\ddot{x}}
  = \underbrace{\pdiff{L[x, \dot{x}, t]}{x}}_{-\partial_{x}V(x, t)}
\end{gather*}
where $\delta S[x]$ is a functional derivative.  I.e. solutions to Newton's law $x=q(t)$
that satisfy the boundary conditions $q(t_i) = x_i$ and $q(t_f) = x_f$ satisfy:
\begin{gather*}
  S[q(t) + \lambda \delta(t)] = S[q] + O(\lambda^2)
\end{gather*}
for small $\lambda$ and all deviations $\delta(t)$ that preserve the boundary conditions
$\delta(t_i) = \delta(t_f) = 0$.

### Example

Consider a particle falling in a gravitational potential $V = mgh$ where $m$ is the mass
of the particle and $g$ is the acceleration due to gravity.  Newton's law
\begin{gather*}
  m\ddot{h}(t) = -mgh(t)
\end{gather*}
has the general solution, where $h_0 = h(0)$ and $v_0 = h'(0)$:
\begin{gather*}
  h(t) = h_0 + v_0t - \frac{gt^2}{2}.
\end{gather*}

:::{margin}
One could easily do an analytic solution here, but it can be messy.  A quick numerical check
is much faster and allows one to play.  I encourage you to make simple numerical checks
often.
:::
Here we demonstrate numerically that the action is indeed quadratic in $\lambda$ for the
solution to Newton's law, but that something that is not a solution to Newton's law has
a linear dependence on $\lambda$.
```{code-cell}
m = 1.2
g = 9.8
h0 = 10.1
v0 = 2.3
t0 = 0
tf = np.sqrt(h0/g)

t = np.linspace(t0, tf, 1000)
h = h0 + v0*t - g*t**2/2
dh_dt = v0 - g*t

h_wrong = h0 + v0*t - g*t**2/2  # Wrong!  (factor in denomonator)
dh_wrong_dt = v0 - g*t/2

# Put whatever you like here, but keep the prefactor
delta = (t-t0)*(t-tf)*np.exp(t**2)
ddelta_dt = np.gradient(delta, t, edge_order=2)

def get_S(h, dh_dt):
    L = m*dh_dt**2/2 - m*g*h
    return np.trapz(L, t)

lams = np.linspace(0, 1)
Ss = [get_S(h+lam*delta, dh_dt + lam*ddelta_dt) for lam in lams]
Ss_wrong = [get_S(h_wrong+lam*delta, dh_wrong_dt + lam*ddelta_dt) for lam in lams]
fig, ax = plt.subplots()
ax.plot(lams, Ss-Ss[0], label="Solution to Newton's Law")
ax.plot(lams, Ss_wrong-Ss_wrong[0], label="Not a solution to Newton's Law")
ax.legend()
ax.set(xlabel="$\lambda$", ylabel=r"$S[h+\lambda\delta] - S[h]$");
```

## Quantum Mechanics

Please review the solution of a harmonic oscillator (HO) in quantum mechanics
\begin{gather*}
  \op{H} = \frac{\hbar^2}{2m}\op{p}^2 + \frac{m\omega^2}{2}\op{x}^2, \qquad
  [\op{x}, \op{p}] = \I \hbar,
\end{gather*}
so that you are comfortable with the notation of raising and lowering operators
\begin{gather*}
  \op{a} = \sqrt{\frac{m\omega}{2\hbar}}\op{x}  + \I \sqrt{\frac{1}{2m\omega \hbar}}\op{p},\\
  \op{a}^\dagger 
    = \sqrt{\frac{m\omega}{2\hbar}}\op{x} - \I \sqrt{\frac{1}{2m\omega \hbar}}\op{p},\\
  [\op{a}, \op{a}^\dagger] = 1.
\end{gather*}
These allow us to factor the Hamiltonian
\begin{gather*}
  \op{H} = \hbar\omega \left(\op{a}^\dagger \op{a} + \frac{1}{2}\right), \qquad
  \op{H}\ket{n} = \ket{n}E_n, \qquad
  E_n = \hbar\omega(n + \tfrac{1}{2}),
\end{gather*}
and construct eigenstates $\ket{n}$ from
the ground state $\ket{0}$ where $\op{a}\ket{0} = 0$:
\begin{gather*}
  \op{a}^\dagger\ket{n} = \sqrt{n+1}\ket{n+1}, \qquad
  \op{a}\ket{n+1} = \sqrt{n+1}\ket{n}.
\end{gather*}
In quantum field theory the analog of the operator $\op{n} = \op{a}^\dagger\op{a}$ (which has
eigenvalues $n$ here) will count the number of quanta in a state.







(sec:Phonons)=
# Phonons

## Classical Theory

Following the text {cite:p}`Donoghue:2022`, we work out the theory here for phonons on a
string of masses and springs.


(sec:BabyProblem)=
# Zee's Baby Problem




## The QFT Approach