diff --git a/_config.yml b/_config.yml
index 203825c0ac3de425e171ab657e6bdb8ee1cfd0e0..e378081f52d48050f1cd478148c1fef2784b1b0b 100644
--- a/_config.yml
+++ b/_config.yml
@@ -1,18 +1,20 @@
 collections:
-  authors:
+  psets:
+    output: true
+  notes:
     output: true
 
 defaults:
   - scope:
       path: ""
-      type: "authors"
+      type: "psets"
     values:
-      layout: "author"
+      layout: "pset"
   - scope:
       path: ""
-      type: "posts"
+      type: "notes"
     values:
-      layout: "post"
+      layout: "note"
   - scope:
       path: ""
     values:
diff --git a/_data/navigation.yml b/_data/navigation.yml
index cb67c7c63c4338c8bfd2b4737d2293c7860c1b82..268b387831a349f3b9b53883d0129077c03a6a84 100644
--- a/_data/navigation.yml
+++ b/_data/navigation.yml
@@ -2,3 +2,5 @@
   link: /
 - name: Problem Sets
   link: /psets.html
+- name: Notes
+  link: /notes.html
diff --git a/_layouts/post.html b/_layouts/note.html
similarity index 59%
rename from _layouts/post.html
rename to _layouts/note.html
index 0fab084a8253662d25f792b5d18519514ca52cee..b07fd4ac3472f1889a64cae10a9f198d20bcb985 100644
--- a/_layouts/post.html
+++ b/_layouts/note.html
@@ -3,8 +3,4 @@ layout: default
 ---
 <h1>{{ page.title }}</h1>
 
-<p>
-  {{ page.date | date_to_string }}
-</p>
-
 {{ content }}
diff --git a/_layouts/pset.html b/_layouts/pset.html
new file mode 100644
index 0000000000000000000000000000000000000000..b07fd4ac3472f1889a64cae10a9f198d20bcb985
--- /dev/null
+++ b/_layouts/pset.html
@@ -0,0 +1,6 @@
+---
+layout: default
+---
+<h1>{{ page.title }}</h1>
+
+{{ content }}
diff --git a/_notes/ch2.md b/_notes/ch2.md
new file mode 100644
index 0000000000000000000000000000000000000000..d629f62de0eb3d69095374a3a5845f5fcc488b48
--- /dev/null
+++ b/_notes/ch2.md
@@ -0,0 +1,65 @@
+---
+title: Chapter 2 Notes
+---
+
+## Photon Shot Noise
+
+The shot noise derivation in the book assumes a constant charge per particle. It also assumes that
+the energy is proportional to the signal squared. Neither of these is true for photons, so I'm
+curious how things look if we relax these assumptions.
+
+Let's assume that each photon energies are normally distributed i.i.d. random variables. I'll still
+model the arrival times with a Poisson distribution. Instead of electric current (in Amps), the
+train of photon arrivals tells us directly the energy transmission per time (Watts).
+
+$$
+W(t) = \sum_{n = 1}^N q_n \delta(t - t_n)
+$$
+
+where $$q_n$$ is the energy of the $$n$$th photon, and $$t_n$$ its time of arrival. The Fourier
+transform of this function is
+
+$$
+\begin{align*}
+W(f) &= \int_\mathbb{R} e^{-2 \pi i t f} \sum_{n = 1}^N q_n \delta(t - t_n) \mathrm{d} t \\
+&= \sum_{n = 1}^N q_n e^{-2 \pi i t_n f} \\
+\end{align*}
+$$
+
+Each photon contributes one term to this sum. The phase of its contribution is determined by its
+arrival time (and the frequency of interest), while the magnitude is determined by its energy.
+
+We want to know the expected magnitude of this function.
+
+$$
+\begin{align*} \langle |W(f)| \rangle
+&= \left \langle \left| \sum_{n = 1}^N q_n e^{-2 \pi i t_n f} \right| \right \rangle \\
+&= \sqrt{N} \langle q \rangle \end{align*}
+$$
+
+Note: I think this is wrong... Formally the expectation should be taken over sequences of arrival
+energies and times $$(q_1, t_1), \ldots, (q_n, t_N)$$. But in this case we can shortcut the math by
+noticing we've ended up with a 2d Gaussian random walk. The arrival times determine direction
+(phase), and the energies determine the step lengths. For a given frequency, all are uncorrelated in
+expectation.
+
+Let's look at the expected squared magnitude as well, so we can compare with the standard shot noise
+formula. Just note that this is not the spectral power density in this case. It's the spectral
+density of Watts squared which doesn't have any particularly useful physical meaning.
+
+$$
+\begin{align*}
+\left \langle |W(f)|^2 \right \rangle
+&= \left \langle \sum_{n = 1}^N q_n e^{-2 \pi i t_n f}
+                 \sum_{m = 1}^N q_m e^{2 \pi i t_m f} \right \rangle \\
+&= \left \langle \sum_{n = 1}^N \sum_{m = 1}^N q_n q_m e^{-2 \pi i (t_n - t_m) f} \right \rangle \\
+&= \left \langle \sum_{n = 1}^N q_n^2 + \sum_{n < m} 2 q_n q_m
+   \text{Re} \left[ e^{-2 \pi i (t_n - t_m) f} \right] \right \rangle \\
+&= \sum_{n = 1}^N \left \langle q_n^2 \right \rangle + 2 \sum_{n < m} \left \langle q_n q_m
+   \text{Re} \left[ e^{-2 \pi i (t_n - t_m) f} \right] \right \rangle \\
+&= N \left \langle q^2 \right \rangle
+\end{align*}
+$$
+
+The second term drops out because the arrival times are generated by a Poisson process, so the real
+part of the phase differences averages out to zero.
diff --git a/_notes/fourier_transform.md b/_notes/fourier_transform.md
new file mode 100644
index 0000000000000000000000000000000000000000..693676ed05c87ac3deda8cefa0630ab606600dc4
--- /dev/null
+++ b/_notes/fourier_transform.md
@@ -0,0 +1,178 @@
+---
+title: Fourier Transforms
+---
+
+To really make sense of chapter 2 I needed to review the properties of Fourier Transforms. These
+notes are based on my prior knowledge and some helpful websites:
+
+- [Properties of Fourier Transform](http://fourier.eng.hmc.edu/e101/lectures/handout3/node2.html)
+- [symmetry.pdf](https://www.cs.unm.edu/~williams/cs530/symmetry.pdf)
+
+
+## Properties of the Fourier Transform
+
+Note: I'm sloppy with the proofs here since all physical functions will have the nice properties
+that make the relevant operations valid, but I don't always call of these properties out when they
+are used.
+
+
+### Basics
+
+For a function $$f : \mathbb{R} \rightarrow \mathbb{C}$$, I use the definitions
+
+$$
+(\mathcal{F} f)(x) = \int_\mathbb{R} f(x') e^{-2 \pi i x' x} \mathrm{d} x'
+$$
+
+$$
+(\mathcal{F}^{-1} f)(x) = \int_\mathbb{R} f(x') e^{2 \pi i x' x} \mathrm{d} x'
+$$
+
+The Fourier Inversion Theorem states that $$\mathcal{F} \mathcal{F}^{-1} = \mathcal{F}^{-1}
+\mathcal{F} = \mathcal{I}$$ (where $$\mathcal{I}$$ is the identity operator). This holds for the space
+of functions whose Fourier transforms exist and for which both the function and the transform are
+absolutely integrable and continuous. All claims I make about functions should be interpreted to
+apply only to functions in this space.
+
+The Fourier transform is linear:
+
+$$
+\mathcal{F(af + bg)} = a \mathcal{F} f + b \mathcal{F} g
+$$
+
+If you shift everything in the original basis (usually the time or space domain), you pick up a
+phase shift in the transformed (i.e. frequency) basis. This follows from a simple change of
+variables.
+
+$$
+\begin{align*}
+(\mathcal{F} f(x' + x_0))(x)
+&= \int_\mathbb{R} f(x' + x_0) e^{-2 \pi i x' x} \mathrm{d} x' \\
+&= \int_\mathbb{R} f(x') e^{-2 \pi i (x' - x_0) x} \mathrm{d} x' \\
+&= e^{2 \pi i x_0 x} \int_\mathbb{R} f(x') e^{-2 \pi i x' x} \mathrm{d} x' \\
+&= e^{2 \pi i x_0 x} (\mathcal{F} f(x'))(x)
+\end{align*}
+$$
+
+
+### Fourier Flips
+
+The Fourier transform has a number of interesting properties related to the flip operator
+$$(\mathcal{R} f)(x) = f(-x)$$. By definition
+
+$$
+(\mathcal{F}^{-1} f)(x) = \int_\mathbb{R} f(x') e^{2 \pi i x' x} \mathrm{d} x'
+= (\mathcal{R} \mathcal{F} f)(x)
+$$
+
+And by changing the variable of integration from $$x'$$ to $$-x'$$
+
+$$
+\begin{align*}
+(\mathcal{F}^{-1} f)(x)
+&= \int_\mathbb{R} f(x') e^{2 \pi i x' x} \mathrm{d} x' \\
+&= \int_\mathbb{R} f(-x') e^{-2 \pi i x' x} \mathrm{d} x' \\
+&= (\mathcal{F} \mathcal{R} f)(x)
+\end{align*}
+$$
+
+Thus $$\mathcal{F}^{-1} = \mathcal{F} \mathcal{R} = \mathcal{R} \mathcal{F}$$. This means that
+
+$$
+\mathcal{I}
+= \mathcal{F} \mathcal{F}^{-1}
+= \mathcal{F} \mathcal{F} \mathcal{R}
+= \mathcal{R} \mathcal{F} \mathcal{F}
+$$
+
+and
+
+$$
+\mathcal{I}
+= (\mathcal{F} \mathcal{F}^{-1}) (\mathcal{F} \mathcal{F}^{-1})
+= \mathcal{F} \mathcal{F} \mathcal{F} \mathcal{F}
+$$
+
+So the Fourier transform generates cycles of functions. The forward transform takes you one forward,
+while the inverse transform takes you one back.
+
+$$
+f(x)
+\longrightarrow \hat{f}(x)
+\longrightarrow f(-x)
+\longrightarrow \hat{f}(-x)
+\longrightarrow f(x)
+$$
+
+
+### Conjugate Symmetries
+
+Recall that the [conjugate](https://en.wikipedia.org/wiki/Complex_conjugate) of a complex number
+$$a + b i$$ is defined as $$\overline{a + bi} = a - bi$$. Conjugation distributes over addition and
+multiplication. Knowing [Euler's formula](https://en.wikipedia.org/wiki/Euler%27s_formula) and that
+cos(x) is even and sin(x) odd, it's easy to show that $$\overline{e^{ix}} = e^{-ix}$$ for
+real $$x$$.
+
+Let's see what the Fourier transform of the conjugate of a function looks like.
+
+$$
+\begin{align*}
+(\mathcal{F} \overline{f})(x)
+&= \int_\mathbb{R} \overline{f(x')} e^{-2 \pi i x' x} \mathrm{d} x' \\
+&= \int_\mathbb{R} \overline{f(x') e^{2 \pi i x' x}} \mathrm{d} x' \\
+&= \overline{\int_\mathbb{R} f(x') e^{2 \pi i x' x} \mathrm{d} x'} \\
+&= \overline{(\mathcal{F}^{-1} f)(x)}
+\end{align*}
+$$
+
+This implies that $$ \mathcal{F}^{-1} f$$ is the complex conjugate of $$\mathcal{F} \overline{f}$$.
+So if we take the complex conjugate before and after, the forward Fourier transform becomes the
+inverse Fourier transform. Let's redraw the arrow diagram from before as a
+[commutative diagram](https://en.wikipedia.org/wiki/Commutative_diagram) that reflects this:
+
+$$
+\begin{array}{ccccccccc}
+f(x) & \longrightarrow & \hat{f}(x) & \longrightarrow & f(-x) & \longrightarrow & \hat{f}(-x) & \longrightarrow & f(x) \\
+\updownarrow && \updownarrow && \updownarrow && \updownarrow && \updownarrow \\
+\overline{f(x)} & \longleftarrow & \overline{\hat{f}(x)} & \longleftarrow & \overline{f(-x)} & \longleftarrow & \overline{\hat{f}(-x)} & \longleftarrow & \overline{f(x)}
+\end{array}
+$$
+
+In this version horizontal arrows indicate Fourier transforms (and you can always go backward along
+a horizontal arrow by taking the inverse Fourier transform), and vertical arrows indicate complex
+conjugates.
+
+
+### Even More Odd
+
+Recall that a function $$f$$ is even if $$f(-x) = f(x)$$, and odd if $$f(-x) = -f(x)$$. Note that a
+function is real if and only if it's equal to its own complex conjugate, and a function is purely
+imaginary if and only if its complex conjugate is its additive inverse. If any of these properties
+hold, we can deduce some interesting things from the commutative diagram above.
+
+- a function is even $$\iff$$ its Fourier transform is even
+- a function is odd $$\iff$$ its Fourier transform is odd
+- a function is real $$\iff$$ the conjugate of its transform is its inverse transform
+  $$\iff$$ its transform is equal to its flipped and conjugated transform
+- a function is real and even $$\iff$$ its transform is real and even
+- a function is real and odd $$\iff$$ its transform is imaginary and odd
+
+The case where $$f$$ is a real function is used often enough to be worth drawing out explicitly. It
+also helps see the special cases for even and odd real functions. Here the diagonal arrows
+represent Fourier transforms (as always they can be inverted), and the vertical arrow represents the
+complex conjugate (which in this particular case is equivalent to the flip operator).
+
+$$
+\begin{array}{rcccl}
+&& \hat{f}(x) = \overline{\hat{f}(-x)} && \\
+& \nearrow && \searrow & \\
+f(x) && \updownarrow && f(-x) \\
+& \nwarrow && \swarrow & \\
+&& \overline{\hat{f}(x)} = \hat{f}(-x) &&
+\end{array}
+$$
+
+In particular, since conjugation doesn't change magnitude, the magnitude of the transform of a real
+function is an even function. Because of this it's common to see the negative frequencies ignored
+when we're dealing with a real function and only care about the magnitude of the transform (like
+for spectral power analysis).
diff --git a/_posts/2019-02-14-pset1.md b/_psets/1.md
similarity index 91%
rename from _posts/2019-02-14-pset1.md
rename to _psets/1.md
index 65573dd0bb332fd4489de665119ed0c15086ba60..be7280a42e6851ed5701d5da8f6fc0db2ac4d5ee 100644
--- a/_posts/2019-02-14-pset1.md
+++ b/_psets/1.md
@@ -6,6 +6,7 @@ title: Problem Set 1
 
 ### (a)
 
+{:.question}
 How many atoms are there in a yoctomole?
 
 Yocto is the SI prefix for $$10^{-24}$$. So there are
@@ -18,6 +19,7 @@ atoms in a yoctomole.
 
 ### (b)
 
+{:.question}
 How many seconds are there in a nanocentury? Is the value near that of any important constants?
 
 Nano is the SI prefix for $$10^{-9}$$.
@@ -36,6 +38,7 @@ This is relatively close to $$\pi$$.
 
 ## 2.2
 
+{:.question}
 A large data storage system holds on the order of a petabyte. How tall would a 1 petabyte stack of
 CDs be? How does that compare to the height of a tall building?
 
@@ -54,6 +57,7 @@ stack of CDs would be roughly twice as tall as any other human-built structure.
 
 ## 2.3
 
+{:.question}
 If all the atoms in our universe were used to write an enormous binary number, using one atom per
 bit, what would that number be (converted to base 10)?
 
@@ -67,6 +71,7 @@ as $$10^\num{3e79}$$.
 
 ## 2.4
 
+{:.question}
 Compare the gravitational acceleration due to the mass of the Earth at its surface to that produced
 by a 1 kg mass at a distance of 1 m. Express their ratio in decibels.
 
@@ -82,6 +87,7 @@ $$
 
 ### (a)
 
+{:.question}
 Approximately estimate the chemical energy in a ton of TNT. You can assume that nitrogen is the
 primary component; think about what kind of energy is released in a chemical reaction, where it is
 stored, and how much there is.
@@ -107,6 +113,7 @@ This is off by a factor of about 1.5, which isn't so bad for not knowing what th
 
 ### (b)
 
+{:.question}
 Estimate how much uranium would be needed to make a nuclear explosion equal to the energy in a
 chemical explosion in 10,000 tons of TNT (once again, think about where the energy is stored).
 
@@ -121,6 +128,7 @@ uranium, which is about 650 grams.
 
 ### (c)
 
+{:.question}
 Compare this to the *rest mass energy* $$E = mc^2$$ of that amount of material (Chapter 15), which
 gives the maximum amount of energy that could be liberated from it.
 
@@ -176,20 +184,24 @@ makes it the world's most valuable substance by mass.)
 
 ### (a)
 
+{:.question}
 What is the approximate de Broglie wavelength of a thrown baseball?
 
 A baseball weighs $$1.45^{-1} \si{kg}$$. A major league fastball is around 100 miles per hour, so
 160 km per hour, but I'll just go with 100 km per hour since we're not all professional pitchers.
 
 $$
-\lambda = \frac{h}{p}
-= \frac{\num{6.626e-34} \si{J.s}}{\num{1.45e-1} \si{kg} \cdot 10^2 \si{km/hour}}
-\cdot \frac{\si{hour}}{60 \si{minutes}} \cdot \frac{\si{minute}}{60 \si{s}}
-= \num{1.3e-35} \si{m}
+\begin{align*}
+\lambda &= \frac{h}{p} \\
+&= \frac{\num{6.626e-34} \si{J.s}}{\num{1.45e-1} \si{kg} \cdot 10^2 \si{km/hour}}
+   \cdot \frac{\si{hour}}{60 \si{minutes}} \cdot \frac{\si{minute}}{60 \si{s}} \\
+&= \num{1.3e-35} \si{m}
+\end{align*}
 $$
 
 ### (b)
 
+{:.question}
 Of a molecule of nitrogen gas at room temperature and pressure? (This requires either the result of
 Section 3.4.2, or dimensional analysis.)
 
@@ -207,15 +219,18 @@ $$
 Thus
 
 $$
-\lambda = \frac{h}{p}
-= \frac{h}{\sqrt{3 m k T}}
-= \frac{\num{6.626e-34} \si{J.s}}{\sqrt{3 \cdot \num{4.6e-26} \si{kg}
-  \cdot \num{1.38e-23} \si{J/K} \cdot 300 \si{K}}}
-= \num{2.7e-11} \si{m}
+\begin{align*}
+\lambda &= \frac{h}{p} \\
+&= \frac{h}{\sqrt{3 m k T}} \\
+&= \frac{\num{6.626e-34} \si{J.s}}{\sqrt{3 \cdot \num{4.6e-26} \si{kg}
+   \cdot \num{1.38e-23} \si{J/K} \cdot 300 \si{K}}} \\
+&= \num{2.7e-11} \si{m}
+\end{align*}
 $$
 
 ### (c)
 
+{:.question}
 What is the typical distance between the molecules in this gas?
 
 The typical distance can be estimated based on the number of particles per volume. If the volume is
@@ -223,33 +238,39 @@ $$V$$, then we can arrange $$n$$ particles on a cubic lattice with a spacing of
 By the ideal gas law, $$V/n = RT/P$$, so the typical distance is
 
 $$
+\begin{align*}
 \left( \frac{V}{n} \right) ^ \frac{1}{3}
-= \left( \frac{R T}{P} \right) ^ \frac{1}{3}
-= \left( \frac{8.3 \si{J/mol/K \cdot 300 K}}{10^5 \si{Pa}}
-  \cdot \frac{\si{mol}}{\num{6.022e23}} \right) ^ \frac{1}{3}
-= \num{3.5e-9} \si{m}
+&= \left( \frac{R T}{P} \right) ^ \frac{1}{3} \\
+&= \left( \frac{8.3 \si{J/mol/K \cdot 300 K}}{10^5 \si{Pa}}
+   \cdot \frac{\si{mol}}{\num{6.022e23}} \right) ^ \frac{1}{3} \\
+&= \num{3.5e-9} \si{m}
+\end{align*}
 $$
 
 ### (d)
 
+{:.question}
 If the volume of the gas is kept constant as it is cooled, at what temperature does the wavelength
 become comparable to the distance between the molecules?
 
 The typical spacing depends on the volume of the gas and number of molecules, both of which are
-fixed in this scenario. So we just need to solve
+fixed in this scenario. So we just need to solve $$\num{3.5e-9} \si{m} = \lambda = h / \sqrt{3 m k
+T}$$ for $$T$$.
 
 $$
-\num{3.5e-9} \si{m} = \lambda = \frac{h}{\sqrt{3 m k T}}
-\implies T = \frac{h^2}{3 m k \lambda^2}
-= \frac{(\num{6.626e-34} \si{J.s})^2}{3 \cdot \num{4.6e-26} \si{kg}
-  \cdot \num{1.38e-23} \si{J/K} \cdot (\num{3.5e-9} \si{m})^2}
-= 0.02 K
+\begin{align*}
+T &= \frac{h^2}{3 m k \lambda^2} \\
+&= \frac{(\num{6.626e-34} \si{J.s})^2}{3 \cdot \num{4.6e-26} \si{kg}
+   \cdot \num{1.38e-23} \si{J/K} \cdot (\num{3.5e-9} \si{m})^2} \\
+&= 0.02 K
+\end{align*}
 $$
 
 ## (2.7)
 
 ### (a)
 
+{:.question}
 The potential energy of a mass m a distance r from a mass $$M$$ is $$−GMm/r$$. What is the escape
 velocity required to climb out of that potential?
 
@@ -262,6 +283,7 @@ $$
 
 ### (b)
 
+{:.question}
 Since nothing can travel faster than the speed of light (Chapter 15), what is the radius within
 which nothing can escape from the mass?
 
@@ -276,6 +298,7 @@ Schwarzschild radius.
 
 ### (c)
 
+{:.question}
 If the rest energy of a mass $$M$$ is converted into a photon, what is its wavelength?
 
 $$
@@ -284,6 +307,7 @@ $$
 
 ### (d)
 
+{:.question}
 For what mass does its equivalent wavelength equal the size within which light cannot escape?
 
 $$
@@ -292,6 +316,7 @@ $$
 
 ### (e)
 
+{:.question}
 What is the corresponding size?
 
 Plugging the above formula for $$M$$ into $$\lambda = h/(M c)$$, we find that
@@ -302,6 +327,7 @@ $$
 
 ### (f)
 
+{:.question}
 What is the energy?
 
 $$
@@ -310,6 +336,7 @@ $$
 
 ### (g)
 
+{:.question}
 What is the period?
 
 $$\tau = 1/f$$ and $$\lambda = c / f = c \tau$$, so
@@ -321,12 +348,14 @@ $$
 
 ## 2.8
 
+{:.question}
 Consider a pyramid of height H and a square base of side length L. A sphere is placed so that its
 center is at the center of the square at the base of the pyramid, and so that it is tangent to all
 of the edges of the pyramid (intersecting each edge at just one point).
 
 ### (a)
 
+{:.question}
 How high is the pyramid in terms of L?
 
 To fit within the square base, the radius of the sphere must be $$L / 2$$. Now imagine rotating the
@@ -338,6 +367,7 @@ lot more tedious.)
 
 ### (b)
 
+{:.question}
 What is the volume of the space common to the sphere and the pyramid?
 
 The sphere is tangent to three points on each wall of the pyramid (the midpoints of each edge). The
diff --git a/_psets/2.md b/_psets/2.md
new file mode 100644
index 0000000000000000000000000000000000000000..c2ca630409a7e0d3ce86150dcb5450c697321aaa
--- /dev/null
+++ b/_psets/2.md
@@ -0,0 +1,431 @@
+---
+title: Problem Set 2
+---
+
+## 3.1
+
+### (a)
+
+{:.question}
+Derive equation (3.16) (i.e. the Poisson probability density function) from the binomial
+distribution and Stirling’s approximation.
+
+Fix any (positive, real) $$\lambda$$. For any $$n$$, the binomial probability mass function for
+$$n$$ trials with probability of success $$p = \lambda / n$$ is
+
+$$
+\begin{align*}
+p(k; n, \lambda/n)
+&= {n \choose k} \left( \frac{\lambda}{n} \right)^k \left( 1 - \frac{\lambda}{n} \right)^{n - k} \\
+&= \frac{n (n - 1) \cdots (n - k + 1)}{k!} \left( \frac{\lambda}{n} \right)^k
+   \left( 1 - \frac{\lambda}{n} \right)^{n - k} \\
+&= \frac{n^k + \mathcal{O}\left( n^{k - 1} \right) }{k!} \left( \frac{\lambda}{n} \right)^k
+   \left( 1 - \frac{\lambda}{n} \right)^{n - k} \\
+&= \frac{\lambda^k}{k!} \left( 1 - \frac{\lambda}{n} \right)^{n - k} +
+   \frac{\mathcal{O}\left( n^{k - 1} \right) }{k!} \left( \frac{\lambda}{n} \right)^k
+   \left( 1 - \frac{\lambda}{n} \right)^{n - k}
+\end{align*}
+$$
+
+As $$n$$ approaches infinity, the second term goes to zero because the $$n^{-k}$$ dominates
+$$\mathcal{O}\left( n^{k - 1} \right)$$ (and the last factor is always between 0 and 1). The first
+term approaches $$\lambda^k e^{-\lambda}/k!$$ which is the Poisson distribution.
+
+### (b)
+
+{:.question}
+Use it to derive equation (3.18) ($$\langle k (k - 1) \cdots (k - m + 1) \rangle = \lambda^m$$) when
+k follows the Poisson distribution with average number of events $$\lambda$$.
+
+First, let's verify that $$\langle k \rangle = \lambda$$. This will serve as the base case for an
+inductive argument.
+
+$$
+\begin{align*}
+\langle k \rangle &= \sum_{k = 0}^\infty k \frac{\lambda^k e^{-\lambda}}{k!} \\
+&= \lambda \sum_{k = 1}^\infty \frac{\lambda^{k - 1} e^{-\lambda}}{(k - 1)!} \\
+&= \lambda \sum_{k = 0}^\infty \frac{\lambda^k e^{-\lambda}}{k!} \\
+&= \lambda
+\end{align*}
+$$
+
+Now assume that $$\langle k (k - 1) \cdots (k - m + 1) \rangle = \lambda^m$$.
+
+$$
+\begin{align*}
+\langle k (k - 1) \cdots (k - m + 1) (k - m) \rangle
+&= \sum_{k = 0}^\infty k \cdots (k - m) \frac{\lambda^k e^{-\lambda}}{k!} \\
+&= \lambda \sum_{k = 1}^\infty (k - 1) \cdots (k - m)
+   \frac{\lambda^{k - 1} e^{-\lambda}}{(k - 1)!} \\
+&= \lambda \sum_{k = 0}^\infty k \cdots (k - m + 1) \frac{\lambda^k e^{-\lambda}}{k!} \\
+&= \lambda \lambda^m \\
+&= \lambda^{m + 1}
+\end{align*}
+$$
+
+### (c)
+
+{:.question}
+Use (b) to derive equation (3.19): $$\sigma / \langle k \rangle = 1 / \sqrt{\lambda}$$.
+
+To compute $$\sigma$$, we need to know what $$\langle k^2 \rangle$$ is. It can be found using the
+same trick we relied on for the last problem.
+
+$$
+\begin{align*}
+\langle k^2 \rangle
+&= \sum_{k = 0}^\infty k^2 \frac{\lambda^k e^{-\lambda}}{k!} \\
+&= \lambda \sum_{k = 1}^\infty k \frac{\lambda^{k - 1} e^{-\lambda}}{(k - 1)!} \\
+&= \lambda \sum_{k = 0}^\infty (k + 1) \frac{\lambda^k e^{-\lambda}}{k!} \\
+&= \lambda \left( \sum_{k = 0}^\infty k \frac{\lambda^k e^{-\lambda}}{k!} +
+                  \sum_{k = 0}^\infty \frac{\lambda^k e^{-\lambda}}{k!} \right) \\
+&= \lambda (\lambda + 1)
+\end{align*}
+$$
+
+Thus $$\sigma^2 = \langle k^2 \rangle - \langle k \rangle^2 = \lambda (\lambda + 1) - \lambda^2 =
+\lambda$$, and $$\sigma / \langle k \rangle = \sqrt{\lambda} / \lambda = 1 / \sqrt{\lambda}$$.
+
+
+## (3.2)
+
+{:.question}
+Assume that photons are generated by a light source independently and randomly with an average rate
+N per second. How many must be counted by a photodetector per second to be able to determine the
+rate to within 1%? To within 1 part per million? How many watts do these cases correspond to for
+visible light?
+
+Note: in the previous problem I used $$\lambda$$ as the expected number of events of a Poisson process.
+Here I'll use $$N$$ to avoid confusion with wavelength.
+
+Since the photons are generated independently and with a constant average rate, it's reasonable to
+model their creation as a Poisson process. We have already found that $$\sigma = \sqrt{N}$$. For
+large $$N$$, the Poisson distribution is very close to the normal distribution. So about two thirds
+of the probability mass lies between $$N - \sigma$$ and $$N + \sigma$$. Thus if $$\sigma \leq 0.01
+N$$, in any given second it's more likely than not that the number of photons emitted is within one
+percent of the true mean. Thus we'd need $$\sqrt{N} \leq 0.01 N$$, i.e. $$N \geq 10^4$$. To have the
+same probability that the number of observed photons is within $$10^{-6} N$$ of the true value, we
+need $$N \geq 10^{12}$$.
+
+The wavelength of visible light is about $$\num{500e-9} \si{m}$$, so the energy of each photon will
+be
+
+$$
+\begin{align*}
+E &= \frac{h c}{\lambda} \\
+&= \frac{\num{6.626e-34} \si{J.s} \cdot \num{3e8} \si{m/s}} {\num{500e-9} \si{m}} \\
+&= \num{3.8e-19} \si{J}
+\end{align*}
+$$
+
+Thus $$10^4$$ photons per second is $$\num{3.8e-15} \si{W}$$, and $$10^{12}$$ photons per second is
+$$\num{3.8e-7} \si{W}$$.
+
+Some caveats about this answer: the Poisson distribution is strictly greater than zero on its whole
+domain, so one can never be *certain* that they have found the correct value within any error
+bounds. That is, even if the real value of $$N$$ is one, there's still a (vanishingly) slight
+chance you'll see a million. The best we can do is find an answer that's [probably approximately
+correct](https://en.wikipedia.org/wiki/Probably_approximately_correct_learning).
+
+Second, I only found limits on the probability that the number of emitted photons is near the true
+(known) average. It's a more subtle question to ask for the probability that a given estimate is
+within one percent of the true (latent) average. That could be tackled with confidence intervals,
+p-values, Bayes factors, or direct posterior estimation, though none would be quite so simple. And
+as an unapologetic Bayesian I'm required to point out that my confidence I've determined $$N$$ with
+some precision depends on my prior for $$N$$. This would be very real if we were, for instance,
+counting photons in a lab module for this course. I would be more confident in my answer if I
+measured something close to $$\num{1e12}$$ photons, than if I measured something like
+$$\num{3.64e12}$$, since I think it's more likely that the instructor would choose a nice round
+number with lots of zeros.
+
+
+## (3.3)
+
+{:.question}
+Consider an audio amplifier with a 20 kHz bandwidth.
+
+### (a)
+
+{:.question}
+If it is driven by a voltage source at room temperature with a source impedance of 10kΩ how large
+must the input voltage be for the SNR with respect to the source Johnson noise to be 20 dB?
+
+$$
+\begin{align*}
+\langle V_\text{noise}^2 \rangle &= 4 k T R \Delta f \\
+&= 4 \cdot \num{1.38e-23} \si{J/K} \cdot 300 \si{K} \cdot 10^4 \si{\ohm} \cdot \num{20e3} \si{Hz} \\
+&= \num{3.3e-12} \si{V^2}
+\end{align*}
+$$
+
+So for the input signal to be 20dB louder, we need
+
+$$
+10 \log_{10} \left( \frac{V_\text{signal}^2}{\num{3.3e-12}} \right) = 20 \\
+V_\text{signal}^2 = 10^2 \cdot \num{3.3e-12} \si{V^2} = \num{3.3e-10} \si{V^2}
+$$
+
+Note that I use a factor of 10 since $$V^2$$ is proportional to power. This signal strength would be
+achieved, for instance, by a sinusoid that ranges between $$-\num{6.6e-10}$$ and $$\num{6.6e-10}$$
+volts.
+
+### (b)
+
+{:.question}
+What size capacitor has voltage fluctuations that match the magnitude of this Johnson noise?
+
+Since a farad is equal to a second per ohm, purely dimensional considerations imply that the noise
+energy of the capacitor is approximately $$kT/C$$. Indeed this turns out to be exactly correct. The
+energy in a capacitor is $$C V^2 / 2$$, so by the equipartition theorem, $$kT/2 = \langle C V^2 / 2
+\rangle$$. This implies that $$\langle V^2 \rangle = k T / C$$. Thus the capacitor that matches the
+voltage fluctuations found in part (b) at room temperature has capacitance
+
+$$
+\begin{align*}
+C &= \frac{k T}{\left \langle V^2 \right \rangle} \\
+&= \frac{\num{1.38e-23} \si{J/K} \cdot 300 \si{K}}{\num{3.3e-12} \si{V^2}} \\
+&= \num{1.25e-9} \si{F}
+\end{align*}
+$$
+
+### (c)
+
+{:.question}
+If it is driven by a current source, how large must it be for the RMS shot noise to be equal to 1%
+of that current?
+
+RMS shot noise is the square root of $$2 q I \Delta f$$. So we want $$\sqrt{2 q \Delta f I} = 0.01
+I$$, which is solved by
+
+$$
+\begin{align*}
+I &= \frac{2q \Delta f}{0.01^2} \\
+&= \frac{2 \cdot \num{1.6e-19} \si{C} \cdot 20 \si{kHz}}{10^{-4}} \\
+&= \num{6.4e-11} A
+\end{align*}
+$$
+
+
+## 3.4
+
+{:.question}
+This problem is much harder than the others. Consider a stochastic process $$x(t)$$ that randomly
+switches between x = 0 and x = 1. Let $$\alpha \mathrm{d}t$$ be the probability that it makes a
+transition from 0 to 1 during the interval $$\mathrm{d}t$$ if it starts in x = 0, and let $$\beta
+\mathrm{d}t$$ be the probability that it makes a transition from 1 to 0 during $$\mathrm{d}t$$ if it
+starts in x = 1.
+
+### (a)
+
+{:.question}
+Write a matrix differential equation for the change in time of the probability $$p_0(t)$$ to be in
+the 0 state and the probability $$p_1(t)$$ to be in the 1 state.
+
+$$
+\frac{d}{dt} \begin{bmatrix} p_0(t) \\ p_1(t) \end{bmatrix}
+= \begin{bmatrix} -\alpha & \beta \\ \alpha & -\beta \end{bmatrix}
+\begin{bmatrix} p_0(t) \\ p_1(t) \end{bmatrix}
+$$
+
+### (b)
+
+{:.question}
+Solve this by diagonalizing the 2 × 2 matrix.
+
+Solving a system of ODEs isn't necessary here, since $$p_1(t) = 1 - p_0(t)$$. So we just need to
+solve
+
+$$
+\begin{align*}
+\frac{d}{dt} p_0(t) &= -\alpha p_0(t) + \beta (1 - p_0(t)) \\
+&= \beta -(\alpha + \beta) p_0(t)
+\end{align*}
+$$
+
+Since the derivative is proportional to the function, it's solved generally by an exponential
+
+$$
+p_0(t) = A + B e^{-(\alpha + \beta) t}
+$$
+
+Then
+
+$$
+\begin{align*}
+\frac{d}{dt} p_0(t) &= -(\alpha + \beta) B e^{-(\alpha + \beta) t} \\
+&= (\alpha + \beta) A - (\alpha + \beta) (A + B e^{-(\alpha + \beta) t}) \\
+&= (\alpha + \beta) A - (\alpha + \beta) p_0(t)
+\end{align*}
+$$
+
+which implies that $$A = \beta / (\alpha + \beta)$$. $$B$$ is determined by $$p_0(t)$$:
+
+$$
+p_0(0) = \frac{\beta}{\alpha + \beta} + B
+$$
+
+So putting everything together we have
+
+$$
+\begin{align*}
+p_0(t) &= \frac{\beta}{\alpha + \beta}
+          + \left(p_0(0) - \frac{\beta}{\alpha + \beta} \right) e^{-(\alpha + \beta) t} \\
+p_1(t) &= \frac{\alpha}{\alpha + \beta}
+          - \left(p_0(0) - \frac{\beta}{\alpha + \beta} \right) e^{-(\alpha + \beta) t} \\
+&= \frac{\alpha}{\alpha + \beta}
+   + \left(p_1(0) - \frac{\alpha}{\alpha + \beta} \right) e^{-(\alpha + \beta) t}
+\end{align*}
+$$
+
+### (c)
+
+{:.question}
+Use this solution to find the autocorrelation function $$\langle x(t)x(t + \tau) \rangle$$.
+
+For positive $$\tau$$,
+
+$$
+\begin{align*}
+\langle x(t) x(t + \tau) \rangle
+&= \sum_{i, j \in \{0, 1\} \times \{0, 1\}} p(x(t) = i \cap x(t + \tau) = j) i j \\
+&= p(x(t) = 1 \cap x(t + \tau) = 1) \\
+&= p_1(t + \tau | x(t) = 1) p_1(t) \\
+&= p_1(\tau | p_1(0) = 1) p_1(t) \\
+&= \left( \frac{\alpha}{\alpha + \beta} + \left(1 - \frac{\alpha}{\alpha + \beta} \right)
+   e^{-(\alpha + \beta) \tau} \right) \frac{\alpha}{\alpha + \beta} \\
+&= \left( \frac{\alpha}{\alpha + \beta} + \frac{\beta}{\alpha + \beta}
+   e^{-(\alpha + \beta) \tau} \right) \frac{\alpha}{\alpha + \beta} \\
+&= \frac{\alpha}{(\alpha + \beta)^2} \left( \alpha + \beta e^{-(\alpha + \beta) \tau} \right)
+\end{align*}
+$$
+
+By symmetry this is an even function. That is, $$\langle x(t) x(t + \tau) \rangle = \langle x(t)
+x(t - \tau) \rangle$$.
+
+### (d)
+
+{:.question}
+Use the autocorrelation function to show that the power spectrum is a Lorentzian.
+
+By Wiener-Khinchin,
+
+$$
+\begin{align*}
+S(f) &= \int_\mathbb{R} \frac{\alpha}{(\alpha + \beta)^2}\left( \alpha + \beta e^{-(\alpha + \beta)
+        |\tau|} \right) e^{-2 \pi i f \tau} \mathrm{d} \tau \\
+&= \frac{\alpha \beta}{(\alpha + \beta)^2} \left( \int_\mathbb{R} \frac{\alpha}{\beta} e^{-2 \pi i f
+   \tau} \mathrm{d} \tau + \int_\mathbb{R} e^{-(\alpha + \beta) |\tau| -2 \pi i f \tau} \mathrm{d}
+   \tau \right)
+\end{align*}
+$$
+
+The first integral evaluates to a delta function $$\alpha / \beta \delta(f)$$. The second can be
+broken into a sum of two integrals over the positive and negative halves of $$\mathbb{R}$$:
+
+$$
+\begin{align*}
+\int_0^\infty e^{-\tau ((\alpha + \beta) + 2 \pi i f)} \mathrm{d} \tau
+&= \left[ \frac{-1}{(\alpha + \beta) + 2 \pi i f}
+   e^{-\tau ((\alpha + \beta) + 2 \pi i f)} \right]_0^\infty \\
+&= \frac{1}{(\alpha + \beta) + 2 \pi i f} \\
+\int_{-\infty}^0 e^{\tau ((\alpha + \beta) - 2 \pi i f)} \mathrm{d} \tau
+&= \left[ \frac{1}{(\alpha + \beta) - 2 \pi i f}
+   e^{\tau ((\alpha + \beta) - 2 \pi i f)} \right]_{-\infty}^0 \\
+&= \frac{1}{(\alpha + \beta) - 2 \pi i f} \\
+\end{align*}
+$$
+
+Putting everything together,
+
+$$
+\begin{align*}
+S(f) &= \frac{\alpha \beta}{(\alpha + \beta)^2} \left( \frac{\alpha}{\beta} \delta(f)
+        + \frac{1}{(\alpha + \beta) + 2 \pi i f} + \frac{1}{(\alpha + \beta) - 2 \pi i f} \right) \\
+&= \frac{\alpha \beta}{(\alpha + \beta)^2} \left( \frac{\alpha}{\beta} \delta(f)
+   + \frac{2 (\alpha + \beta)}{(\alpha + \beta)^2 + (2 \pi f)^2} \right) \\
+&= \frac{\alpha^2}{(\alpha + \beta)^2} \delta(f) + \frac{\alpha \beta}{(\alpha + \beta)^2}
+   \frac{2 (\alpha + \beta)}{(\alpha + \beta)^2 + (2 \pi f)^2} \\
+&= \frac{\alpha^2}{(\alpha + \beta)^2} \delta(f) + \frac{\alpha \beta}{(\alpha + \beta)^2}
+   \frac{2 (\alpha + \beta)^{-1}}{1 + \left( \frac{2 \pi f}{\alpha + \beta} \right)^2} \\
+\end{align*}
+$$
+
+Ignoring the delta function, up to a constant factor this is a Lorentzian distribution with
+$$\tau = 1 / (\alpha + \beta)$$.
+
+### (e)
+
+{:.question}
+At what frequency is the magnitude of the Lorentzian reduced by half relative to its low-frequency
+value?
+
+Looking only at the Lorentzian portion,
+
+$$
+S(0) = \frac{2 \alpha \beta}{(\alpha + \beta)^3}
+$$
+
+So we must plug this in and solve for $$f$$:
+
+$$
+\begin{align*}
+\frac{\alpha \beta}{(\alpha + \beta)^2}
+\frac{2 (\alpha + \beta)^{-1}}{1 + \left( \frac{2 \pi f}{\alpha + \beta} \right)^2}
+&= \frac{1}{2} \frac{2 \alpha \beta}{(\alpha + \beta)^3} \\
+\frac{1}{1 + \left( \frac{2 \pi f}{\alpha + \beta} \right)^2} &= \frac{1}{2} \\
+1 &= \left( \frac{2 \pi f}{\alpha + \beta} \right)^2 \\
+f &= \frac{\alpha + \beta}{2 \pi}
+\end{align*}
+$$
+
+### (f)
+
+{:.question}
+For a thermally activated process, show that a flat distribution of barrier energies leads to a
+distribution of switching times $$p(\tau) \propto 1/\tau$$, and in turn to $$S(f) \propto 1/f$$.
+
+We are assuming the distribution of barrier energies $$p(E)$$ is constant. According to (3.37), for
+a thermally activated process the characteristic switching time is a function of the energy: $$\tau
+= \tau_0 e^{E/kT}$$. So to get the distribution $$p(\tau)$$, we just need to transform $$p(E)$$
+accordingly. In particular,
+
+$$
+p(\tau) = p(E) \frac{k T}{\tau}
+$$
+
+Let's prove that this is the case. Given any random variable $$X$$, and monotonic function $$f$$,
+let $$Y = f(X)$$. Then the cumulative distribution functions are related by
+
+$$
+\begin{align*}
+p(Y \leq y) &= p(f(X) \leq y) \\ &= p(X \leq f^{-1}(y))
+\end{align*}
+$$
+
+By the fundamental theorem of calculus, the probability density function is the derivative of the
+cumulative distribution function. So by employing the chain rule we find that
+
+$$
+\begin{align*}
+p(y) &= \frac{d}{dy} p(Y \leq y) \\
+&= \frac{d}{dy} p(X \leq f^{-1}(y)) \\
+&= p(f^{-1}(y)) \frac{d}{dy} f^{-1}(y)
+\end{align*}
+$$
+
+So our result above for $$p(\tau)$$ follows because $$E = k T \log(\tau / \tau_0)$$, so $$dE / d\tau
+= k T / \tau$$.
+
+Now we must show that $$S(f) \propto 1/f$$.
+
+$$
+\begin{align*}
+S(f) &= \int_0^\infty p(\tau) \frac{2 \tau}{1 + (2 \pi f \tau)^2} \mathrm{d} \tau \\
+&= \int_0^\infty \frac{2 p(E) k T}{1 + (2 \pi f \tau)^2} \mathrm{d} \tau \\
+&= \frac{1}{2 \pi f} \int_0^\infty \frac{2 p(E) k T}{1 + \tau^2} \mathrm{d} \tau \\
+&= \frac{p(E) k T}{\pi f} \int_0^\infty \frac{1}{1 + \tau^2} \mathrm{d} \tau \\
+&= \frac{p(E) k T}{\pi f} \int_0^{\pi/2} \frac{sec^2{\theta}}{1 + tan^2(\theta)} \mathrm{d} \theta \\
+&= \frac{p(E) k T}{\pi f} \int_0^{\pi/2} \mathrm{d} \theta \\
+&= \frac{p(E) k T}{2 f}
+\end{align*}
+$$
diff --git a/_sass/main.scss b/_sass/main.scss
index 9ecbc411044de9c0a2a0ce4c9e1a10bd49e941e9..7a73c536382c304709101a13fbd72d05ab6f9df9 100644
--- a/_sass/main.scss
+++ b/_sass/main.scss
@@ -1,3 +1,34 @@
+body {
+    max-width: 650px;
+    line-height: 1.6;
+    font-size: 18px;
+    color: #333;
+    background-color: #fdfdfd;
+    margin: 40px auto;
+    padding: 0 10px;
+}
+
+h1, h2, h3 {
+    line-height: 1.2;
+}
+
+a {
+    color: #3070B9;
+    text-decoration: none;
+}
+
+a:hover {
+    text-decoration: underline;
+}
+
+a:visited {
+    color: #306590;
+}
+
 .current {
   font-weight: bold;
 }
+
+.question {
+    font-style: italic;
+}
diff --git a/index.md b/index.md
index 5866ab86fd93548c40d2758518ed10adcd7a75da..a90f52a5a567b8ac17c8be2130fd8b47556caac0 100644
--- a/index.md
+++ b/index.md
@@ -4,4 +4,4 @@ title: Home
 
 # MAS 6.244 Physics of Information Technology 2019
 
-This is Erik Strand's as yet totally unstyled tracking page.
+This is Erik Strand's as yet mostly content-free tracking page.
diff --git a/notes.html b/notes.html
new file mode 100644
index 0000000000000000000000000000000000000000..ff57d2ebd4bc33ef9fc032bc1c24a32d2afd6410
--- /dev/null
+++ b/notes.html
@@ -0,0 +1,12 @@
+---
+title: Notes
+---
+<h1>Notes</h1>
+
+<ul>
+  {% for note in site.notes %}
+    <li>
+      <h2><a href="{{ note.url | real_relative_url }}">{{ note.title }}</a></h2>
+    </li>
+  {% endfor %}
+</ul>
diff --git a/psets.html b/psets.html
index 15284eae3e9a20138b083f940e756cf4c3c5a934..1efc5d6ebcbb054db8dbf2c66b0448886400cb8e 100644
--- a/psets.html
+++ b/psets.html
@@ -4,10 +4,9 @@ title: Problem Sets
 <h1>Problem Sets</h1>
 
 <ul>
-  {% for post in site.posts %}
+  {% for pset in site.psets %}
     <li>
-      <h2><a href="{{ post.url | real_relative_url }}">{{ post.title }}</a></h2>
-      <p>{{ post.excerpt }}</p>
+      <h2><a href="{{ pset.url | real_relative_url }}">{{ pset.title }}</a></h2>
     </li>
   {% endfor %}
 </ul>