diff --git a/_psets/2.md b/_psets/2.md index 16b433cab4b1705b9a12ba5b8e2114f05f0d99ad..90ad9f8b11f6c0a100638d2269f8b6411776133d 100644 --- a/_psets/2.md +++ b/_psets/2.md @@ -111,9 +111,11 @@ The wavelength of visible light is about $$\num{500e-9} \si{m}$$, so the energy be $$ -E = \frac{h c}{\lambda} = \frac{\num{6.626e-34} \si{J.s} \cdot \num{3e8} \si{m/s}} - {\num{500e-9} \si{m}} -= \num{3.8e-19} \si{J} +\begin{align*} +E &= \frac{h c}{\lambda} \\ +&= \frac{\num{6.626e-34} \si{J.s} \cdot \num{3e8} \si{m/s}} {\num{500e-9} \si{m}} \\ +&= \num{3.8e-19} \si{J} +\end{align*} $$ Thus $$10^4$$ photons per second is $$\num{3.8e-15} \si{W}$$, and $$10^{12}$$ photons per second is @@ -206,6 +208,7 @@ $$ ## 3.4 +{:.question} This problem is much harder than the others. Consider a stochastic process $$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 @@ -218,29 +221,204 @@ starts in x = 1. 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 $$hx(t)x(t + \tau)i$$. +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 $$S(f) \propto 1/\nu$$. +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 $$dE / d\tau += k T / \tau$$. + + + +$$ +\begin{align*} +\end{align*} +$$