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Moved out homework solutions to the other repo

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 # Exercise Solutions for Lecture 1 ## 1a) First we derive the expression for the reflectance R(x). The general form of a sinusoidal is $$R(x) = A + B \cos(kx)$$ In this case we know the period is $\frac1\lambda$ so $k=\frac{4\pi}{\lambda}$. Applying the conditions $R_{max}=0.5$, $R_{min}=0$ we get $A=B=\frac14$, so $$R(x) = \frac14 (1+\cos(\frac{4\pi}{\lambda}x))$$ Therefore $$\left|\frac{dR}{dx}\right| = \frac\pi\lambda \sin(\frac{4\pi}{\lambda}x)$$ The maximum slope is $\left|\frac{dR}{dx}\right|_{max}=\frac\pi\lambda$, when the sine is equal to 1, or equivalently when $$\frac{4\pi}{\lambda}x_n = n\frac\pi2 \ , \ \forall n\in \mathrm{Z} \\ x_n = n \frac\lambda8$$ At $x_1$ the reflectance is $R_0 = R(x_1) = \frac14$. ## 1b) The responsivity $R_{xp}$ is defined as $$R_{xp} = \frac{dP}{dx}$$ We can write the output power $P$ as $$P = P_{in} R(x)$$ Using the results from section 1a) we can approximate the reflectance $R(x)$ by doing a a Taylor expansion around the point of maximum slope $x_1$ $$R(x) \approx R(x_1) + \frac{dR}{dx}(x_1) \ x = \frac14 + \frac\pi\lambda x$$ We can then write $$P \approx P_{in} \left( \frac14 + \frac\pi\lambda x \right)$$ so $$R_{xp} =\frac{dP}{dx}= \frac{\pi P_{in}}{\lambda}$$ ## 1c) In the lecture we saw that the Power Spectral Density of the reflected laser beam is given by $$S_{pp} = P \hbar \omega_0$$ From b) , we know that at the point of maximum slope $P=P_{in}R_0$ so $$S_{pp} =P_{in}R_0 \hbar \omega_0$$ Then we have that $$S_{xx} = \frac{S_{pp}}{R_{xp}^2} = \frac{P_{in}R_0 \hbar \omega_0 }{P_{in}^2 \frac{\pi^2}{\lambda^2}}$$ using $\omega_0 = \frac{2\pi c}{\lambda}$ $$S_{xx} = 2\frac{R_0 \hbar c\lambda}{P_{in}\pi} \approx 1,8 \cdot 10^{-29} \text{m^2/Hz}$$ so the sensitivity is $\sqrt{S_{xx}}= 4,24\cdot 10^{-14} \ m/\sqrt{Hz}$ Is this reasonable? An Atomic Force Microscope, which is not shot noise limited, can have sensitivities of 2 $\text{fm}/\sqrt{\text{Hz}}$. The Laser Interferometer Gravitational-Wave Observatory (LIGO), which is shot noise limited, has a sensitivity of $10^{-19}\text{m}/\sqrt{\text{Hz}}$. Ligo uses a 2 Watt laser but uses tricks in designing their interferometer differently to get better sensitivity. ## 1d) Measuring for a time $T_m$ would give us an error $$\sigma_x = \sqrt{\frac{S_{xx}}{T_m}}$$ If we want to detect a displacement equal to the diameter of a proton ($D_p \approx 10^{-15}m$), the error has to at least equal to that displacement. We would therefore have to measure for a time $$T_m = \frac{S_{xx}}{\sigma_x^2} = 50,7 \text{ ms}$$
 # Homework 3 ## Exercise 1 ### a) $$S_{xx}^{th}(\omega) = \frac{k_B T \gamma }{m\omega_0^2} \frac{1}{(\omega-\omega_0)^2 + (\gamma/2)^2}$$ Peak value at $\omega = \omega_0$ $$S_{xx}^{th}(\omega_0) = \frac{4k_B T }{m\omega_0^2\gamma} = \frac{4k_B TQ }{m\omega_0^3} = 3,38 \cdot 10^{-26} \ m^2/Hz$$ where we used $Q = \frac{\omega_0}{\gamma}$. The sensitivity is then $$\sqrt{S_{xx}^{th}(\omega_0)} = 183 \ fm/\sqrt{Hz}$$ ![](figures/2a_thermal_sensitivity.png) # ### b) If the signal to noise ratio of our spectral noise peak $S_{xx}^{th}$ is $3,38\cdot 10^{-26} \ m^2/Hz$ then our detector noise level should be $3,38\cdot 10^{-28} \ m^2/Hz$ (Note: strange concept: the signal to noise ratio of detecting noise...) This would require a sensitivity $$\sqrt{S_{xx}^{det}} = \frac{S^{th}_{xx}}{10} = 18 \ \text{fm}$$ As we saw om the lecture 2 homework, this should not be too hard for a laser interferometer, for examples. # ## Exercise 2 ### a) ![](figures/2a_circuit.png) In parallel, impedances add in inverse $$\frac1Z = \frac1{Z_R} + \frac1{Z_L} + \frac1{Z_C} = \frac1R + i\omega C + \frac1{i\omega L}$$ Working out the above expression a bit we get \begin{aligned} Z &= \frac1{ \frac1R + i\left( \omega C - \frac1{\omega L}\right)} \frac{\frac1R - i(\omega L - \frac1{\omega C})}{\frac1R - i(\omega L - \frac1{\omega C})} \\ &= \frac{\frac1R - i(\omega L - \frac1{\omega C})}{\frac1{R^2} + (\omega C- \frac1{\omega L})^2} \end{aligned} $$\text{Re}(Z) = \frac{\frac1R}{\frac1{R^2} + (\omega C- \frac1{\omega L})^2}$$ Using $\Gamma = \frac1{RC}$ and $\omega_0^2=\frac1{LC}$: \begin{aligned} \text{Re}(Z) &= \frac{\frac1R}{\frac1{R^2} + \frac{C^2}{\omega^2}(\omega ^2- \frac1{C L})^2} \\ &=\frac{\omega^2}{RC^2} \frac{\frac1R}{\frac{\omega^2}{R^2C^2} + (\omega ^2- \frac1{C L})^2} \\ &= R\frac{\omega^2}{R^2C^2} \frac1{\omega^2\Gamma^2 + (\omega ^2- \omega_0^2)} \\ &= R \frac{\omega^2 \Gamma^2}{\omega^2 \Gamma^2+ (\omega ^2- \omega_0^2)} \end{aligned} (ok, a big pain but now in a reasonable form...). The PSD of the cavity thermal noise is then $$S_{vv}(\omega) = 4 k_B T R \ \frac{ \omega^2 \Gamma^2}{\omega^2 \Gamma^2+ (\omega ^2- \omega_0^2)}$$ ### 2b) At resonance: $$S_{vv}(\omega_0) = 4 k_B T R$$ Note: this is the PSD that would be produced by the (equivalent) resistor in parallel with my LC circuit that I need to produce the quoted quality factor. In fact , a nice way to interpret the noise out of the cavity is that it corresponds to the Johnson noise of the resistor filtered by the LC cavity. Another way of seeing it is the following: at $\omega_0$, $\frac1{Z_C}+\frac1{Z_L} = 0$ and so $Z(\omega_0)=R$. How big is R? $$R = \frac{Q}{\omega_0 C} = \frac{Q}{C}\sqrt{LC} = Q \sqrt{\frac{L}{C}} = 3.16 \ k\Omega$$ This gives: $$S_{vv} = 4 \cdot (1.38 \cdot 10^{-23} \ \frac{J}{K}) \cdot (300 \ K) \cdot (3.16 \cdot 10^{3} \Omega) = 5.23 \cdot 10^{-17} \ V^2/Hz$$ (recall that $1\ \Omega = 1 \ V^2 \cdot J ^{-1} \cdot s^{-1}$). Then $$\sqrt{S_{vv}} = 7.23 \ nV/ \sqrt{Hz}$$ This should be easily measurable by my amplifier with a sensitivity of $84 \ pV/\sqrt{Hz}$. \ No newline at end of file
 # Homework 4 ## Exercise 1 The expectation value of the energy is: $$\langle E \rangle = \langle \hat{H} \rangle = \frac12 E_0 + \frac12 E_1 = \frac12 \left( \frac12 + \frac32 \right) \hbar \omega = \hbar \omega$$ Now we calculate the "classical" energy $$E_{classical} = \frac12 m \omega^2 \langle \hat{x} \rangle + \frac1{2m} \langle \hat{p} \rangle$$ Using the definition of the position and momentum operators: $$\hat{x} = x_{ZPF} (\hat{a} + \hat{a}^{\dagger})$$ $$\hat{p} = \frac{-i\hbar}{2x_{ZPF}} (\hat{a} - \hat{a}^{\dagger})$$ \begin{aligned} \langle \hat{x}\rangle &= \frac{x_{PZF}}{2} \left( \langle 0| + \langle 1| \right) \left( \hat{a} + \hat{a}^{\dagger} \right) \left( | 0\rangle + |1\rangle \right) \\ &= \frac{x_{PZF}}{2} \left( \langle 0|0\rangle + \langle 1|1\rangle \right) \\ &= x_{PZF} \end{aligned} \begin{aligned} \langle \hat{p}\rangle &= \frac{-i\hbar}{4x_{ZPF}} \left( \langle 0| + \langle 1| \right) \left( \hat{a}^{\dagger} - \hat{a} \right) \left( | 0\rangle + |1\rangle \right) \\ &= \frac{-i\hbar}{4x_{ZPF}}\left( \langle 0|0\rangle - \langle 1|1\rangle \right) \\ &= 0 \end{aligned} with $x_{PZF} = \sqrt{\frac{\hbar}{2m\omega}}$. Therefore $$E_{classical} = \frac12 m \omega^2 x_{PZF}^2 = \frac{\hbar \omega}{4}$$ This state has $\frac{\hbar \omega}{4}$ "classical" contribution to its energy, and a contribution $\frac{3\hbar \omega}{4}$ to its energy induced by quantum fluctuations. # ## Exercise 2 ### a) The probability of having a number of photons $n$ is given by $$P_n = |c_n|^2$$ with $$c_n = \frac{ e^{ -\frac{|\alpha|^2}{2} } \alpha^n}{\sqrt{n!}}$$ $$|c_n|^2 = \frac{ e^{ -|\alpha|^2 } |\alpha|^{2n}}{n!}$$ Therefore $P_n$ follows a Poisson distribution with $\mu = |\alpha|^2$ $$P_n = \frac{ e^{ -|\alpha|^2 } |\alpha|^{2n}}{n!} = \frac{ e^{ -\mu } \mu^{n}}{n!}$$ with average $\langle n\rangle = \mu = |\alpha|^2$ ### b) \begin{aligned} |\alpha(t)\rangle &=\sum_{n=0}^{\infty} c_{n} e^{-i E_{n} t / \hbar}|n\rangle \\ &=\sum_{n=0}^{\infty} \frac{\alpha^{n}}{\sqrt{n !}} e^{-|\alpha|^{2} / 2} e^{-i\left(n+\frac{1}{2}\right) \omega t}|n\rangle \\ &=e^{-i \omega t / 2} \sum_{n=0}^{\infty} \frac{\left(\alpha e^{-i \omega t}\right)^{n}}{\sqrt{n !}} e^{-|\alpha|^{2} / 2}|n\rangle \end{aligned} Apart from the overall phase factor $e^{-i\omega t/2}$ (which doesn't affect its status as an eigenfunction of $\hat{a}$, or its eigenvalues), $|\alpha(t)\rangle$ is the same as $|\alpha\rangle$ but with eigenvalue $\alpha(t) = e^{-i\omega t}\alpha$. ### c) $$\begin{array}{l} \langle x\rangle=\langle\alpha \mid x \alpha\rangle=\sqrt{\frac{\hbar}{2 m \omega}}\left\langle\alpha \mid\left(\hat{a}^{\dagger}+\hat{a}\right)| \alpha\right\rangle=\sqrt{\frac{\hbar}{2 m \omega}}\left(\left\langle \hat{a} \alpha \mid \alpha\right\rangle+\left\langle\alpha \mid \hat{a} \alpha\right\rangle\right)=\sqrt{\frac{\hbar}{2 m \omega}}\left(\alpha+\alpha^{*}\right) \\ \langle p\rangle=\langle\alpha \mid p \alpha\rangle=i \sqrt{\frac{\hbar m \omega}{2}}\left\langle\alpha \mid\left(\hat{a}^{\dagger}-\hat{a}\right)| \alpha\right\rangle=i \sqrt{\frac{\hbar m \omega}{2}}\left(\left\langle \hat{a} \alpha \mid \alpha\right\rangle-\left\langle\alpha \mid \hat{a} \alpha\right\rangle\right)=-i \sqrt{\frac{\hbar m \omega}{2}}\left(\alpha-\alpha^{*}\right) \end{array}$$ ### d) $$x^{2}=\frac{\hbar}{2 m \omega}\left(\hat{a}^{\dagger 2}+\hat{a}^{\dagger} \hat{a}+\hat{a} \hat{a}^{\dagger}+\hat{a}^{2}\right) =\frac{\hbar}{2 m \omega}\left(\hat{a}^{\dagger 2}+2 \hat{a}^{\dagger} \hat{a}+1+\hat{a}^{2}\right).$$ Where we used that $\hat{a} \hat{a}^{\dagger}=\left[\hat{a}, \hat{a}^{\dagger}\right]+\hat{a}^{\dagger} \hat{a}=1+\hat{a}^{\dagger} \hat{a}$. Then: \begin{aligned} \left\langle x^{2}\right\rangle&=\frac{\hbar}{2 m \omega}\left\langle\alpha \mid\left(\hat{a}^{\dagger 2}+2 \hat{a}^{\dagger} \hat{a}+1+\hat{a}^{2}\right) \alpha\right\rangle=\frac{\hbar}{2 m \omega}\left(\left\langle \hat{a}^{2} \alpha \mid \alpha\right\rangle+2\left\langle \hat{a} \alpha \mid \hat{a} \alpha\right\rangle+\langle\alpha \mid \alpha\rangle+\left\langle\alpha \mid \hat{a}^{2} \alpha\right\rangle\right)\\ &=\frac{\hbar}{2 m \omega}\left[\left(\alpha^{*}\right)^{2}+2\left(\alpha^{*}\right) \alpha+1+\alpha^{2}\right]=\frac{\hbar}{2 m \omega}\left[1+\left(\alpha+\alpha^{*}\right)^{2}\right] \end{aligned} For the momentum $$p^{2}=-\frac{\hbar m \omega}{2}\left(\hat{a}^{\dagger 2}-\hat{a}^{\dagger} \hat{a}-\hat{a} \hat{a}^{\dagger}+\hat{a}^{2}\right)=-\frac{\hbar m \omega}{2}\left(\hat{a}^{\dagger 2}-2 \hat{a}^{\dagger} \hat{a}-1+\hat{a}^{2}\right)$$ $$\left\langle p^{2}\right\rangle=-\frac{\hbar m \omega}{2}\left\langle\alpha \mid\left(\hat{a}^{\dagger 2}-2 \hat{a}^{\dagger} \hat{a}-1+\hat{a}^{2}\right) \alpha\right\rangle=-\frac{\hbar m \omega}{2}\left(\left\langle \hat{a}^{2} \alpha \mid \alpha\right\rangle-2\left\langle \hat{a} \alpha \mid \hat{a} \alpha\right\rangle-\langle\alpha \mid \alpha\rangle+\left\langle\alpha \mid \hat{a}^{2} \alpha\right\rangle\right.\\ =-\frac{\hbar m \omega}{2}\left[\left(\alpha^{*}\right)^{2}-2\left(\alpha^{*}\right) \alpha-1+\alpha^{2}\right]=\frac{\hbar m \omega}{2}\left[1-\left(\alpha-\alpha^{*}\right)^{2}\right]$$ Now we can calculate the uncertainties as $$\sigma_{x}^{2}=\left\langle x^{2}\right\rangle-\langle x\rangle^{2}=\frac{\hbar}{2 m \omega}\left[1+\left(\alpha+\alpha^{*}\right)^{2}-\left(\alpha+\alpha^{*}\right)^{2}\right]=\frac{\hbar}{2 m \omega}$$ $$\sigma_{p}^{2}=\left\langle p^{2}\right\rangle-\langle p\rangle^{2}=\frac{\hbar m \omega}{2}\left[1-\left(\alpha-\alpha^{*}\right)^{2}+\left(\alpha-\alpha^{*}\right)^{2}\right]=\frac{\hbar m \omega}{2}$$ $$\sigma_{x} \sigma_{p}=\sqrt{\frac{\hbar}{2 m \omega}} \sqrt{\frac{\hbar m \omega}{2}}=\frac{\hbar}{2}. \quad \mathrm{QED}$$ $$\ No newline at end of file  # Homework 6 ## Exercise 1 ### a) The Hamiltonian is$$ \hat{H} = -g \mu_B B_0 S_z = \frac{ -g \mu_B B_0 S_z}{2} \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} = \begin{pmatrix} -\frac{\hbar \omega_L}{2} & 0 \\ 0 & \frac{\hbar \omega_L}{2} \end{pmatrix} $$where \omega_L is the Larmor frecuency. For convenience, from now on we write \omega_L =\omega. Note that the states$$|\uparrow\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix} |\downarrow\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} $$are the eigenstates of the Hamiltonian \hat{H}. Therfore the initial state |\psi(0)\rangle = \frac1{\sqrt{2}} \left( |\uparrow\rangle + |\downarrow\rangle\right) will evolve as:$$ \begin{aligned} |\psi(t)\rangle &= \frac1{\sqrt{2}} \left( e^{-iE_{\uparrow}t/\hbar} |\uparrow\rangle + e^{-iE_{\downarrow}t/\hbar} |\downarrow\rangle\right) \\ &= \frac1{\sqrt{2}} \left( e^{i\omega t} |\uparrow\rangle + e^{-i\omega t} |\downarrow\rangle\right) \\ &= \frac1{\sqrt{2}} \begin{pmatrix} e^{i\omega t/2} \\ e^{-i\omega t/2} \end{pmatrix} \end{aligned} $$and therefore the expectation value \langle S_x(t) \rangle is:$$ \begin{aligned} \langle S_x(t) \rangle &= \langle \psi(t)| S_x|\psi(t) \rangle \\ &= \frac\hbar2 \frac12 \left( e^{-i\omega t/2}, \ e^{i\omega t/2}\right) \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} \begin{pmatrix} e^{i\omega t/2} \\ e^{-i\omega t/2} \end{pmatrix} \\ &= \frac\hbar2 \frac{e^{-i\omega t/2} + e^{i\omega t/2}}{2} \\ &= \frac\hbar2 \cos\omega t \end{aligned} $$### b) According to the previous section the average \langle S_x(t_1) \rangle is$$ \langle S_x(t_1) \rangle = -\frac\hbar2 $$But what does this mean? We know that the average of an operator A with eigenvalues \{\lambda_i\} can also be written as$$ \langle A \rangle = \sum_i \text{Prob}(\lambda_i) \cdot \lambda_i $$where Prob(\lambda_i) is the probability of measuring the eigenvalue \lambda_i. In our case, S_x has two eigenvalues \pm\frac\hbar2, so we can write:$$ \langle S_x \rangle = \text{Prob}(+\frac\hbar2) \cdot \frac\hbar2 - \text{Prob}(-\frac\hbar2) \cdot \frac\hbar2 $$If \langle S_x(t_1) \rangle = -\frac\hbar2 this means that \text{Prob}(-\frac\hbar2) = 1 and we will always measure -\frac\hbar2 at t_1. Therefore, no collapse occurs since the system is an eigenstate of S_x and nothing happens to the oscillation. ### c) After the measurement the state collapses to the eigenstate S_x corresponding to the eigenvalue +\frac\hbar2. One can check that this corresponds to the initial state |\psi(t=0)\rangle. Therefore, after the measurement the state collapses to |\psi(t=0)\rangle and the oscillation begins again$$ \langle S_x(t) \rangle = \frac\hbar2 \cos\left( \omega (t-\frac{3\pi}{2\omega})\right)  ### d) ![](figures/6d.png) ### e) Each time we collapse the spin, it randomises the phase of the Larmor precession oscillations (unless it occurs at $\langle S_x\rangle = \frac\hbar2$ or $\langle S_x\rangle = -\frac\hbar2$, in which case there is no uncertainty in the measurement outcome and there is no change in the phase of the oscillations). ### f) It will result in stochastic "dephasing" of the oscillations. ### g) If I have no information about the time of the measurements, then the spin will appear only to be executing a random walk in phase. You might be tempted to think: but what if I measure faster that the random measurements? Then, I am actually measuring very accurately already ... and also constantly collapsing the wavefunction... In 2007 there was a very interesting experiment from the group of Serge Haroche who did exactly that: "Quantum jumps of light recording the birth and death of a photon in a cavity" ([Nature 446 297](https://www.nature.com/articles/nature05589)). ### h) At $\Delta = \pi/\omega$ (or actually $\Delta =n \pi/\omega$) there is no uncertainty in the outcome of the measurement. \ No newline at end of file
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