Commit 74a2244a by ignacio

### Added solutions of homework 4 and 6

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 # 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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