Commit 74a2244a authored by ignacio's avatar ignacio
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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
$$
\sigma_{x}^{2}=\left\langle x^{2}\right\rangle-\langle x\rangle^{2}=\frac{\hbar}{2 m \omega}\left[1+\left(\alpha+\alpha^{*}\right)^{2}-\left(\alpha+\alpha^{*}\right)^{2}\right]=\frac{\hbar}{2 m \omega}
$$
$$
\sigma_{p}^{2}=\left\langle p^{2}\right\rangle-\langle p\rangle^{2}=\frac{\hbar m \omega}{2}\left[1-\left(\alpha-\alpha^{*}\right)^{2}+\left(\alpha-\alpha^{*}\right)^{2}\right]=\frac{\hbar m \omega}{2}
$$
$$
\sigma_{x} \sigma_{p}=\sqrt{\frac{\hbar}{2 m \omega}} \sqrt{\frac{\hbar m \omega}{2}}=\frac{\hbar}{2}. \quad \mathrm{QED}
$$
$$
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# Homework 6
## Exercise 1
### a)
The Hamiltonian is
$$
\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.
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