澳洲代写｜PHYS3034｜ Quantum, Statistical and Comp Physics量子、统计和复合物理 悉尼大学
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课程介绍：
The dynamics of complex systems are often described in terms of how they process information and self-organise; for example regarding how genes store and utilise information, how information is transferred between neurons in undertaking cognitive tasks, and how swarms process information in order to collectively change direction in response to predators. The language of information also underpins many of the central concepts of complex adaptive systems, including order and randomness, self-organisation and emergence. Shannon information theory, which was originally founded to solve problems of data compression and communication, has found contemporary application in how to formalise such notions of information in the world around us and how these notions can be used to understand and guide the dynamics of complex systems. This unit of study introduces information theory in this context of analysis of complex systems, foregrounding empirical analysis using modern software toolkits, and applications in time-series analysis, nonlinear dynamical systems and data science. Students will be introduced to the fundamental measures of entropy and mutual information, as well as dynamical measures for time series analysis and information flow such as transfer entropy, building to higher-level applications such as feature selection in machine learning and network inference. They will gain experience in empirical analysis of complex systems using comprehensive software toolkits, and learn to construct their own analyses to dissect and design the dynamics of self-organisation in applications such as neural imaging analysis, natural and robotic swarm behaviour, characterisation of risk factors for and diagnosis of diseases, and financial market dynamics.
| Attribute | Detail |
|---|---|
| Course Code | PHYS3034 |
| Course Title | Quantum, Statistical and Comp Physics |
| Academic Unit | Physics Academic Operations |
| Session | Semester 1, 2023 |
| Number of Units | 6 |
| Pre-Requisites | 2000-level physics (from the context provided) |
| Course Coordinator | Not explicitly mentioned in the provided text |
| Lecturer | Not explicitly mentioned in the provided text |
Statistical Physics统计物理的案例
For a particular model of a gene in a cell, the probability density that said gene produces a concentration $x$ of proteins during the cell cycle is given by
$$
p(x)=A\left(\frac{x}{b}\right)^N e^{-x / b},
$$
where $b$ is a biological constant with units of concentration, $N$ is a physical constant, and $A$ is a normalization parameter.
(a) (5 points) The concentration of proteins that can be produced ranges from zero to infinite. What must $A$ be in order for Eq.(1) to be normalized?
(b) (5 points) What is the mean of the normalized probability density?
(c) (Removed in Final Version) For this probability distribution, compute the average
$$
\left\langle e^{x / a}\right\rangle
$$
and write the result in terms of an infinite sum over a binomial coefficient. (Note that $e^x=$ $\sum_{j=0}^{\infty} x^j / j$ !.)
(a) Given the range of possible protein production, $x$ can go from 0 to $\infty$. Therefore, for $p(x)$ to be normalized, we must obtain 1 when we integrate the function over this entire domain:
$$
\int_0^{\infty} d x p(x)=1 .
$$
From the definition of the probability density we have
$$
\begin{aligned}
1 & =\int_0^{\infty} d x A\left(\frac{x}{b}\right)^N e^{-x / b} \
& =A \int_0^{\infty} d x\left(\frac{x}{b}\right)^N e^{-x / b} \
& =A \int_0^{\infty} d u b u^N e^{-u} \
& =A b \int_0^{\infty} d u u^N e^{-u},
\end{aligned}
$$
where we changed variables with $u=x / b$ in the third line, and factored the $u$-independent constant out of the integral in the final line. By the integral definition of factorial, we have
$$
N !=\int_0^{\infty} d u u^N e^{-u} .
$$
Therefore, the final line of Eq.(4) becomes
$$
1=A b N !
$$
and we can conclude
$$
A=\frac{1}{b N !}
$$
The normalized probability density is therefore
$$
p(x)=\frac{1}{b N !}\left(\frac{x}{b}\right)^N e^{-x / b} .
$$
(b) The mean of a random variable defined by the probability density $p(x)$ (which has a nonzero domain for $x \in[0, \infty))$ is
$$
\langle x\rangle=\int_0^{\infty} d x x p(x) .
$$
Using Eq.(8) to compute this value, we obtain
$$
\begin{aligned}
\langle x\rangle & =\int_0^{\infty} d x x \frac{1}{b N !}\left(\frac{x}{b}\right)^N e^{-x / b} \
& =\frac{1}{N !} \int_0^{\infty} d x\left(\frac{x}{b}\right)^{N+1} e^{-x / b} \
& =\frac{1}{N !} \int_0^{\infty} d u b u^{N+1} e^{-u} \
& =b \frac{(N+1) !}{N !},
\end{aligned}
$$
where in the third line we performed a change of variables with $u=x / b$ and in the final line we used Eq.(5). By the definition of factorial, we ultimately find
$$
\langle x\rangle=b(N+1)
$$
(c) (Not part of final version of exam) We now seek to compute the average of $e^{x / a}$. Noting the Taylor series definition of the exponential
$$
e^{x / a}=\sum_{j=0}^{\infty} \frac{(x / a)^j}{j !},
$$
we have
$$
\begin{aligned}
\left\langle e^{x / a}\right\rangle & =\left\langle\sum_{j=0}^{\infty} \frac{(x / a)^j}{j !}\right\rangle \
& =\sum_{j=0}^{\infty} \frac{1}{j !} \frac{1}{a^j}\left\langle x^j\right\rangle .
\end{aligned}
$$
Computing $\left\langle x^j\right\rangle$ yields
\begin{aligned}
\left\langle x^j\right\rangle & =\int_0^{\infty} d x x^j \frac{1}{b N !}\left(\frac{x}{b}\right)^N e^{-x / b} \
& =\frac{b^j}{b N !} \int_0^{\infty} d x\left(\frac{x}{b}\right)^{N+j} e^{-x / b} \
& =\frac{b^j}{b N !} \int_0^{\infty} d u b u^{N+j} e^{-u}
\end{aligned}
$$
=b^j \frac{(N+j) !}{N !}
$$
where in the second line we multiplied the numerator and the denominator by $b^j$, in the third line we performed a change of variables $u=x / b$, and in the final line we used Eq.(5). Inserting this result into Eq.(13), we find
$$
\left\langle e^{x / a}\right\rangle=\sum \sum_{j=0}^{\infty} \frac{1}{j !} \frac{1}{a^j} b^j \frac{(N+j) !}{N !}=\sum_{j=0}^{\infty} \frac{(N+j) !}{j ! N !}\left(\frac{b}{a}\right)^j,
$$
or
$$
\left\langle e^{x / a}\right\rangle=\sum_{j=0}^{\infty}\left(\begin{array}{c}
N+j \
j
\end{array}\right)\left(\frac{b}{a}\right)^j .
$$
We note that we could evaluate $\left\langle e^{x / a}\right\rangle$ directly using a change of variables in the argument of the exponential of the distribution. The result would be
$$
\begin{aligned}
\left\langle e^{x / a}\right\rangle & =\frac{1}{b N !} \int_0^{\infty} d x e^{x / a}\left(\frac{x}{b}\right)^N e^{-x / b} \
& =\frac{1}{b N !} \int_0^{\infty} d x\left(\frac{x}{b}\right)^N e^{-(1 / b-1 / a) x} \
& =\frac{1}{b N !} \int_0^{\infty} d u \frac{a b}{a-b}\left(\frac{a u}{a-b}\right)^N e^{-u} \
& =\frac{1}{N !}\left(\frac{a}{a-b}\right)^{N+1} \int_0^{\infty} d u u^N e^{-u} \
& =\frac{1}{(1-b / a)^{N+1}},
\end{aligned}
$$
where in the second line we made the change of variables $u=x(a-b) / a b$. Considering Eq.(16), the result Eq.(17) implies
$$
\sum_{j=0}^{\infty}\left(\begin{array}{c}
N+j \
j
\end{array}\right) q^j=\frac{1}{(1-q)^{N+1}}
$$
Quantum Physics量子物理问题
- True or false questions [ 20 points] No explanations required. Just indicate $\mathrm{T}$ or $\mathrm{F}$ for true or false, respectively.
(1) The operators $\sigma_1 \otimes \sigma_1$ and $\sigma_3 \otimes \sigma_1$ commute.
(2) The operators $\sigma_1 \otimes \sigma_1$ and $\sigma_3 \otimes \sigma_3$ commute.
(3) Let $T \otimes S$ be a linear operator on $V \otimes W$. Then $(T \otimes S)^{\dagger}=S^{\dagger} \otimes T^{\dagger}$.
(4) A linear operator on a finite-dimensional vector space is invertible if it is injective.
(5) Angular momentum conservation prevents a particle with spin $1 / 2$ from decaying into two spin- $1 / 2$ particles.
(6) $\left[\mathbf{L}^2, \hat{x}_i\right]=0$. Here $\hat{x}_i$ is the position operator in any of the three directions.
(7) $\mathbf{r} \cdot \mathbf{p}=\mathbf{p} \cdot \mathbf{r}-3 i \hbar$.
(8) $\mathbf{A} \cdot \mathbf{L}=\mathbf{L} \cdot \mathbf{A}$, when $\mathbf{A}$ is a vector under rotations.
(9) In the hydrogen atom the Runge-Lenz (RL) vector $\mathbf{R}$ satisfies the algebra of angular momentum.
(10) Both classically and quantum mechanically the $R L$ vector $\mathbf{R}$ satisfies $\mathbf{R} \cdot \mathbf{L}=0$.
- Expectation value on a generalized squeezed state [10 points] Consider the general squeezed state $|\alpha, \gamma\rangle=D(\alpha) S(\gamma)|0\rangle$ of the harmonic oscillator at time equal zero (here $\alpha \in \mathbb{C}, \gamma \in \mathbb{R}$ ). Find the expectation value of the number operator $\hat{N}$ in this state. As we let time change, does this expectation value exhibit time dependence?
- $3 \mathrm{D}$ bound state $[10$ points]
A particle of mass $m$ is in a potential $V(r)$ that represents a finite depth spherical well of radius $a$ :
$$
V(r)=\left{\begin{aligned}
-V_0 & \text { for } ra,
\end{aligned}\right.
$$
Here $V_0>0$. For the potential to have bound states it should be deep enough. Derive the inequality that $V_0$ must satisfy in order that the potential have a bound state.
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