## 物理代写|凝聚态物理代写condensed matter physics代考|Dynamical Structure Factor and f -Sum Rule

The dynamical structure factor $S(\vec{q}, \omega)$ has two interpretations. First it can be viewed as a measure of the dynamical correlations in the ground (or, at finite $T$, the equilibrium) state. Namely, it is the mean square value of the density fluctuations of wave vector $\vec{q}$ and frequency $\omega$. Secondly, as we shall see below, $(1 / 2 \pi \hbar) S(\vec{q}, \omega)$ is the density of excited states at wave vector $\vec{q}$ created by the density operator. We have
$$S(\vec{q}, \omega)=\frac{1}{N} \int_{-\infty}^{\infty} d t e^{i \omega t} \sum_{j, k=1}^N\left\langle\left\langle e^{i \vec{q} \cdot \vec{r}j(t)} e^{-i \vec{q} \cdot \vec{r}_k(0)}\right\rangle\right\rangle,$$ which is simply the Fourier transform of the density-density correlation, since $$\hat{\rho}{\vec{q}}(t)=\int d^3 \vec{r} e^{-i \vec{q} \cdot \vec{r}} \sum_{j=1}^N \delta^3\left[\vec{r}-\vec{r}j(t)\right]=\int d^3 \vec{r} e^{-i \vec{q} \cdot \vec{r}} \hat{\rho}(\vec{r}, t) .$$ Thus $$S(\vec{q}, \omega)=\int{-\infty}^{+\infty} d t e^{i \omega t} \Pi(\vec{q}, t),$$
where
$$\Pi(\vec{q}, t) \equiv \frac{1}{N}\left\langle\left\langle\hat{\rho}{-\vec{q}}(t) \hat{\rho}{+\vec{q}}(0)\right\rangle\right\rangle .$$
We can interpret the expression as telling us that the density waves in a crystal act as moving diffraction gratings that scatter and Doppler shift the neutron.

Notice from the definition of the density operator that the effect of $\rho_{+\vec{q}}$ is to boost the momentum of one of the particles (we do not know which one) by $-\hbar \vec{q}$, corresponding to the fact that the neutron has acquired momentum $+\hbar \vec{q}$. We can thus also think of $S(\vec{q}, \omega)$ as the spectral density of excited states that couple to the initial state through the density operator:
\begin{aligned} &S(\vec{q}, \omega)=\frac{1}{Z} \sum_i e^{-\beta E_i} S_i(\vec{q}, \omega) \ &S_i(\vec{q}, \omega)=\frac{1}{N} \sum_f\left|\left\langle i\left|\rho_{\vec{q}}\right| f\right\rangle\right|^2 2 \pi \delta\left[\omega-\left(E_f-E_i\right) / \hbar\right] . \end{aligned}

## 物理代写|凝聚态物理代写condensed matter physics代考|Classical Harmonic Oscillator

It is useful as a warm-up exercise to consider a single classical particle in a 1D harmonic oscillator potential. The Hamiltonian is
$$H=\frac{P^2}{2 M}+\frac{1}{2} M \Omega^2 X^2$$
and we would like to compute
$$\left.\Pi(q, t) \equiv\left\langle e^{i q X(t)} e^{-i q X(0)}\right\rangle\right\rangle .$$
The system is classical, so the time evolution is entirely determined by the initial values of position $X(0) \equiv X_0$ and momentum $P(0) \equiv P_0$. We have the exact solution of the classical equations of motion,

$$X(t)=X_0 \cos (\Omega t)+\frac{P_0}{M \Omega} \sin (\Omega t),$$
which is easily seen to obey the initial conditions. Thus
$$\Pi(q, t)=\left\langle\left\langle e^{i q[\cos (\Omega t)-1] X_0} e^{i \frac{q}{M \Omega} \sin (\Omega t) P_0}\right\rangle\right\rangle .$$
Now, in a classical system $X_0$ and $P_0$ are independent random variables given by the Boltzmann probability distribution
$$P\left(X_0, P_0\right)=\frac{1}{Z} e^{-\beta\left(\frac{P_0^2}{2 M}+\frac{1}{2} M \Omega^2 X_0^2\right)},$$
where
$$Z \equiv \frac{1}{h} \int d X_0 \int d P_0 e^{-\beta\left(\frac{P_0^2}{2 M}+\frac{1}{2} M \Omega^2 X_0^2\right)} .$$
(Recall that $Z$ is dimensionless, so we need a factor of $1 / h$, which is the “area” of a quantum state in phase space. The necessity of this factor was one of the important mysteries of classical physics which was later resolved by the discovery of quantum mechanics.)

Let us return to our original definition of $\Pi(q, t)$ in Eq. (4.69) and make an expansion in powers of $q$ :
\begin{aligned} \Pi(q, t)=& 1+i q[\langle\langle X(t)\rangle-\langle\langle X(0)\rangle\rangle]\ &-\frac{1}{2} q^2\left[\left\langle\left\langle X(t)^2\right\rangle\right\rangle+\left\langle\left\langle X(0)^2\right\rangle\right\rangle\right] \ &\left.+q^2[\langle X(t) X(0)\rangle\rangle\right]+\cdots . \end{aligned}

## 物理代写|凝聚态物理代写condensed matter physics代考|Dynamical Structure Factor and f -Sum Rule

$$S(\vec{q}, \omega)=\frac{1}{N} \int_{-\infty}^{\infty} d t e^{i \omega t} \sum_{j, k=1}^N\left\langle\left\langle e^{i \vec{q} \cdot \vec{r} j(t)} e^{-i \vec{q} \cdot \vec{r}k(0)}\right\rangle\right\rangle,$$ 这只是密度-密度相关性的傅里叶变换，因为 $$\hat{\rho} \vec{q}(t)=\int d^3 \vec{r} e^{-i \vec{q} \cdot \vec{r}} \sum{j=1}^N \delta^3[\vec{r}-\vec{r} j(t)]=\int d^3 \vec{r} e^{-i \vec{q} \cdot \vec{r}} \hat{\rho}(\vec{r}, t) .$$

$$S(\vec{q}, \omega)=\int-\infty^{+\infty} d t e^{i \omega t} \Pi(\vec{q}, t)$$

$$\Pi(\vec{q}, t) \equiv \frac{1}{N}\langle\langle\hat{\rho}-\vec{q}(t) \hat{\rho}+\vec{q}(0)\rangle\rangle .$$

，对应于中子获得动量的事实 $+\hbar \vec{q}$. 因此我们也可以想到 $S(\vec{q}, \omega)$ 作为通过密度算子耦合到初始状态的激发 态的谱密度：
$$S(\vec{q}, \omega)=\frac{1}{Z} \sum_i e^{-\beta E_i} S_i(\vec{q}, \omega) \quad S_i(\vec{q}, \omega)=\frac{1}{N} \sum_f\left|\left\langle i\left|\rho_{\vec{q}}\right| f\right\rangle\right|^2 2 \pi \delta\left[\omega-\left(E_f-E_i\right) / \hbar\right] .$$

## 物理代写|凝聚态物理代写condensed matter physics代考|Classical Harmonic Oscillator

$$H=\frac{P^2}{2 M}+\frac{1}{2} M \Omega^2 X^2$$

$$\left.\Pi(q, t) \equiv\left\langle e^{i q X(t)} e^{-i q X(0)}\right\rangle\right\rangle$$

$$X(t)=X_0 \cos (\Omega t)+\frac{P_0}{M \Omega} \sin (\Omega t)$$

$$\Pi(q, t)=\left\langle\left\langle e^{i q[\cos (\Omega t)-1] X_0} e^{i \frac{q}{M \Omega} \sin (\Omega t) P_0}\right\rangle\right\rangle .$$

$$P\left(X_0, P_0\right)=\frac{1}{Z} e^{-\beta\left(\frac{P_0^2}{2 M}+\frac{1}{2} M \Omega^2 X_0^2\right)},$$

$$Z \equiv \frac{1}{h} \int d X_0 \int d P_0 e^{-\beta\left(\frac{p_0^2}{2 M}+\frac{1}{2} M \Omega^2 X_0^2\right)} .$$
(回顾 $Z$ 是无量纲的，所以我们需要一个因子 $1 / h$ ，它是相空间中量子态的“面积”。这个因素的必要性是经 典物理学的重要谜团之一，后来被量子力学的发现解决了。)

$$\Pi(q, t)=1+i q\left[\langle\langle X(t)\rangle-\langle\langle X(0)\rangle\rangle] \quad-\frac{1}{2} q^2\left[\left\langle\left\langle X(t)^2\right\rangle\right\rangle+\left\langle\left\langle X(0)^2\right\rangle\right\rangle\right]+q^2[\langle X(t) X(0)\rangle\rangle\right]+\cdots .$$

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