# 物理代写|热力学代写thermodynamics代考|AMME2262

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## 物理代写|热力学代写thermodynamics代考|Magnon Baths

The low-lying energy states of spin systems coupled by exchange interactions give rise to quantized spin waves. The spin-wave quanta are known as magnons.

The simplest bath Hamiltonian that gives rise to magnons is that of $N$ identical spin- $S$ particles with nearest-neighbor interactions in a ferromagnetic spin lattice. It has the form,

$$H_{\mathrm{B}}=-J \sum_{j, j^{\prime}} \hat{\boldsymbol{S}}{j} \cdot \hat{\boldsymbol{S}}{j^{\prime}}-2 \mu_{0} \mathcal{B}{0} \sum{j} \hat{S}{j z},$$ where $\hat{\boldsymbol{S}}{j}$ is the $j$ th spin operator, $\hat{\boldsymbol{S}}{j}$ are the spin operators of its nearest neighbors on a lattice, $J$ is the positive exchange integral, $2 \mu{0}$ is the magnetic moment of a particle, and $\mathcal{B}_{0} \geq 0$ is the static magnetic field that aligns the spins along the $\mathrm{z}$ axis.

To study this bath, it is convenient to resort to the Holstein-Primakoff transformation of the spin operators to bosonic creation and annihilation operators $a_{j}^{\dagger}, a_{j}$, which has the form
\begin{aligned} &\hat{S}{j}^{+}=\hat{S}{j x}+i \hat{S}{j y}=(2 S)^{1 / 2}\left(1-\frac{a{j}^{\dagger} a_{j}}{2 S}\right)^{1 / 2} a_{j} \ &\hat{S}{j}^{-}=\hat{S}{j x}-i \hat{S}{j y}=(2 S)^{1 / 2} a{j}^{\dagger}\left(1-\frac{a_{j}^{\dagger} a_{j}}{2 S}\right)^{1 / 2} \ &\hat{S}{j z}=S-a{j}^{\dagger} a_{j} \end{aligned}
We next introduce the collective spin-wave (magnon) operators $a(\boldsymbol{k}), a^{\dagger}(\boldsymbol{k})$, that satisfy
$$a_{j}=\frac{1}{\sqrt{N}} \sum_{k} e^{-i k \cdot x_{j}} a(k),$$
where $\boldsymbol{x}{j}$ is the position vector of the $j$ th spin particle. We can rewrite the bath Hamiltonian in terms of these collective operators as $$H{\mathrm{B}}=H_{\mathrm{M}}+H_{\mathrm{M}}^{(1)} .$$

## 物理代写|热力学代写thermodynamics代考|X X Spin-Chain Bath

A quantum bath consisting of a spin- $1 / 2$ chain with nearest-neighbor exchange interactions is defined as $X X$ chain if it is governed by the Hamiltonian
$$H_{\mathrm{B}}=\hbar J \sum_{i=1}^{N-1}\left(\hat{S}{i}^{+} \hat{S}{i+1}^{-}+\hat{S}{i}^{-} \hat{S}{i+1}^{+}\right) .$$
Here $\hat{S}^{\pm}=\hat{S}^{x} \pm i \hat{S}^{y}$ are spin-1/2 raising and lowering (excitation and de-excitation, respectively) operators and $J$ is the spin-spin coupling strength. This Hamiltonian conserves the total magnetization.

It can be transformed into the Hamiltonian of non-interacting spinless fermions that hop between neighboring sites of the form,
$$H_{\mathrm{B}}=\hbar J \sum_{i=1}^{N-1}\left(c_{i}^{\dagger} c_{i+1}+c_{i+1}^{\dagger} c_{i}\right),$$
by using the Jordan-Wigner transformation,
$$c_{i}=e^{i \pi \sum_{j=1}^{i-1} \hat{s}{j}^{+} \hat{s}{j}^{-} \hat{S}{i}^{-} .}$$ The Hamiltonian (3.50) conserves the number of fermions. It simplifies upon performing the discrete sine transform, $$\hat{f{q}}=\sqrt{\frac{2}{N+1}} \sum_{j=1}^{N} \sin \left(\frac{j q \pi}{N+1}\right) c_{j} \equiv \sum_{j=1}^{N} U_{q j} c_{j} \quad(q=1,2, \ldots, N) .$$
The transform matrix $U=\left{U_{q j}\right}$ is real, Hermitian, and unitary, $U=U^{\dagger}, U^{2}=I$. Correspondingly, the inverse transform of (3.52) is
$$c_{j}=\sum_{q=1}^{N} U_{j q} \hat{f}{q} .$$ Moreover, the unitarity of $U$ ensures that $\hat{f}{q}$ and $\hat{f}_{q}^{\dagger}$ satisfy the commutation relations for the fermionic annihilation and creation operators, respectively.

# 热力学代写

## 物理代写|热力学代写thermodynamics代考|Magnon Baths

$$H_{\mathrm{B}}=-J \sum_{j, j^{\prime}} \hat{\boldsymbol{S}} j \cdot \hat{\boldsymbol{S}} j^{\prime}-2 \mu_{0} \mathcal{B} 0 \sum j \hat{S} j z$$

$$\hat{S} j^{+}=\hat{S} j x+i \hat{S} j y=(2 S)^{1 / 2}\left(1-\frac{a j^{\dagger} a_{j}}{2 S}\right)^{1 / 2} a_{j} \quad \hat{S} j^{-}=\hat{S} j x-i \hat{S} j y=$$

$$a_{j}=\frac{1}{\sqrt{N}} \sum_{k} e^{-i k \cdot x_{j}} a(k)$$

$$H \mathrm{~B}=H_{\mathrm{M}}+H_{\mathrm{M}}^{(1)} .$$

## 物理代写|热力学代写thermodynamics代考|X X Spin-Chain Bath

$$H_{\mathrm{B}}=\hbar J \sum_{i=1}^{N-1}\left(\hat{S} i^{+} \hat{S} i+1^{-}+\hat{S} i^{-} \hat{S} i+1^{+}\right)$$

$$H_{\mathrm{B}}=\hbar J \sum_{i=1}^{N-1}\left(c_{i}^{\dagger} c_{i+1}+c_{i+1}^{\dagger} c_{i}\right)$$

$$c_{i}=e^{i \pi \sum_{j=1}^{i-1} \hat{s} j^{+} \hat{s} j^{-} \hat{S}^{-}} .$$

$$\hat{f} q=\sqrt{\frac{2}{N+1}} \sum_{j=1}^{N} \sin \left(\frac{j q \pi}{N+1}\right) c_{j} \equiv \sum_{j=1}^{N} U_{q j} c_{j} \quad(q=1,2, \ldots, N)$$ 地，(3.52)的逆穾换为
$$c_{j}=\sum_{q=1}^{N} U_{j q} \hat{f} q$$

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## MATLAB代写

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