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# 物理代考| Bosons量子力学代写

## 物理代写

7.1 Bosons
Let us order the available single-particle states as $0,1,2, \cdots, i, j, k, \cdots$. The state vector for the many-particle system in the abstract occupation number space is then just the direct product over all the single-particle states with the number of particles in each mode
$\left|n_{0} n_{1} n_{2} \cdots\right\rangle=\left|n_{0}\right\rangle\left|n_{1}\right\rangle\left|n_{2}\right\rangle \cdots \quad ;$ abstract state vector
The operators in this space are just our previous simple harmonic oscillator operators for each mode, and for bosons, we will now denote these operators by $b$ and $b^{\dagger}$. They satisfy the commutation relations
$\left[b_{i}, b_{j}^{\dagger}\right]=\delta_{i j} \quad ;$ commutation relations (7.2)
All of the properties of these operators follow exactly as in Sec. 6.7. As we have seen, this analysis holds for photons. It also holds for non-relativistic spin-zero systems such as ${ }^{4} \mathrm{He}$ atoms, which are also bosons. Let us here focus on this latter case.

Suppose we have a single-particle operator such as the kinetic energy for this many-body system. We write this operator as
$\hat{T}=\sum_{i} \sum_{j} b_{j}^{\dagger}\langle j|T| i\rangle b_{i} \quad ;$ one-body operator
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Introduction to Quantum Mechanics

• From now on we use a hat over a quantity to indicate an operator in the abstract space, except for the creation and destruction operators
• Here $\langle j|T| i\rangle$ is the appropriate single-particle matrix element for the
$$\langle j|T| i\rangle=\int d^{3} x \psi_{j}^{*}(\vec{x}) T \psi_{i}(\vec{x})$$
$$\hat{T}=\sum_{j}\langle j|T| j\rangle b_{j}^{\dagger} b_{j}=\sum_{j}\langle j|T| j\rangle \hat{n}_{j}$$
• In this last case, the many-body matrix element of the kinetic energy From now on we use a hat over a the abstract space, except for the where this is obvious; Here $\langle j|T| i\rangle$ is the appropriate singeder problem at hand $$\langle j|T| i\rangle=\int d^{3}$$ If $T$ is diagonal, then $$\hat{T}=\sum_{j}\langle j|T| j\rangle b_{j}^{\dagger}$$ In this last case, the many-body operator is $$\left\langle n_{1} n_{2} \cdots|\hat{T}| n_{1} n_{2}\right.$$
$$\left\langle n_{1} n_{2} \cdots|\hat{T}| n_{1} n_{2} \cdots\right\rangle=\sum_{j}\langle j|T| j\rangle n_{j}$$ This just adds up the kinetic
We can rewrite this one-body operator $\hat{T}$ by introducing the nonrelativistic quantum field
\begin{aligned} \hat{\psi}(\vec{x}) & \equiv \sum_{j} \psi_{j}(\vec{x}) b_{j} \quad ; \text { quantum field } \ \hat{\psi}^{\dagger}(\vec{x}) & \equiv \sum_{j} \psi_{j}^{*}(\vec{x}) b_{j}^{\dagger} \end{aligned}
The one-body operator is then
$$\hat{T}=\int d^{3} x \hat{\psi}^{\dagger}(\vec{x}) T \hat{\psi}(\vec{x})$$
Suppose one has a two-body operator, such as the potential between all the pairs
$$V=\frac{1}{2} \sum_{i} \sum_{j} V\left(\left|\vec{x}{i}-\vec{x}{j}\right|\right)$$

## 物理代考

7.1 玻色子

$\left|n_{0} n_{1} n_{2} \cdots\right\rangle=\left|n_{0}\right\rangle\left|n_{1}\right\rangle\left|n_{ 2}\right\rangle \cdots \quad ;$ 抽象状态向量

$\left[b_{i}, b_{j}^{\dagger}\right]=\delta_{i j} \quad ;$ 交换关系 (7.2)

$\hat{T}=\sum_{i} \sum_{j} b_{j}^{\dagger}\langle j|T| i\rangle b_{i} \quad ;$ 单体运算符

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• 从现在开始，我们在数量上使用帽子来表示抽象空间中的运算符，除了创建和销毁运算符
• 这里 $\langle j|T| i\rangle$ 是合适的单粒子矩阵元素
$$\langle j|T| i\rangle=\int d^{3} x \psi_{j}^{*}(\vec{x}) T \psi_{i}(\vec{x})$$
$$\hat{T}=\sum_{j}\langle j|T| j\rangle b_{j}^{\dagger} b_{j}=\sum_{j}\langle j|T| j\rangle \hat{n}_{j}$$
• 在最后一种情况下，动能的多体矩阵元素从现在开始，我们在抽象空间上使用帽子，除非这是显而易见的；这里 $\langle j|T| i\rangle$ 是手头的适当的 singeder 问题 $$\langle j|T| i\rangle=\int d^{3}$$ 如果 $T$ 是对角线，则 $$\hat{T}=\sum_{j}\langle j|T| j\rangle b_{j}^{\dagger}$$ 在最后一种情况下，多体运算符是 $$\left\langle n_{1} n_{2} \cdots|\hat{T}| n_{1} n_{2}\对。$$
$$\left\langle n_{1} n_{2} \cdots|\hat{T}| n_{1} n_{2} \cdots\right\rangle=\sum_{j}\langle j|T| j\rangle n_{j}$$ 这只是增加了动力学
我们可以通过引入非相对论量子场来重写这个单体算子 $\hat{T}$
$$\开始{对齐} \hat{\psi}(\vec{x}) & \equiv \sum_{j} \psi_{j}(\vec{x}) b_{j} \quad ; \text { 量子场 } \ \hat{\psi}^{\dagger}(\vec{x}) & \equiv \sum_{j} \psi_{j}^{*}(\vec{x}) b_{j}^{\dagger } \end{对齐}$$
那么一体算子就是
$$\hat{T}=\int d^{3} x \hat{\psi}^{\dagger}(\vec{x}) T \hat{\psi}(\vec{x})$$
假设一个有一个二体算子，比如所有对之间的势
$$V=\frac{1}{2} \sum_{i} \sum_{j} V\left(\left|\vec{x}{i}-\vec{x}{j}\right|\正确的）$$

Matlab代写