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# 物理代考| Fermions 量子力学代写

## 物理代写

$7.2$ Fermions
There is another class of particles, for example, electrons and nucleons, that show a very different behavior. These are the fermions that obey the Pauli exclusion principle, which states that
One cannot put two identical fermions in to the same state.
We have to go back and build this principle into the fermion operators in the abstract occupation number space. Consider the operators in one mode $\left(a, a^{\dagger}\right)$. Instead of imposing commutation relations, we impose anticommutation relations
$$\left{a, a^{\dagger}\right} \equiv a a^{\dagger}+a^{\dagger} a=1$$
$$\begin{gathered} {a, a}=a a+a a=0 \ \left{a^{\dagger}, a^{\dagger}\right}=a^{\dagger} a^{\dagger}+a^{\dagger} a^{\dagger}=0 \end{gathered}$$
pation number space for fermions are
$(7.22)$
The only challenge now is keeping track of signs, since the operators for different modes anti-commute. One has
\begin{aligned} a_{s}\left|n_{0} \cdots n_{s-1} 1_{s} n_{s+1} \cdots\right\rangle &=(-1)^{S_{s}}\left|n_{0} \cdots n_{s-1} 0_{s} n_{s+1} \cdots\right\rangle \ a_{s}^{\dagger}\left|n_{0} \cdots n_{s-1} 0_{s} n_{s+1} \cdots\right\rangle &=(-1)^{S_{s}}\left|n_{0} \cdots n_{s-1} 1_{s} n_{s+1} \cdots\right\rangle \ S_{s} &=n_{0}+n_{1}+\cdots+n_{s-1} \end{aligned}
The trick is to keep the operators paired until they reach the state on which
the trick is to keep the operators paired until they reach the state on which
For fermions, the field operators now become $$\hat{\psi}(\vec{x}) \equiv \sum_{j} \psi_{j}(\vec{x}) a_{j} \quad ; \text { quantum field }$$
$$\hat{\psi}^{\dagger}(\vec{x}) \equiv \sum_{j} \psi_{j}^{*}(\vec{x}) a_{j}^{\dagger}$$
Critra see below.
${ }^{4}$ For the spin- $1 / 2$ electrons and nucle
little more complicated – see below. $66 \quad$ Introduction to Quantum Mechanics
The kinetic and potential energies in this abstract occupation number space
again take the form $^{5}$
\begin{aligned} &\hat{T}=\int d^{3} x \hat{\psi}^{\dagger}(\vec{x}) T \hat{\psi}(\vec{x}) \ &\hat{V}=\frac{1}{2} \int d^{3} x \int d^{3} y \hat{\psi}^{\dagger}(\vec{y}) \hat{\psi}^{\dagger}(\vec{x}) V(|\vec{x}-\vec{y}|) \hat{\psi}(\vec{x}) \hat{\psi}(\vec{y}) \end{aligned}
and the many-body Schrödinger equation is again
\begin{aligned} i \hbar \frac{\partial}{\partial t}|\Psi(t)\rangle &=\hat{H}|\Psi(t)\rangle & & ; \text { Schrödinger eqn } \ \hat{H} &=\hat{T}+\hat{V} & & \end{aligned}
The quantum many-body problem is discussed in detail in [Fetter and

## 物理代考

$7.2$ 费米子

$$\left{a, a^{\dagger}\right} \equiv aa^{\dagger}+a^{\dagger} a=1$$
$$\开始{聚集} {a, a}=a a+a a=0 \ \left{a^{\dagger}, a^{\dagger}\right}=a^{\dagger} a^{\dagger}+a^{\dagger} a^{\dagger}=0 \结束{聚集}$$

$(7.22)$

$$\开始{对齐} a_{s}\left|n_{0} \cdots n_{s-1} 1_{s} n_{s+1} \cdots\right\rangle &=(-1)^{S_{s}}\left |n_{0} \cdots n_{s-1} 0_{s} n_{s+1} \cdots\right\rangle \ a_{s}^{\dagger}\left|n_{0} \cdots n_{s-1} 0_{s} n_{s+1} \cdots\right\rangle &=(-1)^{S_{ s}}\left|n_{0} \cdots n_{s-1} 1_{s} n_{s+1} \cdots\right\rangle \ S_{s} &=n_{0}+n_{1}+\cdots+n_{s-1} \end{对齐}$$

$$\hat{\psi}^{\dagger}(\vec{x}) \equiv \sum_{j} \psi_{j}^{*}(\vec{x}) a_{j}^{\dagger}$$

${ }^{4}$ 对于自旋- $1 / 2$ 电子和原子核

$$\开始{对齐} &\hat{T}=\int d^{3} x \hat{\psi}^{\dagger}(\vec{x}) T \hat{\psi}(\vec{x}) \ &\hat{V}=\frac{1}{2} \int d^{3} x \int d^{3} y \hat{\psi}^{\dagger}(\vec{y}) \帽子{\psi}^{\dagger}(\vec{x}) V(|\vec{x}-\vec{y}|) \hat{\psi}(\vec{x}) \hat{\ psi}(\vec{y}) \end{对齐}$$

$$\开始{对齐} i \hbar \frac{\partial}{\partial t}|\Psi(t)\rangle &=\hat{H}|\Psi(t)\rangle & & ; \文本 { 薛定谔方程 } \ \hat{H} &=\hat{T}+\hat{V} & & \end{对齐}$$

Matlab代写