物理の方程式備忘録

忘れちゃう数学の関数周りの事項の備忘録

# 長さゲージと速さゲージの間のゲージ変換
## 長さゲージ
$$
\begin{align}
\mathrm{i} \frac{\partial}{\partial t} \tilde{\psi}_i(t)=
\left[\frac{1}{2}\hat{p}^2 + \hat{v} + \hat{x}E(t) \right] \tilde{\psi}_i(t)
\end{align}
$$
$$
\hat{\tilde{\rho}}(t) = 2 \sum_i \tilde{\psi}_i(t) \tilde{\psi}_i^\dagger (t)
$$

$$
\hat{O}(t) = 2\sum_i \left\langle \tilde{\psi}_i(t) \left| \hat{\tilde{O}}(t) \right| \tilde{\psi}_i(t) \right\rangle
$$

$$
\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}t} \hat{\tilde{\rho}} = \left[ 
\left\\{\frac{1}{2}\hat{p}^2 + \hat{v} + \hat{x}E(t) \right\\} , \hat{\tilde{\rho}}\right] + \hat{\tilde{s}}
$$
## ゲージ変換
$$
\psi_i(t) = e^{\mathrm{i}A(t) \hat{x}} \tilde{\psi}_i(t)
$$

## 速度ゲージ
$$
\begin{align}
\mathrm{i} \frac{\partial}{\partial t} \psi_i(t)=
\left[\frac{1}{2}\left(\hat{p} + A(t)\right)^2 + e^{-\mathrm{i}A(t) \hat{x}}\hat{v} e^{+\mathrm{i}A(t) \hat{x}}  \right] \psi_i(t)
\end{align}
$$
$$\hat{\rho}(t) = 2 \sum_i \psi_i(t) \psi_i^\dagger (t) 
= e^{+\mathrm{i}A(t) \hat{x}} \hat{\tilde{\rho}}(t) e^{-\mathrm{i}A(t) \hat{x}}
$$

$$
O(t) = 2\sum_i \left\langle \psi_i(t) \left| \hat{O}(t) \right| \psi_i(t) \right\rangle, \quad \hat{O}(t) = e^{+\mathrm{i}A(t) \hat{x}}\hat{\tilde{O}}(t) e^{-\mathrm{i}A(t) \hat{x}}
$$

$$
\mathrm{i}\frac{\mathrm{d}}{\mathrm{d}t} \hat{\rho} = \left[ 
\left\\{\frac{1}{2}\left(\hat{p} + A(t)\right)^2 + e^{-\mathrm{i}A(t) \hat{x}}\hat{v} e^{+\mathrm{i}A(t) \hat{x}} \right\\} , \hat{\rho}\right] + \hat{s}
$$
$$
\hat{s} = e^{+\mathrm{i}A(t) \hat{x}} \hat{\tilde{s}} e^{-\mathrm{i}A(t) \hat{x}}
$$

# Bloch軌道

$$
\frac{1}{i}\frac{\partial}{\partial k} e^{ikx} u_{bk}(x)= e^{ikx}\left(x +  \frac{1}{i}\frac{\partial}{\partial k}\right)u_{bk}(x)
$$


# Hartree-Fock方程式

## 軌道が一般に両方のスピンの成分を独立して持つとき
$$
E_{\mathrm{tot}} = E_{\mathrm{kin}}+E_{\mathrm{pot}}+E_{\mathrm{MF}}
$$

$s=\uparrow, \downarrow$あるいは$s=\alpha, \beta$の二値。

$$
\begin{align}
E_{\mathrm{kin}}+E_{\mathrm{pot}} = \sum_{ss'} \iint \mathrm{d}^3r \mathrm{d}^3r' \gamma (\mathbf{r}s, \mathbf{r}'s') h(\mathbf{r}'s', \mathbf{r}s) \\\\
\frac{\delta ( E_{\mathrm{kin}}+E_{\mathrm{pot}})}{\delta \gamma(\mathbf{r}s,\mathbf{r}'s')} = \\
h(\mathbf{r}'s', \mathbf{r}s)
\end{align}
$$

$$
\begin{align}
E_{\mathrm{MF}} [\gamma] \\
= \\
E_{\mathrm{H}}[\gamma] + E_{\mathrm{F}}[\gamma], \\\\
\gamma = \sum_i^N \phi_i \phi_i^\dagger, \quad \\
\left\langle \left. \mathbf{r} s \right| \phi_i \right\rangle = \varphi_{i}(\mathbf{r} s)
\end{align}
$$

$$
\begin{align}
E_{\mathrm{F}}[\gamma] \\
= \\
-\frac{1}{2} \sum_{ss'} \iint \mathrm{d}^3r \mathrm{d}^3 r' \gamma(\mathbf{r} s, \mathbf{r}' s') w(\mathbf{r} s, \mathbf{r}' s') \gamma(\mathbf{r}' s', \mathbf{r} s) 
\end{align}
$$

## 軌道が偶数本あって、奇数番目がアップスピンで、偶数番目がダウンスピン、かつ$(n-1)/2$番目と$n/2$番目の軌道関数が同じ形の場合

ポテンシャルエネルギーがスピンに対して対角的な場合
$$
\begin{align}
E_{\mathrm{kin}}+E_{\mathrm{pot}} =  \iint \mathrm{d}^3r \mathrm{d}^3r' \rho (\mathbf{r}, \mathbf{r}') h(\mathbf{r}', \mathbf{r}) \\\\
\frac{\delta (E_{\mathrm{kin}}+E_{\mathrm{pot}})}{\delta \rho(\mathbf{r},\mathbf{r}')} = \\
h(\mathbf{r}', \mathbf{r})
\end{align}
$$

$$
\begin{align}
E_{\mathrm{kin}}+E_{\mathrm{pot}} =  2 \sum_{i} \iint \mathrm{d}^3r \mathrm{d}^3r' \varphi_i (\mathbf{r}) \varphi_i^* (\mathbf{r}') h(\mathbf{r}', \mathbf{r}) \\\\
\frac{\delta (E_{\mathrm{kin}}+E_{\mathrm{pot}})}{\delta \\varphi_i^* (\mathbf{r})} = \\
2 \int \mathrm{d}^3r' h(\mathbf{r}, \mathbf{r}') \varphi_i (\mathbf{r}') 
\end{align}
$$


$$
\begin{align}
E_{\mathrm{MF}} [\rho] \\
= \\
E_{\mathrm{H}}[\rho] + E_{\mathrm{F}}[\rho], \\\\
\rho = 2\sum_i^{N/2} \varphi_i \varphi_i^\dagger, \quad \\
\left\langle \left. \mathbf{r} \right| \varphi_i \right\rangle = \varphi_{i}(\mathbf{r})
\end{align}
$$
粒子間相互作用$w$がスピンに依存しない場合$w(\mathbf{r} s, \mathbf{r}' s') \to w(\mathbf{r}, \mathbf{r}')$
$$
\begin{align}
E_{\mathrm{F}}[\rho] \\
= \\
-\frac{1}{4} \iint \mathrm{d}^3r \mathrm{d}^3 r' \rho(\mathbf{r}, \mathbf{r}') w(\mathbf{r} , \mathbf{r}' ) \rho(\mathbf{r}', \mathbf{r} ) 
\end{align}
$$

$$
\begin{align}
\frac{\delta E_{\mathrm{F}}}{\delta \rho(\mathbf{r}, \mathbf{r}')} \\
= \\
-\frac{1}{2}  w(\mathbf{r} , \mathbf{r}' ) \rho(\mathbf{r}', \mathbf{r} ) 
\end{align}
$$

$$
\begin{align}
E_{\mathrm{F}}[\varphi, \varphi^* ] \\
= \\
-\sum_{ij} \iint \mathrm{d}^3r \mathrm{d}^3 r' \varphi_i(\mathbf{r})\varphi_i^*( \mathbf{r}') w(\mathbf{r} , \mathbf{r}' ) \varphi_j(\mathbf{r}') \varphi_j^{}(\mathbf{r}) 
\end{align}
$$

(なぜか最後の軌道関数に複素共役$\bullet^*$をつけるとバグるので外してある)

$$
\begin{align}
\frac{\delta E_{\mathrm{F}}}{\delta \varphi_i^* (\mathbf{r})} \\
= \\
-2 \sum_j  \varphi_j(\mathbf{r}) \int \mathrm{d}^3r   w(\mathbf{r} , \mathbf{r}' ) \varphi_j^*( \mathbf{r}' ) \varphi_i(\mathbf{r}') \\
= \\
-\int \mathrm{d}^3r'   w(\mathbf{r} , \mathbf{r}' ) \rho(\mathbf{r}, \mathbf{r}' ) \varphi_i(\mathbf{r}') 
\end{align}
$$

$$
\begin{align}
E_{\mathrm{H}}[\rho ] \\
= \\
\frac{1}{2}\sum_{ij} \iint \mathrm{d}^3r \mathrm{d}^3 r' \rho(\mathbf{r}, \mathbf{r}) w(\mathbf{r} , \mathbf{r}' ) \rho(\mathbf{r}', \mathbf{r}') 
\end{align}
$$

$$
\begin{align}
\frac{\delta E_{\mathrm{H}}}{\delta \rho(\mathbf{r}, \mathbf{r}')} \\
= \\
\delta (\mathbf{r}', \mathbf{r})\int \mathrm{d}^3 r''  w(\mathbf{r} , \mathbf{r}'' ) \rho(\mathbf{r}'', \mathbf{r} '') 
\end{align}
$$


$$
\begin{align}
E_{\mathrm{H}}[\varphi, \varphi^* ] \\
= \\
2\sum_{ij} \iint \mathrm{d}^3r \mathrm{d}^3 r' \varphi_i(\mathbf{r})\varphi_i^*( \mathbf{r}) w(\mathbf{r} , \mathbf{r}' ) \varphi_j(\mathbf{r}') \varphi_j^{}(\mathbf{r}') 
\end{align}
$$

(なぜか最後の軌道関数に複素共役$\bullet^*$をつけるとバグるので外してある)

$$
\begin{align}
\frac{\delta E_{\mathrm{H}}}{\delta \varphi_i^* (\mathbf{r})} \\
= \\
4 \varphi_i(\mathbf{r}) \sum_j   \int \mathrm{d}^3r   w(\mathbf{r} , \mathbf{r}' ) \varphi_j( \mathbf{r}' ) \varphi_j^*(\mathbf{r}') \\
= \\
2 \varphi_i(\mathbf{r}) \int \mathrm{d}^3r'   w(\mathbf{r} , \mathbf{r}' ) \rho(\mathbf{r}', \mathbf{r}' ) 
\end{align}
$$

$$
\begin{align}
\frac{\delta E_{\mathrm{tot}}}{\delta \\varphi_i^* (\mathbf{r})} = 2 \left[ \int \mathrm{d}^3r' h(\mathbf{r}, \mathbf{r}') \varphi_i (\mathbf{r}') + \varphi_i(\mathbf{r}) \int \mathrm{d}^3r'   w(\mathbf{r} , \mathbf{r}' ) \rho(\mathbf{r}', \mathbf{r}' ) - \frac{1}{2}\int \mathrm{d}^3r'   w(\mathbf{r} , \mathbf{r}' ) \rho(\mathbf{r}, \mathbf{r}' ) \varphi_i(\mathbf{r}')  \right] \\
= \\
2 \int \mathrm{d}^3r' h_{\mathrm{HF}}(\mathbf{r}, \mathbf{r}') \varphi_i (\mathbf{r}')
\end{align}
$$

$$
\begin{align}
h_{\mathrm{HF}}(\mathbf{r}, \mathbf{r}') = \\
h(\mathbf{r}, \mathbf{r}') + \left[ \int \mathrm{d}^3r''   w(\mathbf{r} , \mathbf{r}' ) \rho(\mathbf{r}'', \mathbf{r}'' )\right]\delta(\mathbf{r},\mathbf{r}') - \frac{1}{2}  w(\mathbf{r} , \mathbf{r}' ) \rho(\mathbf{r}, \mathbf{r}' ) 
\end{align}
$$

$$
\begin{align}
h_{\mathrm{HF}}(\mathbf{r}', \mathbf{r}) = \\
\frac{\delta E_{\mathrm{tot}}}{\delta \rho(\mathbf{r}, \mathbf{r}')} 
\end{align}
$$

# 水素原子の波動関数について
## 規格化された1s軌道関数

$$
\begin{align}
\psi(\mathbf{r}) = \frac{1}{\sqrt{\pi}a_B^3}e^{-r/a_B} \\\\
\left\langle \left. \psi \right| \psi \right\rangle = \int_0^{2\pi} \\! \mathrm{d}\phi \int_0^\pi \\! \mathrm{d}\theta \int_0^\infty \\! \mathrm{d} r \ r^2 \sin \theta |\psi(\mathbf{r})|^2 = 1
\end{align}
$$

Wolfram Alpha code: `integrate(integrate(integrate(r^2*sin(phi)*(1/sqrt(pi*a_0^3)*exp(-r/a_0))^2,r,0,inf),phi,0,pi),theta,0,2*pi)`


### 動径関数に対する半径の期待値
$$
\begin{align}\int_0^{2\pi} \\! \mathrm{d}\phi \int_0^\pi \\! \mathrm{d}\theta \int_0^\infty \\! \mathrm{d} r \ r^3 \sin \theta |\psi(\mathbf{r})|^2 = \frac{3}{2}a_B\end{align}
$$
Wolfram Alpha code: `integrate(integrate(integrate(r*r^2*sin(phi)*(1/sqrt(pi*a_0^3)*exp(-r/a_0))^2,r,0,inf),phi,0,pi),theta,0,2*pi)`

### 平均二乗半径

$$
\begin{align}
\left\langle \mathbf{r}^2 \right\rangle =  \int_0^{2\pi} \\! \mathrm{d}\phi \int_0^\pi \\! \mathrm{d}\theta \int_0^\infty \\! \mathrm{d} r \ r^4 \sin \theta |\psi(\mathbf{r})|^2 = 3 a_B^2
\end{align}
$$
つまり$\sqrt{\left\langle \mathbf{r}^2 \right\rangle} = \sqrt{3}a_B$

Wolfram Alpha code: `integrate(integrate(integrate(r^2*r^2*sin(phi)*(1/sqrt(pi*a_0^3)*exp(-r/a_0))^2,r,0,inf),phi,0,pi),theta,0,2*pi)`

## 水素様原子
https://en.wikipedia.org/wiki/Hydrogen-like_atom

クーロンポテンシャル
$$
V(r;\mu,\epsilon_r)=\frac{1}{4\pi \epsilon_0 \epsilon_r}\frac{1}{r}
$$

半径(Bohr半径: $a_B$)
$$
a(\mu,\epsilon_r)= \frac{\epsilon_r}{\mu}a_B
$$

束縛エネルギー(Hartree: $E_H$)
$$
E_n(\mu, \epsilon_r) = \frac{\mu}{\epsilon_r^2} \left(\frac{1}{2}E_H \right)\frac{1}{n^2}
$$