# 摄动理论

$A=\epsilon^0 A_0 + \epsilon^1 A_1 + \epsilon^2 A_2 + \cdots\,\!$

## 一阶无简并摄动理论

### 一阶本征值修正

$Dg(x)=\lambda g(x) \,\!$(1)

$D=D^{(0)}+\epsilon D^{(1)} \,\!$

$D^{(0)} f^{(0)}_i (x)=\lambda^{(0)}_i f^{(0)}_i (x) \,\!$

$\int f^{(0)}_i (x) f^{(0)}_j (x) \,dx = \delta_{ij}\,\!$

$g(x)=f^{(0)}_n (x) + \mathcal{O}(\epsilon)\,\!$

$\lambda=\lambda^{(0)}_n + \mathcal{O}(\epsilon)\,\!$

$g(x)=\sum_m c_m f^{(0)}_m (x)\,\!$(2)

$\lambda^{(0)}_n c_n+ \epsilon \sum_m c_m \int f^{(0)}_n(x) D^{(1)} f^{(0)}_m(x)\,dx =\lambda c_n\,\!$

$A_{nm} = \lambda^{(0)}_n\delta_{nm} + \epsilon \int f^{(0)}_n(x) D^{(1)} f^{(0)}_m(x)\,dx \,\!$

$\lambda=\lambda^{(0)}_n + \epsilon \int f^{(0)}_n(x) D^{(1)} f^{(0)}_n(x)\,dx \,\!$(3)

$\lambda^{(1)}_n=\int f^{(0)}_n(x) D^{(1)} f^{(0)}_n(x)\,dx \,\!$

### 一阶本征函数修正

$g(x)=f^{(0)}_n(x) + \epsilon f^{(1)}_n(x)\,\!$(4)

$\left(D^{(0)} +\epsilon D^{(1)}\right) \left( f^{(0)}_n(x) + \epsilon f^{(1)}_n(x) \right) = ( \lambda^{(0)}_n+ \epsilon \lambda^{(1)}_n) \left( f^{(0)}_n(x) + \epsilon f^{(1)}_n(x) \right) \,\!$

$D^{(1)} f^{(0)}_n (x) +D^{(0)}f^{(1)}_n (x)=\lambda^{(0)}_n f^{(1)}_n (x) +\lambda^{(1)}_n f^{(0)}_n (x)\,\!$(5)

$f^{(1)}_n (x)=\sum_{i\ne n} C_i f^{(0)}_i (x)\,\!$(6)

$(D^{(1)} - \lambda^{(1)}_n) f^{(0)}_n (x)=\lambda^{(0)}_n \sum_{i\ne n} C_i f^{(0)}_i (x) - D^{(0)}\sum_{i\ne n} C_i f^{(0)}_i (x)=\sum_{i\ne n} (\lambda^{(0)}_n - \lambda^{(0)}_i) C_i f^{(0)}_i (x)\,\!$

$\int\, f^{(0)}_j (x) D^{(1)}f^{(0)}_n (x)\, dx=\sum_{i\ne n} (\lambda^{(0)}_n - \lambda^{(0)}_i) C_i \int\, f^{(0)}_j (x) f^{(0)}_i (x)=(\lambda^{(0)}_n - \lambda^{(0)}_j) C_j \,\!$

$f^{(1)}_n(x) = \sum_{m \ne n} \frac {f^{(0)}_m (x)} {\lambda^{(0)}_n- \lambda^{(0)}_m} \int\, f^{(0)}_m(y) D^{(1)} f^{(0)}_n(y) \,dy\,\!$