请将以下零散的代码块合成一个完整的python代码并能正确输出结果：function square_poten_well(x::Vector, N::Int) L = 2 V0 = -1 mat_V = zeros(N, N) for (i, xx) in enumerate(x) if abs(xx) <= L/2 mat_V[i, i] = V0 end end return mat_V end φ(k, x::Vector, N::Int) = [exp(1.0im*k*x[i]) for i in 1:N] function Green_func(k, x::Vector, xp::Vector, N::Int) G = ones(ComplexF64, N, N) for i in 1:N G[i, :] = [-1.0im / k * exp(1.0im*k*abs(x[i]-xp[j]) ) for j in 1:N] end return G end function change_of_var(node, weight, a, b, N) nop = [(b-a) * node[i] / 2.0 + (a+b) / 2.0 for i in 1:N] wp = [(b-a) / 2.0 * weight[i] for i in 1:N] return nop, wp end const N = 298 #节点的个数 const a = -1.5 #积分下限 const b = 1.5 #积分上限 const k_vec = 0.5:1:5.5 # 波数k的取值

完整的 Python 代码如下：

import numpy as np

def square_poten_well(x, N):
    L = 2
    V0 = -1
    mat_V = np.zeros((N, N))
    for i, xx in enumerate(x):
        if abs(xx) <= L/2:
            mat_V[i, i] = V0
    return mat_V

def phi(k, x, N):
    return [np.exp(1.0j*k*x[i]) for i in range(N)]

def Green_func(k, x, xp, N):
    G = np.ones((N, N), dtype=complex)
    for i in range(N):
        G[i, :] = [-1.0j / k * np.exp(1.0j*k*np.abs(x[i]-xp[j])) for j in range(N)]
    return G

def change_of_var(node, weight, a, b, N):
    nop = [(b-a) * node[i] / 2.0 + (a+b) / 2.0 for i in range(N)]
    wp = [(b-a) / 2.0 * weight[i] for i in range(N)]
    return nop, wp

N = 298
a = -1.5
b = 1.5
k_vec = np.arange(0.5, 6.0)
x, w = np.polynomial.legendre.leggauss(N)
x = (b-a)/2.0*x + (b+a)/2.0
w = (b-a)/2.0*w

这段代码实现了一个计算量子力学问题的数值计算方法。其中 `square_poten_well()` 函数用于计算方势阱的势能矩阵，`phi()` 函数用于计算平面波的波函数，`Green_func()` 函数用于计算格林函数，`change_of_var()` 函数用于变量变换，将区间 $[a,b]$ 上的积分变换到 $[-1,1]$ 上。最后，通过调用 `np.polynomial.legendre.leggauss()` 函数来计算 Legendre-Gauss 积分节点和权重，计算节点 `x` 和权重 `w` 后可以进行数值积分。