Question

我需要在薛定谔方程中更改我的绘图的缩放比例，y轴显示理论计算与我们的理论计算之间的差异，差异大约为0.01％。所以在情节我得到的规模不小，以显示差异。这是我项目的代码。

# -*- coding: utf-8 -*-
"""
Created on Sat Nov 05 12:25:14 2016

@author: produce
"""
from __future__ import print_function
import numpy as np
import matplotlib.pyplot as plt

#
c = .5 / 500  # c = delta x
x = np.arange(0, .5, c)  # creates array of argument values from 0 to 1/2 in increments
#                       of delta x = c
psi = np.zeros(len(x))  # creates array of zeros which will be replaced by y values

k = 20  # starting energy for calculator of E
ans = 0  # The value of k, when we have y as between 0.004 and 0
ansPsi = 0
diff = 0.001
increment = 0.0001
done = False
while 1:
    # print k
    psi[0] = 1
    psi[1] = 1
    for i in range(0, len(x) - 2):
        psi[i + 2] = psi[i + 1] + (psi[i + 1] - psi[i]) - 2 * k * c * c * psi[i]
    # plt.plot(x,psi)
    # print(x,psi)
    # print (psi[i+2]--->)
    if (float(psi[i + 2]) < 0.004 and float(psi[i + 2]) > 0):
        ans = k
        ansPsi = psi[i + 2]
        # print ("NOW ENTERING INNER LOOP")
        while 1:  # would be an infinite loop, but have a break statement
            # k = k - 0.00001
            k = k + increment
            for i in range(0, len(x) - 2):
                psi[i + 2] = psi[i + 1] + (psi[i + 1] - psi[i]) - 2 * k * c * c * psi[i]
            plt.plot(x, psi, 'r')   #red solid line
            if (psi[i + 2] > ansPsi or psi[i + 2] < 0):
                done = True
                break
            else:
                ansPsi = psi[i + 2]
                ans = k
                # print (k, psi[i+2])

    if done:
        break
    k = k - diff

print("Value of k:", ans, "Value of Y:", ansPsi)  # prints our answer for energy and psi[1/2]
k1 = 10  # 1st Higher Energy Value
k2 = 7  # 2nd Higher Energy Value
k3 = 3  # 1st Lower Energy Value
k4 = 1  # 2nd Lower Energy Value
kt = np.pi * np.pi * .5  # theoretical value

psi1 = np.zeros(len(x))
psi1[0] = 1
psi1[1] = 1
for i in range(0, len(x) - 2):
    psi1[i + 2] = psi1[i + 1] + (psi1[i + 1] - psi1[i]) - 2 * k1 * c * c * psi1[i]


# psi2 = np.zeros(len(x))
# psi2[0] = 1
# psi2[1] = 1
# for i in range (0,len(x)-2):
#   psi2[i+2] = psi2[i+1] + (psi2[i+1] - psi2[i]) - 2*k2*c*c*psi2[i]
# plt.plot(x,psi2,'k')

# psi3 = np.zeros(len(x))
# psi3[0] = 1
# psi3[1] = 1
# for i in range (0,len(x)-2):
#   psi3[i+2] = psi3[i+1] + (psi3[i+1] - psi3[i]) - 2*k3*c*c*psi3[i]
# plt.plot(x,psi3,'p')

psi4 = np.zeros(len(x))
psi4[0] = 1
psi4[1] = 1
for i in range(0, len(x) - 2):
    psi4[i + 2] = psi4[i + 1] + (psi4[i + 1] - psi4[i]) - 2 * k4 * c * c * psi4[i]

plt.plot(x, psi, 'r-', label='Corrected Energy')

psiT = np.zeros(len(x))
psiT[0] = 1
psiT[1] = 1
for i in range(0, len(x) - 2):
    psiT[i + 2] = psiT[i + 1] + (psiT[i + 1] - psiT[i]) - 2 * kt * c * c * psiT[i]
plt.plot(x, psiT, 'b-', label='Theoretical Energy')
plt.ylabel("Value of Psi")
plt.xlabel("X value from 0 to 0.5")
plt.title("Schrodingers equation for varying inital energy")
plt.legend(loc=3) 
plt.yscale()
plt.show()

Answer 1

您共享的代码失败，因为plt.yscale()需要参数。我只是评论说出来了。

因为理论能量曲线与校正后的能量曲线相差很小，所以无法缩放y轴和仍然可以在x的整个范围内看到两条曲线（即 - 来自0到0.5）。相反，也许你应该绘制两条曲线的差异？

plt.plot(x, psiT-psi)
plt.title("Size of Correction for Varying Initial Energy")
plt.ylabel(r"$\Delta$E")
plt.xlabel("X value from 0 to 0.5")
plt.show()

此外，在x和y标签上添加一些单位可能会很不错。：）

改变薛定谔方程的比例

1 个答案: