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%matplotlib inline
import matplotlib.pyplot as plt
import numpy as np
import scipy.optimize as opt
from scipy.optimize import minimize
The following potential is often used in Physics and other fields to describe symmetry breaking and is often known as the "hat potential":
$$ V(x) = -a x^2 + b x^4 $$Write a function hat(x,a,b)
that returns the value of this function:
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def hat(x,a,b):
return -a*x**2 + b*x**4
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assert hat(0.0, 1.0, 1.0)==0.0
assert hat(0.0, 1.0, 1.0)==0.0
assert hat(1.0, 10.0, 1.0)==-9.0
Plot this function over the range $x\in\left[-3,3\right]$ with $b=1.0$ and $a=5.0$:
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a = 5.0
b = 1.0
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x = np.linspace(-3,3,100);
plt.figure(figsize=(8,6))
plt.plot(x,hat(x,a,b));
plt.xlabel('x');
plt.ylabel('V(x)');
plt.title('Hat Potential');
plt.tick_params(axis='x',top='off',direction='out');
plt.tick_params(axis='y',right='off',direction='out');
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assert True # leave this to grade the plot
Write code that finds the two local minima of this function for $b=1.0$ and $a=5.0$.
scipy.optimize.minimize
to find the minima. You will have to think carefully about how to get this function to find both minima.
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#Finding the Left Minima
guess = (-2)
results = minimize(hat,guess,args=(a,b),method = 'SLSQP')
xL = results.x
print("Left Minima: x = " + str(xL[0]))
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#Finding the Right Minima
guess = (2)
results = minimize(hat,guess,args=(a,b),method = 'SLSQP')
xR = results.x
print("Right Minima: x = " + str(xR[0]))
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x = np.linspace(-3,3,100);
plt.figure(figsize=(8,6))
plt.plot(x,hat(x,a,b));
plt.xlabel('x');
plt.ylabel('V(x)');
plt.title('Hat Potential with Minimums');
plt.tick_params(axis='x',top='off',direction='out');
plt.tick_params(axis='y',right='off',direction='out');
plt.plot(xL, hat(xL,a,b), marker='o', linestyle='',color='red');
plt.plot(xR, hat(xR,a,b), marker='o', linestyle='',color='red');
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assert True # leave this for grading the plot
To check your numerical results, find the locations of the minima analytically. Show and describe the steps in your derivation using LaTeX equations. Evaluate the location of the minima using the above parameters.
To Find the minima we set $\frac{\partial V(x)}{\partial x}$ = 0
$$ \frac{\partial V(x)}{\partial x} = -2ax + 4bx^3 = 0 $$Divide by x on both sides. We know x = 0 is not a minimum. It is a local maximum.
$$ \frac{\partial V(x)}{\partial x} = -2a + 4bx^2 = 0 $$$$ 2a = 4bx^2 $$$$ x^2 = \frac{2a}{4b} $$$$ x = \pm\sqrt{\frac{2a}{4b}} $$$$ x = \pm \sqrt{\frac{5}{2}} = \pm 1.58114 $$
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