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    Special Function Hermite Polinomial 1

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    dharbiswarup1

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    © All Rights Reserved
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    Special Function Hermite Polinomial 1

    Uploaded by

    dharbiswarup1
    5 views4 pages

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    © © All Rights Reserved

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    Copyright:

    © All Rights Reserved
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    5 views4 pages

    Special Function Hermite Polinomial 1

    Uploaded by

    dharbiswarup1

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    of 4

    Special Function(Hermite Polinomial)

    15/04/2024

    1 Hermite Polinomial
    Hermite Differntial equation is given as :

    d2 y dy
    − 2x + 2ny = 0 (1)
    dx2 dx
    The solution that sattisfies the above equation is known as Hermite Polinomial.The Rodrigue’s
    formula is given as :

    2 dn −x2
    Hn (x) = (−1)n ex e (2)
    dxn
    The Orthogonality Condition for Hermite Polinomial is given as:

    Z +∞
    2
    e−x Hm (x).Hn (x)dx = 0 f or n ̸= m (3)
    −∞

    Z +∞
    2
    e−x Hm (x).Hn (x)dx = 2n n!
    p
    (π) f or n=m (4)
    −∞

    1.1 Program:
    1.1.1 Question1.
    Verify the Orthogonality relation for Hermite Polinomial.

    Source Code:
    import numpy as np
    import math
    import matplotlib.pyplot as plt
    from scipy.special import hermite as H
    from scipy.integrate import quad

    m = int(input('Enter the value m = '))


    n = int(input('Enter the value n = '))
    def f(x): return np.exp(-x**2)*H(n)(x)*H(m)(x)
    x = np.linspace(-10,10,100)

    15
    LHS = quad(f,-np.inf,np.inf)
    if m == n:
    RHS = (2**n)*(math.factorial(n)*np.sqrt(np.pi))
    else:
    RHS = 0
    print('LHS = ',LHS[0],'RHS = ',RHS)
    if abs(LHS[0] - RHS)<0.01:
    print('The Orthogonality relation of Legendre plinomial is verified')
    else:
    print('Not Verified')

    Enter the value m = 3


    Enter the value n = 3
    LHS = 85.07778484346483 RHS = 85.07778484346477
    The Orthogonality relation of Legendre plinomial is verified

    1.1.2 Question 2.
    Verify the following recurrence relation for Hermite Polinomial.
    ′
    Hn+1 (x) = 2xHn (x) − 2nHn−1 (x)

    Source Code:
    [22]: #Recurrence relation for Hermite Polinomial 2nd

    from numpy import*


    from scipy.special import hermite as h
    from matplotlib.pyplot import*
    x=linspace(-1,1,50)
    n= int(input ())
    LHS= h(n+1)(x)
    RHS = 2*x*h(n)(x) -2*n*h(n-1)(x)
    plot(x, LHS, '-', label = 'LHS', color='black')
    plot(x, RHS, '*', label = 'RHS', color='black')
    legend()
    xlabel('position(x)')
    ylabel('$H\'_{n+1}(x)$')
    grid()
    title('Verifying the Recurrence relation for Hermite Polynomial: \n␣
    ,→$H\'_{n+1}(x) = 2xH_{n}(x)-2nH_{n-1}(x)$')

    plot()

    16
    1.1.3 Question 3.
    Consider the wavefunction of 1D Harmonic Oscillator:
     1/4  2
    1 1 −x
    ψn (x) = p exp Hn (x)
    (2n n!) π 2

    mω
    Where Hn (x) is the Hermite polinomial and we consider = 1.Plot the wavefunction for n = 0,
    ℏ
    1 and 2 in a single graph.

    Source Code:
    [23]: import numpy as np
    import math
    import matplotlib.pyplot as plt
    from scipy.special import hermite as H
    from scipy.integrate import quad
    def psi(n,x):
    return 1/np.sqrt((2**n)*math.factorial(n))*(1/np.pi)**(1/4)*np.exp(-x**2/
    ,→2)*H(n)(x)

    x = np.linspace(-5,5,100)
    n = np.array([0,1,2])
    for i in n:
    plt.plot(x,psi(i,x),linestyle=(0,(2,i*i)), label =r'$\psi_n(x) \;
    ,→at\quadn=%f$'%i, color = 'black')

    plt.legend(bbox_to_anchor=(1.00, 1),ncol=1)
    plt.grid()
    plt.title(r'Plot of $\psi_n(x)$ for n = 0,1,2')

    17
    plt.ylabel(r'$\psi_n(x)$')
    plt.xlabel('Position(x)')
    plt.axhline(y=0, linestyle='-')
    plt.show()

    18

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