import sympy as sp
from IPython.display import Image, display

x, a, phi, c1, c2, c3, alpha, beta = sp.symbols("x a, phi C_1 C_2 C_3 alpha beta")
m, hbar, E, k, v0 = sp.symbols("m hbar E k V_0", positive=True)
psi = sp.Function("psi")
V = sp.Function("V")

def get_matches(expr, fn, x):
    A = sp.Wild("A")
    B = sp.Wild("B")
    
    pattern = A * fn(B*x)
    matches = []

    for term in sp.Add.make_args(expr):
        m = term.match(pattern)
        if m:
            matches.append((m[A], m[B]))

    return matches

In the finite square well the two regions outside x < -a and x > a have a constant potential V0.

Now the Schrodinger equation is....

We define another constant $ k = \frac{\hbar^2 }{2 m} $

ks = hbar**2 / (2 * m)
ks

$\displaystyle \frac{\hbar^{2}}{2 m}$

eq = sp.Eq(
    -k * sp.diff(psi(x), x, 2) + V(x)*psi(x),
    E * psi(x)
)
display("The Shröedinger Equation")
display(eq)
display(ks)
eq.subs(k, ks)

'The Shröedinger Equation'

$\displaystyle - k \frac{d^{2}}{d x^{2}} \psi{\left(x \right)} + V{\left(x \right)} \psi{\left(x \right)} = E \psi{\left(x \right)}$

$\displaystyle \frac{\hbar^{2}}{2 m}$

$\displaystyle - \frac{\hbar^{2} \frac{d^{2}}{d x^{2}} \psi{\left(x \right)}}{2 m} + V{\left(x \right)} \psi{\left(x \right)} = E \psi{\left(x \right)}$

potential = v0

eq = sp.Eq(
   -k * sp.diff(psi(x), x, 2) + v0*psi(x),
    E * psi(x)
)
psi1 = sp.dsolve(eq)
display(eq)
display(psi1)
display(str(psi1))

$\displaystyle V_{0} \psi{\left(x \right)} - k \frac{d^{2}}{d x^{2}} \psi{\left(x \right)} = E \psi{\left(x \right)}$

$\displaystyle \psi{\left(x \right)} = C_{1} e^{- \frac{x \sqrt{- E + V_{0}}}{\sqrt{k}}} + C_{2} e^{\frac{x \sqrt{- E + V_{0}}}{\sqrt{k}}}$

'Eq(psi(x), C1*exp(-x*sqrt(-E + V_0)/sqrt(k)) + C2*exp(x*sqrt(-E + V_0)/sqrt(k)))'

matches = get_matches(psi1.rhs, sp.exp, x)
display(matches)
#psi1 = psi1.args[1].args[0]
a1,b1 = matches[1]

psi1 = c1*sp.exp(b1*x)
display(psi1)

[(C1, -sqrt(-E + V_0)/sqrt(k)), (C2, sqrt(-E + V_0)/sqrt(k))]

$\displaystyle C_{1} e^{\frac{x \sqrt{- E + V_{0}}}{\sqrt{k}}}$

While inside the finite square well the potential is zero.

eq = sp.Eq(
    -k * sp.diff(psi(x), x, 2),
    E * psi(x)
)
psi2 = sp.dsolve(eq)
display(eq)
display(psi2)

$\displaystyle - k \frac{d^{2}}{d x^{2}} \psi{\left(x \right)} = E \psi{\left(x \right)}$

$\displaystyle \psi{\left(x \right)} = C_{1} \sin{\left(\frac{\sqrt{E} x}{\sqrt{k}} \right)} + C_{2} \cos{\left(\frac{\sqrt{E} x}{\sqrt{k}} \right)}$

Boundary Conditions

# Here I match the sin function but then I turn it into a sin with a phi phase
# so that I can incorporate both results into one.

matches = get_matches(psi2.rhs, sp.sin, x)
c2 =  sp.symbols("C_2")
display(matches)
a1,b1 = matches[0]
psi2 = c2 * sp.sin(b1*x + phi)
display(psi2)

[(C1, sqrt(E)/sqrt(k))]

$\displaystyle C_{2} \sin{\left(\frac{\sqrt{E} x}{\sqrt{k}} + \phi \right)}$

display(psi1.subs(x,-a/2))
display(psi2.subs(x,-a/2))

$\displaystyle C_{1} e^{- \frac{a \sqrt{- E + V_{0}}}{2 \sqrt{k}}}$

$\displaystyle - C_{2} \sin{\left(\frac{\sqrt{E} a}{2 \sqrt{k}} - \phi \right)}$

sp.diff(psi1, x)

$\displaystyle \frac{C_{1} \sqrt{- E + V_{0}} e^{\frac{x \sqrt{- E + V_{0}}}{\sqrt{k}}}}{\sqrt{k}}$

sp.diff(psi2,x)

$\displaystyle \frac{C_{2} \sqrt{E} \cos{\left(\frac{\sqrt{E} x}{\sqrt{k}} + \phi \right)}}{\sqrt{k}}$

lhs = sp.diff(psi1, x) / psi1
lhs

$\displaystyle \frac{\sqrt{- E + V_{0}}}{\sqrt{k}}$

rhs = sp.simplify(sp.diff(psi2, x)/psi2)
rhs

$\displaystyle \frac{\sqrt{E}}{\sqrt{k} \tan{\left(\frac{\sqrt{E} x}{\sqrt{k}} + \phi \right)}}$

# Because of difficulties getting radicals to combine I have decided to square the left hand
# side an right side of both expressions.
(lhs**2)

$\displaystyle \frac{- E + V_{0}}{k}$

rhs

$\displaystyle \frac{\sqrt{E}}{\sqrt{k} \tan{\left(\frac{\sqrt{E} x}{\sqrt{k}} + \phi \right)}}$

# Actually it turns out the right hand side is less troublesome.
rhs.subs(sp.sqrt(E)/sp.sqrt(k), alpha)

$\displaystyle \frac{\alpha}{\tan{\left(\alpha x + \phi \right)}}$

sp.expand((lhs**2)).subs(E/k, alpha**2).subs(v0/k, -beta**2)

$\displaystyle - \alpha^{2} - \beta^{2}$