Chapter 10 - Position and Momentum

We can start using sympy to handle symbolic math (integrals and other calculus):



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from sympy import *



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init_printing(use_unicode=True)



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# SymPy works better if you specify what letters are symbols:
x, y, z = symbols('x y z', real=True)

# notice we can also put some restrictions on the symbols:
a, c = symbols('a c', nonzero=True, real=True)



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integrate?

There are two ways to use the integrate function. In one line, like integrate(x,(x,0,1)) or by naming an expression and then integrating it over a range:

A = (c*cos((pi*x)/(2.0*a)))**2
A.integrate((x,-a,a),conds='none')

We'll use both, at different times. For longer expressions, the second form can be easier to read and write.

First, just try the following, then we'll re-create some examples in the book.



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integrate(x,(x,0,1))









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$\displaystyle \frac{1}{2}$



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integrate(x**2,(x,0,1))









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$\displaystyle \frac{1}{3}$



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A = (c*cos((pi*x)/(2.0*a)))**2
A.integrate((x,-a,a))









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$\displaystyle a c^{2}$

So this tells us the normalization constant should be $c=\frac{1}{\sqrt{a}}$. Check that it is normalized if we do that:



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psi = 1/sqrt(a)*cos((pi*x)/(2.0*a))  # notice we can name the expression something useful.
B = psi**2
B.integrate( (x,-a,a), conds='none')









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$\displaystyle 1$

Because psi is a real function, we can calculate expectation values by integrating over $x$ or $x^2$ with psi**2:



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C = x*psi**2
C.integrate( (x,-a,a), conds='none')









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$\displaystyle 0$



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D = x**2 * psi**2
E = D.integrate( (x,-a,a), conds='none')
E









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$\displaystyle - \frac{- \frac{a^{3}}{6} + \frac{a^{3}}{\pi^{2}}}{a} + \frac{- \frac{a^{3}}{\pi^{2}} + \frac{a^{3}}{6}}{a}$



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E.simplify()  # this is a useful method!









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$\displaystyle \frac{a^{2} \left(-6 + \pi^{2}\right)}{3 \pi^{2}}$



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E.n()  # the .n() method approximates the numerical part. You can look at the full expression below.









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$\displaystyle 0.130690966048658 a^{2}$

Example 10.2



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h = Symbol('hbar', real=True, positive=True)

Use the diff function to take a derivative of a symbolic expression. For example:



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diff(x**2, x)









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$\displaystyle 2 x$



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# Solution
-1j*h*diff( 1/a*cos((pi*x)/(2*a)) ,x)









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$\displaystyle \frac{0.5 i \pi \hbar \sin{\left(\frac{\pi x}{2 a} \right)}}{a^{2}}$



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# Solution
B1 = (pi*h/(2*a))**2 * (cos((pi*x)/(2*a)))**2
B1.integrate( (x,-a,a), conds='none' )









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$\displaystyle \frac{\pi^{2} \hbar^{2}}{4 a}$

Example 10.3



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p = Symbol('p', real=True)



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# Solution
A = integrate(1/sqrt(2*pi*a*h)*exp(-I*p*x/h)*cos((pi*x)/(2*a)),(x,-a,a), conds='none')



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# Solution
A









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$\displaystyle - \frac{\sqrt{2} \sqrt{\pi} a \hbar^{2}}{\sqrt{a \hbar} \left(4 a^{2} p^{2} e^{\frac{i a p}{\hbar}} - \pi^{2} \hbar^{2} e^{\frac{i a p}{\hbar}}\right)} - \frac{\sqrt{2} \sqrt{\pi} a \hbar^{2}}{\sqrt{a \hbar} \left(4 a^{2} p^{2} e^{- \frac{i a p}{\hbar}} - \pi^{2} \hbar^{2} e^{- \frac{i a p}{\hbar}}\right)}$



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psi_p = sqrt(2*a*pi/h) * 2/(pi**2 - (2*p*a/h)**2) * cos(p*a/h)
psi_p









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$\displaystyle \frac{2 \sqrt{2} \sqrt{\pi} \sqrt{\frac{a}{\hbar}} \cos{\left(\frac{a p}{\hbar} \right)}}{- \frac{4 a^{2} p^{2}}{\hbar^{2}} + \pi^{2}}$



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psi_p == sqrt(2*a*pi/h)*2/(pi**2 - (2*p*a/h)**2) * cos(p*a/h)









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True

Which agrees with the book.

This is about as far as we can go in sympy. Unfortunately, many other momentum integrals choke. There are a few hints to get through the rest here:

Problem 10.3



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x, y, z = symbols('x y z', real=True)
a, c = symbols('a c', nonzero=True, real=True, positive=True)

psi = c*1/(a**2 + x**2)  # define the wavefunction with c constant
int1 = integrate(psi*psi,(x,-oo,oo), conds='none')  # integrate psi^2
solutions = solve(int1 - 1,c)  # solve for c, this returns a list of solutions
c2 = simplify(solutions[0])  # simplify the solution for c:
c2









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$\displaystyle \frac{\sqrt{2} a^{\frac{3}{2}}}{\sqrt{\pi}}$



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psi2 = c2/c*psi
psi2









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$\displaystyle \frac{\sqrt{2} a^{\frac{3}{2}}}{\sqrt{\pi} \left(a^{2} + x^{2}\right)}$



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integrate(psi2 * x * psi2,(x,-oo,oo))









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$\displaystyle 0$



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integrate(psi2 * x**2 * psi2,(x,-oo,oo))









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$\displaystyle a^{2}$

So $\Delta x^2 = a^2 - 0^2$ therefore $\Delta x = a$

Problem 10.17:

Now find the momentum representation of the state from 10.3



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p = symbols('p', nonzero=True, real=True, positive=True)
B = integrate(sqrt(1/(2*pi*h))*exp(-I*p*x/h)*psi2,(x,-oo,oo))
B









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$\displaystyle - \frac{\sqrt{a} \left(2 \left(- i \operatorname{Shi}{\left(\frac{a p}{\hbar} \right)} - \frac{\pi}{2}\right) \cosh{\left(\frac{a p}{\hbar} \right)} + 2 i \left(\operatorname{Chi}\left(\frac{a p}{\hbar}\right) - \frac{i \pi}{2}\right) \sinh{\left(\frac{a p}{\hbar} \right)}\right)}{2 \pi \sqrt{\hbar}} - \frac{\sqrt{a} \left(2 \left(i \operatorname{Shi}{\left(\frac{a p}{\hbar} \right)} - \frac{\pi}{2}\right) \cosh{\left(\frac{a p}{\hbar} \right)} - 2 i \left(\operatorname{Chi}\left(\frac{a p}{\hbar}\right) + \frac{i \pi}{2}\right) \sinh{\left(\frac{a p}{\hbar} \right)}\right)}{2 \pi \sqrt{\hbar}}$



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B.simplify()









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$\displaystyle \frac{\sqrt{a} e^{- \frac{a p}{\hbar}}}{\sqrt{\hbar}}$

This agrees with the book after we notice that we had to force $p$ to be positive in order to get the integral to converge. The book has $|p|$ in the argument of the exponent to reflect this constraint.



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