Non degenerate perturbation example

From David J. Griffiths "Introduction to Quantum Mechanics" Problem 6.3

"Two identical bosons are placed in an infinite square well. They interact weakly with one another, via the potential $V(x_{1},x_{2})=-aV_{0}\delta (x_{1}-x_{2})$, where $V_{0}$is a constant with the dimensions of energy, and a is the width of the well).

a)First, ignoring the interaction between the particles, find the ground state and the first excited state -- both the wave function and the associated energies.

b) Use first-order perturbation theory to estimate the effect of the particle-particle interaction on the energies of the ground state and the first excited state.

For the infinite square well the unperturbed energy is given by $E_{n}^{0}={\frac {n^{2}\pi ^{2}\hbar ^{2}}{2ma^{2}}}$and the ground state wave function is given by $\Psi _{n}^{0}={\sqrt {\frac {2}{a}}}sin\left({\frac {n\pi x}{a}}\right)$.

a)

For the case of identical bosons with no interaction between them the ground state is given by the following: $\psi _{11}^{0}(x)=\psi _{1}^{0}(x_{1})\psi _{1}^{0}(x_{2})$

$\psi _{11}^{0}(x)={\sqrt {\frac {2}{a}}}sin\left({\frac {\pi x_{1}}{a}}\right){\sqrt {\frac {2}{a}}}sin\left({\frac {\pi x_{2}}{a}}\right)$

While the energy is given by $E_{1total}^{0}=E_{1}^{0}(x_{1})+E_{1}^{0}(x_{2})$.

$E_{1}^{0}=2E_{1}^{0}=(2){\frac {\pi ^{2}\hbar ^{2}}{2ma^{2}}}={\frac {\pi ^{2}\hbar ^{2}}{ma^{2}}}$

The first excited state is given by $\psi _{12}^{0}(x)={\frac {1}{\sqrt {2}}}\left[\psi _{1}^{0}(x_{1})\psi _{2}^{0}(x_{2})+\psi _{1}^{0}(x_{2})\psi _{2}^{0}(x_{1})\right]$

$\psi _{12}^{0}={\frac {1}{\sqrt {2}}}\left[{\sqrt {\frac {2}{a}}}sin\left({\frac {\pi x_{1}}{a}}\right){\sqrt {\frac {2}{a}}}sin\left({\frac {2\pi x_{2}}{a}}\right)+{\sqrt {\frac {2}{a}}}sin\left({\frac {2\pi x_{1}}{a}}\right){\sqrt {\frac {2}{a}}}sin\left({\frac {\pi x_{2}}{a}}\right)\right]$

$\psi _{12}^{0}={\frac {\sqrt {2}}{a}}\left[sin\left({\frac {\pi x_{1}}{a}}\right)sin\left({\frac {2\pi x_{2}}{a}}\right)+sin\left({\frac {2\pi x_{1}}{a}}\right)sin\left({\frac {\pi x_{2}}{a}}\right)\right]$

While the energy is given by $E_{2}^{0}=E_{2}^{0}(x_{1})+E_{2}^{0}(x_{2})$

$E_{2}^{0}={\frac {\pi ^{2}\hbar ^{2}}{2ma^{2}}}+{\frac {4\pi ^{2}\hbar ^{2}}{2ma^{2}}}={\frac {5\pi ^{2}\hbar ^{2}}{2ma^{2}}}$

b)

The first order perturbation, the energy correction is given by $E_{n}^{1}=\langle \psi _{n}^{1}|H'|\psi _{n}^{1}\rangle$ for the ground state.

The interaction potential given in the problem is the perturbed part of the Hamiltonian. Therefore the first order perturbation in integral form for the ground state is given by:

$E_{1}^{1}=\int \psi _{1}^{1}[-aV_{0}\delta (x_{1}-x_{2})]\psi _{1}^{1}dx$, where $\psi _{1}^{1}={\frac {2}{a}}sin\left({\frac {\pi x_{1}}{a}}\right)sin\left({\frac {\pi x_{2}}{a}}\right)$

$E_{1}^{1}=\int _{0}^{a}\int _{0}^{a}\left({\frac {2}{a}}\right)^{2}sin^{2}\left({\frac {\pi x_{1}}{a}}\right)sin^{2}\left({\frac {\pi x_{2}}{a}}\right)[-aV_{0}\delta (x_{1}-x_{2})]dx_{1}dx_{2}$

$E_{1}^{1}=-aV_{0}\left({\frac {2}{a}}\right)^{2}\int _{0}^{a}sin^{4}\left({\frac {\pi x}{a}}\right)dx$

Then do a substitution such that $y={\frac {\pi x}{a}}$

$E_{1}^{1}=-aV_{0}\left({\frac {2}{a}}\right)^{2}{\frac {a}{\pi }}\int _{0}^{\pi }sin^{4}ydy$

The integral of $sin^{4}(y)$ is given by ${\frac {1}{32}}\left(12y-8sin(2y)+sin(4y)\right)$. Since $y=\pi$ in this case both sine terms go to zero and the integrand yields ${\frac {3\pi }{8}}$.

Therefore the first order correction to the energy is $E_{1}^{1}=-{\frac {3}{2}}V_{0}$

As for the first order perturbation to the energy for the first excited state the wavefunction is given as

$\psi _{12}^{0}={\frac {\sqrt {2}}{a}}\left[sin\left({\frac {\pi x_{1}}{a}}\right)sin\left({\frac {2\pi x_{2}}{a}}\right)+sin\left({\frac {2\pi x_{1}}{a}}\right)sin\left({\frac {\pi x_{2}}{a}}\right)\right]$

Note that if this system was for fermions then this wavefunction would be zero since there would be a minus sign in place of the plus sign.

$E_{2}^{1}=-aV_{0}\int _{0}^{a}\int _{0}^{a}\left[{\frac {\sqrt {2}}{a}}\left(sin\left({\frac {\pi x_{1}}{a}}\right)sin\left({\frac {2\pi x_{2}}{a}}\right)+sin\left({\frac {2\pi x_{1}}{a}}\right)sin\left({\frac {\pi x_{2}}{a}}\right)\right)\right]^{2}\delta (x_{1}-x_{2})dx_{1}dx_{2}$

$E_{2}^{1}=-aV_{0}\left({\frac {2}{a^{2}}}\right)\int _{0}^{a}\left[sin\left({\frac {\pi x}{a}}\right)sin\left({\frac {2\pi x}{a}}\right)+sin\left({\frac {2\pi x}{a}}\right)sin\left({\frac {\pi x}{a}}\right)\right]^{2}dx$

$E_{2}^{1}=-{\frac {2V_{0}}{a}}\int _{0}^{a}\left[4sin^{2}\left({\frac {\pi x}{a}}\right)sin^{2}\left({\frac {2\pi x}{a}}\right)\right]dx$

Use the same substitution as before $y={\frac {\pi x}{a}}$

$E_{2}^{1}=-{\frac {8V_{0}}{a}}\left({\frac {a}{\pi }}\right)\int _{0}^{\pi }[sin^{2}(y)sin^{2}(2y)dy]$

use the identity $sin(2y)=2cos(y)sin(y)$ such that $sin^{2}(2y)=4cos^{2}(y)sin^{2}(y)$

$E_{2}^{1}=-{\frac {8V_{0}}{\pi }}4\int _{0}^{\pi }sin^{2}(y)sin^{2}(y)cos^{2}(y)=-{\frac {32V_{0}}{\pi }}\int _{0}^{\pi }(sin^{4}(y)-sin^{6}(y))dy$

The $sin^{4}(y)$ integrand is the same as before and yields ${\frac {3\pi }{8}}$. The $sin^{6}(y)$ integrand is ${\frac {5y}{16}}-{\frac {15}{64}}sin(2y)+{\frac {3}{64}}sin(4y)-{\frac {1}{192}}sin(6y)$ where all the sine terms are zero so it just yields ${\frac {5\pi }{16}}$.

$E_{2}^{1}=-{\frac {32V_{0}}{\pi }}\left({\frac {3\pi }{8}}-{\frac {5\pi }{16}}\right)=-2V_{0}$