Particle in Infinite Square Well

We are given an unnormalized wavefunction $\psi(x) = x(a-x)$ for a particle in a one-dimensional infinite square well with walls at $x=0$ and $x=a$. The energy eigenfunctions for this system are given by $\phi_n(x) = \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi x}{a}\right)$, for $n=1, 2, 3, \ldots$.

The given wavefunction $\psi(x)$ can be expanded as a linear combination of these energy eigenfunctions:

$$ \psi(x) = \sum_{n=1}^\infty c_n \phi_n(x) $$

According to Parseval's theorem (or the completeness relation for orthonormal basis), the integral of the squared magnitude of the wavefunction is related to the sum of the squared magnitudes of the expansion coefficients:

$$ \int_0^a {\left| {\psi \left( x \right)} \right|^2} dx = \sum_{n=1}^\infty {\left| {{c_n}} \right|^2} $$

The coefficients $c_n$ are found by the projection of $\psi(x)$ onto $\phi_n(x)$:

$$ c_n = \int_0^a \phi_n^*(x) \psi(x) dx $$

Since the energy eigenfunctions $\phi_n(x)$ are real, $\phi_n^*(x) = \phi_n(x)$.

Substitute the expressions for $\psi(x)$ and $\phi_n(x)$:

$$ c_n = \int_0^a \sqrt{\frac{2}{a}} \sin\left(\frac{n\pi x}{a}\right) \cdot x(a-x) dx $$

$$ c_n = \sqrt{\frac{2}{a}} \int_0^a (ax - x^2) \sin\left(\frac{n\pi x}{a}\right) dx $$

We can split this into two integrals:

$$ c_n = \sqrt{\frac{2}{a}} \left[ a \int_0^a x \sin\left(\frac{n\pi x}{a}\right) dx - \int_0^a x^2 \sin\left(\frac{n\pi x}{a}\right) dx \right] $$

Let $k = \frac{n\pi}{a}$. We need to evaluate $\int_0^a x \sin(kx) dx$ and $\int_0^a x^2 \sin(kx) dx$. Using integration by parts:

For $\int x \sin(kx) dx$:

• Let $u=x$, $dv=\sin(kx) dx$. Then $du=dx$, $v=-\frac{1}{k}\cos(kx)$.
• $\int_0^a x \sin(kx) dx = \left[-\frac{x}{k}\cos(kx)\right]_0^a - \int_0^a -\frac{1}{k}\cos(kx) dx$
• $= \left(-\frac{a}{k}\cos(ka) - 0\right) + \frac{1}{k}\left[\frac{1}{k}\sin(kx)\right]_0^a$
• $= -\frac{a}{k}\cos(ka) + \frac{1}{k^2}\sin(ka)$.
• With $k=\frac{n\pi}{a}$, $ka=n\pi$. $\cos(n\pi)=(-1)^n$, $\sin(n\pi)=0$.
• $\int_0^a x \sin\left(\frac{n\pi x}{a}\right) dx = -\frac{a}{n\pi/a}(-1)^n + \frac{1}{(n\pi/a)^2}(0) = -\frac{a^2}{n\pi}(-1)^n = \frac{a^2}{n\pi}(-1)^{n+1}$.

For $\int x^2 \sin(kx) dx$:

• Let $u=x^2$, $dv=\sin(kx) dx$. Then $du=2x dx$, $v=-\frac{1}{k}\cos(kx)$.
• $\int_0^a x^2 \sin(kx) dx = \left[-\frac{x^2}{k}\cos(kx)\right]_0^a - \int_0^a -\frac{2x}{k}\cos(kx) dx$
• $= \left(-\frac{a^2}{k}\cos(ka) - 0\right) + \frac{2}{k}\int_0^a x\cos(kx) dx$.

Now evaluate $\int_0^a x\cos(kx) dx$:

• Let $u=x$, $dv=\cos(kx) dx$. Then $du=dx$, $v=\frac{1}{k}\sin(kx)$.
• $\int_0^a x\cos(kx) dx = \left[\frac{x}{k}\sin(kx)\right]_0^a - \int_0^a \frac{1}{k}\sin(kx) dx$
• $= \left(\frac{a}{k}\sin(ka) - 0\right) - \frac{1}{k}\left[-\frac{1}{k}\cos(kx)\right]_0^a$
• $= \frac{a}{k}\sin(ka) + \frac{1}{k^2}[\cos(ka) - \cos(0)]$
• $= \frac{a}{k}\sin(ka) + \frac{1}{k^2}(\cos(ka) - 1)$.

Substitute back into the $\int_0^a x^2 \sin(kx) dx$ integral:

• $\int_0^a x^2 \sin(kx) dx = -\frac{a^2}{k}\cos(ka) + \frac{2}{k}\left[\frac{a}{k}\sin(ka) + \frac{1}{k^2}(\cos(ka) - 1)\right]$
• $= -\frac{a^2}{k}\cos(ka) + \frac{2a}{k^2}\sin(ka) + \frac{2}{k^3}(\cos(ka) - 1)$.
• With $k=\frac{n\pi}{a}$, $ka=n\pi$, $\cos(n\pi)=(-1)^n$, $\sin(n\pi)=0$.
• $\int_0^a x^2 \sin\left(\frac{n\pi x}{a}\right) dx = -\frac{a^2}{n\pi/a}(-1)^n + \frac{2a}{(n\pi/a)^2}(0) + \frac{2}{(n\pi/a)^3}((-1)^n - 1)$
• $= -\frac{a^3}{n\pi}(-1)^n + \frac{2a^3}{n^3\pi^3}((-1)^n - 1) = \frac{a^3}{n\pi}(-1)^{n+1} + \frac{2a^3}{n^3\pi^3}((-1)^n - 1)$.

Now substitute these results back into the expression for $c_n$:

$$ c_n = \sqrt{\frac{2}{a}} \left[ a \left(\frac{a^2}{n\pi}(-1)^{n+1}\right) - \left(\frac{a^3}{n\pi}(-1)^{n+1} + \frac{2a^3}{n^3\pi^3}((-1)^n - 1)\right) \right] $$

$$ c_n = \sqrt{\frac{2}{a}} \left[ \frac{a^3}{n\pi}(-1)^{n+1} - \frac{a^3}{n\pi}(-1)^{n+1} - \frac{2a^3}{n^3\pi^3}((-1)^n - 1) \right] $$

The first two terms cancel:

$$ c_n = \sqrt{\frac{2}{a}} \left[ - \frac{2a^3}{n^3\pi^3}((-1)^n - 1) \right] $$

Now consider the term $((-1)^n - 1)$:

• If $n$ is even, $n=2m$: $((-1)^{2m} - 1) = 1 - 1 = 0$. So $c_n = 0$ for even $n$.
• If $n$ is odd, $n=2m-1$: $((-1)^{2m-1} - 1) = -1 - 1 = -2$.

Thus, only coefficients for odd $n$ are non-zero. Let $n = 2m-1$ where $m = 1, 2, 3, \ldots$.

For odd $n = 2m-1$:

$$ c_{2m-1} = \sqrt{\frac{2}{a}} \left[ - \frac{2a^3}{(2m-1)^3\pi^3}(-2) \right] = \sqrt{\frac{2}{a}} \frac{4a^3}{(2m-1)^3\pi^3} $$

Now calculate $|c_n|^2 = |c_{2m-1}|^2$ for odd $n$:

$$ |c_{2m-1}|^2 = \left( \sqrt{\frac{2}{a}} \frac{4a^3}{(2m-1)^3\pi^3} \right)^2 = \frac{2}{a} \frac{16a^6}{(2m-1)^6\pi^6} = \frac{32a^5}{(2m-1)^6\pi^6} $$

The integral $\int_0^a |\psi(x)|^2 dx$ is equal to the sum of these squared coefficients:

$$ \int_0^a |\psi(x)|^2 dx = \sum_{n=1, \text{odd}}^\infty |c_n|^2 = \sum_{m=1}^\infty |c_{2m-1}|^2 $$

$$ \int_0^a |\psi(x)|^2 dx = \sum_{m=1}^\infty \frac{32a^5}{(2m-1)^6\pi^6} $$

$$ \int_0^a |\psi(x)|^2 dx = \frac{32a^5}{\pi^6} \sum_{m=1}^\infty \frac{1}{(2m-1)^6} $$

Replacing the summation index $m$ with $n$, we get:

$$ \int_0^a |\psi(x)|^2 dx = \frac{32a^5}{\pi^6} \sum_{n=1}^\infty \frac{1}{(2n-1)^6} $$

The integral $\int_0^a |\psi(x)|^2 dx$ is proportional to the series $\sum_{n=1}^\infty \frac{1}{(2n-1)^6}$, which can also be written as $\sum_{n=1}^\infty (2n-1)^{-6}$.

Let's briefly check the direct integration of $|\psi(x)|^2$:

$$ \int_0^a |x(a-x)|^2 dx = \int_0^a (ax - x^2)^2 dx = \int_0^a (a^2 x^2 - 2ax^3 + x^4) dx $$

$$ = \left[ \frac{a^2 x^3}{3} - \frac{2ax^4}{4} + \frac{x^5}{5} \right]_0^a = \frac{a^2 a^3}{3} - \frac{a a^4}{2} + \frac{a^5}{5} $$

$$ = \frac{a^5}{3} - \frac{a^5}{2} + \frac{a^5}{5} = a^5 \left(\frac{10 - 15 + 6}{30}\right) = \frac{a^5}{30} $$

So we have $\frac{a^5}{30} = \frac{32a^5}{\pi^6} \sum_{n=1}^\infty \frac{1}{(2n-1)^6}$. This explicitly shows the proportionality and implies the sum value is $\frac{\pi^6}{960}$.

Therefore, $\int_0^a |\psi(x)|^2 dx$ is proportional to the infinite series $\sum_{n = 1}^\infty (2n - 1)^{ - 6}$.

The provided integral identities involving 't' were not directly used in this integration by parts approach.