maximum position uncertainty of particle in a box

Question

I want to verify mathematically

for wave function $\psi(x)$ satisfying

$\psi(x)=0$ for $\lvert x \rvert \ge \frac{L}{2} $ and

$\int_{- \frac{L}{2}}^{\frac{L}{2}} \lvert \psi(x) \rvert ^2 dx = 1 $

Below holds

$\int_{-\frac{L}{2}}^{\frac{L}{2}} x^2 {\lvert \psi(x)\rvert}^2 dx - \left( \int_{-\frac{L}{2}}^{\frac{L}{2}} x {\lvert \psi(x) \rvert}^2 dx \right) ^2 \le \frac{L^2}{4} $

I know that the equality holds for ${\lvert \psi(x) \rvert}^2 = \frac{1}{2} \delta (x - \frac{L}{2}) +\frac{1}{2} \delta (x + \frac{L}{2}) $

But.. I dont konw how to establish the inequality

Thanks in advance for any help

It seems that you need to solve for the eigenfunctions first. When you do that, then try computing $\langle X^2\rangle - \langle X\rangle^2$ on $\psi_n(x)$. — Mnifldz, Mar 24 '15 at 05:36
@Mnifldz This holds for any shape of the potential within the requirements of the OP. I don't think you will be able to find the eigenfunctions explicitly. — Rol, Mar 24 '15 at 08:06
Note that only the probability density $|\psi(x)|^2$ is relevant. This result is known as Popoviciu's inequality on variances, see also here. — Oscar Cunningham, Mar 24 '15 at 11:50

Rol · Answer 1 · 2015-03-24T11:20:06.397

Given that $\lvert x \rvert \le \frac{L}{2}$ in the range of the integral, $ x^2 \le \frac{L^2}{4}$ and thus

$ \int_{-\frac{L}{2}}^{\frac{L}{2}} x^2 {\lvert \psi(x)\rvert}^2 dx \le \frac{L^2}{4} \overbrace{\int_{-\frac{L}{2}}^{\frac{L}{2}} {\lvert \psi(x)\rvert}^2 dx}^{1} \le \frac{L^2}{4} . $

Substracting any positive number, like $\left( \int_{-\frac{L}{2}}^{\frac{L}{2}} x {\lvert \psi(x) \rvert}^2 dx \right) ^2$, from the LHS will preserve the inequality.

The equality holds iff

$ \int_{-\frac{L}{2}}^{\frac{L}{2}} x^2 {\lvert \psi(x)\rvert}^2 dx = \frac{L^2}{4}$

and

$\int_{-\frac{L}{2}}^{\frac{L}{2}} x {\lvert \psi(x) \rvert}^2 dx = 0.$

I wonder if the function you gave is the only case of equality.

maximum position uncertainty of particle in a box

1 Answers1