Question

The uncertainty in a particle's momentum in an infinite well in the general case of arbitrary $\mathit{n}$is given by$\frac{\mathbf{n}\mathbf{\pi }\mathbf{h}}{\mathbf{L}}$ .

Short answer

For $n=0$the uncertainty vanishes but is perfectly finite for all other values. Thus, so long as$\Delta x^2$ is large enough (it is), the uncertainty principle is perfectly satisfied for all $n>0$.

Step-by-step solution

The concept and the formula used.

Heisenberg's uncertainty principle states that it is impossible to measure or calculate exactly, both the position and the momentum of an object.

Consider, energy $E=0$. Then, the momentum of the state must satisfy$E=\frac{p^2}{2m}$. Now, there are two solutions for$P$ corresponding to positive and negative momentum. The uncertainty can thus be calculated as follows:

${\mathrm{\Delta P}}^2=⟨{\mathrm{P}}^2⟩-{⟨\mathrm{P}⟩}^2$

$=2mE$

$=\frac{{\mathrm{n}}^2{\mathrm{\pi }}^2{\mathrm{\hslash }}^2}{{\mathrm{L}}^2}$

$\mathrm{\Delta P}=\frac{\mathrm{n\pi }\mathrm{\hslash }}{\mathrm{L}}$

Conclusion

Clearly, for$n=0$ the uncertainty vanishes but is perfectly finite for all other values. Thus, so long as$\Delta x^2$ is large enough (it is), the uncertainty principle is perfectly satisfied for all$n>0$ .