Question

We say that the ground state for the particle in a box has nonzero energy. What goes wrong with $\mathbf{\Psi }$in equation 5.16 if n = 0 ?

Short answer

If n = 0 in equation of ${\mathrm{\Psi }}_{\mathrm{n}}\left(\mathrm{x}\right)=\sqrt{\frac{2}{\mathrm{L}}}\sin \left(\frac{\mathrm{n\tau \tau x}}{\mathrm{L}}\right)$, this would make locating our particle difficult, meaning that it no longer exists.

Step-by-step solution

Calculating Ψn(x) for n=0

The equation is given $\mathrm{\Psi }=5.16$wheren n = 0

The equation in question says that for 0<x<L

${\mathrm{\Psi }}_{\mathrm{n}}\left(\mathrm{x}\right)=\sqrt{\frac{2}{\mathrm{L}}}\sin \left(\frac{\mathrm{n\tau \tau x}}{\mathrm{L}}\right)$

If n = 0 ,this equation gives us ${\mathrm{\Psi }}_{\mathrm{n}}=0$. This would make it impossible to locate our particle, implying that the particle no longer exists.

Conclusion

If a square well potential had a ground state with zero energy, it would suggest that the particle would be at rest inside the square well (due to its zero energy), which would obviously violate Heisenberg's uncertainty principle.

Therefore, It is not possible to locate the particle if n = 0