Does the n=0 State in a Quantum Box Violate the Uncertainty Principle?

[pivoxa15]

How would you go about this question?

Show that by allowing the state n=0 for a particle in a 1D box will violoate the uncertainty principle, delta(x)delta(p)>=h(bar)/2

I have tried to substitute all sorts of different relationships but do seem to get anywhere. I have showed that E=0 for a ground state electron but can't seem to relate it to the uncertainty principle.

 

[dextercioby]

You just have to compute the uncertainties for each variable, x and p_{x}, knowing the wavefunction.

Daniel.

 

[pivoxa15]

You just have to compute the uncertainties for each variable, x and p_{x}, knowing the wavefunction.

Daniel.

If we allow n=0, the wavefunction will cease to exist hence the particle will cease to exist. Hence momentum and position of a particle does not exist or could you say 0. Hence any change in the n=0 state, the particle will continue to cease to exist. In this way the HU principle will not be satisfied. But my argument is pretty vague. Is there a quantifiable way to express this?

 

[dextercioby]

What's the energy spectrum for the particle...?

Daniel.

 

[pivoxa15]

What do you mean by the energy spectrum?

Do you mean the energy levels?
At n=0, Energy=0.
At other levels, Energy=n^2(pie)^2(hbar)^2/(2mL^2)

But how does it relate to the UC?