(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

Consider a particle of mass m in a vee-shaped potential whose analytic form is

V(x)= -bx (x<=0)

V(x)= bx (x>=0)

Use what is known about the uncertainty principle and the simple harmonic oscillator to show that the lowest state energy is ((hbar)^2(b)^2/m)^(1/3). Show that this funky result has the correct dimensions.

2. The attempt at a solution

The width of the well is 2bx. I used delta p = hbar/2bx, and then subbed into E=p^2/2m but the b is in the denominator, and I have no clue where the 1/3 power comes from.

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# Particle in Potential Well

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