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.