Homework 1, solutions

Problem 1, solution

Assume the uncertainty in the position of the electron is Dx about x=0.  Then the uncertainty in its momentum is Dp=h/Dx about p=0.

The kinetic energy is on the order of T=(Dp)2/2m and the potential energy is on the order of V=(1/2)mw2(Dx)2.  The energy is E=T+V»h2/(2m(Dx)2)+(1/2)mw2(Dx)2.

The ground state has the lowest energy, so let us minimize E with respect to Dx.

DE/d(Dx) = -h2/(m(Dx)3)+mw2Dx = 0,  (Dx)4 = h2/(m2w2)

Inserting (Dx)2 = h/(mw) into the equation for E yields E = hw.   The ground state energy of the harmonic oscillator is on the order of hw.

 

Problem 2, solution

 

 

 The probability of finding the particle with momentum between p and p+dp  is

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Problem 3, solution

The electron energy is E=10eV=1.6´10-18J.

Its speed is v=(2E/m)1/2=1.88´106m/s.

It travels 2km in ~10-3s.

The initial uncertainty in its position is Dx=10-9m.

Dp=h/Dx,  Dv=h/(mDx)=1.1´105m/s.

After 10-3s the uncertainty in its position is on the order of  Dv10-3s=110m.