🚀 go-pugleaf

            +        +
            +        +
   V(x)=OO  + V(x)=0 +  V(x)=OO (infinite potential to the right
            +        +           and left on the well)
            ++++++++++

           x=0      x=a

 Suppose you have a particle (1-D) in an infinite square well ....
assume also that the state of the system is the ground state of the
box, so |Psi> = sqrt (1/a) sin kx. If BOTH walls disolve at t=0,
the probability distributions of energy and momentum can be calculated
from Q(E)=|<En|Psi>|^2 and taking advantage from the fact that the
Hamiltonian and Momentum operators conmute for this particular new
state. All is joy and happines :-)

                
Now, consider the particle in the box problem again. Now at time t=0 
only the left side of the wall vanishes. The energy distribution can
be calculated using the formula above. How do you calculate the momentum
distribution????? - [H,P] =! 0 anymore .... -this is an old homework
problem I never solved ... thanks in advance



            +        
            +        
   V(x)=OO  + V(x)=0             
            +                             
            ++++++++++++++++++++++++++++++++++++

           x=0









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  $^$ "I perceive the Universe as  ^^^^^^   Jose Luis Paredes             $^$
  $^$  a single equation .... and  ^^^^^^   Physicist by choice !!        $^$
  $^$  it's so simple!"            ^^^^^^   University of Houston         $^$
  $^$  Lt. Barkley -Star Trek TNG  ^^^^^^   Jose@crcc.uh.edu              $^$
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