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A particle in two boxes

  1. Jun 9, 2008 #1
    Suppose we have a particle in a 1-dimensional box, such that the particle is in its lowest energy state. The energy of a particle in a 1-dimensional box is E = h^2*n^2/(8*m*L^2). Therefore, if the particle is in its lowest energy level, n = 1, and the box has a length of d, then E = h^2/(8*m*d^2). Now suppose we divide the box into two boxes, A and B, using an impenetratable barrier so that each new box has a length of d/2. Now, if the particle is found to be in box A, the minimum energy it can have is n = 1, where E = h^2*n^2/(8*m*L^2) and therefore, E= h^2/(8*m*(d/2)^2) = 4*h^2/(8*m*d^2). This is the same for the particle being found in box B by symmetry. How is it possible that the energy has increased by simply adding a dividing barrier.
  2. jcsd
  3. Jun 9, 2008 #2


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    Science Advisor

    The process of insertion of the barrier would require energy. That energy would be transmitted to the particle. The whole process can be simulated by a time-dependent potential. It is known that a time-dependent potential corresponds to a non-conserved energy due to an external energy source.
  4. Jun 9, 2008 #3
    To Me

    Well Gentleman To Me The Energy Of Particle Is Decreased On Applying Barrier,bcoz We Have Restricted Particle To Finite Area.if We Further Introduce Separation,then Again Energy Is Decreased.so What We Are Doing,is Just Limiting Its Range Of Freeness.
    Have You Studied Gibb's Paradox,it Also Tell The Same Thing But Regarding Entropy.
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