WKB Approximation Problem

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  E6510: Finding energy levels using WKB Submitted by: Elad Cohen, Ran PeiserThe problem: Find, using the WKB approximation, the energy levels of a one dimensional problem with a potentialof   V   ( x ) =  sign ( α ) · x α , x ≥ 0.For what values of   α  is there a floor?In particular, discuss the cases of   α  = 2 , 0 , − 1 The solution: According to the WKB approximation:   x 2 x 1 P   ( x ) dx  =  π    n +  12  . For  x 1 ,x 2 - the turning pointsin the energy  E  n . P  ( x ) =   2 m [ E  n − V    ( x )] =   2 m [ E  n − sign ( α ) · x α ]We will notice that for  α >  0,  E  n  >  0. And for  α <  0,  E  n  <  0 (In this case  E  n  >  0 is allowed, butthese are not bound states). x 1  = 0  x 2  = ( sign ( α ) · E  n ) 1 /α We will refer to  sign ( α ) as  S   from now on.    x 2 x 1 P  ( x ) dx  = √  2 m    ( SE  n ) 1 /α 0   E  n − S  · x α dx  = √  2 m ( SE  n ) 12    ( SE  n ) 1 /α 0   S  −  x α E  n dx  == √  2 m ( SE  n ) 12    ( SE  n ) 1 /α 0   S  − S    x ( SE  n ) 1 /α  α dx  == √  2 m ( SE  n ) 12 +  1 α    10   S  (1 − t α ) dt We will define the integral:  C   ( α ) =   10   S  (1 − t α ) dt  and we shall get:    x 2 x 1 P  ( x ) dx  = √  2 m ( SE  n ) 12 +  1 α C  ( α ) =  π    n + 12  Extracting  E  n  from the expression we get a solution for all  α  = 0: E  n  =  S   π    n +  12  √  2 mC   ( α )   2 αα +2 The condition for floor to exist is the convergence of the integral defining  C   ( α ).The reason is when the integral converges, it means that until certain energy, there is a finite area inthe phase domain, and therefore a finite number of allowed states, hence, there has to be a ground1  state (floor). When the integral does not converge, there is infinite area in the phase domain, whichmeans infinite number of states, hence, no ground state. Integrand { C   ( α ) } =   S  (1 − t α )  ≈  t α 2  α <  01  α >  0When  t → 0 . Therefore, the integral always converges for  α >  0, but for  α <  0 the integral only converges for α > − 2. Conclusion: There is a floor for each   α > − 2 . α  = 2 ,  Harmonic Oscilator C   (2) =  π 4 E  n  = 2   2 m    n +  12  = 2   ω  n +  12  1 =  12 mω 2  ω  =   2 m  We got the expected result with an additional factor of 2, coming from the fact we only regardedthe positive half of the x axis. α  = − 1 ,  Hydrogen Atom C   ( − 1) =  π 2 E  n  = −  √  2   ( n + 12 ) √  m  − 2 = −  m 2   2 ( n + 12 ) 2 We got the spectrum of the hydrogen atom, except the  12  which was added to the denominator.For the case of   α  = 0 we will solve separately. V    ( x ) = 0  P   ( x ) = √    2 mE  n   L 0  P   ( x ) dx  =   L 0 √  2 mE  n dx  =  L √  2 mE  n  =  π    n +  12     E  n  =  π 2   2 ( n + 12 ) 2 2 mL 2 We got the spectrum of an infinite potential well with soft boundary conditions.  E  n  =  E  n − E  n − 1  =  π 2   2 2 mL 2  n +  12  2 −  n −  12  2  =  π 2   2 nmL 2 lim L →∞  E  n  = 0Meaning that in the limit of   L →∞  the gap between neighboring energy levels is reduced to zero,and we get a continues spectrum.Of course there is a floor since the energy has to be positive which fits the ’floor condition’ fromthe general case.2
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