2001AugQualQUANT - Ben Sauerwine Practice for Qualifying...

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Ben Sauerwine Practice for Qualifying Exams Thanks to Yossef and Shiang Yong for their input in this problem. Problem Source: CMU August 2001 Qualifying Exam Electron in a spherical well An electron is in a spherical well of radius a and depth , i.e., the non-relativistic Hamiltonian is 0 V V m p H + = 2 2 v (1) with m = the mass of an electron and a r V a r V V > = = 0 0 (2) In this problem, we take 1 2 = π h . The solutions of the Schrodinger equation are of the form () ( ) s lm Y r R χ φ θ , , where ( ) r R is the radial wave equation, () , lm Y is a spherical harmonic, and s is a non-relativistic spinor. ( ) r R is a solution to the equation 0 1 2 1 2 2 2 = + + R r l l V E m dr dR r dr d r (3) where E is the energy of the state. Note that in spherical coordinates, the operator has the form 2 ( φθ , 1 1 2 2 2 2 Ω + = r dr dR r dr d r ) (4) where Ω is an operator in the angular variables. The differential equation satisfied by spherical Bessel or Hankel functions is 0 1 1 1 2 2 2 = + + x F x l l dx x dF x dx d x l l (5)
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(a) Explain what properties of the spherical harmonics are used to obtain equation (3) starting with the Hamiltonian given in equation (1). Note the form of the operator given above. 2 () [] 0 2 2 2 2 2 = + = + = ψ ψψ V E m E H V m H h Next, separation of variables is assumed so that: [ ]() ( ) 0 , 2 2 = + φθ lm Y r R V E m Now when one expands the operator, one gets 2 ( ) ( ) 0 , 2 , 1 1 2 2 2 = + Ω + lm Y r R V E m r dr d r dr d r This is particularly convenient, since the spherical harmonics are orthogonal eigenfunctions of the spherical harmonic differential equation, here written φ θ , Ω , with eigenvalue . At this point, it is clear that by allowing the spherical harmonic differential equation to commute with the radial portion and act on the spherical harmonic, then dividing out the spherical harmonic portion will give (3) exactly. ( 1 + l l ) (b) For our spherical well problem, what conditions must ( ) r R satisfy at: (i) 0 = r Since this wave function must be continuous in space, I see that it must without a doubt have a derivative of zero at the origin or otherwise the spherical harmonics approaching from either side could force a discontinuity as one approaches from either side: otherwise constra no l if R int 0 0 0 ' = = Second, I see that a discontinuity could be imposed by the spherical harmonics if the spherical harmonic is not spherically symmetric: otherwise finite R l if R 0 0 0 0 = (ii) a r = At this boundary, the wave function must be continuous:
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() () () ( ) a R a R and a R a R a r a r a r a r < > < > = = ' ' (iii) r The wave function must be normalizable () 0 = R (c) The solutions for for bound electrons are spherical Bessel functions r R ( r j l ) α inside the well and spherical Hankel functions ( ) r h l β outside the well.
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2001AugQualQUANT - Ben Sauerwine Practice for Qualifying...

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