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Unformatted text preview: Solutions to Problem Set 4 Physics 341 by: Callum Quigley 1 Delta-Function Potentials (i) and (ii) See midterm solutions (iii) Away from x = ± a the potential vanishes, so for bound states ( E < 0), the wavefunctions will go like (superpositions of) ψ ( x ) ’ e ± κx with energy E =- ~ 2 κ 2 2 m . More precisely, ψ n ( x ) = A n e κx , x ≤ - a B n e κx + C n e- κx ,- a ≤ x ≤ a D n e- κx , x ≥ a (1.1) where n labels the different possible energy eigenstates. Imposing continuity of ψ ( x ) at x = ± a we have A n = B n + C n e 2 κa (1.2) D n = B n e 2 κa + C n . However, as in the case of a single delta-function, ψ ( x ) is discontinuous at x = ± a . Inte- grating the Schr¨odinger equation around those points yields lim → ( ψ ( ± a + )- ψ ( ± a- )) = 2 mg ~ 2 ψ ( ± a ) ≡ - κ ψ ( ± a ) . (1.3) Note that κ = 2 m | g | ~ 2 > 0. Plugging in our ansatz for ψ ( x ) leads to the constraints A n = κ κ ( A n- B n + C n e 2 κa ) (1.4) D n = κ κ ( D n- C n + B n e 2 κa ) . Combining (1.2) and (1.4), and performing a little algebra, we obtain the following transcen- dental relation on the allowed energies e- 2 κa = ± 2 κ κ- 1 (1.5) 1 which also constrains B n = ± C n . Thus, an even and an odd solution are possible ψ ± ( x ) = A ± e κx , x ≤ - a B ± ( e κx ± e- κx ) ,- a ≤ x ≤ a ± A ± e- κx , x ≥ a . (1.6) Notice, however, that for κ a < 1 the odd solution disappears from the spectrum. This can be seen as follows: we know κ = 0 is a trivial solution of e- 2 κa = 1- 2 κ/κ , now consider the derivatives of both sides at 0. If the RHS has a the greater slope, then we are guaranteed that the two curves intersect again with κ > 0. Otherwise, no solution exists (with κ > 0). In particular, as a → 0 the energies must satisfy 1 = ± 2 κ κ- 1 . (1.7) For the even solution, this basically reduces to the single delta-function potential result: κ = κ (though now g/ra 2 g ). In terms of the energy E + =- 2 m | g | ~ 2 . (1.8) For the odd solution, there is only the trivial solution κ = E- = 0. This limit agrees with our intuition, since a → 0 reduces to parts ( i ) and ( ii ) (with the modified g ). 2 Transmission Through Multiple Barriers Note: I should begin by pointing out that when I discussed this problem with the professor, the intention was to find k ’s (in the region of each barrier) which are not integer multiples of each other, as opposed to the energies (away from the barriers). Unfortunately, this isn’t how the question was posed. I’ll attempt to answer both “interpretations” here.how the question was posed....
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