hw211a - Physics 512 Homework Set #11 Due Friday, April 4...

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Physics 512 Winter 2003 Homework Set #11 – Due Friday, April 4 1. This is similar to Sakurai, Chapter 6, Problem 7. Two identical spin- 1 2 fermions are placed in a one-dimensional inFnite square well of size L V ( x )= ± ,x< 0or x>L ; 0 , 0 <x<L We assume that the two particles interact mutually via a δ -function potential given by V int ( x 1 ,x 2 λδ ( x 1 x 2 ) ±ind the (complete, not just ground state) eigenenergies of this system, including the Frst order energy shift from V int . Take into account both triplet and singlet spin combinations. The single particle states have the familiar form φ n ( x r 2 L sin nπx L ,E n = ¯ h 2 π 2 2 mL 2 n 2 where n =1 , 2 , 3 ,... . Hence for identical fermions, we may form the singlet and triplet combinations ψ s = 1 2 ² φ n 1 ( x 1 ) φ n 2 ( x 2 )+ φ n 2 ( x 1 ) φ n 1 ( x 2 ) ³ χ singlet ψ t = 1 2 ² φ n 1 ( x 1 ) φ n 2 ( x 2 ) φ n 2 ( x 1 ) φ n 1 ( x 2 ) ³ χ triplet all with unperturbed energy E (0) n 1 ,n 2 = ¯ h 2 π 2 2 mL 2 ( n 2 1 + n 2 2 ) Note that only the singlet combination survives when n 1 = n 2 . Furthermore, in this case, the normalization factor 1 / 2 must be replaced by 1 / 2 (since φ n 1 and φ n 2 are no longer orthogonal when n 1 = n 2 ). Applying ±rst order time independent perturbation theory with V int , we ±nd E (1) n 1 ,n 2 = h ψ | V int | ψ i = λ 2 ´ 2 L µ 2 Z L 0 Z L 0 h sin n 1 πx 1 L sin n 2 2 L ± sin n 2 1 L sin n 1 2 L i 2 × δ ( x 1 x 2 ) dx 1 dx 2 = 2 λ L 2 Z L 0 h sin n 1 1 L sin n 2 1 L ± sin n 2 1 L sin n 1 1 L i 2 dx 1 1
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This clearly vanishes for the space antisymmetric combination. Hence only the singlet combination picks up an energy shift at Frst order. We Fnd E (1) n 1 ,n 2 = 8 λ L 2 Z L 0 sin 2 n 1 πx 1 L sin 2 n 2 1 L dx 1 = 2 λ L 2 Z L 0 ± 1 cos 2 n 1 1 L ²± 1 cos 2 n 2 1 L ² dx 1 = 2 λ L 2 Z L 0 ³ 1 cos 2 n 1 1 L cos 2 n 2 1 L + 1 2 cos 2( n 1 + n 2 ) 1 L + 1 2 cos 2( n 1 n 2 ) 1 L ´ dx 1 = 2 λ L (1 + 1 2 δ n 1 ,n 2 ) (since n 1 and n 2 are positive integers and we are integrating over complete periods of cos ). Reintroducing a factor of 1 / 2 for the case when n 1 = n 2 (to correct for the normalization), we Fnd the Frst order energy shifts E (1) n 1 ,n 2 = µ 2 λ/L, n 1 6 = n 2 3 λ/ 2 L, n 1 = n 2 ( singlet only ) Thus the low-lying energy spectrum looks like with V int E (0) n 1 n 2 2 1 2 2 3 1 1 1 2 5 8 10 s s s s t t (where t and s denote triplet and singlet states). It is straightforward to understand why only the singlet states feel the perturbation; for the triplet, the space antisymmetric wave- function vanishes when x 1 = x 2 , and as a result is una±ected by V int .
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This note was uploaded on 02/15/2011 for the course PHYS 512 taught by Professor Unknow during the Winter '03 term at Cornell College.

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hw211a - Physics 512 Homework Set #11 Due Friday, April 4...

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