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# 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 infinite square well of size L V ( x ) = , x < 0 or 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 ) Find the (complete, not just ground state) eigenenergies of this system, including the first 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 ) = 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 φ n 1 ( x 1 ) φ n 2 ( x 2 ) + φ n 2 ( x 1 ) φ n 1 ( x 2 ) χ singlet ψ t = 1 2 φ n 1 ( x 1 ) φ n 2 ( x 2 ) φ n 2 ( x 1 ) φ n 1 ( x 2 ) 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 first order time independent perturbation theory with V int , we find E (1) n 1 ,n 2 = ψ | V int | ψ = λ 2 2 L 2 L 0 L 0 sin n 1 πx 1 L sin n 2 πx 2 L ± sin n 2 πx 1 L sin n 1 πx 2 L 2 × δ ( x 1 x 2 ) dx 1 dx 2 = 2 λ L 2 L 0 sin n 1 πx 1 L sin n 2 πx 1 L ± sin n 2 πx 1 L sin n 1 πx 1 L 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 first order. We find E (1) n 1 ,n 2 = 8 λ L 2 L 0 sin 2 n 1 πx 1 L sin 2 n 2 πx 1 L dx 1 = 2 λ L 2 L 0 1 cos 2 n 1 πx 1 L 1 cos 2 n 2 πx 1 L dx 1 = 2 λ L 2 L 0 1 cos 2 n 1 πx 1 L cos 2 n 2 πx 1 L + 1 2 cos 2( n 1 + n 2 ) πx 1 L + 1 2 cos 2( n 1 n 2 ) πx 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 find the first order energy shifts E (1) n 1 ,n 2 = 2 λ/L, n 1 = 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 unaffected by V int .
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hw211a - Physics 512 Homework Set#11 Due Friday April 4...

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