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

This preview shows pages 1–3. Sign up to view the full content.

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

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
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 .
This is the end of the preview. Sign up to access the rest of the document.

## 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.

### Page1 / 7

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

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online