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# solution2 - 4 Use the variational method with wave...

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4. Use the variational method with wave functions of the form φ a ( r )= r a π 2 1 r 2 + a 2 and φ a ( r e r/a π a 3 to estimate the ground state energy of a particle subject to a Coulomb potential V ( r e 2 /r. After verifying that the wave function is properly normalized in 3D we evaluate ~ φ = φ 0 ˆ r φ 0 = r a π 2 2 r ( r 2 + a 2 ) 2 from which, with u = r/a, 1 2 m h P 2 i α = ~ 2 2 m Z d 3 r | ~ φ | 2 = ~ 2 2 m Z d 3 r ¯ ¯ φ 0 ¯ ¯ 2 = ~ 2 2 m 4 a π 2 Z d 3 r r 2 ( r 2 + a 2 ) 4 = 2 ~ 2 a m π 2 (4 π ) Z 0 r 4 ( r 2 + a 2 ) 4 dr = 8 ~ 2 m π a 2 Z 0 u 4 (1 + u 2 ) 4 du = ~ 2 4 ma 2 Similarly, we f nd h V i α = e 2 Z d 3 rr 1 | φ | 2 = 4 π e 2 a π 2 Z 0 r ( r 2 + a 2 ) 2 dr = 4 e 2 π a Z 0 u (1 + u 2 ) 2 du = 2 π e 2 a . Setting h H i a = ~ 2 4 ma 2 2 π e 2 a h H i a a =0= 1 2 μ ~ 2 π +4 e 2 ma ma 3 π we f nd ~ 2 4 ma 2 = e 2 π a a = ~ 2 π 4 me 2 From this value of a, we obtain the following estimate (upper bound) for the ground state energy: E var 0 = ~ 2 4 ma 2 2 π e 2 a = e 2 π a = 8 π 2 me 4 2 ~ 2 Repeating this with the normalized wave function φ = e r/a / π a 3 , f nd that ~ φ = a 1 φ ˆ r ,so 1 2 m h P 2 i α = ~ 2 2 m Z d 3 r | ~ φ | 2 = ~ 2 2 ma 2 Z d 3 r | φ | 2 = ~ 2 2 ma 2 . Then, with h V i α = e 2 Z d 3 1 | φ | 2 = 4 π e 2 π a 3 Z 0 re r/a dr = e 2 a Z 0 ue u du = e 2 a , we set h H i a = ~ 2 2 ma 2 e 2 a h H i a a ~ 2 + e 2 ma ma 3 . and deduce ~ 2 2 ma 2 = e 2 2 a a = ~ 2 me 2 . Thus, E 0 = ~ 2 2 ma 2 e 2 a = e 2 2 a = me 4 2 ~ 2 , which is the exact result for the problem. The Lorentizian wave function gives an estimate equal to ¡ 8 / π 2 ¢ E 0 0 .

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solution2 - 4 Use the variational method with wave...

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