130_notes.dvi

=

[

−a^2 ̄h^2

2 m

+

1

2

mω^2

]

∞∫

−∞

x^2 e−ax

2

dx

∞∫

−∞

e−ax^2 dx

+

[

̄h^2 a

2 m

]

∫∞

−∞

e−ax

2

dx=

√

π

a

=

√

πa−^1 /^2

−

∫∞

−∞

x^2 e−ax

2

dx=

√

π

(

−

1

2

)

a−^3 /^2

∫∞

−∞

x^2 e−ax

2

dx=

1

2

√

π

a^3

=

1

2 a

√

π

a

E′=

[

−a ̄h^2

4 m

+

1

4 a

mω^2

]

+

̄h^2 a

2 m

=

1

4 a

mω^2 +

̄h^2

4 m

a

∂E′

∂a

=

−mω^2

4 a^2

+

̄h^2

4 m

= 0

4 a^2 ̄h^2 = 4m^2 ω^2

a=

mω

̄h

ψ=e−

mω2 ̄hx 2

E′=

mω^2

4

̄h

mω

+

̄h^2

4 m

mω

̄h

=

1

4

̄hω+

1

4

̄hω

OK.

25.7 Derivations and Computations

25.7.1 Calculation of the ground state energy shift

To calculate the first order correction to the He ground state energy, we gotta do this integral.

∆Egs=〈u 0 |V|u 0 〉=

∫

d^3 r 1 d^3 r 2 |φ 100 (~r 1 )|^2 |φ 100 (~r 2 )|^2

e^2

|~r 1 −~r 2 |

First, plug in the Hydrogen ground state wave function (twice).

∆Egs=

[

1

4 π

4

(

Z

a 0

) 3 ]^2

e^2

∫∞

0

r^21 dr 1 e−^2 Zr^1 /a^0

∫∞

0

r 22 dr 1 e−^2 Zr^2 /a^0

∫

dΩ 1

∫

dΩ 2

1

|~r 1 −~r 2 |