Today’s Outline - February 17, 2014csrri.iit.edu/~segre/phys406/14S/lecture_09.pdfToday’s Outline - February 17, 2014 Hydrogen molecule ion Problem 7.14 Problem 7.15 Midterm Exam

Today’s Outline - February 17, 2014

• Hydrogen molecule ion

• Problem 7.14

• Problem 7.15

Midterm Exam #1Monday, February 24, 2014Covers through HW #04

Tutoring sessions:Thursday & Friday, 12:00–13:45, 121 E1

C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 1 / 12

Hydrogen molecule ion trial function

The Hamiltonian for the hydrogen molecule ion has two protons and oneelectron. The vectors between the electron and the individual protons arelabeled as ~r1 and ~r2.

H = − ~2

2m∇2 − e

2

4π�0

(1

r1+

1

r2

)The trial solution is a linear combination of atomic orbitals centered abouteach of the protons which must be normalized first to get A.

ψ ≡ A [ψ0(r1) + ψ0(r2)] =A√πa3

[e−r1/a + e−r2/a

]1 =

∫|ψ|2d3~r

= |A|2[∫|ψ0(r1)|2d3~r +

∫|ψ0(r2)|2d3~r + 2

∫ψ0(r1)ψ0(r2)d

3~r

]= |A|2

[1 + 1

+ 2

∫ψ0(r1)ψ0(r2)d

3~r

]= 2|A|2 [1 + I ]




H = − ~2

2m∇2 − e

2

4π�0

(1

r1+

1

r2

)

The trial solution is a linear combination of atomic orbitals centered abouteach of the protons which must be normalized first to get A.

ψ ≡ A [ψ0(r1) + ψ0(r2)] =A√πa3

[e−r1/a + e−r2/a

]1 =

∫|ψ|2d3~r

= |A|2[∫|ψ0(r1)|2d3~r +

∫|ψ0(r2)|2d3~r + 2

∫ψ0(r1)ψ0(r2)d

3~r

]= |A|2

[1 + 1

+ 2

∫ψ0(r1)ψ0(r2)d

3~r

]= 2|A|2 [1 + I ]




H = − ~2

2m∇2 − e

2

4π�0

(1

r1+

1

r2

)The trial solution is a linear combination of atomic orbitals centered abouteach of the protons

which must be normalized first to get A.

ψ ≡ A [ψ0(r1) + ψ0(r2)] =A√πa3

[e−r1/a + e−r2/a

]1 =

∫|ψ|2d3~r

= |A|2[∫|ψ0(r1)|2d3~r +

∫|ψ0(r2)|2d3~r + 2

∫ψ0(r1)ψ0(r2)d

3~r

]= |A|2

[1 + 1

+ 2

∫ψ0(r1)ψ0(r2)d

3~r

]= 2|A|2 [1 + I ]




H = − ~2

2m∇2 − e

2

4π�0

(1

r1+

1

r2



ψ ≡ A [ψ0(r1) + ψ0(r2)]

=A√πa3

[e−r1/a + e−r2/a

]1 =

∫|ψ|2d3~r

= |A|2[∫|ψ0(r1)|2d3~r +

∫|ψ0(r2)|2d3~r + 2

∫ψ0(r1)ψ0(r2)d

3~r

]= |A|2

[1 + 1

+ 2

∫ψ0(r1)ψ0(r2)d

3~r

]= 2|A|2 [1 + I ]




H = − ~2

2m∇2 − e

2

4π�0

(1

r1+

1

r2



ψ ≡ A [ψ0(r1) + ψ0(r2)] =A√πa3

[e−r1/a + e−r2/a

]

1 =

∫|ψ|2d3~r

= |A|2[∫|ψ0(r1)|2d3~r +

∫|ψ0(r2)|2d3~r + 2

∫ψ0(r1)ψ0(r2)d

3~r

]= |A|2

[1 + 1

+ 2

∫ψ0(r1)ψ0(r2)d

3~r

]= 2|A|2 [1 + I ]




H = − ~2

2m∇2 − e

2

4π�0

(1

r1+

1

r2


ψ ≡ A [ψ0(r1) + ψ0(r2)] =A√πa3

[e−r1/a + e−r2/a

]

1 =

∫|ψ|2d3~r

= |A|2[∫|ψ0(r1)|2d3~r +

∫|ψ0(r2)|2d3~r + 2

∫ψ0(r1)ψ0(r2)d

3~r

]= |A|2

[1 + 1

+ 2

∫ψ0(r1)ψ0(r2)d

3~r

]= 2|A|2 [1 + I ]




H = − ~2

2m∇2 − e

2

4π�0

(1

r1+

1

r2


ψ ≡ A [ψ0(r1) + ψ0(r2)] =A√πa3

[e−r1/a + e−r2/a

]1 =

∫|ψ|2d3~r

= |A|2[∫|ψ0(r1)|2d3~r +

∫|ψ0(r2)|2d3~r + 2

∫ψ0(r1)ψ0(r2)d

3~r

]= |A|2

[1 + 1

+ 2

∫ψ0(r1)ψ0(r2)d

3~r

]= 2|A|2 [1 + I ]




H = − ~2

2m∇2 − e

2

4π�0

(1

r1+

1

r2


ψ ≡ A [ψ0(r1) + ψ0(r2)] =A√πa3

[e−r1/a + e−r2/a

]1 =

∫|ψ|2d3~r

= |A|2[∫|ψ0(r1)|2d3~r +

∫|ψ0(r2)|2d3~r + 2

∫ψ0(r1)ψ0(r2)d

3~r

]

= |A|2[

1 + 1

+ 2

∫ψ0(r1)ψ0(r2)d

3~r

]= 2|A|2 [1 + I ]




H = − ~2

2m∇2 − e

2

4π�0

(1

r1+

1

r2


ψ ≡ A [ψ0(r1) + ψ0(r2)] =A√πa3

[e−r1/a + e−r2/a

]1 =

∫|ψ|2d3~r

= |A|2[∫|ψ0(r1)|2d3~r +

∫|ψ0(r2)|2d3~r + 2

∫ψ0(r1)ψ0(r2)d

3~r

]= |A|2

[1 + 1

+ 2

∫ψ0(r1)ψ0(r2)d

3~r

]

= 2|A|2 [1 + I ]




H = − ~2

2m∇2 − e

2

4π�0

(1

r1+

1

r2


ψ ≡ A [ψ0(r1) + ψ0(r2)] =A√πa3

[e−r1/a + e−r2/a

]1 =

∫|ψ|2d3~r

= |A|2[∫|ψ0(r1)|2d3~r +

∫|ψ0(r2)|2d3~r + 2

∫ψ0(r1)ψ0(r2)d

3~r

]= |A|2

[1 + 1 + 2

∫ψ0(r1)ψ0(r2)d

3~r

]

= 2|A|2 [1 + I ]




H = − ~2

2m∇2 − e

2

4π�0

(1

r1+

1

r2


ψ ≡ A [ψ0(r1) + ψ0(r2)] =A√πa3

[e−r1/a + e−r2/a

]1 =

∫|ψ|2d3~r

= |A|2[∫|ψ0(r1)|2d3~r +

∫|ψ0(r2)|2d3~r + 2

∫ψ0(r1)ψ0(r2)d

3~r

]= |A|2

[1 + 1 + 2

∫ψ0(r1)ψ0(r2)d

3~r

]= 2|A|2 [1 + I ]


Hydrogen molecule ion normalization

I =1

πa3

∫e−(r1+r2)/ar2 sin θ dr dθ dφ

R

r2

r1θ

z

y

x

2

1

The ideal coordinate system for this integral isto align the vector between the protons alongthe z-axis which gives

r1 ≡ r , r2 =√

r2 + R2 − 2rR cos θ

The φ integration gives 2π

I =2π

πa3

∫ ∞0

∫ π0

e−r/ae−√r2+R2−2rR cos θ/ar2 sin θ dr dθ

Let’s look at the θ integral first...



I =1

πa3


R

r2

r1θ

z

y

x

2

1


r1 ≡ r ,

r2 =√



I =2π

πa3

∫ ∞0

∫ π0





I =1

πa3


R

r2

r1θ

z

y

x

2

1


r1 ≡ r , r2 =√



I =2π

πa3

∫ ∞0

∫ π0




The overlap integral

The θ integral can be simplified byusing the substitution

y ≡√r2 + R2 − 2rR cos θ

d(y2) = 2y dy

= 2rR sin θ dθ

Iθ =

∫ π0

e−√r2+R2−2rR cos θ/a sin θ dθ

=1

rR

∫ r+R|r−R|

ye−y/a dy

=1

rR

[−aye−y/a

∣∣∣r+R|r−R|

+ a

∫ r+R|r−R|

e−y/a dy

]=

a

rR

[−(r + R)e−(r+R)/a + |r − R|e−|r−R|/a

− ae−(r+R)/a + ae−|r−R|/a

]= − a

rR

[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a

]The overlap integral thus becomes

I = − 2Ra2

∫ ∞0

r[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a

]e−r/a dr




y ≡√r2 + R2 − 2rR cos θ

d(y2) = 2y dy = 2rR sin θ dθ

Iθ =

∫ π0

e−√r2+R2−2rR cos θ/a sin θ dθ

=1

rR

∫ r+R|r−R|

ye−y/a dy

=1

rR

[−aye−y/a

∣∣∣r+R|r−R|

+ a

∫ r+R|r−R|

e−y/a dy

]=

a

rR

[−(r + R)e−(r+R)/a + |r − R|e−|r−R|/a


]= − a

rR

[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a


I = − 2Ra2

∫ ∞0


]e−r/a dr




y ≡√r2 + R2 − 2rR cos θ


Iθ =

∫ π0

e−√r2+R2−2rR cos θ/a sin θ dθ =

1

rR

∫ r+R|r−R|

ye−y/a dy

=1

rR

[−aye−y/a

∣∣∣r+R|r−R|

+ a

∫ r+R|r−R|

e−y/a dy

]=

a

rR

[−(r + R)e−(r+R)/a + |r − R|e−|r−R|/a


]= − a

rR

[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a


I = − 2Ra2

∫ ∞0


]e−r/a dr




y ≡√r2 + R2 − 2rR cos θ


Iθ =

∫ π0


1

rR

∫ r+R|r−R|

ye−y/a dy

=1

rR

[−aye−y/a

∣∣∣r+R|r−R|

+ a

∫ r+R|r−R|

e−y/a dy

]

=a

rR

[−(r + R)e−(r+R)/a + |r − R|e−|r−R|/a


]= − a

rR

[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a


I = − 2Ra2

∫ ∞0


]e−r/a dr




y ≡√r2 + R2 − 2rR cos θ


Iθ =

∫ π0


1

rR

∫ r+R|r−R|

ye−y/a dy

=1

rR

[−aye−y/a

∣∣∣r+R|r−R|

+ a

∫ r+R|r−R|

e−y/a dy

]=

a

rR

[−(r + R)e−(r+R)/a + |r − R|e−|r−R|/a


]

= − arR

[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a


I = − 2Ra2

∫ ∞0


]e−r/a dr




y ≡√r2 + R2 − 2rR cos θ


Iθ =

∫ π0


1

rR

∫ r+R|r−R|

ye−y/a dy

=1

rR

[−aye−y/a

∣∣∣r+R|r−R|

+ a

∫ r+R|r−R|

e−y/a dy

]=

a

rR

[−(r + R)e−(r+R)/a + |r − R|e−|r−R|/a − ae−(r+R)/a + ae−|r−R|/a

]

= − arR

[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a


I = − 2Ra2

∫ ∞0


]e−r/a dr




y ≡√r2 + R2 − 2rR cos θ


Iθ =

∫ π0


1

rR

∫ r+R|r−R|

ye−y/a dy

=1

rR

[−aye−y/a

∣∣∣r+R|r−R|

+ a

∫ r+R|r−R|

e−y/a dy

]=

a

rR


]= − a

rR

[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a

]

The overlap integral thus becomes

I = − 2Ra2

∫ ∞0


]e−r/a dr




y ≡√r2 + R2 − 2rR cos θ


Iθ =

∫ π0


1

rR

∫ r+R|r−R|

ye−y/a dy

=1

rR

[−aye−y/a

∣∣∣r+R|r−R|

+ a

∫ r+R|r−R|

e−y/a dy

]=

a

rR


]= − a

rR

[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a


I = − 2Ra2

∫ ∞0


]e−r/a dr


More integration...

I = − 2Ra2

∫ ∞0


]e−r/a dr

= − 2Ra2

{

e−R/a∫ ∞0

[r2 + r(R + a)

]e−2r/a dr

−e−R/a∫ R0

[r(R + a)− r2

]dr − eR/a

∫ ∞R

[r2 − r(R − a)

]e−2r/a dr

}= − 2

Ra2

{e−R/a

[−2a

3

8− a

2

4(R + a)

]e−2r/a

∣∣∣∣∞0

−e−R/a[r2

2(R + a)− r

3

3

∣∣∣∣R0

− eR/a[−a

2r2 − 2a

2

4r − 2a

3

8+

a

2r(R − a) + a

2

4(R − a)

]e−2r/a

∣∣∣∣∞R

}= − 2

Ra2e−R/a

[

a3

4+

a2R

4+

a3

4−R

3

2− aR

2

2+

R3

3

−aR2

2− a

2R

2− a

3

4+

aR2

2− a

2R

2+

a2R

4− a

3

4

]


More integration...

I = − 2Ra2

∫ ∞0


]e−r/a dr

= − 2Ra2

{

e−R/a∫ ∞0

[r2 + r(R + a)

]e−2r/a dr

−e−R/a∫ R0

[r(R + a)− r2

]dr − eR/a

∫ ∞R

[r2 − r(R − a)

]e−2r/a dr

}

= − 2Ra2

{e−R/a

[−2a

3

8− a

2

4(R + a)

]e−2r/a

∣∣∣∣∞0

−e−R/a[r2

2(R + a)− r

3

3

∣∣∣∣R0

− eR/a[−a

2r2 − 2a

2

4r − 2a

3

8+

a

2r(R − a) + a

2

4(R − a)

]e−2r/a

∣∣∣∣∞R

}= − 2

Ra2e−R/a

[

a3

4+

a2R

4+

a3

4−R

3

2− aR

2

2+

R3

3

−aR2

2− a

2R

2− a

3

4+

aR2

2− a

2R

2+

a2R

4− a

3

4

]


More integration...

I = − 2Ra2

∫ ∞0


]e−r/a dr

= − 2Ra2

{e−R/a

∫ ∞0

[r2 + r(R + a)

]e−2r/a dr

−e−R/a∫ R0

[r(R + a)− r2

]dr − eR/a

∫ ∞R

[r2 − r(R − a)

]e−2r/a dr

}

= − 2Ra2

{e−R/a

[−2a

3

8− a

2

4(R + a)

]e−2r/a

∣∣∣∣∞0

−e−R/a[r2

2(R + a)− r

3

3

∣∣∣∣R0

− eR/a[−a

2r2 − 2a

2

4r − 2a

3

8+

a

2r(R − a) + a

2

4(R − a)

]e−2r/a

∣∣∣∣∞R

}= − 2

Ra2e−R/a

[

a3

4+

a2R

4+

a3

4−R

3

2− aR

2

2+

R3

3

−aR2

2− a

2R

2− a

3

4+

aR2

2− a

2R

2+

a2R

4− a

3

4

]


More integration...

I = − 2Ra2

∫ ∞0


]e−r/a dr

= − 2Ra2

{e−R/a

∫ ∞0

[r2 + r(R + a)

]e−2r/a dr

−e−R/a∫ R0

[r(R + a)− r2

]dr

− eR/a∫ ∞R

[r2 − r(R − a)

]e−2r/a dr

}

= − 2Ra2

{e−R/a

[−2a

3

8− a

2

4(R + a)

]e−2r/a

∣∣∣∣∞0

−e−R/a[r2

2(R + a)− r

3

3

∣∣∣∣R0

− eR/a[−a

2r2 − 2a

2

4r − 2a

3

8+

a

2r(R − a) + a

2

4(R − a)

]e−2r/a

∣∣∣∣∞R

}= − 2

Ra2e−R/a

[

a3

4+

a2R

4+

a3

4−R

3

2− aR

2

2+

R3

3

−aR2

2− a

2R

2− a

3

4+

aR2

2− a

2R

2+

a2R

4− a

3

4

]


More integration...

I = − 2Ra2

∫ ∞0


]e−r/a dr

= − 2Ra2

{e−R/a

∫ ∞0

[r2 + r(R + a)

]e−2r/a dr

−e−R/a∫ R0

[r(R + a)− r2

]dr − eR/a

∫ ∞R

[r2 − r(R − a)

]e−2r/a dr

}

= − 2Ra2

{e−R/a

[−2a

3

8− a

2

4(R + a)

]e−2r/a

∣∣∣∣∞0

−e−R/a[r2

2(R + a)− r

3

3

∣∣∣∣R0

− eR/a[−a

2r2 − 2a

2

4r − 2a

3

8+

a

2r(R − a) + a

2

4(R − a)

]e−2r/a

∣∣∣∣∞R

}= − 2

Ra2e−R/a

[

a3

4+

a2R

4+

a3

4−R

3

2− aR

2

2+

R3

3

−aR2

2− a

2R

2− a

3

4+

aR2

2− a

2R

2+

a2R

4− a

3

4

]


More integration...

I = − 2Ra2

∫ ∞0


]e−r/a dr

= − 2Ra2

{e−R/a

∫ ∞0

[r2 + r(R + a)

]e−2r/a dr

−e−R/a∫ R0

[r(R + a)− r2

]dr − eR/a

∫ ∞R

[r2 − r(R − a)

]e−2r/a dr

}= − 2

Ra2

{e−R/a

[−2a

3

8− a

2

4(R + a)

]e−2r/a

∣∣∣∣∞0

−e−R/a[r2

2(R + a)− r

3

3

∣∣∣∣R0

− eR/a[−a

2r2 − 2a

2

4r − 2a

3

8+

a

2r(R − a) + a

2

4(R − a)

]e−2r/a

∣∣∣∣∞R

}

= − 2Ra2

e−R/a[

a3

4+

a2R

4+

a3

4−R

3

2− aR

2

2+

R3

3

−aR2

2− a

2R

2− a

3

4+

aR2

2− a

2R

2+

a2R

4− a

3

4

]


More integration...

I = − 2Ra2

∫ ∞0


]e−r/a dr

= − 2Ra2

{e−R/a

∫ ∞0

[r2 + r(R + a)

]e−2r/a dr

−e−R/a∫ R0

[r(R + a)− r2

]dr − eR/a

∫ ∞R

[r2 − r(R − a)

]e−2r/a dr

}= − 2

Ra2

{e−R/a

[−2a

3

8− a

2

4(R + a)

]e−2r/a

∣∣∣∣∞0

−e−R/a[r2

2(R + a)− r

3

3

∣∣∣∣R0

− eR/a[−a

2r2 − 2a

2

4r − 2a

3

8+

a

2r(R − a) + a

2

4(R − a)

]e−2r/a

∣∣∣∣∞R

}= − 2

Ra2e−R/a

[

a3

4+

a2R

4+

a3

4−R

3

2− aR

2

2+

R3

3

−aR2

2− a

2R

2− a

3

4+

aR2

2− a

2R

2+

a2R

4− a

3

4

]C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 5 / 12

More integration...

I = − 2Ra2

∫ ∞0


]e−r/a dr

= − 2Ra2

{e−R/a

∫ ∞0

[r2 + r(R + a)

]e−2r/a dr

−e−R/a∫ R0

[r(R + a)− r2

]dr − eR/a

∫ ∞R

[r2 − r(R − a)

]e−2r/a dr

}= − 2

Ra2

{e−R/a

[−2a

3

8− a

2

4(R + a)

]e−2r/a

∣∣∣∣∞0

−e−R/a[r2

2(R + a)− r

3

3

∣∣∣∣R0

− eR/a[−a

2r2 − 2a

2

4r − 2a

3

8+

a

2r(R − a) + a

2

4(R − a)

]e−2r/a

∣∣∣∣∞R

}= − 2

Ra2e−R/a

[a3

4+

a2R

4+

a3

4

−R3

2− aR

2

2+

R3

3

−aR2

2− a

2R

2− a

3

4+

aR2

2− a

2R

2+

a2R

4− a

3

4


More integration...

I = − 2Ra2

∫ ∞0


]e−r/a dr

= − 2Ra2

{e−R/a

∫ ∞0

[r2 + r(R + a)

]e−2r/a dr

−e−R/a∫ R0

[r(R + a)− r2

]dr − eR/a

∫ ∞R

[r2 − r(R − a)

]e−2r/a dr

}= − 2

Ra2

{e−R/a

[−2a

3

8− a

2

4(R + a)

]e−2r/a

∣∣∣∣∞0

−e−R/a[r2

2(R + a)− r

3

3

∣∣∣∣R0

− eR/a[−a

2r2 − 2a

2

4r − 2a

3

8+

a

2r(R − a) + a

2

4(R − a)

]e−2r/a

∣∣∣∣∞R

}= − 2

Ra2e−R/a

[a3

4+

a2R

4+

a3

4−R

3

2− aR

2

2+

R3

3

−aR2

2− a

2R

2− a

3

4+

aR2

2− a

2R

2+

a2R

4− a

3

4


Normalization, finally!

Gathering all the like terms

and the normalization con-stant is expressed in terms ofthe overlap integral

Now to calculate the expec-tation value of the Hamilto-nian

I =2

Ra2e−R/a

[a2R

2+

aR2

2+

R3

6

]= e−R/a

[1 +

(R

a

)+

1

3

(R

a

)2]|A|2 = 1

2(1 + I )

Hψ =

[− ~

2

2m∇2 − e

2

4π�0

(1

r1+

1

r2

)]A [ψ0(r1) + ψ0(r2)]

= − ~2

2m∇2Aψ0(r1)−

~2

2m∇2Aψ0(r2)

− A e2

4π�0

(1

r1ψ0(r1) +

1

r1ψ0(r2) +

1

r2ψ0(r1) +

1

r2ψ0(r2)

)= E1ψ0(r1) + E1ψ0(r2)− A

e2

4π�0

[1

r1ψ0(r2) +

1

r2ψ0(r1)

]






I =2

Ra2e−R/a

[a2R

2+

aR2

2+

R3

6

]

= e−R/a

[1 +

(R

a

)+

1

3

(R

a

)2]|A|2 = 1

2(1 + I )

Hψ =

[− ~

2

2m∇2 − e

2

4π�0

(1

r1+

1

r2

)]A [ψ0(r1) + ψ0(r2)]

= − ~2

2m∇2Aψ0(r1)−

~2

2m∇2Aψ0(r2)

− A e2

4π�0

(1

r1ψ0(r1) +

1

r1ψ0(r2) +

1

r2ψ0(r1) +

1

r2ψ0(r2)

)= E1ψ0(r1) + E1ψ0(r2)− A

e2

4π�0

[1

r1ψ0(r2) +

1

r2ψ0(r1)

]






I =2

Ra2e−R/a

[a2R

2+

aR2

2+

R3

6

]= e−R/a

[1 +

(R

a

)+

1

3

(R

a

)2]

|A|2 = 12(1 + I )

Hψ =

[− ~

2

2m∇2 − e

2

4π�0

(1

r1+

1

r2

)]A [ψ0(r1) + ψ0(r2)]

= − ~2

2m∇2Aψ0(r1)−

~2

2m∇2Aψ0(r2)

− A e2

4π�0

(1

r1ψ0(r1) +

1

r1ψ0(r2) +

1

r2ψ0(r1) +

1

r2ψ0(r2)

)= E1ψ0(r1) + E1ψ0(r2)− A

e2

4π�0

[1

r1ψ0(r2) +

1

r2ψ0(r1)

]






I =2

Ra2e−R/a

[a2R

2+

aR2

2+

R3

6

]= e−R/a

[1 +

(R

a

)+

1

3

(R

a

)2]|A|2 = 1

2(1 + I )

Hψ =

[− ~

2

2m∇2 − e

2

4π�0

(1

r1+

1

r2

)]A [ψ0(r1) + ψ0(r2)]

= − ~2

2m∇2Aψ0(r1)−

~2

2m∇2Aψ0(r2)

− A e2

4π�0

(1

r1ψ0(r1) +

1

r1ψ0(r2) +

1

r2ψ0(r1) +

1

r2ψ0(r2)

)= E1ψ0(r1) + E1ψ0(r2)− A

e2

4π�0

[1

r1ψ0(r2) +

1

r2ψ0(r1)

]






I =2

Ra2e−R/a

[a2R

2+

aR2

2+

R3

6

]= e−R/a

[1 +

(R

a

)+

1

3

(R

a

)2]|A|2 = 1

2(1 + I )

Hψ =

[− ~

2

2m∇2 − e

2

4π�0

(1

r1+

1

r2

)]A [ψ0(r1) + ψ0(r2)]

= − ~2

2m∇2Aψ0(r1)−

~2

2m∇2Aψ0(r2)

− A e2

4π�0

(1

r1ψ0(r1) +

1

r1ψ0(r2) +

1

r2ψ0(r1) +

1

r2ψ0(r2)

)

= E1ψ0(r1) + E1ψ0(r2)− Ae2

4π�0

[1

r1ψ0(r2) +

1

r2ψ0(r1)

]






I =2

Ra2e−R/a

[a2R

2+

aR2

2+

R3

6

]= e−R/a

[1 +

(R

a

)+

1

3

(R

a

)2]|A|2 = 1

2(1 + I )

Hψ =

[− ~

2

2m∇2 − e

2

4π�0

(1

r1+

1

r2

)]A [ψ0(r1) + ψ0(r2)]

= − ~2

2m∇2Aψ0(r1)−

~2

2m∇2Aψ0(r2)

− A e2

4π�0

(1

r1ψ0(r1) +

1

r1ψ0(r2) +

1

r2ψ0(r1) +

1

r2ψ0(r2)

)= E1ψ0(r1) + E1ψ0(r2)− A

e2

4π�0

[1

r1ψ0(r2) +

1

r2ψ0(r1)


Expectation value of H

〈H〉 = E1 − 2|A|2e2

4π�0

[〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉+〈ψ0(r1) ∣∣∣∣ 1r1

∣∣∣∣ψ0(r2)〉]

The second term is called thedirect integral

and the third term is the ex-change integral

The final expression for theexpectation value of theHamiltonian becomes

We also have to include theproton-proton repulsion en-ergy

D ≡ a〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉

=a

R−(

1 +a

R

)e−2R/a

X ≡ a〈ψ0(r1)

∣∣∣∣ 1r1∣∣∣∣ψ0(r2)〉

=

(1 +

R

a

)e−R/a

〈H〉 =[

1 + 2D + X

1 + I

]E1

Vpp =e2

4π�0

1

R= −2a

RE1



〈H〉 = E1 − 2|A|2e2

4π�0

[〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉+〈ψ0(r1) ∣∣∣∣ 1r1

∣∣∣∣ψ0(r2)〉]The second term is called thedirect integral




D ≡ a〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉

=a

R−(

1 +a

R

)e−2R/a

X ≡ a〈ψ0(r1)

∣∣∣∣ 1r1∣∣∣∣ψ0(r2)〉

=

(1 +

R

a

)e−R/a

〈H〉 =[

1 + 2D + X

1 + I

]E1

Vpp =e2

4π�0

1

R= −2a

RE1


Direct & exchange integrals

D = a

〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉

= a

〈ψ0(r2)

∣∣∣∣ 1r1∣∣∣∣ψ0(r2)〉

= a1

πa3

∫e−2r2/a

1

r1d3r =

1

πa2

∫r e−

2a

√r2+R2−2rR cos θ sin θ dr dθ dφ

=2π

πa2

∫ ∞0

r

[∫ π0

e−2a

√r2+R2−2rR cos θ sin θ dθ

]dr

X =

〈ψ0(r1)

∣∣∣∣ 1r1∣∣∣∣ψ0(r2)〉 = a 1πa3

∫e−r1/ae−r2/a

1

r1d3r

=1

πa2

∫r e−r/ae−

2a


=2π

πa2

∫ ∞0

r e−r/a[∫ π

0e−

2a


]dr



D = a

〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉 = a〈ψ0(r2) ∣∣∣∣ 1r1

∣∣∣∣ψ0(r2)〉

= a1

πa3

∫e−2r2/a

1

r1d3r =

1

πa2

∫r e−

2a


=2π

πa2

∫ ∞0

r

[∫ π0

e−2a


]dr

X =

〈ψ0(r1)

∣∣∣∣ 1r1∣∣∣∣ψ0(r2)〉 = a 1πa3

∫e−r1/ae−r2/a

1

r1d3r

=1

πa2

∫r e−r/ae−

2a


=2π

πa2

∫ ∞0

r e−r/a[∫ π

0e−

2a


]dr



D = a

〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉 = a〈ψ0(r2) ∣∣∣∣ 1r1

∣∣∣∣ψ0(r2)〉= a

1

πa3

∫e−2r2/a

1

r1d3r

=1

πa2

∫r e−

2a


=2π

πa2

∫ ∞0

r

[∫ π0

e−2a


]dr

X =

〈ψ0(r1)

∣∣∣∣ 1r1∣∣∣∣ψ0(r2)〉 = a 1πa3

∫e−r1/ae−r2/a

1

r1d3r

=1

πa2

∫r e−r/ae−

2a


=2π

πa2

∫ ∞0

r e−r/a[∫ π

0e−

2a


]dr



D = a

〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉 = a〈ψ0(r2) ∣∣∣∣ 1r1

∣∣∣∣ψ0(r2)〉= a

1

πa3

∫e−2r2/a

1

r1d3r =

1

πa2

∫r e−

2a


=2π

πa2

∫ ∞0

r

[∫ π0

e−2a


]dr

X =

〈ψ0(r1)

∣∣∣∣ 1r1∣∣∣∣ψ0(r2)〉 = a 1πa3

∫e−r1/ae−r2/a

1

r1d3r

=1

πa2

∫r e−r/ae−

2a


=2π

πa2

∫ ∞0

r e−r/a[∫ π

0e−

2a


]dr



D = a

〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉 = a〈ψ0(r2) ∣∣∣∣ 1r1

∣∣∣∣ψ0(r2)〉= a

1

πa3

∫e−2r2/a

1

r1d3r =

1

πa2

∫r e−

2a


=2π

πa2

∫ ∞0

r

[∫ π0

e−2a


]dr

X =

〈ψ0(r1)

∣∣∣∣ 1r1∣∣∣∣ψ0(r2)〉

= a1

πa3

∫e−r1/ae−r2/a

1

r1d3r

=1

πa2

∫r e−r/ae−

2a


=2π

πa2

∫ ∞0

r e−r/a[∫ π

0e−

2a


]dr



D = a

〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉 = a〈ψ0(r2) ∣∣∣∣ 1r1

∣∣∣∣ψ0(r2)〉= a

1

πa3

∫e−2r2/a

1

r1d3r =

1

πa2

∫r e−

2a


=2π

πa2

∫ ∞0

r

[∫ π0

e−2a


]dr

X =

〈ψ0(r1)

∣∣∣∣ 1r1∣∣∣∣ψ0(r2)〉 = a 1πa3

∫e−r1/ae−r2/a

1

r1d3r

=1

πa2

∫r e−r/ae−

2a


=2π

πa2

∫ ∞0

r e−r/a[∫ π

0e−

2a


]dr



〈Hψ〉 = E1 − 2|A|2e2

4π�0

[〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉+〈ψ0(r1) ∣∣∣∣ 1r1





D ≡ a〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉

=a

R−(

1 +a

R

)e−2R/a

X ≡ a〈ψ0(r1)

∣∣∣∣ 1r1∣∣∣∣ψ0(r2)〉

=

(1 +

R

a

)e−R/a

〈H〉 =[

1 + 2D + X

1 + I

]E1

Vpp =e2

4π�0

1

R= −2a

RE1



〈Hψ〉 = E1 − 2|A|2e2

4π�0

[〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉+〈ψ0(r1) ∣∣∣∣ 1r1





D ≡ a〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉

=a

R−(

1 +a

R

)e−2R/a

X ≡ a〈ψ0(r1)

∣∣∣∣ 1r1∣∣∣∣ψ0(r2)〉

=

(1 +

R

a

)e−R/a

〈H〉 =[

1 + 2D + X

1 + I

]E1

Vpp =e2

4π�0

1

R

= −2aRE1



〈Hψ〉 = E1 − 2|A|2e2

4π�0

[〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉+〈ψ0(r1) ∣∣∣∣ 1r1





D ≡ a〈ψ0(r1)

∣∣∣∣ 1r2∣∣∣∣ψ0(r1)〉

=a

R−(

1 +a

R

)e−2R/a

X ≡ a〈ψ0(r1)

∣∣∣∣ 1r1∣∣∣∣ψ0(r2)〉

=

(1 +

R

a

)e−R/a

〈H〉 =[

1 + 2D + X

1 + I

]E1

Vpp =e2

4π�0

1

R= −2a

RE1


Energy upper bound

〈H〉 =

1 + 2 aR − (1 + aR ) e−2R/a + (1 + Ra ) e−R/a1 + e−R/a

[1 +

(Ra

)+ 13

(Ra

)2] − 2aRE1

Substituting x ≡ R/a, the total expectation value becomes

F (x) =〈H〉−E1

= −1 + 2x

{(1− 23x

2)e−x + (1 + x)e−2x

1 + (1 + x + 13x2)e−x

}

This can be plotted andthe equilbrium distance andbinding energy extracted

Req = 2.4a = 1.3Å → 1.06Å

Ebind = 1.8eV → 2.8eV

A stable molecule with theelectron shared between thetwo protons.

-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0 1 2 3 4 5 6

F(x

)

x

Req

E1


Energy upper bound

〈H〉 =


[1 +

(Ra

)+ 13

(Ra

)2] − 2aRE1


F (x) =〈H〉−E1

= −1 + 2x

{(1− 23x

2)e−x + (1 + x)e−2x

1 + (1 + x + 13x2)e−x

}This can be plotted andthe equilbrium distance andbinding energy extracted

Req = 2.4a = 1.3Å → 1.06Å



-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0 1 2 3 4 5 6

F(x

)

x

Req

E1


Energy upper bound

〈H〉 =


[1 +

(Ra

)+ 13

(Ra

)2] − 2aRE1


F (x) =〈H〉−E1

= −1 + 2x

{(1− 23x

2)e−x + (1 + x)e−2x

1 + (1 + x + 13x2)e−x

}

This can be plotted andthe equilbrium distance andbinding energy extracted

Req = 2.4a = 1.3Å → 1.06Å



-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0 1 2 3 4 5 6

F(x

)

x

Req

E1


Energy upper bound

〈H〉 =


[1 +

(Ra

)+ 13

(Ra

)2] − 2aRE1


F (x) =〈H〉−E1

= −1 + 2x

{(1− 23x

2)e−x + (1 + x)e−2x

1 + (1 + x + 13x2)e−x

}This can be plotted andthe equilbrium distance andbinding energy extracted

Req = 2.4a = 1.3Å → 1.06Å



-1.2

-1

-0.8

-0.6

-0.4

-0.2

0

0 1 2 3 4 5 6

F(x

)

x

Req

E1


Problem 7.14

If the photon had a nonzero mass (mγ 6= 0), the Coulomb potential wouldbe replaced with the Yukawa potential,

V (~r) = − e2

4π�0

e−µr

r

where µ = mγc/~. Estimate the binding energy of a “hydrogen” atomwith this potential. Assume µa� 1.


Problem 7.15

Suppose you are given a quantum system whose Hamiltonian H0 admitsjust two eigenstates, ψa (with energy Ea) and ψb (with energy Eb). Theyare orthogonal, normalized and non-degenerate (assume Ea < Eb). Nowturn on a perturbation H ′, with the following matrix elements:〈

H ′〉

=

(0 hh 0

), h = constant

a. Find the exact eigenvalues of the perturbing Hamiltonian.

b. Estimate the energies of the perturbed system using second-orderperturbation theory.

c. Estimate the ground state energy of the pertirbed system using thevariational principle with a trial wavefuction of the form

ψ = (cosφ)ψa + (sinφ)ψb

where φ is an adjustable parameter.

d. Compare the answers to the sections above.


PreambleToday's Outline - February 17, 2014

Variational MethodHydrogen molecule ion trial functionHydrogen molecule ion normalizationThe overlap integralMore integration...Normalization, finally!Expectation value of HDirect & exchange integralsExpectation value of HEnergy upper boundProblem 7.14Problem 7.15

Documents

Today’s Outline - February 17, 2014csrri.iit.edu/~segre/phys406/14S/lecture_09.pdfToday’s Outline - February 17, 2014 Hydrogen molecule ion Problem 7.14 Problem 7.15 Midterm Exam