Today’s Outline - February 17, 2014csrri.iit.edu/~segre/phys406/14S/lecture_09.pdfToday’s...
Transcript of Today’s Outline - February 17, 2014csrri.iit.edu/~segre/phys406/14S/lecture_09.pdfToday’s...
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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
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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
-
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
-
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
-
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
-
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 ]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 2 / 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 ]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 2 / 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 ]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 2 / 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 ]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 2 / 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 ]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 2 / 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 ]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 2 / 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 ]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 2 / 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 ]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 2 / 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 ]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 2 / 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 ]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 2 / 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 ]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 2 / 12
-
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...
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 3 / 12
-
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...
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 3 / 12
-
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...
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 3 / 12
-
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...
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 3 / 12
-
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...
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 3 / 12
-
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...
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 3 / 12
-
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...
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 3 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 4 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 4 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 4 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 4 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 4 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 4 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 4 / 12
-
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
]
= − arR
[(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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 4 / 12
-
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
]
= − arR
[(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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 4 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 4 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 4 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 4 / 12
-
More integration...
I = − 2Ra2
∫ ∞0
r[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a
]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
r[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a
]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
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 5 / 12
-
More integration...
I = − 2Ra2
∫ ∞0
r[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a
]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
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 5 / 12
-
More integration...
I = − 2Ra2
∫ ∞0
r[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a
]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
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 5 / 12
-
More integration...
I = − 2Ra2
∫ ∞0
r[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a
]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
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 5 / 12
-
More integration...
I = − 2Ra2
∫ ∞0
r[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a
]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
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 5 / 12
-
More integration...
I = − 2Ra2
∫ ∞0
r[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a
]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
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 5 / 12
-
More integration...
I = − 2Ra2
∫ ∞0
r[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a
]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
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 5 / 12
-
More integration...
I = − 2Ra2
∫ ∞0
r[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a
]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
r[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a
]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
]C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 5 / 12
-
More integration...
I = − 2Ra2
∫ ∞0
r[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a
]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
r[(r + R + a)e−(r+R)/a − (|r − R|+ a)e−|r−R|/a
]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
-
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)
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 6 / 12
-
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)
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 6 / 12
-
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 = 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)
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 6 / 12
-
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 = 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)
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 6 / 12
-
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)
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 6 / 12
-
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)
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 6 / 12
-
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)
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 6 / 12
-
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)− Ae2
4π�0
[1
r1ψ0(r2) +
1
r2ψ0(r1)
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 6 / 12
-
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)− Ae2
4π�0
[1
r1ψ0(r2) +
1
r2ψ0(r1)
]
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 6 / 12
-
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)
]C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 6 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 7 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 7 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 7 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 7 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 7 / 12
-
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
√r2+R2−2rR cos θ sin θ dr dθ dφ
=2π
πa2
∫ ∞0
r e−r/a[∫ π
0e−
2a
√r2+R2−2rR cos θ sin θ dθ
]dr
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 8 / 12
-
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
√r2+R2−2rR cos θ sin θ dr dθ dφ
=2π
πa2
∫ ∞0
r e−r/a[∫ π
0e−
2a
√r2+R2−2rR cos θ sin θ dθ
]dr
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 8 / 12
-
Direct & exchange integrals
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
√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
√r2+R2−2rR cos θ sin θ dr dθ dφ
=2π
πa2
∫ ∞0
r e−r/a[∫ π
0e−
2a
√r2+R2−2rR cos θ sin θ dθ
]dr
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 8 / 12
-
Direct & exchange integrals
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
√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
√r2+R2−2rR cos θ sin θ dr dθ dφ
=2π
πa2
∫ ∞0
r e−r/a[∫ π
0e−
2a
√r2+R2−2rR cos θ sin θ dθ
]dr
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 8 / 12
-
Direct & exchange integrals
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
√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
√r2+R2−2rR cos θ sin θ dr dθ dφ
=2π
πa2
∫ ∞0
r e−r/a[∫ π
0e−
2a
√r2+R2−2rR cos θ sin θ dθ
]dr
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 8 / 12
-
Direct & exchange integrals
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
√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)〉
= a1
πa3
∫e−r1/ae−r2/a
1
r1d3r
=1
πa2
∫r e−r/ae−
2a
√r2+R2−2rR cos θ sin θ dr dθ dφ
=2π
πa2
∫ ∞0
r e−r/a[∫ π
0e−
2a
√r2+R2−2rR cos θ sin θ dθ
]dr
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 8 / 12
-
Direct & exchange integrals
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
√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
√r2+R2−2rR cos θ sin θ dr dθ dφ
=2π
πa2
∫ ∞0
r e−r/a[∫ π
0e−
2a
√r2+R2−2rR cos θ sin θ dθ
]dr
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 8 / 12
-
Direct & exchange integrals
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
√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
√r2+R2−2rR cos θ sin θ dr dθ dφ
=2π
πa2
∫ ∞0
r e−r/a[∫ π
0e−
2a
√r2+R2−2rR cos θ sin θ dθ
]dr
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 8 / 12
-
Direct & exchange integrals
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
√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
√r2+R2−2rR cos θ sin θ dr dθ dφ
=2π
πa2
∫ ∞0
r e−r/a[∫ π
0e−
2a
√r2+R2−2rR cos θ sin θ dθ
]dr
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 8 / 12
-
Expectation value of H
〈Hψ〉 = 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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 9 / 12
-
Expectation value of H
〈Hψ〉 = 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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 9 / 12
-
Expectation value of H
〈Hψ〉 = 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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 9 / 12
-
Expectation value of H
〈Hψ〉 = 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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 9 / 12
-
Expectation value of H
〈Hψ〉 = 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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 9 / 12
-
Expectation value of H
〈Hψ〉 = 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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 9 / 12
-
Expectation value of H
〈Hψ〉 = 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
= −2aRE1
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 9 / 12
-
Expectation value of H
〈Hψ〉 = 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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 9 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 10 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 10 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 10 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 10 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 10 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 10 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 10 / 12
-
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
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 10 / 12
-
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.
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 11 / 12
-
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.
C. Segre (IIT) PHYS 406 - Spring 2014 February 17, 2014 12 / 12
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