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Unformatted text preview: VIII. The Variational
Principle Griffiths Chapter 7 1 A. The Variational Principle
The variational principle is another approximation method for quantum
mechanics. Perturbation theory allowed us to find approximate wave functions and energies
for systems that we couldn’t solve the
Schrödinger equation for exactly.
2 A. The Variational Principle The variational principle only tells us about
the energy of a system, and only of the
ground state. Furthermore, we can only find an upper
bound on the ground state energy. On the other hand, the variational principle
is easy to understand and use, and the ground state energy of a system is often the most important thing we want to know.
3 A. The Variational Principle 1. Statement of the Principle
The ground state energy of a system is,
according to the variational principle,
Egs ≤ ψ H ψ
ground state
expectation value of H, 〈H〉 See board VIII.A.1 4 A. The Variational Principle 2. An Example
Griffiths Example 7.1: Find an upper bound on the ground
state energy for the one dimensional harmonic oscillator. 2 d 2
1
H=−
+ mω 2 x2
2m dx2 2 Solution: Egs 1
= ω
2 What luck! Equal to the exact result. Used a Gaussian trial function with free parameter and then
M INIMIZED with respect to the free parameter. See board VIII.A.2 5 A. The Variational Principle 3. Another Example
Griffiths Example 7.2: Find an upper bound on the ground state energy for the delta function potential.
2 d 2
H=−
− αδ (x)
2
2m dx Solution: Egs mα 2
≤− 2
π For exact result, replace π w ith 2. Negative for
b ound state Used a Gaussian trial function again with a free parameter
a nd then MINIMIZED with respect to the free parameter.
See board VIII.A.3
6 A. The Variational Principle 4. A Third Example
Griffiths Example 7.3: Find an upper bound on the ground state energy for the one dimensional square well. where Solution: Egs 2 d 2
H=−
+ V (x)
2
2m dx
0 , if 0 ≤ x ≤ a
V (x) =
∞,
otherwise
122
For exact result, replace 12 with π2.
≤
2
2ma Could NOT use a Gaussian trial function here.
It “leaks” outside the potential – it is nonzero in the infinity region of the potential See board VIII.A.4 7 B. The Ground State of Helium
The helium atom consists of two electrons and a
nucleus of two protons (as well as two neutrons
which we can ignore in our model).
2
e2
2
2
H=−
(∇1 + ∇2 ) −
2m
4π0
kinetic energy
o f electron 1 kinetic
e nergy of
e lectron 2 2 protons in the
nucleus 2
2
1
+
−
r1
r2
r1 − r2  Coulomb potential Coulomb potential
b et ween electron b et ween electron
1 a nd the nucleus 2 a nd the nucleus Coulomb potential
b et ween electrons There is no analytical solution for the time
independent Schrödinger equation for the
energy levels of the electrons in a helium atom.
8 B. The Ground State of Helium 1. The Simplest Approximate Solution
We can ignore the electronelectron Coulomb
interaction term altogether
e
2
2
H=−
(∇1 + ∇2 ) −
2m
4π0
2 2 2
2
1
+
−
r1
r2
r1 − r2  The Schrödinger equation is now separable between r1 and r2. The ground state energy in this approximation is Egs = 109 eV.
The experimentally measured ground state energy of helium is 78.975 eV
n ot a very good match The ground state wave function in this approximation
is just the product of two hydrogen wave functions. 8 −2(r1 +r2 )/a
e
ψgs (r1 , r2 ) =
3
See board VIII.B.1
πa
9 B. The Ground State of Helium 2. A Better Approximate Solution 8 −2(r1 +r2 )/a
e
ψgs (r1 , r2 ) =
3
πa
We can use the simplest approximation wave function as
a trial function in the variational method.
This gives us H = 8E1 + Vee expectation
v alue of the
H amiltonian ground state
e nergy of the
h ydrogen atom expectation value of the
e lectronelectron Coulomb
i nteraction term
Vee = e2
4π0 8
π a3 2 e−4(r1 +r2 )/a
dr1 dr2
r1 − r2  The ground state energy in this approximation is Egs = 75 eV.
The experimentally measured ground state energy of helium is 78.975 eV
See board VIII.B.2 Not bad.
10 B. The Ground State of Helium 3. An Even Better Solution We can use the simplest approximation wave function as
a trial function in the variational method, but with a different effective nuclear charge – Z instead of 2.
The logic is that the trial function ignores electron electron interactions completely. We now suppose that the presence of another electron partially “screens” the
interaction between the first electron and the nucleus.
In other words, the first electron doesn’t see a bare nucleus of charge +2, but rather a nucleus surrounded
by a cloud of negative charge that is the second
electron. 11 B. The Ground State of Helium Therefore we should use a trial wave function
of the form
Z 3 −Z (r1 +r2 )/a
ψgs (r1 , r2 ) =
e
3
πa
compare with
23 ψgs (r1 , r2 ) = Ide nt al
ic ! If we use this wave function, it is more convenient
to write our Hamiltonian
e
(∇2 + ∇2 ) −
2
2m 1
4π0 2
e2
2
2
H=−
(∇ + ∇2 ) −
2m 1
4π0 H=− as 8 −2(r1 +r2 )/a
e
3
πa 2 2 2
2
1
+
−
r1
r2
r1 − r2  Z
Z
+
r1
r2 12 e2
+
4π0
(Z − 2) (Z − 2)
1
+
+
r1
r2
r1 − r2  Just adding and subtracting Z/r terms B. The Ground State of Helium The expectation value of the Hamiltonian will be
H = 2Z 2 E1 + 2(Z − 2)
e
1
+ Vee
4π0
r
2 expectation value with
respect to hydrogenic wave
f unction ψ100 e xcept with
c harge Z – see Griffiths
e quation 6.55 1 Z
r = same as before but
w ith Z instead of 2
Vee = − 5Z
E1
4 a The expectation value of the Hamiltonian will be
H = [−2Z 2 + (27/4)Z ]E1 13 B. The Ground State of Helium Minimizing with respect to Z gives us the ground
state energy in this approximation of Egs = 77.5 eV.
The experimentally measured ground state energy of helium is 78.975 eV
Within 2%
See board VIII.B.3 Griffiths points out that we could continue adding
refinements to the trial function in this way, with
more adjustable parameters, and get closer to the
measured result.
But he gets bored and stops.
We’ll stop too.
14 C. The Hydrogen Molecule Ion
We have looked at atomic hydrogen and atomic
helium. Now let’s look at molecular hydrogen, H2, but ionized
so that it only has one electron (since that’s easier). We can use the variational principle to not only find a
bound on the ground state energy, but also to determine whether such an ionized molecule will be
stable at all – as opposed to forming a hydrogen
atom and a free proton. 15 C. The Hydrogen Molecule Ion The hydrogen molecule ion is shown
schematically in Griffiths figure 7.5. The Hamiltonian for this system is
e
2
H=−
∇−
2m
4π0
2 2 16 1
1
+
r1
r2 C. The Hydrogen Molecule Ion The Hamiltonian for this system is
e
2
H=−
∇−
2m
4π0
2 2 1
1
+
r1
r2 We could use the hydrogen atom ground state wave
function as our trial function in the variational
principle. ψ0 (r) = √ 1
π a3 e−r/a Better yet, we should let the electron have an equal
chance of being associated with either proton. 17 C. The Hydrogen Molecule Ion ψ0 (r) = √ 1
π a3 e−r/a Better yet, we should let the electron have an equal
chance of being associated with either proton.
Let’s use a combination of hydrogen atom ground
state wave functions, one associated with each
proton. hydrogen atom wave function
a ssociated with 1st proton normalization factor hydrogen atom wave function
a ssociated with 2nd proton ψ = A[ψ0 (r1 ) + ψ0 (r2 )] LCAO (Linear Combination of Atomic Orbitals) technique
18 C. The Hydrogen Molecule Ion 1. Normalization
Normalizing this trial function gives us a value for A.
1
A =
2(1 + I )
2 where the “overlap integral” I is
I = e−R/a 1 + R
a + 1
3 R
a 2 and R is the distance between the two protons.
(a is still the Bohr radius).
See board VIII.C.1
19 C. The Hydrogen Molecule Ion 2. Expectation Value Next, we calculate the expectation value of the
Hamiltonian. H = E1 − 2A2
e
1
1
aψ0 (r1 ) ψ0 (r1 ) + aψ0 (r1 ) ψ0 (r2 )
4π0 a
r2
r1 ≡ ≡ See board VIII.C.2 2 D: Direct
i ntegral X: Exchange
i ntegral a
a −2R/a
D=
− 1+
e
R
R X= 1+ See Grifﬁths problem 7.8 20 R
a e−R/a C. The Hydrogen Molecule Ion Therefore the expectation of the Hamiltonian is
(D + X )
H = 1 + 2
E1
1+I But this is for a given distance R between the two
protons.
Since we don’t know what this separation should be,
we should add the potential energy for the protonproton separation.
Vpp e2 1
2a
=
= − E1
4π0 R
R
21 C. The Hydrogen Molecule Ion Combining all of this, Griffiths gives us the upper
bound for the total energy of the system, in units of
E1 and expressed as a function of x = R/a.
F (x) = −1 + 2
x (1 − (2/3)x )e + (1 + x)e
1 + (1 + x + (1/3)x2 )e−x
2 22 −x −2x C. The Hydrogen Molecule Ion We see that indeed ionic molecular hydrogen can
form instead of neutral atomic hydrogen plus a free
proton because there is a minimum below the binding
energy of neutral atomic hydrogen.
The upper bound on the ground state energy for
ionic molecular hydrogen is 1.8eV. experimental value is 2.8eV The predicted separation of the protons is 2.4a, or
1.3Å. experimental value is 1.06Å
23 The End
Quantum Physics 25 2 ...
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This note was uploaded on 09/28/2011 for the course PHYS 334 taught by Professor Resch during the Spring '08 term at Waterloo.
 Spring '08
 RESCH
 mechanics

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