Download Quantum Mechanics Lecture 30 Dr. Mauro Ferreira

Quantum Mechanics
Lecture 30
Dr. Mauro Ferreira
E-mail: [email protected]
Room 2.49, Lloyd Institute
Suppose we want to calculate the ground state energy Egs of a
system described by the Hamiltonian H, but we’re unable to
solve the Schroedinger equation.
The variational method
Pick any arbitrary normalized wave function
the quantity |Ĥ| ⇥
Egs ⇥ ⇤ |Ĥ| ⌅
⇤H⌅
and calculate
Egs ⇥ ⇤ |Ĥ| ⌅
⇤H⌅
Proof: We start by writing as a linear combination of the
(unknown) eigenfunctions of H.
=
cn
where
n
Ĥ
n
= En
n
n
Because
1=
is normalized:
| ⇥=
m|
cm
m
cn
n
n⇥
=
cm cn
m
n
m|
n⇥ =
n
|cn |2
Meanwhile
Ĥ⇥ =
m |Ĥ|
cm
m
cn
n
n⇥
But by definition: Egs
⇥H⇤
Egs
n
|cn |2 = Egs
=
cm En cn
m
En
Q.E.D.
n
m| n⇥
=
n
En |cn |2
By minimizing <H>
we’re certain to approach
the true ground state
Suppose that we want to find out the ground state for the
one-dimensional harmonic oscillator
2 2
d
m⇥ 2 x2
+
=E
2
2m dx
2
We pick as a trial wave function:
(x) = A e
A is determined by normalization: A =
H⇥ = T ⇥ + V ⇥
⇥T ⇤ =
2
2m
2
|A|
m 2 2
V⇥=
|A|
2
2
⇤
+⇥
dx e
bx
2
⇥
+⇥
dx e
⇥
b m 2
H⇥ =
+
2m
8b
2bx2
d2
e
2
dx
bx
2
⇥
2b
bx2
⇥1/4
2
b
=
2m
m 2
x =
8b
2
Now all we need to do is to
minimize <H>
2
b m 2
H⇥ =
+
2m
8b
2
d
⇥H⇤ =
db
2m
m
b=
2
H⇥min
1
=
2
m 2
=0
2
8b
min
= Ae
m
2
x2
If a solution is known, the so-called ladder operators generate
new solutions that are higher or lower in the energy spectrum.
(â )
0
=0
(Ground state)
1
d
[
+ m⇥x]
2 m⇥ dx
d 0
=
dx
d
0
m⇥
=
x
0
0 (x)
= Ae
=0
0
m⇥
m
2
0
x2
m⇥ 2
dx x ⇥ ln 0 =
x +C
2
m⇥ 1
)4
Normalization: A = (
gs
lecture 8
To obtain the energy E0 associated with the state ψ0, we write
the Schroedinger Eq. in the form
1
⇥(â+ â + ) 0 = E0 0
2
and make use of the fact that (â )
E0 =
2
0
= 0 to conclude that
Egs
Once the ground state is known, we can obtain the excited
states by repeatedly applying the ladder operator
n (x)
= An (â+ )
n
0 (x)
In general, one can show that An =
1
En = (n + )
2
1
n!
lecture 8
2
b m 2
H⇥ =
+
2m
8b
2
d
⇥H⇤ =
db
2m
m 2
=0
2
8b
m
b=
2
H⇥min
min
1
=
2
H⇥min = Egs
min
=
gs
= Ae
m
2
x2
Suppose that we want to find out the ground state for a
delta-function potential
2
d2 ⇤
2m dx2
⇥(x) ⇤ = E⇤
We pick as a trial wave function:
(x) = A e
2b
A is determined by normalization: A =
H⇥ = T ⇥ + V ⇥
⇥T ⇤ =
2
2m
2
|A|
⇤
+⇥
dx e
⇥
+⇥
⇥V ⇤ =
2
|A|
2
b
⇥H⇤ =
2m
bx
2
dx e
⇥
2b
⇥
2bx2
d2
e
2
dx
⇥(x) =
bx
2
⇥
2
⇥1/4
b
=
2m
⇥
2b
⇤
bx2
same as before
2b
⇥
2
b
⇥H⇤ =
2m
2
d
⇥H⇤ =
db
2m
⇥H⇤min =
m
⇥
⌅
2
2
2⇥b
=0
2m2 2
b=
⇥ 4
The Hamiltonian contains a single bound state given by
Ĥ (x) = E (x)
where
m
⇥(x) =
and
m
2
E=
1
e
m |x|/
2
2
Egs
2
(eigenfunction)
(eigenvalue)
(x)
0.75
0.5
0.25
-7
-6
-5
-4
-3
-2
-1
0
1
2
3
4
5
6
x
7
-0.25
lecture 10
2b
⇥
2
b
⇥H⇤ =
2m
2
d
⇥H⇤ =
db
2m
⇥H⇤min =
m
⇥
H⇥min > Egs
⌅
2⇥b
=0
2m2 2
b=
⇥ 4
2
2
error ≈ 57%
Let’s now use the variational method to estimate the ground state
of the hydrogen atom
We pick as a trial wave function: ⇤(r, , ⇥) = A e
r/
1
A is determined by normalization: A =
⇥
3
H⇥ = T ⇥ + V ⇥
T⇥ =
⇥V ⇤ =
⇥H⇤ =
2
2m
e2
2
2
2m
e2
2
⇥ ⇤H⌅min =
m e4
2 2
d
⇥
⇤H⌅ =
d
2
m
3
+
e2
2
=0
Hydrogen atom
V (r) =
scattering
states
e2 1
4⇥ 0 r
Radial equation:
r
bound
states
2
2
d u
+
2
2m dr
1
r
2
e
4⇥
⇥
1
⌅(⌅ + 1)
+
u = Eu
2
2m
r
0 r
2
Vef f (x)
Vef f =
2
e2 1
⌅(⌅ + 1)
+
4⇥ 0 r 2m
r2
x
=0
=1
=2
=3
lecture 21
cj+1 =
2(j + ⇥ + 1)
0
(j + 1)(j + 2⇥ + 2)
⇥
Recursive formula for the coefficients
cj
We can obtain all coefficients in terms of the initial one (co).
The only way this expansion is compatible with the previous
asymptotic analysis is if the series terminates at a finite value jmax.
cjmax +1 = 0
⇥ 2(jmax + ⇥ + 1)
0
=0
Defining the so-called principal quantum number n as
n
E=
jmax + + 1
2 2
⇥
=
2m
⇥ En =
0
me4
8⇤ 2 20 2
13.6 eV
n2
⌅20
Egs
= 2n
⇥ En =
⇤
where n = 1, 2, 3, ...
m
2 2
e2
4⇥ 0
⇥2 ⌅
1
n2
lecture 21
Let’s now use the variational method to estimate the ground state
of the hydrogen atom
We pick as a trial wave function: ⇤(r, , ⇥) = A e
r/
1
A is determined by normalization: A =
⇥
3
H⇥ = T ⇥ + V ⇥
T⇥ =
⇥V ⇤ =
⇥H⇤ =
2
2m
e2
2
2
2m
e2
2
⇥ ⇤H⌅min =
m e4
2 2
d
⇥
⇤H⌅ =
d
2
m
3
+
e2
2
=0
H⇥min = Egs
Solution to the infinite well is
1.5
⇥gs =
⇤
1
2
sin
L
x⇥
L
Egs =
0.5
0
0.5
2 2
2mL2
1
Trial wave function:
= A x (L
No variation
parameter required
x)
A is determined by normalization: A =
30
L5
H⇥ = T ⇥ + V ⇥
H⇥ = T ⇥ =
2
5 2
A2 L3
=
6m
mL2
error ≈ 1.3%