Quantum Theory of Hydrogen
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Transcript Quantum Theory of Hydrogen
Quantum Theory of Hydrogen
shrödinger's equation for hydrogen
separation of variables
“A facility for quotations covers the absence of original thought.”—
Dorothy L. Sayers
Chapter 6
Quantum Theory of the Hydrogen Atom
6.1 Schrödinger's Equation for the Hydrogen Atom
We have discovered a "new" theory—quantum mechanics as
exemplified by Schrödinger's equation.
We tested it on three simple model systems in Chapter 5.
We ought to test it on something a bit more “realistic,” like an
atom.
What’s the simplest atom you can think of?
“A physicist is an atom’s way of knowing about atoms.”—biochemist George
Wald
The hydrogen atom is the simplest physical system containing
interaction potentials (i.e., not just an isolated particle).
Simple: one proton, one electron, and the electrostatic
(Coulomb) potential that holds them together.
The potential energy in this case is just
e2
V=4 ε 0r
(the attractive potential between charges of +e and –e,
separated by a distance r).
This is a stationary state potential (no time dependence). We
could just plug it in to Schrödinger’s equation to get
2m
+ 2
2
x
2
e2
E -
4
ε
r
0
= 0
The potential looks quite simple, but it is a function of r, not x
or (xyz). What can we do about that?
x
x2 + y 2 + z2 = r 2 ⇒ r =
2m
+ 2 E- 2
x
4 ε0
e
2
2
+ y 2 + z2
2
x 2 + y 2 + z2
= 0
Lots of luck solving that!
(To say nothing of the fact that we shouldn’t be using the 1D Schrödinger
equation for a 3D problem.)
We need to let the symmetry of the problem dictate our
mathematical approach.
The spherically symmetric potential “tells” us to use spherical
polar coordinates!
http://hyperphysics.phy-astr.gsu.edu/hbase/sphc.html
In spherical polar coordinates, r is
the length of the radius vector
from the origin to a point (xyz)
r=
x
2
point
(xyz)
+ y 2 + z2 ,
is the angle between the
vector and the +z axis
z
-1
θ = cos
x 2 + y 2 + z2
radius
,
and is the angle between the projection of the radius vector
onto the xy plane and the +x axis
-1 y
= tan .
x
The equations on the previous slide tell us how to express
(r) in terms of (xyz). We can also express (xyz) in terms of
(r):
x = r sin θ cos , y = r sin θ sin , z = r cos θ.
Now we can re-write the 3D Schrödinger equation
2m
+ 2 + 2 + 2 (E- V ) = 0
2
x
y
z
2
2
2
in three dimensions, and in spherical polar coordinates, as
1 2
1
r
+ 2
sinθ
2
r r r r sinθ θ
θ
1 2 2m
+ 2 2
+ 2 (E- V ) = 0.
2
r sin θ
If we plug in our potential V and multiply both sides by r2sin2,
we get
2
sin θ r
+ sinθ
sinθ
r r
θ
θ
2 2mr 2 sin2θ e2
+ 2+
+ E = 0 .
2
4 ε0r
2
This looks nasty! We had better have a little talk about this
equation!
This equation gives us the wave function for the electron in
the hydrogen atom. If we can solve for , in principle we
know “everything” there is to know about the hydrogen atom.
When we solved Schrödinger's equation in one dimension, we
found that one quantum number was necessary to describe
our systems.
For example, in the Bohr atom, the electron moves in an orbit,
but we need only one parameter to specify its position in the
fixed orbit, so we only need one quantum number.
Here, in three dimensions and with three boundary conditions,
we will find that we need three quantum numbers to describe
our electron.
At the end of this section Beiser tells what the quantum
numbers for the hydrogen atom are, and gives their possible
values. I’ll skip that for now, because until we see where they
come from and what they mean, they aren't of much use to
us.
Another comment: we are really solving Schrödinger's
equation for the electron in a hydrogen atom, aren't we.
Nevertheless, we talk about solving the "hydrogen atom,"
because our solution will provide us with much of what we
need to know about hydrogen.
6.2 Separation of Variables
We now “solve” the hydrogen atom.
Here are some math activities, arranged from “fun” to “ugh:”
solving linear algebraic equations
solving coupled algebraic equations (e.g. xy together)
solving linear differential equations
solving coupled differential equations (e.g. derivatives
mixed together)
We have a coupled linear differential equation to solve. Maybe
if we are clever, like we were with the tunneling calculation, we
can make the problem easier.
A big improvement would be to uncouple the variables.
Stated more mathematically, when we have an equation like
the one above, we like to see if we can "separate" the
variables; i.e., "split" the equation into different parts, with
only one variable in each part.
Our problem will be much simplified IF we can write
(r,θ, ) = R(r) (θ) () = R
I’ve never known you (as a class) to shy away from leaps of
logic, so how about if we assume that and see where it leads
us?
Let’s assume
ψ(r,θ,φ) = R Θ Φ
and see where it takes us.
If our assumption works, in the orderly world of mathematics
we know it must have been right.
With this assumption, the partial derivatives in Schrödinger's
equation become
∂
∂R
dR
=
=
∂r
∂r
dr
∂
∂
d
= R
= R
∂θ
∂θ
dθ
∂
∂
d
= R
= R
∂
∂
d
The partial derivatives become full derivatives because R, ,
and depend on r, , and only.
To separate variables, plug = R into Schrödinger's
equation and divide by R. The result is
sin2θ d 2 dR sinθ d
d
r
+
sinθ
R dr dr
dθ
dθ
1 d2 2mr 2 sin2θ e2
+
+
+ E = 0 .
2
dφ
4ε0r
We have separated out the variable! The term
1 d 2Φ
Φ d
is a function of only. Let's put it over on the right hand side
of the equation. This gives us…
sin2θ d 2 dR sinθ d
d
r
+
sinθ
R dr dr
dθ
dθ
2mr 2 sin2θ e2
1 d2
+
+ E = .
2
d
4ε0r
This equation has the form
f(r,θ) = g()
f is a function of r and only, and g is a function of only.
“How can this be? The RHS has only in it (but no r and ),
and the LHS has only r and in it (but no ).” And LHS=RHS?
Yup, you heard right!
“And yet you’re telling me LHS = RHS. I repeat, how can this
be?”
Only one way!
f(r,θ) = a constant, independent or r,θ, and = g()
“Are you telling me everything is just a constant?”
Absolutely not! It’s just that the particular combination of
terms on the LHS happens to add up to a constant, which is
the same as the constant given by the particular combination
of terms on the RHS.
sin2θ d 2 dR sinθ d
d
r
+
sinθ
R dr dr
dθ
dθ
2mr 2 sin2θ e2
1 d2
+
+ E = constant = .
2
d
4ε0r
This is really good. We've taken the one nasty equation in r,
and separated it into two equations, one in r, and the other in
only. Do you think maybe we can separate the r part…
sin2θ d 2 dR sinθ d
d
r
+
sinθ
R dr dr
dθ
dθ
2mr 2 sin2θ e2
1 d2
+
+ E = constant = .
2
d
4ε0r
It turns out (although we won't do the math in this course)
that the “constant” must be the square of an integer.
If not, our differential equations have no solution.
Thus, we can write the RHS of this equation as
1 d2
= m2 .
d
Kind of hard to see, but
that’s a lowercase script ℓ
Where did this mℓ come from? It’s an integer. We just
“happened” to give it that “name.”
The LHS of our big Schrödinger equation also must equal mℓ2.
If we set the LHS equal to mℓ2, divide by sin2, and rearrange,
we get
m2
1 d 2 dR 2mr e
1
d
d
r
+
+
E
=
sinθ
.
2
2
R dr dr
dθ
4ε0r
sin θ sinθ dθ
2
2
Once again we have separated variables. The LHS is a
function of r only, and the RHS is a function of only.
Again, the only way to satisfy this equation is for LHS=a
constant=RHS.
Solution of the resulting differential equations will result in
restrictions on this constant. In this case, the constant must
equal an integer times the next larger integer: ℓ(ℓ+1).
We have taken our initial differential equation and split into 3.
Here are the pieces, rewritten slightly:
1 d2
= m2
d
1
d
d
sinθ
+
sinθ dθ
dθ
m2
=0
+1 2
sin θ
1 d 2 dR 2m e 2
+ E r
+ 2
2
r dr dr
4ε0r
+1
R=0
2
r
“You sure went to a lot of trouble, and all you’ve done is
replaced one differential equation by three.”
But in the “one” the variables were coupled, and in the “three”
the variables are separated. Huge improvement!
The first attempt to LOG into ARPANET (precursor to the
Internet), fall 1969 (http://www.netvalley.com/intval1.html):