Transcript orbital quantum number

Quantum Theory of Hydrogen
quantum numbers
(I will not finish this lecture 03-02-2005)
“When it comes to atoms, language can be used only as in poetry. The
poet, too, is not nearly so concerned with describing facts as with
creating images.”—Neils Bohr
Picking up where we left off… we are “solving the hydrogen
atom.
We “separated” our 3D Schrödinger equation into 3 singlevariable differential equations…
6.3 Quantum Numbers
We briefly saw two quantum numbers in the previous lecture:
mℓ and ℓ. They were “separation constants,” and you were
told that the differential equations for hydrogen could be
solved only if they took on integral values.
In this section, we find out what the allowed values of mℓ and
ℓ are.
We find the first quantum number by solving the differential
equation for .
d2
2
+
m
 = 0
2
d
That equation should look familiar to you; you've seen it a
number of times before. It has solutions which are sines and
cosines, or complex exponentials. We write the general
jm 
solution
  = Ae
.
 
We will get the constant A by normalization.
Now, because  and +2 represent a single point in space,
we must have
Ae
jm 
= Ae
j m   + 2 
This happens only for mℓ = 0, 1, 2, 3, ...
.
For reasons which are not yet obvious, mℓ is called the
magnetic quantum number.
Our differential equation for  is
1
d 
d  
 sinθ
+
sinθ dθ 
dθ  


m2 
=0
+1 2
sin θ 

It involves the term




m2 
 .
+1 2
sin θ 

It turns out that from differential equations that the equation
for  can be solved only if ℓ is an integer greater than or equal
to the absolute value of mℓ.
ℓ is another quantum number, called the orbital quantum
number, and the requirement on ℓ can be restated as
mℓ = 0, 1, 2, 3, ..., ℓ.
I’ll summarize OSE’s
in a bit.
Finally, the radial differential equation is

1 d  2 dR   2m  e 2
+ E r
 + 2 
2
r dr  dr  
 4ε0r


+1 
R=0
2
r

It can be solved only for energies E which satisfy the same
condition as we found on the energies for the Bohr atom:
me 4
En = 322 ε0
2
E1
1
=
, n = 1, 2, 3...
2
2
n
n
n is called the principal quantum number.
Here’s the differential equation for R again:

1 d  2 dR   2m  e 2
+ E r
 + 2 
2
r dr  dr  
 4ε0r


+1 
R=0
2
r

Note that the product ℓ(ℓ+1) shows up in the equation for R,
and n comes out of solving this equation. Another math
requirement for valid solutions is that n(ℓ+1).
We can express the requirement that n=1,2,3,… and n(ℓ+1)
as a condition on ℓ:
ℓ = 0, 1, 2, ..., (n-1).
Summarizing our quantum numbers:
n = 1, 2, 3, ...
ℓ = 0, 1, 2, ..., (n-1)
mℓ = 0, 1, 2, 3, ..., ℓ
We summarize this section by noting that solutions to the
Schrödinger equation for the hydrogen atom must be of the
form
 = Rnℓ ℓmℓ mℓ ,
with conditions on the quantum numbers n, ℓ, and mℓ as
discussed above.
We aren't going to go any further with our solutions to the
Schrödinger equation, other than to note that they are wellknown, and Beiser tabulates some of them in Table 6.1.
Note how Table 6.1 is set up. For n=1, the only allowed
possibilities are ℓ=mℓ=0. For this case, Beiser lists the three
solutions R, , and .
For n=2, ℓ can be either 0 or 1. If ℓ=0 then mℓ=0. If ℓ=1 then
mℓ=0 and mℓ=1 are allowed. The solutions for mℓ=1 are the
same. Beiser tabulates the three solutions.
Here's an example. Suppose we have an electron with a
principal quantum number n=3 (corresponding to the second
excited state of the Bohr hydrogen atom) and orbital and
magnetic quantum numbers ℓ=2 and mℓ=-1.
Then, according to table 6.1,
1 ± j
   =
e
2
 θ =
15
sin θ cos θ
2
and
4
r 2 - r / 3a0
R r  =
e
.
2
3/2
81 30 a 0 a0
The wave function  is the product of all three of those
functions.
I wouldn't care to calculate and plot the wave function by
hand, but with Mathcad the problem is rather easy.
With the wave functions in Table 6.1, you can calculate all
sorts of fun stuff, like ground state energies (see the example
on page 207), excited state energies, expectation values,
probabilities, etc.
6.4 Principal Quantum Number
Our result for the allowed values of the principal quantum
number n and, and dependence of the electron energy En on
n, turn out to be exactly the same as for the Bohr model.
Is this just luck or was Bohr on to something deeper?
In fact, it is not just luck. Both results depend on the wave
nature of the electron.
The Bohr model is, however, unable to provide additional
details which the full quantum mechanical solution does.
Electron energies in the hydrogen atom are quantized, and
they are negative numbers:
me 4
En = 322 ε0
2
E1
1
= 2 , n = 1, 2, 3...
2
n
n
It is true that any positive energy may lead to a solution to
Schrödinger's equation…
…but a positive energy means the electron is not bound, so we
don't have a electron in the hydrogen atom.
The only possible negative (bound electron) energies are those
given by the equation above.
None of this information is new; we have seen it all before.
6.5 Orbital Quantum Number
I've already written down the differential equation for R(r)
several times. Sigh… I guess I’d better repeat it again.

1 d  2 dR   2m  e 2
+ E r
 + 2 
2
r dr  dr  
 4ε0r


+1 
R=0
2
r

On the next slide, I’ll write it in an alternate form, which uses:
E=K+V
K = Kradial + Korbital
V = - e2/ ( 4 0 r ) .
The alternate form is:
1 d  2 dR 
2m 
r
 + 2 K radial + K orbital 2
r dr  dr 

+1 
R=0
2
2mr 
2

Ohhhhh nooooo…
The above equation is supposed to have only r in it. But Korbital
depends on tangential velocity, so it seems to have angular
dependence in it!
I have two choices: throw out the last lecture and a half and
start over, or…
1 d  2 dR 
2m 
r
 + 2 K radial + K orbital 2
r dr  dr 

+1 
R=0
2
2mr 
2

…somehow make Korbital just “go away.”
Let’s see… both Kradial and Korbital are positive numbers, so they
can’t somehow cancel each other out…
…but if the radial equation is to really have only radial
dependence in it, it clearly cannot have an orbital kinetic
energy term.
The only way to make Korbital "go away" from this equation is
to have
K orbital =
2

+1
2mr
2
.
Some review from Physics 23.
Orbital kinetic energy:
K orbital
Orbital angular momentum:
1
2
= m v orbital
2
L orbital = r m v orbital
Just as we can write kinetic energy in terms of momentum, we
can write orbital kinetic energy in terms of orbital angular
momentum:
K orbital
L2
=
2 m r2
Combining our two equations for Korbital (the one on this slide,
and the one on the previous slide), we find that…
2
K orbital =

+1
2 m r2
L2
=
2 m r2
which gives us
L=

+1
.
Because the orbital quantum number ℓ is quantized, the
electron’s orbital angular momentum L (a vector) is also
quantized.
In fact, L is quantized in (non-integral) multiples of ħ.
The lowest possible L is zero, when ℓ=0.
Huh? Are you trying to tell me an orbiting electron in
hydrogen can have zero angular momentum?
The ground-state hydrogen electron has zero angular
momentum! It cannot be orbiting in any classical
sense!
L=
The lowest possible nonzero L is
2
.
Here’s a table from the good old days of spectroscopy,
showing how we label angular momentum states:
ℓ=
0
s
1
p
2
d
3
f
4
g
5
h
6
i
Know how to use this table!
How would you label an electron with n=3, ℓ=2?
ℓ
n
It would be a 3d electron. Note that a 3d electron can have
any one of several possible allowed mℓ's.
See table 6.2, page 210 of Beiser for designation of atomic
states.
6.6 Magnetic Quantum Number
Adding to our house of cards, we have come
up with an orbital quantum number (and
found an orbiting electron that doesn’t “go
around”).
Regardless, an electron in a hydrogen atom is in an “orbit.” It
has an angular momentum vector L (a vector).
We found the magnitude of L above:
What else does any vector have?
L=

+1
.
A direction, of course. What does the direction of L in an
isolated hydrogen atom mean?
Not much! (Actually, nothing, unless you specify a coordinate
system.)
But an orbiting electron is like a current in a loop, which gives
rise to a magnetic field, and can interact with an external
magnetic field.
An external magnetic field therefore gives us a meaningful
reference for specifying the direction of the electron orbital
angular momentum vector in the hydrogen atom.
By convention, we put our hydrogen atom's z-axis along the
direction of the applied magnetic field B.
Then mℓ gives the component of L in the direction of B:
Lz = m
.
This is called “space
quantization.” See your text.
Huh? Where did this come from?
Straight out of nowhere. Not from anything we've done here!
Let’s just accept it as true. If you really want to see, come to
my office some time and we’ll dig out my graduate text on
quantum mechanics…
OK, so now we know how to calculate a hydrogen atom
electron’s total angular momentum, and it’s z-component, the
part parallel to an applied magnetic field B.
Thinking back to Physics 24, if you had a loop of wire capable
of rotating, and the loop carried a current, and was placed in
an external magnetic field, what would the loop do?
The current loop would
experience a torque, and,
in the absence of external
forces, rotate until the
torque became zero.
In the process, the magnetic field due to the current loop
would line up with the external magnetic field.
What does quantum mechanics say about the “same” scenario
in the hydrogen atom?
Or, in the language of the electron in hydrogen, can L ever be
parallel to B?
Answer:
L=

+1
Lz = m
Because mℓ is at most equal to ℓ, then Lz=mℓħ is always less
than L.
Huh? Does this mean that my current loop in the previous
slide can never exactly “line up” its magnetic field with the
external magnetic field?
Quantum mechanics says “yes.” However, the quantum
numbers will be so large that you will never see the difference
between “almost” and “exactly” aligned.
Suppose we place a hydrogen atom in a magnetic field. Is Lz
always the maximum possible (i.e., does the hydrogen atom
always try to "line up" with the field)?
Answer: not necessarily. “Space
quantization.” See figure 6.4.
Understand this figure for test/quiz
questions!
The figure to the right is (intentionally) not fullylabeled. Also, I used a circle instead of a semicircle
due to the lack of a semicircle tool in Powerpoint.
Here's why Lz is always less than L.
If L could point exactly along the z axis (magnetic field axis)
then the electron orbit would lie exactly in the xy plane and
the uncertainty in the z position coordinate would be zero.
The momentum uncertainty in the z direction would then be
infinite. This is intolerable for an electron in an atom.
The tilt of L with respect to the z-axis lets us satisfy the
uncertainty principle.
October 24, 2003, go here next.
(The final 3 slides of lecture 21 are duplicated at the start of lecture 22.)
6.7 Electron Probability Density
Recall that the volume element in spherical polar coordinates is
dV = r 2 sin θ dr dθ d .
Thus, the electron probability density in hydrogen is
P(r,θ, ) dV =  dV = RR   r 2 sin θ dr dθ d
*
dV
P(r,θ, ) dV = RR r 2 dr   sin θ dθ   d
P(r,θ, ) dV = P(r) dr  P(θ) dθ P() d
P(r) dr = R*R r2 dr
(the probability of finding the
electron within infinitesimal dr
centered at r)
We often wish to calculate the probability of finding the
electron in some volume element in space:
Probability =  P r,θ,   dV .
Because  is separable, and R, , and  are orthonormal,* we
can write the triple integral as three one-dimensional integrals.
Probability  r  =
Probability(θ) =
Probability() =
 P r  dr
 P(θ) dθ
 P() d
*As one consequence of this, the R, , and  in table 6.1 are independently
normalized, and the product wave function is also normalized.
I am using my own personal shorthand notation that
Probability(r) means “the probability of finding the electron
within some dr centered at r,” etc.
Important note: in spherical polar coordinates, 0  r  , 0 
  , and 0    2. It makes no sense to calculate
probabilities outside these regions.
When you calculate <r>, the integral goes from 0 to , NOT
from - to !
Some terminology and important notes (I discuss only the
radial part, but it applies to the angular parts too)…