Download 3D quantum mechanics, hydrogen atom

Physics 228 Today:
Ch 41: 1-3:
3D quantum mechanics, hydrogen atom
Website: Sakai 01:750:228 or www.physics.rutgers.edu/ugrad/228
Happy April Fools Day
Monday, April 1, 2013
Example / Worked Problems
What is the ratio of the energy of the 3rd to the
2nd excited states of a harmonic oscillator?
The harmonic oscillator energy levels are En = (n+½)ħω, with
ω2 = k'/m. The ground state is n = 0, the second excited state
is n = 2, and the third excited state is n = 3.
Thus: r = E3/E2 = (3+½)/(2+½) = 7/5.
What energy photon is given off when an electron in
an infinite 1-D square well transitions from the 2nd
excited state to the ground state?
The infinite square well energy levels are En = n2ħ2π2/2mL2. The
ground state is n = 1, and the second excited state is n = 3.
Thus the photon or transition energy is
Eγ = En=3 - En=1 = (32-12)ħ2π2/2mL2 = 8ħ2π2/2mL2 = 4ħ2π2/mL2.
Monday, April 1, 2013
Example / Worked Problems
What is the ratio of the energy of the 3rd to the
2nd excited states of a harmonic oscillator?
The ground state is n = 0.
What energy photon is given off when an electron in
an infinite 1-D square well transitions from the 2nd
excited state to the ground state?
The ground state is n = 1.
While we use n to denote the levels, note that for some
potentials we use n = 0 for the ground state, while for other
potentials we use n = 1 for the ground state. On a test, we
will expect you to know n of the ground state.
Monday, April 1, 2013
Example / Worked Problems II
What is the momentum of an electron in the 1st
excited state of an infinite square well potential?
The infinite square well energy levels are En = n2ħ2π2/2mL2. The first
excited state is n = 2.
The only way we know how to get the momentum is from the
energy. The kinetic energy in this state is E2 = 4ħ2π2/2mL2 = 2ħ2π2/
mL2. So we expect the magnitude of the momentum is p = (2mE)½ =
2πħ/L = h/L.
But is this the answer to the question? You need to be careful about
how it is asked. If the "expectation value" of the momentum is
requested, this is the signed momentum, not the magnitude, and the
answer is 0. Since the particle in a box isn't going anywhere, its
average momentum must be 0, although the average magnitude of
the momentum is non-0. Be careful about whether the average
momentum or the magnitude of the momentum is requested.
If you have this on a test, and are not sure, ASK!
Monday, April 1, 2013
3D Quantum Mechanics
When we go from 1D to 3D, we need to take
the 3D derivative. We need vector calculus
since the derivative can be different in
different directions. We use the gradient
operator:
∂
∂
∂
�
∇=
x̂ +
ŷ +
ẑ
∂x
∂y
∂z
The time-independent S.E. now becomes
2
−� 2
∇ ψ(�x) + U (x)ψ(�x) = Eψ(�x)
2m
or, for a 3D square well potential we can write
�
�
−�2 2
∇ ψ(�x) = E − U (�x) ψ(�x)
2m
where inside the box U0 = 0, and outside the box U0 = ∞.
Monday, April 1, 2013
Infinite 3D Square Well
Let's consider a simple cubical box, each side
of length L, and the box extending from 0 to L
in x, y, and z directions.
Again, as in the 1D case, outside the box U = ∞, so the
differential equation is solved if the wave function ψ(x) = 0. Thus
the particle is in the box and we require ψ(x) = 0 at x (or y or z)
= 0 or x (or y or z) = L. In the box U(x) = 0, so S.E. is simply:
−�2 2
∇ ψ(�x) = Eψ(�x)
2m
We again need a wave function that has the same form as its
2nd derivative, and the right boundary conditions. It is similar
to the 1D case, and we can guess the result:
ψ(x) = (2/L)3/2 sin(qπx/L) sin(rπy/L) sin(sπz/L)
Monday, April 1, 2013
Note the new
indices: q in x, r in y,
and s in z directions.
Infinite 3D Square Well
Energy Levels
In the 1D case, we had k = qπ/L, ψ(x) = (2/L)1/2 sin(qπx/L), and Eq =
q2ħ2π2/2mL2, with q = 1, 2, 3... What happens now with the 3D case
where ψ(x) = (2/L)3/2 sin(qπx/L) sin(rπy/L) sin(sπz/L)?
The three directions are independent, so we will have kx = qπ/L, ψx(x)
= (2/L)1/2sin(qπx/L), and Eqx = q2ħ2π2/2mL2, with q = 1, 2, 3 ... and
similar in the y and z directions. Then
Eqrs = Eq + Er + Es = (q2+r2+s2)ħ2π2/2mL2.
The g.s. will have q=r=s=1, and E111 = 3ħ2π2/2mL2.
Monday, April 1, 2013
iClicker
Eqrs = Eq + Er + Es = (q2+r2+s2)ħ2π2/2mL2.
The g.s. will have q=r=s=1, and E111 = 3ħ2π2/2mL2.
What is the energy of the 1st excited state?
How many of them are there?
a) E2 = Egs, 1
b) E2 = 2Egs, 1
c) E2 = 4Egs, 1
d) E2 = 2Egs, 3
e) E2 = 4Egs, 3
Monday, April 1, 2013
Degenerate States
States of the same energy are called "degenerate states".
Is is possible to break the degeneracy by having all 3 sides
of the box different lengths, so that
Eqrs = Eqx + Ery + Esz = (q2+r2+s2)ħ2π2/2mL2 →
Eqrs = Eqx + Ery + Esz = (q2/Lx2 +r2/Ly2 +s2/Lz2) ħ2π2/2m.
Monday, April 1, 2013
Probability Distributions
Again in 3D QM, as in the 1D case, the probability of finding a
particle at some point x0, y0, z0 in space is given by ψ*ψ = |ψ|2 at
that point.
Note an odd feature of QM: the particle in, e.g., the ψ2,1,1 state
can never be found in the x=L/2 plane, but it can be on one side
or the other, and one expects moves between the two sides. It
crosses the plane, but it is never found there!
Monday, April 1, 2013
A 3D spherical box
A complication, not covered in the textbook, is to have a 3D
spherical box. Now using x, y, z makes the problem more difficult
since the, e.g., x limits depend on the y and z values. Instead we
use spherical coordinates, r, θ, φ. The 3D Laplacian operator
operating on ψ makes S.E., slightly rearranged, of the form:
2
�
�
�
�
∂
∂
ψ
1
∂
∂ψ
1
∂ψ
1
−2mE
2
2
∇ ψ(�x) = 2
r
+ 2
sin θ
+ 2 2
=
ψ
2
2
r ∂r
∂r
r sin θ ∂θ
∂θ
�
r sin θ ∂φ
It is natural to guess the solution is of the form ψ(r,θ,φ) =
R(r)Θ(θ)Φ(φ). Since each term differentiates only one of the three
functions, we come up with 3 independent equations, like in the
3D cubical box case. The radial function R needs to go to 0 at the
edge of the box, r = r0, but not at the center, r=0. The angular
dependences are usually written as "spherical harmonics" Ylm(θ,φ),
which you might have encountered elsewhere. The are of the form
of Legendre polynomials times e±imφ.
(Actually solving the equations is too much math for us.)
Monday, April 1, 2013
The Hydrogen Atom
For hydrogen, like the spherical
box, we have spherical
symmetry, but the potential
energy is the Coulomb potential
energy between opposite
charges:
U(r) = -e2/4πε0r.
The radial wave function R(r)
must go to 0 at large r, since the
electrons are bound to the
protons, but does not necessarily
go to 0 at the origin. The
angular functions must be finite
and periodic in φ, with a period
of 2π. The wave functions are
characterized by 3 quantum
numbers ...
Monday, April 1, 2013
The Principal Quantum Number
n = 1, 2, 3, ...: The wave function is of the form of a polynomial
series in r (∑i=0,...n-1 ciri) (some ci are 0) times an exponential
exp(-r/na0). Here a0 is the Bohr radius, 0.053 nm. The resulting
energies are the same as in the Bohr model, En = -13.6 eV / n2.
n=2
n=3
examples of
wave functions
n=1
We refer to the n=1, 2, 3... sets of states as the K, L, M, N, ... shells.
Monday, April 1, 2013
The Orbital Angular Momentum Quantum Number
l = 0, 1, ... n-1: (Or if you fix l, then n = l, l+1, l+2, ...) Formally, the
orbital angular momentum in quantum mechanics is not actually lħ,
but is rather L = [l(l+1)]½ħ. But we commonly refer to:
l=
0
1
2
3
4
5 ...
as an
s
p
d
f
g
h .... state
√2
√6
√12
√20
(or L/ħ = 0
√30)
and say the orbital angular momentum is quantized to integral
multiples of ħ.
In the Bohr model each value of n describes a circular orbit with a
particle speed and radius and orbital angular momentum. With the
S.E., each orbital (each particular value of n) has n different
possible degenerate orbital angular momenta states.
Monday, April 1, 2013
The Magnetic
Quantum Number
ml = 0, ±1, ... ±l: "magnetic" or "zcomponent of angular momentum"
quantum number. The z-projection of the
orbital angular momentum has magnitude
mlħ.
The azimuthal direction of the angular
momentum L is arbitrary, and if, e.g.,
there is a magnetic field in the z
direction the angular momentum will
precess about the z direction.
In the semiclassical picture shown, the
angular momentum of an l = 2 state with
ml = +2 makes an angle of cosθ = 2/√6 =
0.816 ➭ θ = 35o with the z-axis. For ml =
+1, cosθ = 1/√6 = 0.408 ➭ θ = 66o.
Monday, April 1, 2013
Summary of Hydrogen States
States within the same shell are degenerate.
Shell
K
L
"
M
"
"
N
etc...
Monday, April 1, 2013
n
1
2
2
3
3
3
4
l
0
0
1
0
1
2
0
ml
0
0
0, ±1
0
0, ±1
0, ±1, ±2
0
notation
1s
2s
2p
3s
3p
3d
4s
iClicker
How many degenerate states are there in the n=3 shell?
(If you know about the spin of the electron, ignore it.)
Shell
K
L
"
M
"
"
N
etc...
Monday, April 1, 2013
n
1
2
2
3
3
3
4
l
0
0
1
0
1
2
0
ml
0
0
0, ±1
0
0, ±1
0, ±1, ±2
0
notation
1s
2s
2p
3s
3p
3d
4s
a) 1
b) 5
c) 9
d) 12
e) 14
More probability
distributions
In the case of xyz coordinate systems,
the probability is ψ*ψ because the
volume element is constant.
But for spherical coordinates systems,
the volume integral uses:
dV = r2dr dcosθ dφ
so the probability distribution uses
r2ψ*ψ in the plots to the right. Thus
all the distributions go to 0 at the
origin.
If we just showed ψ or ψ*ψ, you
would see the s orbits do not go to 0
at the origin, but the p, d, f... orbits
do. Why is this?
Monday, April 1, 2013
More probability
distributions
sphere
Monday, April 1, 2013
To indicate the 3D nature of the distributions,
we can show them in the xz plane.
Rotate the image around the z axis for the
3D view.
sphere +
donut
dumbbell
shell
iClicker
What angle does the angular momentum make with the zaxis in the semi-classical picture for the ml = 0 2p state?
a) There is no 2p state, you need at least 3p
b) |cosθ| = 1/√2
c) |sinθ| = 1/√2
d) 0o
e) 90o
Monday, April 1, 2013