Download Atomic and Molecular Physics for Physicists Ben-Gurion University of the Negev

Ben-Gurion University of the Negev
Atomic
Atomic and
and Molecular
Molecular Physics
Physics
for
for Physicists
Physicists
Ron Folman
Chapter 7:
The Hydrogen atom: spin, fine-structure, hyperfine structure
(Stern-Gerlach)
Main Reference: Open University (unit 8)
Copies made from Modern Physics, Paul A. Tipler
and Ralph A. Llewellyn
W.H. Freeman and company
Exercises:
Dudi Moravchik.
www.bgu.ac.il/atomchip
www.bgu.ac.il/nanofabrication
www.bgu.ac.il/nanocenter
Following the square and harmonic
potentials in 1d, we now turn to a
3d electric potential which is only dependent
on the distance r from the origin.
The total magnetic moment of the atom is:
µ= [e/(2m) lz + e/(m) sz + e/(2M) iz] =
µ[eħ/(2m) ml + eħ/(m) ms + eħ/(2M) mi]=
µB [ml + 2ms] + µN [mi]
Example I:
Calculating the
properties of an
atomic wave function
Question: show that the potential
Energy of the Hydrogen ground state
is twice the total energy.
Answer:
note: r0 is the bohr radius (i.e. n=1)
in MKS (q=e)
rn= 4πε0 n2ħ2 / (me2)=
n2ε0h2/πme2
En= - 1/n2 me4 /(8ε02h2)=
-1/n2r0 e2/8πε0=> E1= - e2/8πε0r0
E1= - q2/2r0
<V>=2E1
What is the classic radius of an electron with energy E0?
What is the probability to find it in the larger radius?
(note that the answer in the exam solution is not correct for this and the previous)
Classical radius: mv^2/rc=F=q^2/rc^2 → q^2/rc=mv^2 → q^2/2rc=EK=<EK>
Quantum: <EK>=E0-<V>= q^2/2r0
Hence: r0=rc
For the probability, integrate from r0 to infinity over psi square!
Example II:
Spin properties of a wave function
ψ n ,l ,m,ms
1
1
3
=
ψ 2,1, −1,1 / 2 + ψ 2,1,1,1 / 2 − ψ 2,1,1, −1 / 2
5
5
5
Question: Show that it is normalized and find the average z component of the
orbital and spin angular momentums. What is the uncertainty in the first?
Answer:
2
Summary of 3 external B fields
B’>0
B=Constant
1. Stern-Gerlach exp.
1. Zeeman effect
2. Force acting on a
neutral atom. Number
of discrete force
projections on z, depends
on the value of µz.
2. Energy splitting
Number of discrete energy
levels depends on the
number of projections on z,
i.e. different values of µz.
B=0
1. Fine structure
Splitting caused by
Spin-orbit coupling.
2. No force on the
atom.
3. No force.
The total magnetic moment of the atom is:
µz = [e/(2m) lz + e/(m) sz + e/(2M) iz] = [eħ/(2m) ml + eħ/(m) ms + eħ/(2M) mi]=
µB [ml + 2ms] + µN [mi]
In this course we ignore the last term which is caused by the Nuclear spin and which
gives rise to the Hyperfine structure (i.e. final state splitting for B=0 case).
Exercise examples
B=0
Question: The fine structure splitting of the 2P3/2 and
the 2P1/2 levels in hydrogen is 4.5 10-5 eV. From this
estimate the magnetic field the 2P electron in hydrogen
experiences (the field originates from the movement
of the proton in the frame of the electron).
Answer: µ = µB [ml + 2ms] and E=µB
In the frame of the electron we are interested only in
the electron spin and hence: µ = µB [2ms]= 2µB (+/- ½)
Therefore: ∆E= 2µBB
And B= ∆E/2µB = 4.5 10-5 / (2x 5.79 10-5) = 0.39T
B=Constant
Question: One of the main absorption lines of the Rubidium atom is at 6.8GHz
(this micro wave transition may be used also for atomic clocks).
Assume that the maximum JZ of the higher level is 2.5 and for the lower level its 1.5.
What will the photon spectrum look like (assume that a photon can only change
JZ by 0 or +/-1).
Answer: JZ for the higher level is -2.5, -1.5, -0.5, 0.5, 1.5, 2.5 and for the lower level
Its -1.5, -0.5, 0.5, 1.5. Let us for example take the transition from JZ=2.5 in the higher
level to JZ=1.5 in the lower level. As the Zeeman energy change is µB, the energy
for the higher level changed by Bµ=B µB h [2+2x1/2] and for the lower level by
Bµ=B µB h [1+2x1/2], a new photon will appear in the spectrum with an frequency
of 6.8 109 + {B µB h [3-2]}/h Hz, when µB is given in units of Hz/G (1.4 10^6)
What is the magnetic field on the sun???
B’>0
Stern-Gerlach experiment
Question: what will be the distance between the 2 points on the screen if the SG device
is 10cm long, and if the distance to the screen is 10 times larger, and if B(z)=Az.
Answer: according to the given field, F=2 µB sZ A and therefore the transverse velocity
which the passing particle acquires is +/- F/m t and the distance between the two hitting
points is 2 F/m t T, where t is the time it takes the particle to pass the magnetic field and
T is the time it takes it to arrive at the screen. The final distance between the points is 2
F/m L/V 10L/V (if one neglects the transverse distance inside the SG machine) and
2* F/m L/V 10L/V + 2* 1/2 F/m (L/V)2 (without neglecting the latter).
Question: what would happen in the SG exp. if they would take an atom in level l=1?
Answer: As the magnetic moment contribution of the electron depends on its total
angular momentum J, and as JZ=LZ+SZ, and as, when calculating the distances and
therefore the forces one has to take into account that g for the orbital motion is gL=1
while for the spin is gS=2, we will have the following forces acting on the atoms:
F(LZ=+1, SZ=+1/2), F(LZ=+0, SZ=+1/2), F(LZ=-1, SZ=+1/2),
F(LZ=+1, SZ=-1/2), F(LZ=+0, SZ=-1/2), F(LZ=-1, SZ=-1/2),
We then find F(LZ=+1, SZ=+1/2)=A µB (LZ*gL+ SZ*gS)=
A µB (LZ+ 2SZ)=2AµB, and in the same way:
F(LZ=0, SZ=+1/2)= AµB, F(LZ=-1, SZ=+1/2)=0, F(LZ=+1, SZ=-1/2)=0,
F(LZ=0, SZ=-1/2)= -AµB, and F(LZ=-1, SZ=-1/2)= -2AµB.
We therefore expect to see 5 separate hits on the screen.
Last example: We have said that for each n level, l=0,1,…n-1 (l is the orbital angular
momentum). This means 2l+1 projections on the z axis (noted by ml). As the Pauli
exclusion principle states that two fermions (spin half particles) cannot be at the exact
same quantum state if they are at the same position, each quantum level of the atom
(denoted by n, l, m) can only hold 2
electrons (as they have two possibilities
for their spin).
Therefore, the total number of electrons
each shell (n,l level) can hold is 2(2l+1).