Transcript Lecture 2

a from the quantum Hall effect
The everyday Hall effect: A current density jx in the
presence of a perpendicular magnetic field Bz induces an
electric field Ey which is perpendicular to both. From
balance of forces F=qvB and F=qE the field is equal to
vB where v is the average velocity of charge carriers.
The Hall resistance R is defined to be the ratio of
induced field to applied current; R=E/j=vB/Nvq=B/Nq.
So the normal Hall resistance is a measure of N, the
density of charge carriers in the sample. It is sensitive to
material type, impurities, defects, temperature, etc.
The quantum Hall effect: If electrons
are confined in a thin layer at low
temperature in a high magnetic field the
Hall resistance vs. B rises in a series of
steps which are quantised at levels given
by R=h/ie2 where i is an integer.
quantum Hall continued
The quantum Hall effect was discovered by von Klitzing in 1980 (Nobel prize 1985).
It was totally unexpected and initially unexplained.
A partial explanation: A perpendicular
magnetic field splits the states in a 2D electron
gas into “Landau levels”. The number of
current carrying states in each level is eB/h.
The position of the Fermi level relative to the
Landau levels changes with B. So the number
of charge carriers is equal to the number of
filled Landau levels, i, times eB/h. Plugging
this into the classical expression for R gives
R=h/ie2.
From our point of view the important features are;
•The resistance can be precisely measured ( 1 in 108 ).
•It is simply related to fundamental constants. R=h/ie2, a= e2/2e0hc so R=1/2ie0ca. Both e0
and c are constants without errors: e0= 1/m0c2, m0=4p . 10-7 (NA-2) and c=299,792,458
(ms-1) by definition.
•The measurement is done at very low energy so higher order corrections are negligible.
B.I.Halperin. Scientific American 1986. D.R.Leadley http://www.warwick.ac.uk/~phsbm/qhe.htm
GF from the muon lifetime.
Before electro-weak unification, muon decay ( and
all other weak interactions ) were described by an
effective 4-fermion coupling of strength GF.
We now know that weak decays take place
via an intermediate W boson like this :
However, the Fermi constant GF is still one of
the most accurately measurable parts of the
standard model. So it is taken as one of the
primary constants. It is related to the other
“more fundamental” constants by
GF=sqrt(2).g2/8.mW2 . Note g is related to
other electroweak parameters through
sinqw=e/g and cosqw=mW/mZ.
The muon lifetime is calculated
to be
1/tm= GF2 mm5/192p3 + h.o.c.
The muon lifetime experiment
An accurate measurement of the muon lifetime was made back in 1984 at
Saclay, Paris. Not improved since. Because GF is accurate enough for present tests of SM?
Method: A muon source is built up by stopping 140 MeV/c p+ ( and
some m+ ) during a 3 ms burst of the linac. A positron from muon decay is
detected as a triple coincidence in one of the 6 scintillator telescopes.
Some time after the end of each burst a clock is started and the time
distribution of positron signals is recorded.
Some questions
How do you stop a 140 MeV/c pion or muon ?
Any charged particle passing through matter
loses energy by ionisation or excitation of some
of the atoms that it passes. If it hits a thick
enough piece of matter it will slow down and
stop unless something else happens first (e.g..
nuclear interaction, bremsstrahlung or the particle itself
decays).
The energy loss rate, -dE/dx, of a pion of
bg=1 in aluminium is about 2.5 (MeV g-1 cm2)
and is rapidly rising as the momentum falls.
The energy loss in a thickness t of the sulphur
target will be more than;
DE = t (cm) . 2.5 (MeV g-1cm2) . rsulphur(g cm-3)
Sulphur has a density of 2.1 so we can say that
11 cm will stop the pions, probably half that is
enough. ( K.E. = sqrt(p2+m2)-m = 58 MeV )
Where do the muons from pion decay go ?
What happens when they stop?
99.99% decay p  mnn with a
lifetime of 26 ns.
In 2 body decay of M, the energy of decay products in the rest frame of M is E 1 =
(M2 – m22 + m12)/2M. mp =139.6 MeV, mm = 105.7 MeV so Em = 109.8 MeV. They
will stop within a cm or so from where the pion decayed.
Possible sources of error
Backgrounds. Cosmic and Beam-induced:
The arrival time of cosmic rays will be random so they will simply contribute a flat
background which is easy to subtract. The cosmic background level was measured by
taking data at long times after the linac burst.
The beam particles can create other states with short lifetimes. Their decays may trigger
the telescope and mimic a positron from muon decay. They will be correlated with
beam bursts in the same way as the muons, so are a potentially a serious bias to the
muon lifetime. The paper states that there was only one significant background of this
type ( X-rays from thermal neutron capture ) and that it was fitted with a single
exponential with a time constant of 160 ms.
Random coincidences:
If two decays occur within the resolving time of the electronics then only one of them
will be seen. This is more likely to happen when decays are happening frequently – it is
proportional to instantaneous rate. It is taken into account in the fit:
R(t) = R0 [ exp(-lt) + A r exp(-2lt) + B ]
effect, B=background.
l = muon lifetime, r=rate, A=coincidence
Errors due to muon polarisation
The muons which are produced from pion decay in the target are unpolarised, but some muons
reach the target after pion decay in flight and these ones are partially polarised (Why?). The
direction of the positron from muon decay is correlated with the direction of the muon spin, so
if the telescope efficiency is not isotropic a change of spin direction leads to a change of
counting rate and therefore a lifetime error. This effect is minimised by
•Using sulphur for the target which rapidly depolarises the muons (Why?). Some muons will
stop in other materials so this does not entirely eliminate the problem.
•Reducing the magnetic field in the target region with Helmholtz coils to 0.1 G ( ie nearly
cancelling Earth’s field ). It is the change of spin direction by precession that is harmful.
•Measuring the lifetime in each of the 6 telescopes independently and checking that all give
the same result.
Lifetime result
The result is tm+= 2197.078  0.073 ns.
This is converted into GF including corrections of up to (me / mm )8 and (mm / mW )2 and a2.
It gives, together with a similar measurements of tm- , the result
GF/(c)3 = 1.16637(1)  10-5 GeV-2
mZ from LEP lineshape scan
Dr Loebinger has already told how
the mass of the Z boson was
measured very accurately at LEP-1
by scanning the energy of the
accelerator across the resonance
and measuring the cross section at
each point.
The result is mZ=91187.5  2.1
MeV. The uncertainty is mainly 1.7
MeV due to uncertainty of the
beam energy and common to all
LEP experiments. There is also a
common 0.3 MeV due to
theoretical uncertainty of initial
state radiation.