Roberto De Renzi DiFeST, Department of Physics and Earth … · 2017. 5. 15. · Roberto De Renzi...

12/05/2014 De Renzi - ISIS Muon Training

Introduction to µSR

Roberto De Renzi DiFeST, Department of Physics and Earth Sciences University of ParmaItaly

1896 1987 2014

Setup of the first spectrometer at ISIS, MuSR


Introduction to µSR

• Muon history– The charged particles– Anti-matter– The neutrinos– Parity violation

• How it works– Production– Spin polarization– Transport– Implantation– Detection

• Few examples• Summary


J.J. Thomson: the electron

1896 1955 2014

B

Electrons orbit around BThomson measures the ratio :a light (lepton) q


E. Rutherford: the nucleus

1911 1955 2014

particles througha gold foil scatter atlarge angles

An even better cake!

Bohr atom

E Rutherford


E. Rutherford: the proton

1918 1955 2014

particles throughN2 scatter hydrogennuclei

Let's call them protons!

E Rutherford


P.A.M. Dirac, C. Anderson: the positron

1928 1955 2014

From relativity, quadratic energy form

Dirac predicted the electron sea

E 2=p2 c 2+m2 c 4

E =√p2 c 2+m2 c 4 E =−√p2 c 2+m2 c 4e- ?

E

e-

e-

0

B

No, it's an anti-electron

1932

PAM Dirac

CD Anderson


Antimatter

1932 1955 2014

B

p ⇔ p̄p + p -

e ⇔ ēe - e +

proton

electron

antiproton

antielectron = positron

Dirac's E


The name is my idea, yo!Nuclear massdidn't add up!

J. Chadwick: the neutron

1932 1955 2014

particles from Po on Boproduce unknown radiation

Po

N2

N

n

Neutral particle with mass mnc2

= 938 MeV = 1.0014 mpc

2

= 1800 mec2

heavy (baryon) J Chadwick

E Rutherford


Pauli suggested a neutralparticle for β decays

1930 1955 2014

Beta decay at rest, if it were a 2-body decayproducts would have fixed energies

614 C → 7

14 N +e−

e−714 N

There must bean additional

neutralparticle,

the neutron!

Instead they have an energy spectrum 3-body decay!

?

W Pauli


Pauli suggested a neutrino

1930 1955 2014

After Chadwick'sdiscovery of theneutron let's call

it neutrino

1932614 C → 7

14 N +e−+ν̄ e

There must bean additional

neutralparticle,

the neutron!

Instead they have an energy spectrum

Beta decay at rest, if it were a 2-body decayproducts would have fixed energies

614 C → 7

14 N +e−

e−714 N

W Pauli

E Fermi


H. Yukawa: the meson

1935 1955 2014

Nuclei are made of n and p+. What force binds them with finite range?Coulomb force = exchange of photons

In analogy

e- hν p+ mν = 0

n meson p+ mc2 ~ 150 MeV

V (r )∝1r

V (q )∝ 1q2

V (r )∝1r

e− h r

√2mc V (q )∝ 1

q 2+2m2 c 2

h2

Mass justifies screening, finite range

0.5 140 980 Mev leptons mesons baryons

H Yukawa


C. Anderson, S Neddermayer: mesotron

1936 1955 2014

Within two years a new particle with that mass (~) is found. C.D. Anderson calls it mesotron

100 < mc2 < 150 MeV

withcosmic ray balloons

τ ~ 2 μs I measured itin 1941

However its decay is a bit slow

and it has spin S=1/2

But I don't know that yet

CD AndersonBB Rossi

VF Hess


M. Conversi: not Yukawa's meson?

1946 1955 2014

Furthermore the mesotron does not interactstrongly enough with matter.

μ ⇔ μ̄μ- μ+

Fe C

µ+ 0.67 ± 0.07 0.36 ± 0.05

µ- 0.03 ± 0.03 0.27 ± 0.03

μ- + p+ → n + νμ

M Conversi


π

e

μ

μ

C. Powell G. Occhialini: Two particles, pion and muon

1947 1955 2014

Three tracks in a photographic emulsions at Mt Chacaltaya (5600 m).

The π is Yukawa's meson

mπ = 140 MeV/c2

τπ = 26 ns S = 0

The μ is a lepton (a heavier electron)

mμ = 106 MeV/c2

τμ = 2200 ns S = ½

C Powell

G Occhialini


π

e

μ

μ

C. Powell G. Occhialini: Two particles, pion and muon

1947 1955 2014

Three tracks in a photographic emulsions at Mt Chacaltaya (5600 m).

The π is Yukawa's meson

mπ = 140 MeV/c2

τπ = 26 ns S = 0

The μ is a lepton (a heavier electron)

mμ = 106 MeV/c2

τμ = 2200 ns S = ½

Hey, there's somethingmissing here

W Pauli

C Powell

G Occhialini


So what is missing?

1947 1955 2014

Hey, there's somethingmissing there

Linear momentum conservation!

πe μμ

μμ

W Pauli


π

eμ

μ

Pion and muon, bothweak decays

1947 1955 2014

π+ → μ++ νe

μ+ → e ++ν̄μ+νe

νe

νμ

νe

Matter Antimattere - ē=e+

p - p̄=p -

n n̄π- π0=π̄0 π̄=π

+

μ- μ̄=μ+


Recognitions

Incidentally: Nobel prize for

● 1906 J.J. Thomson Physics● 1908 E. Rutherford Chemistry● 1933 P.A.M Dirac and E. Schrödinger Physics● 1935 J. Chadwick Physics● 1936 C.D. Anderson, V.F. Hess Physics● 1938 E. Fermi Physics● 1945 W. Pauli Physics● 1949 H. Yukawa Physics● 1950 C.F. Powell Physics


uup

ccharm

ttop

ggluon

ddown

sstrange

bbottom

photon

e μ τ WW boson

νe

νμ

ντ

Z0Z

0 boson

What we know today

1955 2014

Three families

Qua

rks

Lep

tons

Un ifi

e d f o

r ces

BaryonsMesons

p=uud n=ddu

π=u d̄ π0=u ūd d̄


So we now have everything

We can produce pions

and they produce muons

However μSR needs another ingredient to work....

p+ + n → n + n + π+

π+ → μ+ + νμ


Lee and Yang: parity violation

In certain interactions (e.g. magnetic) parity is broken

Weak interactions violate parity

i.e. the mirror imagedoes not exist in nature

1957 2014

CN Yang

TD Lee


Parity violation

weak interactions violate parity

1957 2014

Madame CB Wu demonstrated that

Only right-handed anti-neutrinosand left-handed neutrinosexist in nature

TD Lee and CN Yang gotthe 1957 Nobel prize

Anisotropicdecay

CB Wu


Parity violation

weak interactions violate parity

1957 2014

Also Garwin Lederman & Weinrich showed that

μ+ → e+ + ν̄ μ + νe

Nµ

=- 1 Ne = -1 +1 = 0

Nµ = -1


Let's sum up

Accelerate protons to Ek > 280 MeV ~ 2mπc2, to impinge on a target

p+ n


Pion production

Accelerate protons to > 280 MeV and impinge them on a target

n

n

π

p++n → n + n + π+

Lots of pions


Pion decay

Pions that decay at rest on the surface of the target

π

τπ=26 ns


Parity violation

Remember! The pion is S=0 Two body decay

π+ → μ+ + νμ

Sµ=½Sν=½

100% spin polarized muon beamsthanks to parity violation

τπ=26 ns

π


Energy and momentum

(c = 1)

mπ=√mμ2+p2+p

p=m π

2−mμ2

2m π= 29.8 MeV/c E μ=

mμ2+mπ

2

2m π= 109.8 MeV/c2

β=vc= p

E μ= 29.8

109.8≈0.271

√mπ2+p π2=√mμ2+pμ2+√mν2+pν2 Energy pμ=−p ν=p

pp π

and momentum

conservation

I.e.

Hence

E μ ,k=√mμ2+p2−mμ= 4.12 MeV/c2And the muon kinetic energy is


Muon beam transport

Quadrupole Dipole

Quadrupole pair

brings muonsto stop ina sample(mostly at aninterstitial site)


Let's introduce Muonium

MuO

Mu = µ+ + e-

1s

Bound state, light isotope of H

Paramagnetic

In matter it most oftenbinds to other ions formingcovalent bonds

Diamagnetic


Thermalization

4 MeV

Electron scattering(ionization) 10-10 s

2-3 keV

Muonium formation(e- capture/loss + collisions) 10-12 s200 eV few eV


Muon decay

Average lifetime τμ = 2.2 μs

e+

μ+ → e+ + νe + ν̄μ

N e− tτμ


Muon decay

Three body decay

e+

μ+ → e+ + νe + ν̄μTakes place like this:

μ+ν̄μνe

Emax ~ ½mµc2

~ 50 MeV

μ+

e+

ν̄μνe

(by parity violation this does not take place)


Muon decay

Three body decay

e+

μ+ → e+ + νe + ν̄μ

ν̄μ

νe

Emin = 0

μ+

or like this:


Energy distribution in the muon decay

Positron distribution

with asymmetry

and

probability of emission

P (x ,θ)=1+A (x )cosθ

E (x )= 2 x2

3−2 x

A (x )=2 x −13−2 x


Asymmetry of the muon decay

E =E max

P (θ)∝1+cosθ

average over all energies

P (θ)∝1+ 13

cosθ

Probability of e+ emission

Sµ Sµ


No spin dynamics

F B

Sµ

Asymmetry


Spin dynamics: precession

m=γℏS

Magnetic moment

ℏ d Sdt

=m×B loc

ω m [−sinω t sinθcosω t sinθ0 ]=m (−γB loc )[−sinωt sinθcosω t sinθ

0 ]http://www.fis.unipr.it/~derenzi/dispense/pmwiki.php?n=NMR.SpinPrecession

Larmor

B = B ẑ

x̂

ẑŷ

θ

m(0)ẑ

x̂

Semiclassical dynamics

m (t )=m [cosωt sinθsinω t sinθcosθ ]


Transverse field µSR

B

No spin dynamics ω=−γB loc

γ2π

=135.5 MHz/T

Spin precession at the Larmorfrequency


e.g. in a magnetic material

B locα=∑

i

D iαβ m i

β

α ,β=x , y , z

Local field

ω=−γB locLarmor frequency

B loc

dipolar Fermi contact

+ Ac m1α

∝m magneticmoment


MuSR nowadays

µ


PSI GPS: another workhorse

µ

x

y

x

y

z


Example: Antiferromagnetic YBa2Cu3O6+x


The first magnet ever:Fe3O4

A spinel ferrimagnetwith the metal-insulator Verwey transition


Examples

MnSi helimagnet

site determined by DFT


Where?

J-PARC

TRIUMFISIS

PSI


Summary

S γ m τ

½ 135.5 MHz/T 105.66 MeV 2.197 µs

γe/206.8 (*) 206.8 me

3.18 γp mp/9

* The anomalous electron g (QED corrections) is 2.0023193043615(5) cfr. the anomalous muon g 2.0023318414(1)

γ=g e2m

Muon properties

θ

Sμ

B


Bibliography

Particle hystory survey● D. Griffith, John Wiley, New York, 1987, Ch. 1

μSR● A. Schenck, Adam Hilger, Bristol 1986

● A. Yaouanc, P. Dalmas de Reotier, Oxford Univ. Press, 2011. - 486 p

● S.J. Blundell Contemporary Physics 40, 175 (1999) http://arxiv.org/abs/cond-mat/0207699

● Some private notes athttp://www.fis.unipr.it/~derenzi/dispense/pmwiki.php?n=MuSR.MuSR

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Roberto De Renzi DiFeST, Department of Physics and Earth … · 2017. 5. 15. · Roberto De Renzi...

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