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PHY211 - KTI Herbstsemester 2019 Exercise Sheet 1 O. Steinkamp

A. Mathad, D. Lancierini Issued: 18.09.2019 https://www.physik.uzh.ch/de/lehre/PHY211/HS2019. Discussed: 20.09.2017 html

Lab experiment: : Measurements on Positronium is a composed by an and its antiparticle, a . The physical properties of this bound state can be described by borrowing the formalism used to solve the spectrum of the . Recall that when inserting in the time independent Schrödinger equation:

  m · m − ~ ∇2 + V ψ(x, y, z) = Eψ(x, y, z) with µ = p e 2µ mp + me a central potential of Coulombian form the normalizable solutions occur Figure 1: The positron- only if the binding energy E assumes discrete values of the form: ium system

2 ~ En = − 2 2 (1) 2µa0n

2 2 Where n is the and and a0 = ~ /µe the Bohr radius.

Exercise 1: Positronium binding energy and annihilation

a) Derive the binding energy for the positronium . Can it be regarded as a relativistic bound state? Motivate your answer.

b) When the electron and the positron coincide, they annihilate to produce at least two photons. Using relativistic momentum conservation show that the annihilation to one photon is not kine- matically allowed.

c) What energy will the two photons be emitted at?

SOLUTION: For convenience let’s express the electron charge in CGS units

2 2 eMKS eCGS = 4π0 = (1.602 × 10−19C)2(8.988 × 109Nm2/C2) = 2.307 × 10−28J · m = 14.42 eVÅ the Bohr radius for the positronium bound state becomes: 2 2 2~ 2 (hc) a0 = 2 = 2 2 2 e me 4π mec e 1 (1.240 × 104 eVÅ)2 = 2π2 (0.511 × 106eV)(14.42 eVÅ) = 1.057Å

The reduced mass µ for the positronium system is µ = me/2. Note that the Bohr radius for the positronium is roughly double the one for the , this can also be deduced from the fact that the reduced mass for the hydrogen atom is to a very good approximation the electron mass, i.e.

mpm˙ e mpm˙ e mpme ¨m¨pm˙ e µ = =   −−−−−→   ∼ me m + m me me p e mp 1 + m¨p 1 + mp ¨ mp

a) Using the values found and eq. (1) we are ready to derive the binding energy of the ground state of the positronium: 2 ~ En = − 2 2 2µa0n 2 2 ~ (hc) = − 2 2 = 2 2 2 2 mea0n 4π (mec )(1.057Å) n 6.822 eV = −−→n=1 6.822 eV n2

Now since the total rest mass of the positronium is 2me = 1022 keV and is six orders of magnitude greater than the binding energy of the system, it can be treated non-relativistically.

In order to show that the process e+ + e− → γ is not kinematically allowed and positronium must annihilate to at least two photons we make use of conservation of four-momentum. In units in which ~ = c = 1 the four momentum of a particle is expressed as:

p = (E, px, py, pz)

Where E is its energy and pi is the i-th spatial component of its three-momentum. If the particle is real (or equivalently on-shell) it satisfies Einstein’s dispersion relation:

2 2 2 2 2 2 p = E − px − py − pz = m | {z } q Note: this is equivalent to E = m2 + ~p 2 Where m is its rest mass.

b) For the process e+ + e− → γ, conservation of four momentum implies:

pe− + pe+ = pγ

pe− = pγ − pe+ , now square both sides to get: (2) 2 2 me = 0 + me − 2pe+ · pγ → pe+ · pγ = 0 Now pe+ · pγ is a dot product between two four vectors: pe+ = (E, px, py, pz) and pγ = (ω, kx, ky, kz). Since it’s a scalar, it is invariant with respect to the reference frame where it’s evaluated and, for simplicity, we chose to evaluate it in the electron’s rest frame where ~p = 0. In RF this reference frame it reads: p + = (m , 0, 0, 0) and the scalar product becomes p + · p = m ω. e e e γ e Conservation of four-momentum (2) requires this scalar product to be 0, and since me 6= 0, it must be that ω = 0 and thus a non-zero energy photon would violate it. c) e+e− → γγ positronium annihilation occurs when the electron and the positron coincide, i.e. when they’re both at rest in their center of mass frames. Due to momentum conservation the two photons will be emitted back-to-back with equal and opposite momentum and will share evenly the energy available of 1022 keV, i.e. two 511 keV photons will be emitted. Exercise 2: Ortho and para positronium The ground state of the positronium can be found in two configurations depending on the relative orientations of the spins of the electron and the positron. The state with antiparallel spins is known as para-positronium while the one with parallel spins is referred to as ortho-positronium. Use internet search to discuss the properties of the two positronium ground states:

a) What physics principle allows only even/odd number of photons for the decay of the para/ortho positronium?

b) Which one of the two do you expect to have longer lifetime?

c) What causes the longer lived bound state to have a lower lifetime in material than in vacuum?

SOLUTION: The full wavefunction of positronium can be considered as the product of the orbital and a spin vector:

Ψ = ψ |S, S i (3) n,l,m,S,Sz n,l,m z

The spin component of the total wavefunction can be decomposed in linear combinations of the spin wavefunctions of the individual particles:

 |S = 1,S = 1i = | ↑i| ↑i  z  1 |S = 1,Sz = 0i = √ (| ↑i| ↓i + | ↓i| ↑i) (4)  2  |S = 1,Sz = −1i = | ↓i| ↓i

n |S = 0,S = 0i = √1 (| ↑i| ↓i − | ↓i| ↑i) z 2

Where for example the state | ↑i| ↑i is the product of a positron spin state with Sz = +1/2 and an electron spin state with Sz = +1/2 which produces an eigenstate of total spin S = 1 and z component Sz = 1. The first three expressions above describe the totally symmetric triplet state referred to as ortho-positronium (o-Ps), the last expression is the totally antisymmetric singlet state referred to as para-positronium (p-Ps). In order to characterise the decays of the two positronium states we need to consider the operation of spatial reflection ~r → −~r referred to as parity P , and that of charge conjugation parity C, i.e. the exchange of particles and antiparticles. The positronium wave function (3) is an eigenstate of parity, its eigenvalue is given by the product of the parities of the individual wavefunctions: the obital wavefunction is an eigenstate of parity with eigenvalue (−1)l while the parity of the spin vector is the product of the parity of the individual particle’s spin vector (intrinsic parity) which is conventionally opposite for particles and antiparticles.

P|Ψ(~r)i = |Ψ(−~r)i = (−1)l · (−1)(+1) |Ψ(~r)i | {z } | {z } Angular momentum Intrinsic parity Positronium is also an eigenstate of charge conjugation C and in this case the operation is equivalent to the application of parity and echange of spin labels on the right hand side of eq. 4 Show that the described operation produces a sign change of −(−1)S to the total wavefunction A system of n-photons is an eigenstate of charge conjugation parity with eigenvalue

C|n-photonsi = (−1)n|n-photonsi

Where n is the number of emitted photons. Considering that the electron and positron overlap when l = 0 and that the electromagnetic force, which mediates this process, conserves both parity and charge conjugation parity it must be that:

(−1)S = (−1)n

a) And S even/odd decays to an even/odd number of photons The decay widths for the two processes e+e− → γγ and e+e− → γγγ are found to be proportional to:

! 2 ~ Γ(p-Ps → 2γ) ∝ 4πα 2 (5) mecvrel

! 16 2 3 ~ Γ(o-Ps → 3γ) ∝ (π − 9)α 2 (6) 9 mecvrel b) With α ∼ 1/137 fine structure constant. Since the decay time is inversely proportional to the decay width, τ ∝ Γ−1 the extra α factor in (6) causes the ortho-positronium decay time to be longer than its para counterpart

c) When in immersed in material, the bounded positron of ortho-positronium can annihilate in 2 γ with an electron which is not its partner electron. This effect, which is called pick-off annihilation, lowers the decay time of the o-Ps in materials wrt when in vacuum. Lab experiment: : Angular correlation of successive gamma decays The conservation of angular momentum provides information on the structure of decaying nuclei and plays a controlling role in the gamma ray emission. From a schematic viewpoint, a stationary nucleus in a definite quantum mechanical state makes a transition to a lower energy state during gamma decay and emits a single photon. Both the initial and final states of the nucleus will have a definite angular momentum and parity thus the photon emission process must connect the two states and conserve both parity and an- gular momentum. The angular momenta of the initial and final states nuclei, labelled as Ji and Jf satisfy the relation: Figure 2: the gamma decay cas- 60 cade after the 27Co undergoes beta

|Ji − Jf | ≤ ` ≤ Ji + Jf (7) decay

Where ` is the angular momentum carried away by the photon in the emission. The order of the multipole of the photon is a quantification of the amount of angular momentum carried by the photon, i.e. a photon with ` = 1, 2 ... units of angular momentum is called a di/quadru ... -pole photon. The angular distribution of the emitted radiation from a single photon is isotropic since the nuclei are oriented at random. Anisotropic angular distributions can be observed if a preferred direction is established when a successive photon is emitted in a cascade from the same nucleus. The observed angular correlations are analysed in terms of a power series of cosθ, W (θ), normalized such that W (θ = 90◦) = 1

k X 2i W (θ) = 1 + aicos θ (8) i=1

Where 2k is the order of the lowest multipole in the cascade. A further restriction on the number of terms in W (θ) is ai = 0 for i > Jf

Exercise 3: On the angular correlation of successive gamma decays

60 a) Use non-relativistic kinematics to calculate the recoil kinetic energy of the 28Ni for each gamma emission in fig. 2

b) Use equations (7) and (8) to determine the maximum order in cosθ for the angular distributions W (θ) of the cascade transitions to the ground state with angular momenta:

Ji = 0 → Jf = 2 → Jgs = 0

and

Ji = 1 → Jf = 1 → Jgs = 0

c) In which cases do you expect the gamma ray cascade to be isotropic? SOLUTION:

a) Conservation of energy requires that

2 ∗ 2 (m0c ) = m0c + Eγ + Tr

2 ∗ 2 Where (m0c ) is the rest mass of the excited nucleus and (m0c ) the one of the ground state, Eγ the photon energy and Tr the recoil kinetic energy of the nucleus after gamma emission. The nucleus must recoil in the opposite direction of the emitted photon thus pr = pγ. The classical expression for kinetic energy Tr = pr/2m0 then reads:

2 2 p p =E /c , E =E E r r r r γ γ (9) Tr = −−−−−−−−−−−→ 2 2m0 2m0c

60 2 For 28Ni, m0c = 57.94 amu · 931.49 MeV / amu = 52923.61 MeV and sobstituting in eq. (9):

(1.17 MeV )2 T = = 12.93 eV r,1 2(52923.61 MeV ) (1.33 MeV )2 T = = 16.71 eV r,2 2(52923.61 MeV )

b) For the transition Ji = 0 → Jf = 2 → Jgs = 0,

` = 2

the order of the multipole is 2` = 4 and the angular distribution W (θ) will be described by a 4-th order polynomial in cosθ:

2 X 2i 2 4 W (θ) = 1 + aicos θ = 1 + a1cos θ + a2cos θ i=1

For the transition Ji = 1 → Jf = 1 → Jgs = 0,

0 < ` ≤ 2

but since Jf = 1 and i = 0, ∀i > Jf the angular distribution W (θ) will be described by a 2-nd order polynomial in cosθ:

1 X 2i 2 W (θ) = 1 + aicos θ = 1 + a1cos θ i=1

c) The special cases in which Jf = 0 or Jf = 1/2 are such that ai = 0, ∀i ≥ 1 and thus the cascade emission will be isotropic, i.e. W (θ) = 1