# Spin And Parity Assignments For Kids

In quantum mechanics, a **parity transformation** (also called **parity inversion**) is the flip in the sign of *one*spatialcoordinate. In three dimensions, it is also often described by the simultaneous flip in the sign of all three spatial coordinates (a point reflection):

It can also be thought of as a test for chirality of a physical phenomenon, in that a parity inversion transforms a phenomenon into its mirror image. A parity transformation on something achiral, on the other hand, can be viewed as an identity transformation. All fundamental interactions of elementary particles, with the exception of the weak interaction, are symmetric under parity. The weak interaction is chiral and thus provides a means for probing chirality in physics. In interactions that are symmetric under parity, such as electromagnetism in atomic and molecular physics, parity serves as a powerful controlling principle underlying quantum transitions.

A matrix representation of **P** (in any number of dimensions) has determinant equal to −1, and hence is distinct from a rotation, which has a determinant equal to 1. In a two-dimensional plane, a simultaneous flip of all coordinates in sign is *not* a parity transformation; it is the same as a 180°-rotation.

## Simple symmetry relations[edit]

Under rotations, classical geometrical objects can be classified into scalars, vectors, and tensors of higher rank. In classical physics, physical configurations need to transform under representations of every symmetry group.

Quantum theory predicts that states in a Hilbert space do not need to transform under representations of the group of rotations, but only under projective representations. The word *projective* refers to the fact that if one projects out the phase of each state, where we recall that the overall phase of a quantum state is not an observable, then a projective representation reduces to an ordinary representation. All representations are also projective representations, but the converse is not true, therefore the projective representation condition on quantum states is weaker than the representation condition on classical states.

The projective representations of any group are isomorphic to the ordinary representations of a central extension of the group. For example, projective representations of the 3-dimensional rotation group, which is the special orthogonal group SO(3), are ordinary representations of the special unitary group SU(2) (see Representation theory of SU(2)). Projective representations of the rotation group that are not representations are called spinors, and so quantum states may transform not only as tensors but also as spinors.

If one adds to this a classification by parity, these can be extended, for example, into notions of

*scalars*(*P*= +1) and*pseudoscalars*(*P*= −1) which are rotationally invariant.*vectors*(*P*= −1) and*axial vectors*(also called*pseudovectors*) (*P*= +1) which both transform as vectors under rotation.

One can define **reflections** such as

which also have negative determinant and form a valid parity transformation. Then, combining them with rotations (or successively performing *x*-, *y*-, and *z*-reflections) one can recover the particular parity transformation defined earlier. The first parity transformation given does not work in an even number of dimensions, though, because it results in a positive determinant. In odd number of dimensions only the latter example of a parity transformation (or any reflection of an odd number of coordinates) can be used.

Parity forms the abelian group due to the relation . All Abelian groups have only one-dimensional irreducible representations. For , there are two irreducible representations: one is even under parity, , the other is odd, . These are useful in quantum mechanics. However, as is elaborated below, in quantum mechanics states need not transform under actual representations of parity but only under projective representations and so in principle a parity transformation may rotate a state by any phase.

## Classical mechanics[edit]

Newton's equation of motion (if the mass is constant) equates two vectors, and hence is invariant under parity. The law of gravity also involves only vectors and is also, therefore, invariant under parity.

However, angular momentum is an axial vector,

- ,
- .

In classical electrodynamics, the charge density is a scalar, the electric field, , and current are vectors, but the magnetic field, is an axial vector. However, Maxwell's equations are invariant under parity because the curl of an axial vector is a vector.

## Effect of spatial inversion on some variables of classical physics[edit]

### Even[edit]

Classical variables, predominantly scalar quantities, which do not change upon spatial inversion include:

- , the time when an event occurs
- , the mass of a particle
- , the energy of the particle
- , power (rate of work done)
- , the electric charge density
- , the electric potential (voltage)
- , energy density of the electromagnetic field
- , the angular momentum of a particle (both orbital and spin) (axial vector)
- , the magnetic field (axial vector)
- , the auxiliary magnetic field
- , the magnetization
- Maxwell stress tensor.
- All masses, charges, coupling constants, and other physical constants, except those associated with the weak force

### Odd[edit]

Classical variables, predominantly vector quantities, which have their sign flipped by spatial inversion include:

- , the helicity
- , the magnetic flux
- , the position of a particle in three-space
- , the velocity of a particle
- , the acceleration of the particle
- , the linear momentum of a particle
- , the force exerted on a particle
- , the electric current density
- , the electric field
- , the electric displacement field
- , the electric polarization
- , the electromagnetic vector potential
- , Poynting vector.

## Quantum mechanics[edit]

### Possible eigenvalues[edit]

In quantum mechanics, spacetime transformations act on quantum states. The parity transformation, , is a unitary operator, in general acting on a state as follows: .

One must then have , since an overall phase is unobservable. The operator , which reverses the parity of a state twice, leaves the spacetime invariant, and so is an internal symmetry which rotates its eigenstates by phases . If is an element of a continuous U(1) symmetry group of phase rotations, then is part of this U(1) and so is also a symmetry. In particular, we can define , which is also a symmetry, and so we can choose to call our parity operator, instead of . Note that and so has eigenvalues . However, when no such symmetry group exists, it may be that all parity transformations have some eigenvalues which are phases other than .

For electronic wavefunctions, even states are usually indicated by a subscript g for *gerade* (German: even) and odd states by a subscript u for *ungerade* (German: odd). For example, the lowest energy level of the hydrogen molecule ion (H_{2}^{+}) is labelled and the next-closest (higher) energy level is labelled .^{[1]}

### Consequences of parity symmetry[edit]

When parity generates the Abelian group ℤ_{2}, one can always take linear combinations of quantum states such that they are either even or odd under parity (see the figure). Thus the parity of such states is ±1. The parity of a multiparticle state is the product of the parities of each state; in other words parity is a multiplicative quantum number.

In quantum mechanics, Hamiltonians are invariant (symmetric) under a parity transformation if commutes with the Hamiltonian. In non-relativistic quantum mechanics, this happens for any potential which is scalar, i.e., , hence the potential is spherically symmetric. The following facts can be easily proven:

- If and have the same parity, then where is the position operator.
- For a state of orbital angular momentum with z-axis projection , then .
- If , then atomic dipole transitions only occur between states of opposite parity.
^{[2]} - If , then a non-degenerate eigenstate of is also an eigenstate of the parity operator; i.e., a non-degenerate eigenfunction of is either invariant to or is changed in sign by .

Some of the non-degenerate eigenfunctions of are unaffected (invariant) by parity

## Nuclear Physics *PHY303*

## Problems

### Constants

Charge of the electrone = 1.6 10^{-19}C

Mass of the electronm_{e}= 9.11 10^{-31}kg = 511 keV/c^{2}

Mass of the protonm_{p}= 1.673 10^{-27}kg = 938.272 MeV/c^{2}

Mass of the neutronm_{n}= 1.675 10^{-27}kg = 939.566 MeV/c^{2}

Planck constanth = 6.626 10^{-34}J s = 4.136 10^{-15}eV s

Boltzmann constantk = 1.38 10^{-23}J K^{-1}= 8.617 10^{-5}eV K^{-1}

Speed of light in free spacec = 3.00 10^{8}m s^{-1}

Permittivity of free space_{0}= 8.85 10^{-12}F m^{-1}

Permeability of free spaceµ_{0}= 4 10^{-7}H m^{-1}

Avogadro constantN_{A}= 6.02 10^{26}kg-mol^{-1}

Rydberg constantR = 1.10 10^{7}m^{-1}

Bohr magnetonµ_{B}= 9.27 10^{-24}J T^{-1}

Nuclear magnetonµ_{N}= 5.0508 10^{-27}J T^{-1}= 3.1525 10^{-14}MeV T^{-1}

Fine structure constant= 1/137

### Useful data

Unified atomic mass unit1u = 1.66 10^{-27}kg = 931.502 MeV/c^{2}

Energy conversion1 eV = 1.6 10^{-19}J

Year in seconds1 yr = 3.16 10^{7}s

Atmospheric pressure1 atmosphere = 1.01 10^{5}N m^{-2}

Acceleration due to gravity on Earth's surfaceg = 9.81 m s^{-2}1 gram moleculeatSTPoccupies22.4 litres

**Look at the Lecture notes for Chapter 1 or the constants and useful data.**

- In an experiment carried out with a beam of thermal neutrons it is found that on traversing a

**2mm**thick foil of

**, some**

^{197}Au**70%**of the neutrons are removed. What is the total thermal neutron cross-section for this isotope of gold? Comment on the result of the cross-section measurement in the light of the fact that the radius of a gold nucleus is

**6.5 x 10**. (Density of gold:

^{-15}m**19300 kg m**)

^{-3}A beam of low energy neutrons, intensity

**10**traverses a foil of

^{6}s^{-1}**, thickness**

^{235}U**0.15 kg m**. If the fission cross-section is

^{-2}**2 x 10**find the rate of fissions induced in the foil by the neutrons. In soving this problem you can neglect the reduction of the beam as it passes through the foil ("thin foil" condition). By also doing the full calculation show that this approximation is justified. Carry this further by devising a criterion that should be satisfied if the "thin foil" approximation is to be used.

^{-26}m^{2}The nucleus

**decays to its mirror nucleus**

^{27}Si_{14}**by positron emission with a maximum energy of**

^{27}Al_{13}**3.48 MeV**. Find the difference in the coulomb energy between the two nuclei and hence estimate the value of

**r**in the expression for the nuclear radius

_{0}**R = r**.

_{0}x A^{1/3}For the nuclide

**the neutron and proton separation energies are**

^{16}O**15.7**and

**12.2 MeV**respectively. Estimate the radius of this nucleus assuming that the particles are removed from its surface and that the difference in separation energies is due to the Coulomb potential energy of the proton.

Estimate the angle at which the first diffraction minimum occurs when

**alpha particles**of kinetic energy

**100 MeV**are scattered by a nucleus of

**. Assume that the nucleus behaves as an impenetrable disc.**

^{56}FeFraunhofer diffraction by a circular disc with diameter

**D**produces a ring shaped pattern. The first minimum appears at an angle of

**1.22 /D**where is the wavelength of the light.

Calculate the kinetic energy of the alpha particle emitted in the process

**. The binding energy of the alpha particle is**

^{235}U +^{231}Th**28.3 MeV**and you may assume the following values (

**in MeV**) for the five coefficients in the semi-empirical expression for the binding energy of heavier nuclei: volume

**15.5**; surface

**16.8**; Coulomb

**0.72**; asymmetry

**23**; pairing

**34**.

**If you're stuck look at the**

**solutions**

**Look at the Lecture notes for Chapter 2 or the constants and useful data.**

- A simple quantum mechanical model of the deuteron is based on a spherically symmetric potential well given by

By insering the Laplacian operator for polar coordinates into the Schrödinger equation

show that the following function is a solution:

where

Include in your answer explicit expressions for and .

Use the continuity and normalisation conditions to find expressions for the coefficients and in the deuteron wave functions given in the previous question. Given that the binding energy of the deuteron is

**2.2 MeV**and assuming

**D**and

**b**are

**35 MeV**and

**2 x 10**respectively,estimate for what fraction of the time the separation of the neutron and the proton is greater than this value of

^{-15}m**b**.

The small binding energy of the deuteron indicates that the maximum of lies only just inside the radius of the well. Use this information to estimate the value of if is approximately

**2 x 10**.

^{-15}mFor a spherically symmetric potential well of the above given radius what would be the approximate depth if the binding energy of the deuteron were

**10 MeV**? Estimate the value of the magnitude of below which no bound state will exist.

Using the values

**µ**and

_{d(L= 0)}= 0.8798 µ_{N}**µ**find the percentage admixture of the (

_{d(L= 2)}= 0.3101 µ_{N}**) state if the deuteron magnetic momemt were given by**

*L*= 2**µ**. Note that, as indicated in the lecture,

_{d}= 0.8325 µ_{N}**a**.

_{0}^{2}+ a_{2}^{2}= 1On the basis of the

**Yukawa potential**discussed in the lecture, what would be the mass of the exchanged particle if the 'range' of the potential were

**10**?

^{-14}m**If you're stuck look at the**

**solutions**

**Look at the Lecture notes for Chapter 3 or the constants and useful data.**

- Given that the ordering of the nuclear levels is

1s ; _{1/2}1p ; _{3/2}1p ; _{1/2}1d ; _{5/2}1d ; _{3/2}2s ; _{1/2}1f ; _{7/2}2p ; _{3/2}1f_{5/2} |

justify the following ground state spin and parity assignments

; ^{3}He_{2}(1/2 +) ; ^{20}Ne_{10}(0 +) ; ^{27}Al_{13}(5/2 +) ; ^{41}Sc_{21}(7/2 -).^{69}Ga_{31}(3/2 -) |

Estimate the separation of the

**1p**and

_{1/2}**1d**energy levels for nuclei with mass number

_{5/2}**A ~ 16**given the following information:

the ordering of the lowest nuclear energy levels is | 1s ; _{1/2}1p ; _{3/2}1p ; _{1/2}1d ; _{5/2}1d_{3/2} |

and the total binding energy for the oxygen isotopes is | ^{15}O . . . . . . 111.96 MeV^{16}O . . . . . . 127.62 MeV^{17}O . . . . . . 131.76 MeV |

Find the configuration of the protons and neutrons in the incomplete shells and hence the ground state

**spin**and

**parity**assignments for the following nuclei

; ^{7}Li_{3} ; ^{23}Na_{11} ; ^{33}S_{16}^{41}Sc_{21} |

under the assumption that the ordering of the lowest single particle nuclear energy levels is

1s ; _{1/2}1p ; _{3/2}1p ; _{1/2}1d ; _{5/2}2s ; _{1/2}1d ; _{3/2}1f ; _{7/2}2p._{3/2} |

In this model the first excited states can be produced either

- by excitation of the unpaired nucleon into the next

**higher**subshell, orby pairing this nucleon with another excited from the next

**lower**subshell.

The observed nuclear moments of

**are:**

^{209}Bi_{83}I = 9/2 ; µ = + 4.1 µ ; _{N}Q = - 0.4 x 10^{-28} m^{2} |

Determine the expected values for these moments according to the simple Shell model and comment on any significant differences. Why would you expect

**to have an unusually low cross section (**

^{209}Bi**~ 0.003 barn**) for the capture of

**1 MeV**neutrons, compared with the average for a heavy nuclide which is about

**0.1 barn**?

The nuclide

**has a low lying level with spin and parity**

^{42}Sc_{21}**7**and an excitation energy of

^{+}**618 keV**. What is the likely Shell model configuration for this state and its expected magnetic moment?

Consider a permanently deformed nucleus whose shape is represented by an ellipsoid of revolution the surface of which is described by

where

**b**is the deformation parameter and the spherical harmonic

The quadrupole moment of a classical point charge

**e**is of the form

**e(3z**. Assuming that the charge of the nucleus

^{2}- r^{2})**Ze**is spread uniformly inside the ellipsoidal surface, obtain an expression for the electric quadrupole moment of the deformed nucleus.

**If you're stuck look at the**

**solutions**

**Look at the Lecture notes for Chapter 4 or the constants and useful data.**

- The isotope

**is a positron emitter, decaying to an excited state of**

^{14}O_{8}**. The gamma rays from this latter have an energy of**

^{14}N_{7}**2.313 MeV**and the maximum energy of the positrons is

**1.835 MeV**. The mass of

**is**

^{14}N_{7}**14.003074 u**and that of the electron is

**0.000549 u**. Write the equation for the decay of the oxygen isotope and sketch an energy level diagram for the process. Given that one unified mass unit (

**u**) is equal to

**931.502 MeV/c**find the mass of

^{2}**.**

^{14}O_{8}Given the following binding energies:

, ^{4}He_{2}B = 28.3 MeV ; , ^{235}U_{92}B = 1739.7 MeV ; , ^{231}Pa_{91}B = 1716.3 MeV |

and that the five terms which make up the binding energy in the mass formula have the values (in

**MeV**): volume,

**13.8**; surface,

**13.0**; coulomb,

**0.58**; asymmetry,

**20.0**; pairing,

**33.5**; show that the following decay sequence is possible

(^{235}U_{92}alpha decay) (^{231}Th_{90}beta decay) .^{231}Pa_{91} |

Calculate the energy of the alpha particle and the maximum energy of the decay electron.

The daughter nucleus of a given alpha emitter has several accessible excited states and so the kinetic energy of an emitted particle can have one of several possible values. For a particular heavy nucleus these values in

**MeV**are:

5.545 ; 5.513 ; 5.486 ; 5.469 ; 5.443 ; 5.417 ; 5.389. |

It is also observed that the daughter nuclei produced emit gamma rays with one of the following energies (in

**keV**)

26 ; 33 ; 43 ; 56 ; 60 ; 99 ; 103 ; 125. |

Use this information to sketch a decay scheme indicating the energy levels and marking the gamma ray transitions. You may assume that the most energetic alpha particle leaves the daughter nucleus in its ground state.

The binding energy of an alpha particle is

**28.3 MeV**. On the basis of the semi empirical mass formula, estimate from which mass number

**A**onwards alpha decay is energetically allowed for all nuclei.

If the neutrino has a mass

**m**rather than the zero value assumed in the lecture show that the phase space factor in the expression for the beta decay rate obtained there has to be multiplied by

**[1 - m**. Further show this means that the slope of the electron energy spectrum as

^{2}c^{4}/(T_{0}- T)^{2}]^{1/2}**T**approaches

**T**now becomes infinite rather than zero, as is the case for a massless neutrino.

_{0}A free neutron decays into a proton, electron and antineutrino. Assuming the latter to be massless and the original neutron to be at rest, calculate the maximum momentum that could be carried off by the electron and compare this with the maximum momentum which the antineutrino could have.

**If you're stuck look at the**

**solutions**

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In the case of difficulties with this course contact Prof Neil Spooner.If you find this material at all helpful please let me know!

© 1999 - FH Combley, 2005 - CN Booth, 2009 - NJC Spooner

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