Spin And Parity Assignments For Kids

In quantum mechanics, a parity transformation (also called parity inversion) is the flip in the sign of onespatialcoordinate. 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]


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


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 (H2+) is labelled and the next-closest (higher) energy level is labelled .[1]

Consequences of parity symmetry[edit]

When parity generates the Abelian group ℤ2, 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

Two dimensional representations of parity are given by a pair of quantum states which go into each other under parity. However, this representation can always be reduced to linear combinations of states, each of which is either even or odd under parity. One says that all irreducible representations of parity are one-dimensional.

Nuclear Physics PHY303



Charge of the electron e = 1.6 10-19 C
Mass of the electron me = 9.11 10-31 kg = 511 keV/c2
Mass of the proton mp = 1.673 10-27 kg = 938.272 MeV/c2
Mass of the neutron mn = 1.675 10-27 kg = 939.566 MeV/c2
Planck constant h = 6.626 10-34 J s = 4.136 10-15 eV s
Boltzmann constant k = 1.38 10-23 J K-1 = 8.617 10-5 eV K-1
Speed of light in free space c = 3.00 108 m s-1
Permittivity of free space 0 = 8.85 10-12 F m-1
Permeability of free space µ0 = 4 10-7 H m-1
Avogadro constant NA = 6.02 1026 kg-mol-1
Rydberg constant R = 1.10 107 m-1
Bohr magneton µB = 9.27 10-24 J T-1
Nuclear magneton µN = 5.0508 10-27 J T-1 = 3.1525 10-14 MeV T-1
Fine structure constant = 1/137

Useful data

Unified atomic mass unit 1u = 1.66 10-27 kg = 931.502 MeV/c2
Energy conversion 1 eV = 1.6 10-19 J
Year in seconds 1 yr = 3.16 107 s
Atmospheric pressure 1 atmosphere = 1.01 105 N m-2
Acceleration due to gravity on Earth's surface g = 9.81 m s-2
1 gram molecule at STP occupies 22.4 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 197Au, some 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 10-15 m. (Density of gold: 19300 kg m-3)

    A beam of low energy neutrons, intensity 106 s-1 traverses a foil of 235U, thickness 0.15 kg m-2. If the fission cross-section is 2 x 10-26 m2 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.

    The nucleus 27Si14 decays to its mirror nucleus 27Al13 by positron emission with a maximum energy of 3.48 MeV. Find the difference in the coulomb energy between the two nuclei and hence estimate the value of r0 in the expression for the nuclear radius R = r0 x A1/3.

    For the nuclide 16O the neutron and proton separation energies are 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 56Fe. Assume that the nucleus behaves as an impenetrable disc.
    Fraunhofer 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 235U + 231Th. The binding energy of the alpha particle is 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:


    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 10-15 m respectively,estimate for what fraction of the time the separation of the neutron and the proton is greater than this value of 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 m.

    For 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 µd(L= 0) = 0.8798 µN and µd(L= 2) = 0.3101 µN find the percentage admixture of the (L= 2) state if the deuteron magnetic momemt were given by µd = 0.8325 µN. Note that, as indicated in the lecture, a02 + a22 = 1.

    On 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

    1s1/2 ; 1p3/2 ; 1p1/2 ; 1d5/2 ; 1d3/2 ; 2s1/2 ; 1f7/2 ; 2p3/2 ; 1f5/2

    justify the following ground state spin and parity assignments

    3He2(1/2 +) ; 20Ne10(0 +) ; 27Al13(5/2 +) ; 41Sc21(7/2 -) ; 69Ga31(3/2 -).

    Estimate the separation of the 1p1/2 and 1d5/2 energy levels for nuclei with mass number A ~ 16 given the following information:

    the ordering of the lowest nuclear energy levels is1s1/2 ; 1p3/2 ; 1p1/2 ; 1d5/2 ; 1d3/2
    and the total binding energy for the oxygen isotopes is15O . . . . . . 111.96 MeV
    16O . . . . . . 127.62 MeV
    17O . . . . . . 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

    7Li3 ; 23Na11 ; 33S16 ; 41Sc21

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

    1s1/2 ; 1p3/2 ; 1p1/2 ; 1d5/2 ; 2s1/2 ; 1d3/2 ; 1f7/2 ; 2p3/2.

    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.
    Determine the spin and parity for these two types of excited state for each of the four given nuclides.

    The observed nuclear moments of 209Bi83 are:

    I = 9/2 ; µ = + 4.1 µN ; Q = - 0.4 x 10-28 m2

    Determine the expected values for these moments according to the simple Shell model and comment on any significant differences. Why would you expect 209Bi to have an unusually low cross section ( ~ 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 42Sc21 has a low lying level with spin and parity 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(3z2 - r2). Assuming that the charge of the nucleus 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 14O8 is a positron emitter, decaying to an excited state of 14N7. The gamma rays from this latter have an energy of 2.313 MeV and the maximum energy of the positrons is 1.835 MeV. The mass of 14N7 is 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/c2 find the mass of 14O8.

    Given the following binding energies:

    4He2, B = 28.3 MeV ; 235U92, B = 1739.7 MeV ; 231Pa91, 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

    235U92 (alpha decay) 231Th90 (beta decay) 231Pa91.

    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 - m2c4/(T0 - T)2]1/2. Further show this means that the slope of the electron energy spectrum as T approaches T0 now becomes infinite rather than zero, as is the case for a massless neutrino.

    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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