IB DP Physics: HL复习笔记7.3.4 Particle Conservation Laws

Particle Conservation Laws


  • Electric charge must be conserved during particle interactions
    • This means charge must have the same value overall before and after the interaction
  • Charge is found in quarks, leptons and exchange particles and their anti-matter counterparts

Baryon Number

  • The baryon number, B, is the number of baryons in an interaction
  • B depends on whether the particle is a baryon, anti-baryon or neither
    • Baryons have a baryon number B = +1
    • Anti-baryons have a baryon number B = –1
    • Particles that are not baryons have a baryon number B = 0
  • Baryon number is a quantum number and is conserved in all interactions
    • This is one of the indicators for whether an interaction is able to occur or not

Lepton number

  • The lepton number, L, is the number of leptons in an interaction
  • L depends on whether the particle is a lepton, anti-lepton or neither
    • Leptons have a lepton number L = +1
    • Anti-leptons have a lepton number L = –1
    • Particles that are not leptons have a lepton number L = 0
  • Lepton number is a quantum number and is conserved in all interactions
    • This is one of the indicators for whether an interaction is able to occur or not

Conservation Laws

  • All particle interactions must obey a set of conservation laws. These are conservation of:
    • Charge, Q
    • Baryon number, B
    • Lepton Number, L
    • Strangeness, S
    • Energy (or mass-energy)
    • Momentum
  • However, strangeness does not need to be conserved in weak interactions. It can change by either 0, +1 or –1
  • Quantum numbers such as QB, L and S can only take discrete values (ie. 0, +1, –1, 1/2)
  • To know whether a particle interaction can occur, check whether each quantum number is equal on both sides of the equation
    • If even one of them, apart from strangeness in weak interactions, is not conserved then the interaction cannot occur


Example of a working out what is conserved in Kaon decay. This decay must be through the weak interaction since S is not conserved


  • Strangeness, S, like baryon and lepton number, is a quantum number
  • Strangeness is conserved in every interaction except the weak interaction
  • This means that strange particles are always produced in pairs (e.g. K+ and K–)
  • S depends on whether the particle contains a strange quark, anti-strange quark, or no strange quarks
    • Particles with an anti-strange quark have S = +1
    • Particle with a strange quark have S = –1
    • Particles with no strange quark have S = 0


Only particles with a strange or anti-strange quark have a strangeness of +1 or –1

  • Strangeness can change by 0, +1 or –1 in weak interactions

Strange Particles

  • Strange particles are particles that include a strange or anti-strange quark
  • An example of these are kaons
  • Strange particles always:
    • Are produced through the strong interaction
    • Decay through the weak interaction
  • An example of a kaon production could be:


Kaons are produced through the strong interaction. This is shown by the gluon exchange particle.

  • An example of kaon decay could be:


Kaons decay via the weak interaction. This is shown by the W+ boson.

Worked Example

The lambda nought particle Λ0 is has a quark composition uds.

Show, in terms of the conservation of charge, strangeness, baryon number and lepton number whether the following interaction is permitted:2.3.5-Conservation-Laws-Worked-Example

Step 1: Determine conservation of charge, Q

    • The Λ0 has a charge of 0
    • The pion π has a charge of –1
    • The positron e+ has a charge of +1
    • The electron neutrino νe has a charge of 0

0 = –1 + 1 + 0

    • Therefore, charge is conserved

Step 2: Determine conservation of strangeness, S

    • Λ0 has an s quark, so must have a strangeness of –1
    • None of the particles on the right hand side of the decay has a strange quark

–1 = 0 + 0 + 0

    • Therefore, strangeness is not conserved

Step 3: Determine conservation of baryon number, B

    • Λ0 is a baryon since it has 3 quarks, so must have a baryon number of +1
    • The pion π is a meson so has a baryon number 0
    • The positron is a lepton so has a baryon number 0
    • The electron neutrino νe is a lepton so has a baryon number 0

+1 = 0 + 0 + 0

    • Therefore, baryon number is not conserved

Step 4: Determine conservation of lepton number, L

    • Λ0 is a baryon, so must have a lepton number of 0
    • The pion π is a meson so has a lepton number of 0
    • The positron e+ is an anti-lepton so has a lepton number of –1
    • The electron neutrino νe is a lepton so has a lepton number of +1

0 = 0 + (–1) + 1

    • Therefore, lepton number is conserved

Step 5: Conclusion

    • Since the baryon number is not conserved, this interaction is not permitted

Worked Example

The equation for β decay is2.3.3-Beta-Minus-Equation

Using the quark model of beta decay, prove that the charge is conserved in this decay.


Worked Example

The sigma baryon has a quark structure of suu. It decays to produce a proton and pion as shown in the equation below2.2.6-Sigma-Baryon-Decay-EquationProve that this decay is via the weak interaction.

Step 1: Determine the strangeness, S of each particle

    • Since sigma baryon has one s quark, it has S = –1
    • The proton and pion has no strange particles, so they have S = 0

Step 2: Determine strangeness, S on both sides of the equation

    • The sigma baryon has a S = –1 but the meson and proton have a S = 0

–1 = 0 + 0

Step 3: Comment on the conservation of strangeness

    • Since S is not conserved on both sides of the decay equation (only changed by –1), this decay is via the weak interaction
    • This is because S is conserved in all other types of interaction (strong and EM), but isn't always conserved in weak interactions