Chapter 4.2 by Manu Verma

Chapter 4 Symmetries And Conservation Laws

Chapter 4 4.1 Why Conservation And Symmetry 4.2 Angular Momentum 4.3 Flavour Symmetry Isospin 4.4 Parity and Charge Conjugation 4.5 Interactive Exercise

4.2 Angular Momentum

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Angular Momentum Angular momentum is a central concept There are two kinds of angular momenta. Orbital: Relative angular Momentum. This has a classical analog. Subject to quantum conditions, a particle can be in any orbital angular momentum state.

Fig 2: Spin

Fig 1: orbital

Spin: Intrinsic Angular Momentum. This is an intrinsic property i.e. attribute to a given species of particle. Spin appears to have a clear classical analog. However consider a proton of radius 1 fm and calculate velocity at equator for spin = 침/2 In relativity, cannot untangle intrinsic or orbital angular momentum This is exists

why

intrinsic

Fig 3: A sphere that flies by an observer in a straight line is seen from different directions (labeled 1 through 4) as time goes by. The surface coordinate grid shows that this straight line flyby is accompanied by an apparent rotation. Dayalbagh Educational Institute

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Fig 4 Quantization of orbital Angular Momentum

Classically = r x p

↑p

Takes continuous _ values, since r x p are quantized, we can expect same of ℒ _

|ℒ| can only take certain values. _ ℒ can only assume certain orientations w.r.t. a defined direction in space. Usually the direction is defined by a magnetic field.

Stern and Gerlach performed classic experiment on “Space Quantization.” In the “differential operator” representation of quantum mechanics, we represent momentum by a differential operator

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In this representation orbital angular momentum is also represented by a_ differential operator. For example, the z-component of ℒ

Just as a particle has a part of its wave function corresponding to its momentum, it has a part corresponding to angular momentum. For a particle in a well defined state of momentum

iħ∇ Momentum operator

Eigenstate of momentum

Eigenvalue of momentum

Eigenstate of momentum

Same thing goes for angular momentum

Angular momentum operator

Angular momentum eigenstate ℓ→ magnitude M→3-component

Value of total Angular momentum corresponding to ℓ

For z component, or orientation

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A measurement of the magnitude of the angular momentum yields a value ℓ(ℓ+1)ħ2 ℓ= 0, 1, 2 … … For a given value of ℓ a measurement of the z-component ℓZ always gives mℓħ mℓ = -ℓ, -ℓ+1, … -1, 0, +1... ℓ-1, ℓ Note that for a given magnitude of angular momentum there are (2ℓ+1) Values possible for the orientation. Quantization of orbital angular momentum leads to integral values of ℓ, so (2ℓ+1) is odd. An experimental surprise was to see that some atoms had lines which split into even no of states Magnetic field on No field (2ℓ+1) Degenerate states

odd no. of spectral lines Fig 5(a)

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No magnetic field

Magnetic field on 2 = even

Fig 5(b)

2 orientation; (2ℓ+1) = 2; ℓ= ½ ½ integral spin We interpret this by saying that there is some attribute of an electron that makes it able to take only two orientations with respect to a defend direction in space Behaves like intrinsic ½ integral angular momentum What do we mean by: Behaves like angular momentum? Quantum mechanical operators obey Commutation relations

When a commutator is zero, the two quantities can be simultaneously measured

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So for angular momentum

Means that we can only measure the magnitude of angular momentum and one of its components the set of commutation relations define a Group algebra related to a: symmetry Since both orbital and spin angular momentum operators obey the same algebra. They can be identified with the same physical attribute Angular momentum Spin angular momentum obeys the same eigenvalue equations as any other kind of angular momentum

Since a particle can have both spin and orbital angular momentum, we also have the concept of Total Angular Momentum. Also a system of two particles with spin, will have some value of total angular momentum Fig 6(a)

â&#x20AC;&#x153;classical mnemonicsâ&#x20AC;? Dayalbagh Educational Institute

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“Total angular momentum” Total angular momentum will also obey the eigenvalue equations

The total angular momentum can take on both integral and ½ integral values J = 0, ½, 1, 3/2, 2... For each value of J, M has (2J+1) values Note that any kind of particle can be put into any orbital angular momentum quantum state. However, the spin angular momentum is an intrinsic property of the particle

Angular Momentum In quantum mechanics, it is impossible to measure all three components of orbital angular momentum simultaneously: we can at best measure L2 along with one component LZ. Also these measurements can only return fixed values: L2 gives ℓ (ℓ+1) ħ2 ; ℓ = 0,1,2,3... Lz gives mℓ ħ ; mℓ = -ℓ, - ℓ+1, ..., 1, 0, 1, ...ℓ (2ℓ +1) possible values. Represented by 'kets': | ℓ, mℓ›. Dayalbagh Educational Institute

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Similarly, spin angular momentum measures give S2 gives S (S+1) ħ2, Sz gives msħ,

S= 0, , 1, ,...

ms = -S,..., S.

Represented by Example: A particle can have any value for orbital angular momentum ℓ, but a fixed value of S. For a pion, kaon : spin = 0 Electron, proton, neutron, quark S = ρ, Ψ, Gluon, photon, S = 1 Δ, Л, S =  Example: Write down the total angular momentum for

1, ℓ = 3 j = 4 (J2= 20 ħ2); j = 3 (J2= 12 ħ2); j = 2 (J2= 6 ħ2).

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Addition Of Angular Momenta For systems with several sources of angular momentum, one needs some way of adding the various contributions in order to get the Total Angular Momentum J = J1 + J2 Classically → just add all 3-components Quantum mechanics → This procedure would be meaningless, since the 3-components do not commute. One cannot add them if one does not know the values! One is obliged to work only with Magnitude and

commute

One component For states |J1〉 |J2〉 Consider states

| j1 m1〉 | j2 m2〉

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The magnitude of the angular momenta do not add. Classically one can think of a vector diagram J1 m1 = - ½

J1 m1 = + ½

J J2 m2 = + 1

J2 m2 = +1

m = m1 + m2 = (1+1/2 = 3/2) = (1-1/2 = 1/2) Fig 7(a), (b)

From this example one can see that adding j1 = 1 j2 = ½ can give us either a state of total angular momentum j= 3/2 or j = ½ If the two contributing angular momenta are parallel the magnitudes add. If they are anti parallel they subtract. In general there is a probability of getting intermediate values: Get every j from ( j1 + j2) down to | j1 - j2 | in integer steps j | j1 -j2|, | j1 -j2 | +1, …, ( j1+j2)-1, ( j1 -j2) So it is possible to enter what total angular momentum states are possible. But we do not know how to find the relative probabilities of different states turning up in any given case.

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In general, if we have two states They combine according to the rule

Contributing states

√ of probability of |j, m〉 occurring

Total State

Clebsch-Gordan Coefficients Take our vector example of _ J= ℒ +S This corresponds to Combining with From adding the z-values, see that resulting states could be:

What are the probabilities of various states? Total j J m m m1m2 Contributing

m1m2

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Clebsch Gordan Coeff

Table 1

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Table 2

Parallel → only one combined state

If orbital is anti-parallel to spin the probability of

|3/2, ½〉 is 1/3 |½ , ½〉 is 2/3 Note that this is a m = +½ state which can come from j=3/2 or j= ½ The remaining combinations are:

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Example 1: a quark and an anti quark are found together, in a state of zero orbital angular momentum, to form a meson, what are the possible values of the meson's spin?

Solution 1: Quark and anti quark carry spin ½, so total spin= 1 or 0. The spin 0 combination is called pseudo scalar meson (π, k, ŋ, ŋ2). The spin 1 combination is called vector scalar meson (ρ, k*,Φ, ω). Add three angular momenta, we combine two of them initially and then add on the third momentum.

Example 2: What happens in combining three quarks in a zero orbital angular momentum, to the spins?

Solution 2:

The two quarks combine to give the total

angular momentum of 1 or 0. Adding the third quark gives  or ½

and ½. Thus the

baryons have a spin , ½. The s=  is called the decuplet baryon and s=½ is called octet. Rule: To compute total angular momenta j from j 2, we define

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are called clebsch: Gordan coefficients. These coefficients tell us the probability of getting j (j + 1)ħ2, for any j value. The method is to measure J2 on a system consisting of two angular momentum states and , the probability is square of corresponding C-G coefficient. Convention: The spin combination is called the triplet, it is symmetric under interchange of particles 1 and 2, Whereas the spin -0 combination is called 'singlet', which is antisymmetric under interchange.

Example 3: An electron occupies the orbital state |2,-1› and the spin state | ›. We want to find values of J2 and probability of each value:

Solution 3: The j

values are

From C-G table (table 3.3.1), for j1= 2, j2= ½ and m1= -1, m2= ½ gives

Hence probability of state Dayalbagh Educational Institute

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Table: Clebsch -Gordan(C-G) coefficients

and probability of state Note: Total probability is 1.

Example 4:

The two spin states combine to give spin 1

and spin 0. Find the C-G decomposition for these states:

Solution 4:

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So the three spin states are

Where the spin 0 state is

The above equations can be read from C-G table directly, i.e.

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Spin ½: The proton, neutron, electron, all quarks, all leptons carry spin ½. A particle with spin ½ have mS=½ ('spin up') [↑] or mS= -½ ('spin down') [↓]. This has another representation by two-component column vectors or spinor:

A particle of spin ½ can only exist in one or other of these two states, but the most general state of a spin ½ particle is the linear combination are complex numbers. |α|2 is the probability that measurement of Sz would give the value of +½ħ, |β|2 is the probability of getting -½ ħ. Thus |α|2 +|β|2 =1. To measure components of S, we associate 2 x 2 matrix:

The eigenvalues of Ŝx are ± ħ/2 and corresponding normalized eigen vectors are

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An arbitrary spinner can be expressed as a linear combination of these eigen vectors:

Where

The probability that a measurement of SX will give the value ½ ħ is |a|2, value -½ħ is |b|2.

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Example 5:

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Suppose we measure (Sy)2 on a particle in the

state α . What is the value and probability of each? β

Solution 5: The matrix responding (S )2 is y

and every spinor is an eigen vector of Ŝy2 with eigenvalue . Similarly, for ŜX2,ŜZ2, as every spinor is an eigenstate of Ŝ2= ŜX2+ Ŝy2+ ŜZ2, with eigenvalue

Every spinor is an eigenvector of

with eigenvalue ħ2/4

Thus one is certain to measure ħ2/4 The same goes for So every spinor is an eigenstate of

½ Just S2= s(s+1) ħ2

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Fig 9

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Example 6: Compare the production rates for Solution 6:

π+ + p → Δ++ π- + p → Δo

Since pions have I = 1 and nucleons have I = 

we can calculate the rates from the 1 x  Clebsch-Gordan Coefficients Remember: these coefficients tell you how to combine quantized angular momentum (c.f. Electron orbital and spin AM ) In terms of Iz, π+ + p = (+1, +) This can only combine to give (,+) In Terms of Iz, π- + p = (-1, +) This can combine to give (,-) or (,-)

Table 1

Using the table, we can construct the general wave function for these (notice √ sign)

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Remember, the probability of obtaining eigenvalue is the coefficient squared, so

particular

Remember, in terms of (I, Iz) Δ++ ≡

(,+)

Δo

(,-)

≡

and as isospin is conserved in the strong interaction π+ + p → Δ++

100% of the time

π- + p → Δo

33% of the time

(because the decay in Δ0 is forbidden when π- + p is in the (,-) state)

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The Pauli spin matrices are defined: So that

Spinors occupy an intermediate position between scalars and vectors. Under a rotation of co-ordinate axes, the component of a vector change, similarly, the component of a spinor transform as:

Where U(θ) is the 2x2 matrix

U(θ) is a matrix of determinant 1: all such rotation matrices constitute the group SU(2). Hence,

spin

particles

transform

under

rotation

according to the 2-dimensional representation of SU(2). Similarly, particles of spin 1, described by vectors, belong to the 3-dimensional representation of SU(2). Spin  particles, are described by a 4-component object, transform under the 4-dimensional representation of SU(2). SU(2) is the same group as SO(3), the group of rotations in 3-dimensions. Dayalbagh Educational Institute

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different

spin,

belong

different

representation of the rotation group.

Spin and statistics Integral angular momentum has a fairly clear classical analog. ½ integral does not. In fact its existence is a consequence of relativistic quantum mechanics Particles with integral and ½ integral spins behave in profoundly different ways Consider two identical particles (1) + (2) they are at different space points and have same J, but different Jz One can imagine an operator which exchanges the position of the particles In nature all wave functions for identical particles are one of two types |1, 2〉 = +|2, 1〉 symmetric wave function integer spins Bose-Einstein statistics bosons. |1, 2〉 = -|2, 1〉 anti symmetric wave function ½ integer spins fermi-Dirac statistics fermions What does “statistics” refer to?

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Statistics

Assume two particles have exactly the same quantity numbers |1, 2〉 Now if we exchange the two particles, we have state |2, 1〉

Bosons

But, there is no observable difference |1, 2〉 = |2, 1〉 However, for fermions |1, 2〉 = -|2, 1〉

Fermions

These equalities only satisfied if |1, 2〉 = |2, 1〉= 0 Pauli Exclusion → no two fermions can be in identical quantum states

Classification Of Particles By Spin Boson (integer spin)

Fermions (half-integer spin)

Spin 0

Spin 1

Spin 

Spin

Mediators

Quarks/ leptons

Pseudo scalar mesons

Vector mesons

Baryon octet

Baryon decuplet

Table 3

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