Symmetries Flashcards

1
Q

symmetries

When are reference frames equivalent?

A

If (1) the set of all possible experimental outcomes, and (2) the dynamical laws
governing the temporal evolution of physical systems are the same for the two experimenters.

In other words: the experimenter will not be able to tell, neither by the mere results of the experiments nor by the physical laws that they can infer, in which of the two frames they have been placed.

Mathematically: the Hilbert space H and the associated space of rays H, corresponding to the possible states that they can assign to the system, must be the same for O and O′

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2
Q

symmetries

What’s an invertible mapping?

A

We can define a mapping M from the space of rays H = {R} to itself, which maps the ray R that observer O assigns to the state of the system to the eay R′ that observer O′ assigns to the same state of the system.

If the two observers are equivalent, and there is a map from O to O′, then there must also be an inverse mapping from O′ to O, for otherwise the two frames would be distinguishable and therefore not equivalent.

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3
Q

symmetries

What’s Wigner’s theorem? What does it imply?

A

Any invertible transformation M on the space of rays H, M : HH that conserves probabilities can be implemented as a transformation M on the space of vectors H, M : HH, that is either linear and unitary or antilinear and antiunitary. In other words, any such M can be obtained as the natural mapping between rays following from a unitary or antiunitary mapping M between vectors.

  • implication: it’s enough to look at unitary and antiunitary mappings of the Hilbert space of the system onto itself when relating two equivalent observers
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4
Q

symmetries

What is antilinearity and antiunitarity?

A

Antilinearity: an antilin. operator acting on a linear combination of elements from the Hilbert space produce complex conjugates of the coefficients

Antiunitarity: an antilinear operator that is norm-preserving

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5
Q

symmetries

What are the consequences of equivalent reference frames for the temporal evolution of the system?

A

We want the dynamical evolution of a physical system to be governed by the same laws for both observers, or stated differently we want that the equations of motion have the same form for both observer (i.e. they are invariant in form). This entails that the transformed of the evolved of the state vector of the system is equal to the evolved of the transformed of this vector: MU(t)ψ(0) = U(t)Mψ(0)

So if the two observers are equivalent, they must see the same Hamiltonian, so for M to be a symmetry it must commute/anticommute with the Hamiltonian for unitary/antiunitary operators.

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6
Q

symmetries

How can the group structrure of symmetries be described?

A

The symmetry transformations of a physical system form a group. This holds for both time-dependent and independent transformations.

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7
Q

symmetries

What are continuous symmetries? What’s the physical meaning of one-parameter families?

A

A symmetry is continuous if it consists
of a family of symmetry transformations parameterised by a set of continuous variables
(otherwise it’s discrete).

The representatives of the transformations on the Hilbert space of the system,
M = M(α), depend on continuous parameters and if they are connected to the identity, as in M(α0) = 1 for some α0, M must be unitary.

For one-parameter unitary operators, the operator Q (derivative of M) is a self-adjoint operator commuting with H, and so a conserved physical quantity that can be diagonalised simultaneously with the Hamiltonian. Examples: energy and momentum, associated with the symmetry under temporal and spatial translations; angular momentum, associated with the symmetry under rotations

The above is true for multiple parameters as well.

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8
Q

Discrete symmetries

What does parity do? Why is it a unitary operator?

A

It consists in the change of the sign of all the spatial coordinates of our reference frame, while leaving time unchanged. Generally this means that all the components of the momentum of a particle will change sign, while angular momenta (and spin in particular) will remain unchanged: P|p, s(z) ; α〉 = η(α)|− p, s(z) ; α〉.

  • η(α): intrinsic parity (necessary phase factor for generality)
  • P is a unitary op.: if it were to be antiunitary ({P, H} = 0), applying HP to ψ(E), the result would be negative energy
  • so [P, H] = 0, so they can be diagonalized simultaniously
  • using the energy and momentum eigenstates: P|E, l, l(z); α〉 = (−1)^l|E, l, l(z); α〉
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9
Q

Discrete symmetries

What are the properties of intrinsic parity? How to determine it?

A

In general, we can choose arbitrarily one intrinsic parity for each conserved charge, and since strong and electromagnetic interactions conserve each individual particle type, we can fix the intrinsic parities of all quarks and all negatively charged leptons to 1.

  • does not change the physical content of parity transformation
  • for truly neutral particles, it can’t be determined through a phase transformation
  • η(α)η(¯α) = +/–1 for bosons/fermions
  • for antiquarks and positively charged leptons it’s −1
  • η(α)^2 = 1 for self-conjugate bosons
  • determining η(α): through S matrix for scattering and decay processes (conservation of angular momentumn limits the possible combinations of S and l) or through experimental/theoretical considerations
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10
Q

Discrete symmetries

What is the charge conjugation operator? What are its properties?

A

It consists in exchanging particles with the corresponding antiparticles, keeping momenta and spins unchanged: Cp, s(z); α〉 = ξ(α)|p, s(z); ¯α〉.

  • ξ(α) is a phase included for generality
  • unitary operator (same reasoning as for parity): [C, H] = 0
  • changing to antiparticles leads to the changing sign of all the charges, like electric charge (Q), baryon (B), lepton (L) and lepton family (L(l)) numbers
  • since magnetic momentum depends on the charge, it’s also changed
  • for the charges {C, O} = 0
  • ξ(α)ξ(¯α) = 1, both for bosons and fermion, so C^2 = 1 (it’s just a phase transformation)
  • the value of the phase is only relevant for self-conjugate neutral particles
  • violations of charge conjugation symmetry come only from the weak interactions
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11
Q

Discrete symmetries

What is the time reversal operator? What does it do?

A

It reverses the arrow of time, under such a transformation, the signs of both the momentum components and the spin components change: T|p, s(z); α〉 = ζ(α,s(z))|−p, −s(z); α〉.

  • the phase factor depends on s(z) and the particle type: ζ(α,s(z)) = (−1)^(s−s(z)) ζ(α)
  • antiunitary operator: {T, iH} = 0, [T, H] = 0
  • as a consequence of antiunitarity, the residual phase ζ(α) has no physical meaning, since it can be reabsorbed in a redefinition of the particle states
  • physical consequences: T(†)Ω±T = Ω∓, T(†)ST = S(†), S(Tf,Ti) = S(if), M(Tf,Ti) = M(if)
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12
Q

Discrete symmetries

What is the CPT theorem and what are its implications?

A

It is a general theorem of quantum field theory, the CPT theorem, that for any Lorentz- and translation-invariant theory of local quantum fields, the antiunitary transformation Θ = CPT is a symmetry: Θ|p, s(z); α〉 = ξ(α)η(α)ζ(α,s(z))|p, −s(z) ;¯α〉 = θ(α,sz)|p, −s(z); ¯α〉.

Implications:

  • particles and antiparticles must have the same mass
  • unstable particles: the lifetime is the same as that of the corresponding antiparticle
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13
Q

Continuous symmetries: rotations

How to describe a particle at rest with rotations?

A

It is an experimental fact that a massive particle in its rest frame can be in more than one physical state, i.e., the “particle is at rest” state characterised by vanishing spatial momentum p = 0 is not unique: any rotation of the coordinate system still yields a frame where the particle is at rest.

  • the resting states of a particle form a Hilbert space from which two observers can assign vectors that are related by rotations
  • the two observers are related by a unitary operator
  • to the operator there are unitary matrices assigned: U(R)|n〉 = ∑(n′) D(n′n)(R)|n′〉 —» D(n′n)(R) = 〈n′|U(R)|n〉
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14
Q

Continuous symmetries: rotations

What is the representation of SO(3)? What’s the goal with it?

A

A mapping from a group G to the space of invertible n × n complex matrices GL(C, n), associating a matrix D(g) to each element of G: D: G → GL(C, n), g → D(g).

  • this obeys: D(g1g2) = D(g1)D(g2)
  • acceptable ways to implement the effect of rotations on the state vectors of a physical system are then restricted to the unitary representations of the rotation group that must hold up to a phase: projective representations (D(g1g2) = e^(iφ(g1,g2))D(g1)D(g2))
  • unitary reps are always completely reducible

Goal: finding unitary representations of SO(3), so finding all the irreducible projective representations of SO(3)

  • D(g) can be brought to a block diagonal form (because of the complete reducibility) where each block provides an irreducible rep of the group
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15
Q

Continuous symmetries: rotations

What are the properties of the rep. of SO(3) as a Lie group?

A

SO(3) is a Lie group, so it’s a group and also a smooth manifold: their elements g = g(α) can be parameterized in terms of a set of continuous real variables α = (α1, α2, . . . , α(d)), and form a continuous space that is locally identical to the Euclidean space R^d.

  • for (simply) connected Lie groups: the component of a Lie group connected to the identity element can essentially be entirely reconstructed starting from the elements infinitesimally close to the identity (g(α) = 1 + iα(a)L(a) + . . . , L(a) ≡ −i(∂g/∂α(a)∣α=0)
  • matrices L(a) are the (infinitesimal) generators of the groups: [La, Lb] = iC(ab)^c L(c)
  • tangent space: the real linear space spanned by {L(a)} at the identity element
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16
Q

Continuous symmetries: rotations

What are the properties of the Lie algebra of SO(3)?

A

This linear space endowed with the matrix commutator defines the Lie algebra associated with the Lie group.

  • for each group there is a unique Lie algebra
  • reps of a Lie group can be more easily reconstructed from its Lie algebra
  • rep. of a Lie algebra: a linear mapping d(X) = X^a d(L(a)) of its elements X = X^a L(a), X^a ∈ R into a space of matrices that respects commutation relations (Lie homomorphism)
  • a rep. of its algebra can always be obtained by differentiation: d(L(a)) = −i(∂/∂α(a)) D(g(α))|α=0
  • a unitary irrep of the group yields an irrep of the algebra in terms of Hermitean matrices
  • if all the D(g), and so the U(g), leave a subspace invariant, then so will the H(L(a)) = −i(∂U(g)/∂α(a))|α=0
17
Q

Continuous symmetries: rotations

What’s the connection between SO(3) and SU(2)? What’s their Lie algebra?

A

They share the same Lie algebra (SO(3) is connected SU(2) is simply connected). The problem of classifying the physically relevant representations of SO(3) then boils down to classifying the irreducible unitary representations of SU(2).

  • and so the irreducible Hermitean representations of the corresponding algebra
  • all the projective representations of SO(3) are induced by a representation of SU(2)

The Lie algebra:

  • for SO(3): L(a) = −i(∂O/∂α(a))|α=0 , [L(a), L(b)] = iε(abc)L(c)
  • for SU(2): s(a) ≡ σ(a)/2 = −i(∂U/∂α(a))|α=0, [s(a), s(b)] = 1/4 [σ(a), σ(b)] = iε(abc) s(c)

Independently of the choice of basis, both antisymmetric purely imaginary 3×3 matrices and Hermitean traceless 2×2 complex matrices form real linear spaces of dimension 3, both closed under matrix commutators, and so they correspond to the same abstract Lie algebra.

18
Q

Continuous symmetries: rotations

How to obtain the irreps of SU(2)?

A

We look for a set of Hermitean matrices d(s(a)) that obey [d(s(a)), d(s(b))] = iε(abc)d(s(c)).

  • the entries of these matrices: S(a) = H(s(a)) (Hermitean operators acting on a linear space having the entries of d(s(a)) as their matrix elements between basis vectors)
  • introduction of ladder operators: s± = s1 ± i s2 ([s3, s±] = ±s±, [s+, s−] = 2 s3), same for S
  • Casimir operator: S^2 = S1^2 + S2^2 + S3^2
  • diagonalizing S3: there must be an upper and lower limit (otherwise: infinite sequence of eigenvectors)
  • a definite choice of phase: Condon-Shortley convention (S− has positive-definite matrix elements between basis vectors)
19
Q

Continuous symmetries: rotations

What is the result for the irreps of SU(2)?

A

The irreducible representations of SU(2) are labelled by an integer or half-integer number s, that is the spin of the representation.

  • dimension of the rep.: 2s+1
  • the representatives S±,3 of s±,3 act on the basis elements |s, m〉
  • total spin operator s^2: the rep. S^2 is a constant multiple of the identity
  • the value of s, or equivalently the eigenvalue of S^2, entirely characterises the rep
  • the representation with integer s as ordinary reps of SO(3), while half-integer s provides a projective rep
  • similar relations are obeyed by orbital angular momentum
20
Q

Continuous symmetries: rotations

What are the simplest irreps of SU(2)?

A

Trivial representation: 1D, s = 0

  • group rep: D0(U ) = 1, for every U ∈ SU(2)
  • generated by: d0(s) = 0
  • singlet state

Defining/fundamental representation: 2D, s = 1/2

  • group rep: D(1/2)(U) = U
  • generated by: d(1/2)(s) = s = σ/2
  • doublet state (formed by the eigenvectors of s3)

Complex conjugate representation: 2D, s = 1/2

  • group rep: D(¯1/2)(U) = U∗
  • generated by d(¯1/2)(s) = −σ∗/2
  • for SU(2) this is equivalent to the defining rep
  • opposite eigenvalues but same eigenvectors

Adjoint representation: the rep of a Lie group on its own Lie algebra, 3D, s = 1

  • linear mapping from the Lie algebra to itself
  • group rep: D1(U) = Ad(U), Ad(U)X ≡ UXU† (X = X(a)s(a) , X(a) ∈ R), linear superposition of the generators
  • generated by: d1(s) = ad(s), ad(A)X ≡ [A, X] (A is an op. of the algebra)
  • d1(s(a))(cb) = ad(s(a))(cb) = −iε(acb) ≡ (T^(A)(a))(cb)
  • [ad(s(a)), ad(s(b))] = iε(acb) ad(s(c)), same for the matrix components
  • the T matrices are the same as the generators of SO(3)
  • the eigenvectors are the algebra elements s+, s3, s− with eigenvalues 1, 0, −1, respectively
  • triplet state
21
Q

Continuous symmetries: rotations

How to compose representations?

A

The full effect of a rotation can then be seen as rotating the spatial part of the state from |p〉 —» |Rp〉, combined with that of rotating the original spin state from |s, s(z)〉 —» ∑(s(z)′) |s, s(z)′〉 D^(s)(s(z)’ s(z))(R). Altogether: U(R(α)) = e^(iαJ), J = L + S.

  • the goal is: obtaining the eigenvectors of J^2 and J3 from the eigenvectors of L^2, L3, S^2 and S3 (the J ones are conserved)
  • eigenvectors of J^2 with different eigenvalues do not mix with each other under rotations, and will form the bases of separate irreducible representations

Addition of angular momenta: composition of the individual s = 1/2 reps

  • the direct product of two s = 1/2 reps equals the direct sum of the s = 1 and the s = 0 reps, i.e., 1/2 ⊗ 1/2 = 0 ⊕ 1 (2 ⊗ 2 = 1 ⊕ 3 with the dimensions)

For the composition of representations s1 and s2 the general result: s1 ⊗ s2 = |s1 − s2| ⊕ (|s1 − s2| + 1) ⊕ (s1 + s2 − 1) ⊕ (s1 + s2), where each irreducible representation in the sum appears exactly once.