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Lecture 1 – Symmetries & Conservation

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Lecture 1 – Symmetries & Conservation LECTURE 1 – SYMMETRIES & CONSERVATION Contents • Symmetries & Transformations • Transformations in Quantum Mechanics • Generators • Symmetry in Quantum Mechanics • Conservations Laws in Classical Mechanics • Parity Messages • Symmetries give rise to conserved quantities . Symmetries & Conservation Laws Lecture 1, page1 Symmetry & Transformations Systems contain Symmetry if they are unchanged by a Transformation . This symmetry is often due to an absence of an absolute reference and corresponds to the concept of indistinguishability . It will turn out that symmetries are often associated with conserved quantities . Transformations may be: Active: Active • Move object • More physical Passive: • Change “description” Eg. Change Coordinate Frame • More mathematical Passive Symmetries & Conservation Laws Lecture 1, page2 We will consider two classes of Transformation: Space-time : • Translations in (x,t) } Poincaré Transformations • Rotations and Lorentz Boosts } • Parity in (x,t) (Reflections) Internal : associated with quantum numbers Translations: x → 'x = x − ∆ x t → 't = t − ∆ t Rotations (e.g. about z-axis): x → 'x = x cos θz + y sin θz & y → 'y = −x sin θz + y cos θz Lorentz (e.g. along x-axis): x → x' = γ(x − βt) & t → t' = γ(t − βx) Parity: x → x' = −x t → t' = −t For physical laws to be useful, they should exhibit a certain generality, especially under symmetry transformations. In particular, we should expect invariance of the laws to change of the status of the observer – all observers should have the same laws, even if the evaluation of measurables is different. Put differently, the laws of physics applied by different observers should lead to the same observations. It is this principle which led to the formulation of Special Relativity. Symmetries & Conservation Laws Lecture 1, page3 Illustration from Non-Relativistic Mechanics Observer 1 For Observer 1: m Mass m at position x, attached to Spring. Spring fixed to Wall at x = W. Natural length of Spring is L, and spring constant k. Boundary conditions: x = W+L+A & x& = 0 at t = 0. x x=W Observers 1 & 2 are in inertial frames. Observer 3 is in an accelerating frame wrt 1 & 2. Observer 2 Observer 3 Symmetries & Conservation Laws Lecture 1, page4 Observer 1 measures things in a coordinate frame {x}, in which the Wall is stationary. Newton: F = m&x& − k(x − (W + L)) = m&x& Implicitly replace x − (W + L) by ξ,the extension of the Spring, and solve in terms of sin( ωt) and cos( ωt). Solution: x = W + L + A cos( ωt), where ω2 = k / m Observer 2 measures in a coordinate frame {x’}, in which Observer 1 is moving with a uniform velocity: x' = x +∆ + vt . As far as Observer 2 is concerned, − k(x'−(W'+L)) = m&x&' The expression of for the Force (and mass and spring constant) is unchanged. Again, implicitly replace x'−(W'+L) by ξ – this works because the derivative of W' = W +∆ + vt vanishes, allowing us to replace &x&' with ζ&& . Solution: x' = W'+L + A cos( ωt), where ω2 = k / m The frames of the two observers are equally good, there is no way of absolutely distinguishing the frames – this indistinguishability corresponds to a symmetry (or equality) between the two frames – associated with the transformation between the two. The physical predictions of the two observers are identical: the amplitude and frequency are the same, although the description of the position of the mass is different. Symmetries & Conservation Laws Lecture 1, page5 1 2 However, Observer 3 is accelerating: ''x = x + 2 at , the equation of motion assumed by Observer 3, is − k(x''−(W''+L)) = m&x&'' 1 2 This time, one cannot implicitly replace x''−(W''+L) by ξ – because the derivative of W '' = W +2 at does not vanish. The solution is a little more complicated: Solution: x'' = W''+L − ma / k + (A + ma / k)cos( ωt), where ω2 = k / m This is a different solution – the amplitude is different. The reason is that the force was incorrectly attributed as − k(x''−(W''+L)) – as far as Observer 3 is now concerned, there is an additional force ma required to accelerate the Mass (and the Wall) – like a gravitational force. The best way to solve for the motion of the Mass, is to solve in frame {x} and then transform. Why the lack of symmetry between the Observers ? We know that Observer 3 (not 1 & 2) is accelerating because he is forced into the back of his car seat. Newton’s Law is only applicable for observers in inertial frames. One must be careful to understand the validity of given equation under a transformation. Symmetries & Conservation Laws Lecture 1, page6 Transformations in Quantum Mechanics Consider a scalar wavefunction: ψ(x) Make a transformation from one coordinate system to another: x → x' Define the transformed wavefunction in the new frame by: ψ'(x' ) ≡ ψ(x) The intention is that x and x' correspond to the same point in space-time and the wavefunctions ψ' and ψ describe the same event . ψ ψ′ ψ′ ψ x x x x=0 x=1 x=2 x=3 x′ x′ x′=10 x′=110 x′=210 x′=310 At the point where the event happens, the wavefunction has some (relatively) well-defined value. Symmetries & Conservation Laws Lecture 1, page7 Note ψ' will be a different function from ψ: If x' = f(x), and the inverse transformation is f −1, then x = f −1 )'x( . Hence ψ )'x(' ≡ ψ )x( = ψ f( −1 ))'x( so ψ ()' = ψ f( −1()) For example, if ψ(x) = x and x' = exp( x), then ψ'(x' ) ≡ ψ(x) = x = log( x' ). In general, assume that the new wavefunction can be derived from the old one by a transformation of the wavefunction itself: ψ' = Uψ where U is an operator In general, ψ can be expressed as a linear superposition of base states {φi}: ψ = ∑ c iφi In the new description, ψ → ψ', φ → φ', so ψ' = ∑ 'c i φ'i . But since ψ and ψ' correspond to the same states, as do φ and φ', then we would expect 'c i = c i ⇒ ⇒ ψ' = ∑ c iφ'i (U ∑ c iφi ) = ∑ c i (U φi ) This is the definition of a linear operator . Note there are two distinct transformations: a. the transformation describing the change in “description” (coordinate frame): x → x' ≡ f(x) b. the associated transformation of the wavefunction: ψ → ψ' = Uψ Symmetries & Conservation Laws Lecture 1, page8 Furthermore, the overlap between any states ψa and ψb is an observable and should be independent of the description. Using the bra-ket notation for compactness: ψa ↔ | a > , then H H H H ψ'a = Uψa ↔ | 'a >= U | a > and ψ'a = ψa U ↔ < |'a =< a | U where H denotes the Hermitian conjugate (often shown by a dagger). H H The overlap is < |'b 'a >=< b | U U | a > and if this is equal to < b | a > for all ψa and ψb, then U U = I. The transformation of the wavefunctions is Unitary . (But see later.) How do operators transform ? H Consider the observable < b | A | a >≡ ∫ ψ b Aψ a . Want < b'| A'| a' >=< b | A | a > for all ψa and ψb. ⇒< b | UH 'A U | a >=< b | A | a >⇒ UH 'A U = A ⇒ 'A = UAU H If A is invariant, then 'A = A ⇒ UAU H = A ⇒ UA = AU ⇒ [ ,A ]U = 0 Symmetries & Conservation Laws Lecture 1, page9 Lastly, if states { | i > } form an orthonormal basis: • < i|j > = δij • All states | a > in the vector space can be written as a linear superposition | a > = ∑ αi i| > Then the transformed states | i' > = U | i > also form an orthonormal basis: H < 'i|'j > = < |j U U i| > = < i|I|j > = δij and since | i' > is derived from | i > etc, then if i=j, the states | i' > and | j' > must be equal. 'i|>= U i| > = ∑ < |j U i| > j| > Since < j | U | i > is just (the set of coefficients of) a unitary matrix, the matrix can be inverted: i| > = ∑ < |j UH i| > 'j| > Hence any state | a > can be written as: H | a > = ∑ αi i| > = ∑ αi ∑ < |j U i| > 'j| > = ∑βi 'j| > which is the requirement for a basis. Symmetries & Conservation Laws Lecture 1, page10 Generators Under a transformation, ψ → ψ' = Uψ . Assume that the unitary operator can be expressed as: U = exp( iaX ), where a ∈ℜ. What is X ? Naively it is the “log” of U, but this is non-trivial, since we are dealing with operators and functions need to be defined by their power series. X is defined as the Generator of the transformation. Symmetries & Conservation Laws Lecture 1, page11 Exponentiation of Operators Define: 1 2 1 3 ∞ 1 p 0 exp( A) ≡ 1+ A + !2 A + !3 A + ... = ∑p=0 !p A where A = I Then H ∞ 1 p H ∞ 1 H p H exp( A) = (∑p=0 !p A ) = ∑p=0 !p (A ) = exp( A ) If U = exp( iaX ), then it is easy to show U−1 = exp( −iaX ) – it follows from the normal rules for multiplying exponentials of scalars, which in turn can be proved by expanding the exponentials. The result follows easily because all the terms commute. If the U is unitary, then U−1 = UH , hence exp( −iaX ) = (exp( iaX )) H = exp(( iaX )H ) = exp( −iaX H ) Equating terms order by order in a implies XH = X , i.e. X is Hermitian . These results can be derived by • Looking at the series expansions of exponentials and considering Binomial expressions • Considering infinitisimal terms: exp( ε) ≈ 1+ ε for small ε Lim A N • Building finite transformations from a product of infinitisimal ones: exp( A) = N→∞ 1( + N ) So the generator of a Unitary transformation is a Hermitian operator . In QM, Hermitian operators are postulated to correspond to observables.
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