Quantum Mechanics II Classical → Quantum Physics Quantum Mechanics over Phase-Space; Feynman-Hibbs Path-Integrals; Quantization and Anomalies Tristan Hübsch Department of Physics and Astronomy, Howard University, Washington DC http://physics1.howard.edu/~thubsch/
Q Classical → Quantum Physics M II Quantum Mechanics over Phase-Space Recall: Heisenberg’s indeterminacy relations: ∆ x i ∆ p i > ½ ħ , i = x, y, z… Classical mechanics is defined over phase-space ( x i , p j ) Heisenberg relations make phase-space “granular” …albeit not as a chess-board with a fixed tiling E.g. the linear harmonic oscillator ³ / 2 ħω ⁵ / p 2 ħω ⁷ / 2 ħω ⁹ / 2 ħω … etc. … ½ ħω q Phase-space “area” enclosed by an orbit But, ordinary space is an integral multiple of ħ . remains just as continuous . 2
Q Classical → Quantum Physics M II Quantum Mechanics over Phase-Space Classical mechanics is defined over phase-space ( q i , p j ) All observables are real functions, F ( q i , p j ), over phase-space Dynamics is governed by the equations of motion d F d t = ∂ F � H , F ∂ t + � PB …which “converts” to the Heisenberg equations of motion d b d t = ∂ b ⇥ b ⇤ F F H , b ∂ t + 1 F i ¯ h But what of the state operator? There is no analogous dynamical object in classical physics. Also, quantum mechanics results in (amplitudes of) probabilities, not actual orbits, trajectories, positions… 3
Q Classical → Quantum Physics M II Quantum Mechanics over Phase-Space To reproduce quantum mechanics from classical mechanics on Z b b the phase space, we must introduce probability distributions. b Such a distribution, ! Q ( q , p ), must satisfy Z Z d q ρ Q ( q , p ) = h p | b ρ | p i d p ρ Q ( q , p ) = h q | b ρ | q i b Z …as well as Z Z ρ ∗ ( q , p ) = ρ ( q , p ) d p ρ Q ( q , p ) = 1 ρ ( q , p ) > 0 d q Turns out: For any desired quantum state operator, there are infinitely many functions # Q ( q , p ) that satisfy the above equations But, there is no uniformly specified choice for all state operators 1 − 1 There is no universal “quantization” assignment b → ρ Q ( q , p ) ← ρ 4
Two Proof-of-Concept Distributions Kodi Husimi Eugene Wigner
Q Classical → Quantum Physics M II The Wigner Representation b Consider the function over phase space [Wigner, 1932] Z ∞ �b� � � h h q � 1 �b � q + 1 1 d y e ipy /¯ ρ W ( q , p ) : = 2 y i 2 y ρ Reverse-engineered! 2 π ¯ h Constructs a classical distribution from a given � ∞ quantum state operator. Z ∞ � � �b� h h p � 1 �b � p + 1 1 d k e iqk /¯ 2 k i 2 k = ρ 2 π ¯ h � ∞ Then: Z Z Z 1 ∞ 0 0 � � � � �b � q + 1 �b � q i d p e ipy /¯ h 1 d p ρ W ( q , p ) = h q � 1 2 y i = h q d y 2 y ρ ρ 2 π ¯ h � ∞ | {z } � � | {z } h δ ( y ) 2 π ¯ � i �b h Easy: Z ZZ Z b b d q d p ρ W ( q , p ) = d q h q | b ρ | q i q i = Tr [ b ρ ] = 1 ρ † = b ρ ∗ b W ( q , p ) = ρ W ( q , p ) ⇔ ρ 6
Q Classical → Quantum Physics M II The Wigner Representation b b ⇔ However, for two states to be orthogonal ZZ ρ 2 ] 6 1 Tr [ b 0 ! b ] = d q d p ρ 1 W ( q , p ) ρ 2 W ( q , p ) = Tr [ b ρ 1 b ρ 2 ] = Assuming that 0 ≤ ! W ( q , p ) ≤ 1 (appropriate for probabilities) it must be that # 1 and # 2 have no overlap. Then, N orthogonal states ⇒ each ≠ 0 only over 1/ N of phase-space When N → ∞ (as typical), the # ’s would have to be $ -function-like ZZ But, phase-space is granular: distributions cannot be b b pinpointed better than size-2 "ħ “areas” in the phase-space ⇢ 0 � ⇢ � � ( q , p ) 62 A ZZ ZZ ρ i W ( q , p ) = d q d p ρ i W ( q , p ) = 1 A � 1 2 ( q , p ) 2 A ZZ ZZ d q d p ρ 2 1 W = 2 π ¯ h h O i : = d q d p ρ 1 W O ( q , p ) ) → 2 π ¯ h A > 2 π ¯ h A 6 1 So, the only way out is to permit ! W ( q , p ) to be negative …and so fail as a probability distribution. 7
Q Classical → Quantum Physics M II ZZ ZZ The Wigner Representation Nevertheless… 2 π a e � q 2 /4 a 2 1 p p ψ ( q ) = h q | ψ i = ρ = | ψ i h ψ | b q ) e � p 2 / ( 2 4 2 h e � q 2 / ( 2 4 2 p ) 1 ρ W ( q , p ) = 4 q = a 4 p = ¯ h /2 a π ¯ (Only true for the Gaussian!) …turns out non-negative. Time dependence: ⇥ b b⇤ = ⇤ ⇥ b ⇤ + i ⇥ b ⇤ d b ρ d t = i i P 2 , b H , b W , b ρ ρ ρ ¯ h 2 M ¯ h h ¯ ⇥ b b⇤ ⇥b b ⇤ ⇥b b ⇤ b Calculate the 1st term in momentum rep., 2nd in coordinate rep., …then transform into the Wigner representation h ) n − 1 d n W ∂ n ρ W ∂ρ W ∂ t = p ∂ρ W 1 ∂ q + ∑ q ! ( − 1 2 i ¯ d q n ∂ p n M n = odd For the LHO, all ( n ≥ 2)-order derivatives of W ( q ) vanish (Only true for the LHO!) …the equation = classical Liouville equation. 8
Q Classical → Quantum Physics M II The Husimi Representation Main problem: no simultaneous eigenstates | q , p � K. Husimi: use the next-best-thing, such that � ( x � q ) � 2 2 π σ e � + ipx /¯ h 1 p h x | q , p i = 2 σ p Gaussian “lump,” centered at ( q, p ) in the phase-space with the complementary half-widths ∆ q = " and ∆ p = ħ /(2 " ) These functions are not orthogonal (all overlaps ≥ 0) � � but… …and are over-complete: ∫ dq dp| q , p � � q , p | = 2 $ħ Define then � � �b � q , p i 1 ρ H ( q , p ) : = h h q , p ρ 2 π ¯ This equals to a Gaussian smoothing of ! W ( q , p ) 2 Is a true probability density function w/fuzziness controlled by % , and complementary for q and p 9
Q Classical → Quantum Physics M II Feynman-Hibbs Path-Integrals (Section 4.8) ⇣� ⌘ ⇤ b b ⇥ A “triviality”: � � � b � � � h i b h | i h � b � Ψ ( t 0 ) i i = b U ( t , t 0 ) h x | Ψ ( t 0 ) i ) i : = h x U ( t , t 0 ) Ψ ( x , t ) : = h x | Ψ ( t ) i Z � � � � � x 0 ih x 0 � � x 0 d x 0 h x � b � Ψ ( t 0 ) i � b � Ψ ( t 0 ) i = U ( t , t 0 ) = h x U ( t , t 0 ) 1 � Z Z � � � Z Z Z Z Z Z Z Z Z Z x 0 d x 0 G ( x , t ; x 0 , t 0 ) Ψ ( x 0 , t 0 ) � b � � ) i = Z Z Z Z Z …except, Z Z Z Z D[ x 0 ( t ) ] G ( x , t ; x 0 , t 0 ) Ψ ( x 0 , t 0 ) = a sum over all possible (unrestricted) paths x ( t ) that take x 0 ( t 0 ) → x ( t ). …is a NEW kind of integral. x t t 0 t 10
Q Classical → Quantum Physics M II Feynman-Hibbs Path-Integrals (Section 4.8) The “propagator” G ( x , t ; x 0 , t 0 ) concatenates: G ( x , t ; x 0 , t 0 ) = G ( x , t ; x 1 , t 1 )· G ( x 1 , t 1 ; x 0 , t 0 ) Z Z Z Z Z Z Z Z Z = G ( x , t ; x 2 , t 2 )· G ( x 2 , t 2 ; x 1 , t 1 )· G ( x 1 , t 1 ; x 0 , t 0 ) = … etc. Subdividing until each time-interval [ t i , t i+ 1 ] is infinitesimal, � � � h � � � ⇡ h P 2 i 4 � e � i 4 j t [ b 2 M + W ] /¯ � x j i G ( x j + 1 , t j + 1 ; x j , t j ) ⇡ h x j + 1 4 j t : = ( t j + 1 � t j ) � � b � � h ) � � ⇡ h i � e � i 4 j t b P 2 / ( 2 M ¯ � � � x j i e � i 4 j tW ( x j ) /¯ h Z ⇡ h x j + 1 � � Z b � � � h ) � Z � e � i 4 j t b P 2 / ( 2 M ¯ � p ih p | x j i e � i 4 j tW ( x j ) /¯ � � h = d p h x j + 1 Z Z Z d p h x j + 1 | p ih p | x j i e � i 4 j tp 2 / ( 2 M ¯ = e � i 4 j tW ( x j ) /¯ h ) h Z Z Z d p e ip ( x j + 1 � x j ) e � i 4 j tp 2 / ( 2 M ¯ = e � i 4 j tW ( x j ) /¯ h ) h 1 2 π ¯ h i M ( x j + 1 � x j ) 2 ⇢ � h q = e � i 4 j tW ( x j ) /¯ M h 4 j t exp 2 i π ¯ 2¯ h 4 j t 11
Q Classical → Quantum Physics M Z II i M ( x j + 1 � x j ) 2 ⇢ � h q h i = e � i 4 j tW ( x j ) /¯ M Feynman-Hibbs Path-Integrals h 4 j t exp ⇥ ⇤ 2 i π ¯ h 4 j t 2¯ So, N Z Z ∏ G ( x , t ; x 0 , t 0 ) = lim G ( x j + 1 , t j + 1 ; x j , t j ) d x 1 · · · d x N · · · N → ∞ j = 0 � M ⇥ xj + 1 � xj ⇤ 2 N N + 1 h ∑ N i 2 e Z � W ( x j ) Z Z Z Z j = 0 4 j t Z Z Z Z ⇣ ⌘ 2 M 4 jt ∏ ¯ = lim d x k q 2 i π ¯ h 4 k t N ! ∞ k = 0 For infinitesimal subdivisions ⌘ ⇣ h x j + 1 � x j i 2 i 2 M ( x j + 1 � x j ) 2 ⇤ 2 = i → i ⇥ d x i h 4 j t ! h d t h 4 j t 4 j t 2¯ ¯ ¯ d t Z ⇥ m Z Z Z Z Z Z Z Z Z . D[ x ( t ) ] e iS [ x ( t )] /¯ h x 2 − W ( x ) ⇤ S [ x ( t ) ] : = = d t 2 The classical path minimizes S [ x ( t )] by definition …and is the dominant single contribution in the integral …nearby paths are sub-dominant, but there is many of them …even wildly non-classical paths contribute! 12
Q Classical → Quantum Physics M II Quantization and Anomalies Extra! In general, “quantization” is a prescription of assigning a quantum theory to a classical one. Not unique at all [Pauli, 1930’s]: ⇥ b b⇤ ⇤ ⇥ b ⇤ Q 2 b PQ 2 7! b P b b Q b P b Q = b P b h b P = b P b h b Q 2 Q 2 + i ¯ Q 2 + 2 i ¯ Q Q or or ⇥ b b b b⇤ ⇥ b b 7! b b b b Denote by “ ( ” the chosen prescription: b b b P = π ( P ) , Q = π ( Q ) , F = π ( F ( Q , P )) But, classical observables are not independent. b b b So, compute: ⇣� ⌘ ⌘ ⇤ − π ⇥ ⇤ − = anomaly ( b A , b i π ( A ) , π ( B ) B ) A , B h ¯ PB for all Poisson brackets/commutators in a theory and redefine the prescription & until all anomalies cancel—if possible especially for the cases representing symmetries and conservation laws! 13
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