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Canonical Quantization Phys540.nb 27 2 Canonical quantization 2.1. Lagrangian mechanics and canonical quantization Q: How do we quantize a general system? 2.1.1.Lagrangian Lagrangian mechanics is a reformulation of classical mechanics. L = T - V (2.1) Here, the quantity L is called the Lagrangian, where T is the kinetic energy and V is the potential energy. For a pendulum, the kinetic energy is 2 1 ⅆθ 1 9 2 T = m r2 = m r θ (2.2) 2 ⅆt 2 and the potential is V $ - m g r cos 8 (2.3) And thus 9 1 9 2 L:8, 8; $ m r2 8 1 m g r cos 8 (2.4) 2 9 In general, a Lagrangian depends on coordinates (e.g. 8) and their corresponding velocities (e.g. 8) 9 9 9 L(q1, q2 < qn; q1, q2 <qn) (2.5) 9 where qi represent all coordinates in the systems and qi are the velocities. 2.1.2. Equation(s) of motion The equation of motion in Lagrangian mechanics is - =L =L 9 $ (2.6) -t =qi =qi For the pendulum example discussed above, - =L =L 9 $ (2.7) -t =8 =8 So 9 1 2 2 - =L - =T - = m r 8 - 9 .. 2 2 2 (2.8) 9 $ 9 $ 9 $ :m r 8; $ m r 8 -t =8 -t =8 -t =8 -t 28 Phys540.nb and =L =V =m g r cos 8 $ - $ $ -m g r sin8 (2.9) =8 =8 =8 As a result, the EOM is .. m r2 8 $ -m g r sin 8 (2.10) .. m r 8 $ -m g sin 8 (2.11) which recovers the second law of Newton: m a $ F (2.12) 2.1.3. Hamiltonian, canonical coordinates and canonical momenta From a Lagrangian described above, we can define one canonical momentum for every coordinate: =L pi $ 9 (2.13) =qi If the system has n coordinates, we will have n canonical momenta. If the coordinates are the usual 3D Cartesian coordinates, these momenta coincide with the “momentum” we usually used in Newton’s theory. If the coordinates that we used here are angles, then the momenta will be angular momenta L. Hamiltonian: H $ >pi qi - L $ T 1 V (2.14) H is a function of pi and qi For the pendulum example, 9 1 2 2 =L =L =T = m r 8 9 2 2 (2.15) p $ 9 $ 9 $ 9 $ 9 $ m r 8 =8 =8 =8 =8 9 Notice that 8 is just the angular velocity ". Here, this p is in fact m r ", i.e. the angular momentum. 9 9 1 9 2 H $ p 8 - L $ p 8 - m r 8 - m g r cos 8 (2.16) 2 9 9 Because p $ m r2 8, we can replace all 8 by p5m r2 p 1 p 2 p2 p2 p2 H $ p - m r2 - m g r cos 8 $ - - m g r cos 8 $ - m g r cos 8 (2.17) m r2 2 m r2 m r2 2 m r2 2 m r2 This Hamiltonian is indeed a function of p and 8 2.1.4. Canonical quantization In a quantum system, all physical observable are presented by quantum operators, including H, qi and pi. The relation between them remains the same as in classical mechanics. The only ingredient that we need to add here is the “uncertainty relation”: &qi, pj* $ ? 2 %i,j (2.18) &qi, qj* $ &pi, pj* $ 0 (2.19) i.e., we request a coordinate NOT to commute with its canonical momentum. The commutator is assumed to be a constant, ?2 for any canonical pair pi and qi. 2.1.5. Summary Step 1: Start from classical mechanics Step 2: Define coordinates Step 3: Find the canonical momentum for each coordinate Phys540.nb 29 Step 4: Write down the Hamiltonian Step 5: Require canonical commutation relation Example: for the example we considered above, we have only one coordinate and one momentum, and the Hamiltonian looks like p2 H $ - m g r cos q (2.20) 2 m r2 When we quantize this system, we turn everything into quantum operators @ @ p2 H $ - m g r cos q@ (2.21) 2 m r2 And we request &q@, p@* $ ? 2. Q: What is a cos function of a quantum operator? A: We use Taylor expansions x2 x4 cos x $ 1 - 1 1 < (2.22) 2 24 So q@ 2 q@ 4 cos q@ $ 1 - 1 1 < (2.23) 2 24 And thus @ @ @ p2 p2 m g r m g r H $ - m g r cos q@ $ - m g r 1 q@ 2 - q@ 4 1 < (2.24) 2 m r2 2 m r2 2 24 We know that a constant in energy is not important, so we can drop the constant part -m g r @ @ @ p2 p2 m g r m g r H $ - m g r cos q@ $ 1 q@ 2 - q@ 4 1 < (2.25) 2 m r2 2 m r2 2 24 The first two terms here are the Hamiltonian of a harmonic oscillator. And the rest part of the Hamiltonian is known as the non-harmonic part of the potential. 2.2. Creation and annihilation operators 2.2.1. Creation and annihilation operators. Define: 1 x qi - ? pi x A (2.26) ai $ 2 2 1 x qi 1 ? pi x (2.27) ai $ 2 2 A where x is some arbitrary (positive) real number. It is easy to prove that ai and ai are conjugate of each other, :aA;A $ a (2.28) A and it is easy to verify that Bai, ai C $ 1 30 Phys540.nb 1 1 x qi 1 ? pi x qi - ? pi x x A A A ai ai - ai ai $ Bai, ai C $ D , E $ 2 2 2 2 (2.29) x ? ? 1 ? -?F? 2 &qi, qi* 1 &pi, qi* - &qi, pi* 1 &pi, pi* $ - &qi, pi* $ $ 1 2 2 2 2 2 2 2 2x 2 2 More generally, we can prove that 1 if i $ j Ba , a AC $ % $ G (2.30) i j ij 0 if i / j For the example that we considered above, there is only one coordinate and one momentum, q and p, and thus, we only have one creation operator and one annihilation operator x q - 1 ? p x aA $ (2.31) 2 2 x q 1 1 ? p x a $ (2.32) 2 2 2.2.2. Vacuum For simplicity, from now on, we will only focus on the example that discussed above, but please keep in mind that all conclusions can be generalized easily to more complicated Hamiltonians. Here we make two assumptions: Assumption #1: there is one (and only one) quantum state such that a 0H $ 0 (2.33) We call this state the vacuum state. Assumption #2: The states 0〉, a† 0, a† a† 0, a† a† a† 0… form a complete basis of the Hilbert space. Comment: These assumptions turn out to be true for most cases that we study in quantum mechanics. 2.2.3. particle number operator Define particle number operator n $ aA a (2.34) 2.2.4. Hamiltonian: the harmonic part @ @ @ p2 p2 m g r m g r H $ - m g r cos q@ $ 1 q@ 2 - q@ 4 1 < (2.35) 2 m r2 2 m r2 2 24 Lets keep only the first two terms, and ignore all higher order terms of q@ p@ 2 m g r @ @ 2 H0 $ 1 q (2.36) 2 m r2 2 The part that we ignored are @ m g r @ % H $ - q4 1 < (2.37) 24 and H $ H0 1 % H For the harmonic part H0 p@ 2 m g r @ @ 2 H0 $ 1 q (2.38) 2 m r2 2 Phys540.nb 31 Because x q - 1 ? p x aA $ (2.39) 2 2 x q 1 1 ? p x a $ (2.40) 2 2 we know that 2 q $ :aA 1 a; (2.41) 2 x x 2 p $ ? :aA - a; (2.42) 2 Thus, 2 ? x 2 :aA - a; 2 2 p@ 2 m g r m g r 2 x 2 m g r 2 (2.43) @ @ 2 A A 2 A 2 H0 $ 1 q $ 1 :a 1 a; $ - :a - a; 1 :a 1 a; 2 2 2 2 m r 2 2 m r 2 2 x 4 m r 4 x Here :aA - a;2 $ :aA - a; :aA - a; $ aA aA - aA a - a aA 1 a a (2.44) and :aA 1 a;2 $ :aA 1 a; :aA 1 a; $ aA aA 1 aA a 1 a aA 1 a a (2.45) So x 2 m g r 2 x 2 m g r 2 @ A A A A H0 $ - 1 :a a 1 a a; 1 1 :a a 1 a a; (2.46) 4 m r2 4 x 4 m r2 4 x The first term here violate the particle conservation law, i.e. it doesn’t commute with the particle number operator. However, we can get ride of it by choosing the value of x, such that the first term vanish x 2 m g r 2 - 1 $ 0 (2.47) 4 m r2 4 x i.e. x2 $ m2 g r3 (2.48) x $ m2 g r3 $ m r g r (2.49) And thus, our Hamiltonian preserves the particle number, &H0, n* $ 0. x 2 m g r 2 x 2 m g r 2 @ A A A A H0 $ - 1 :a a 1 a a; 1 1 :a a 1 a a ; $ 4 m r2 4 x 4 m r2 4 x 2 m g r 2 2 g g 2 m r g r 1 :aA a 1 a aA; $ 1 :aA a 1 a aA; 2 (2.50) 4 m r 4 m r g r 4 r 4 r g aA a 1 a aA $ 2 r 2 Because Ba, aAC $ a aA - aA a $ 1, a aA $ aA a 1 1 and thus 32 Phys540.nb aA a 1 a aA aA a 1 aA a 1 1 2 aA a 1 1 1 @ A H0 $ 2 g5r $ 2 g5r $ 2 g5r $ 2 g5r a a 1 (2.51) 2 2 2 2 Notice that g5r $ " is actually the angular frequency of the harmonic oscillator, and aA a is in fact the particle number operator n we defined above @ 1 H0 $ 2 " n 1 (2.52) 2 We can understand this Hamiltonian as the following: (1) each of our fictitious particle has energy 2".
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