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 2 T = m r2 = m r θ (2.2) 2 ⅆt 2 and the potential is
V = - m g r cos θ (2.3) And thus
1 2 Lθ, θ = m r2 θ + m g r cos θ (2.4) 2 In general, a Lagrangian depends on coordinates (e.g. θ) and their corresponding velocities (e.g. θ) L(q1, q2 … qn; q1, q2 …qn) (2.5) 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 ⅆ ∂L ∂L = (2.6) ⅆt ∂qi ∂qi For the pendulum example discussed above, ⅆ ∂L ∂L = (2.7) ⅆt ∂θ ∂θ So 1 2 2 ⅆ ∂L ⅆ ∂T ⅆ ∂ m r θ ⅆ .. 2 2 2 (2.8) = = = m r θ = m r θ ⅆt ∂θ ⅆt ∂θ ⅆt ∂θ ⅆt 28 Phys540.nb
and ∂L ∂V ∂m g r cos θ = - = = -m g r sinθ (2.9) ∂θ ∂θ ∂θ As a result, the EOM is .. m r2 θ = -m g r sin θ (2.10) .. m r θ = -m g sin θ (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 = (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 + V (2.14)
H is a function of pi and qi For the pendulum example, 1 2 2 ∂L ∂L ∂T ∂ m r θ 2 2 (2.15) p = = = = = m r θ ∂θ ∂θ ∂θ ∂θ Notice that θ is just the angular velocity ω. Here, this p is in fact m r ω, i.e. the angular momentum.
1 2 H = p θ - L = p θ - m r θ - m g r cos θ (2.16) 2 Because p = m r2 θ, we can replace all θ by p/m r2
p 1 p 2 p2 p2 p2 H = p - m r2 - m g r cos θ = - - m g r cos θ = - m g r cos θ (2.17) m r2 2 m r2 m r2 2 m r2 2 m r2 This Hamiltonian is indeed a function of p and θ
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] = ⅈ ℏ δi,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, ⅈℏ 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 ] = ⅈ ℏ. Q: What is a cos function of a quantum operator? A: We use Taylor expansions
x2 x4 cos x = 1 - + + … (2.22) 2 24 So
q 2 q 4 cos q = 1 - + + … (2.23) 2 24 And thus p2 p2 m g r m g r H = - m g r cos q = - m g r + q 2 - q 4 + … (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 = + q 2 - q 4 + … (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 - ⅈ pi x † (2.26) ai = 2 ℏ
1 x qi + ⅈ pi x (2.27) ai = 2 ℏ † where x is some arbitrary (positive) real number. It is easy to prove that ai and ai are conjugate of each other,
a†† = a (2.28)
† and it is easy to verify that ai, ai = 1 30 Phys540.nb
1 1 x qi + ⅈ pi x qi - ⅈ pi x x † † † ai ai - ai ai = ai, ai = , = 2 ℏ 2 ℏ (2.29) x ⅈ ⅈ 1 ⅈ -ⅈ*ⅈ ℏ [qi, qi] + [pi, qi] - [qi, pi] + [pi, pi] = - [qi, pi] = = 1 2 ℏ 2 ℏ 2 ℏ 2 ℏx ℏ ℏ More generally, we can prove that 1 if i = j a , a † = δ = (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 ⅈ p x a† = (2.31) 2 ℏ
x q + 1 ⅈ p x a = (2.32) 2 ℏ
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 0〉 = 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 = a† a (2.34)
2.2.4. Hamiltonian: the harmonic part p2 p2 m g r m g r H = - m g r cos q = + q 2 - q 4 + … (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 = + q (2.36) 2 m r2 2 The part that we ignored are m g r δ H = - q4 + … (2.37) 24
and H = H0 + δ H
For the harmonic part H0
p 2 m g r 2 H0 = + q (2.38) 2 m r2 2 Phys540.nb 31
Because
x q - 1 ⅈ p x a† = (2.39) 2 ℏ
x q + 1 ⅈ p x a = (2.40) 2 ℏ we know that
ℏ q = a† + a (2.41) 2 x
x ℏ p = ⅈ a† - a (2.42) 2
Thus,
2 ⅈ x ℏ a† - a 2 2 p 2 m g r m g r ℏ x ℏ m g r ℏ (2.43) 2 † † 2 † 2 H0 = + q = + a + a = - a - a + a + a 2 2 2 2 m r 2 2 m r 2 2 x 4 m r 4 x Here
a† - a2 = a† - a a† - a = a† a† - a† a - a a† + a a (2.44) and
a† + a2 = a† + a a† + a = a† a† + a† a + a a† + a a (2.45) So x ℏ m g r ℏ x ℏ m g r ℏ † † † † H0 = - + a a + a a + + a a + 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 ℏ m g r ℏ - + = 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 ℏ m g r ℏ x ℏ m g r ℏ † † † † H0 = - + a a + a a + + a a + a a = 4 m r2 4 x 4 m r2 4 x ℏ m g r ℏ ℏ g g ℏ m r g r + a† a + a a† = + a† a + a a† 2 (2.50) 4 m r 4 m r g r 4 r 4 r g a† a + a a† = ℏ r 2 Because a, a† = a a† - a† a = 1, a a† = a† a + 1 and thus 32 Phys540.nb
a† a + a a† a† a + a† a + 1 2 a† a + 1 1 † H0 = ℏ g/r = ℏ g/r = ℏ g/r = ℏ g/r a a + (2.51) 2 2 2 2
Notice that g/r = ω is actually the angular frequency of the harmonic oscillator, and a† a is in fact the particle number operator n we defined above 1 H0 = ℏ ω n + (2.52) 2 We can understand this Hamiltonian as the following: (1) each of our fictitious particle has energy ℏω. (2) The system also have an zero point energy (energy of the vacuum) ℏω . So the total energy of the system is 2 1 H0 = ℏ ω n + (2.53) 2 Because the energy is carried by these fictitious particles, the total energy is proportional to particle number (up to the zero point energy ℏω/2). Because particle number are quantized integers, the energy is quantized. This is the origin of quantization
2.2.5. Hamiltonian: the non-harmonic part How about the terms that we ignored? We can treat them as interactions between particles (i.e. scatterings).
2.2.6. Waves and particle-wave duality For a harmonic oscillator, we know that we can think about it using these fictitious particles. These fictitious particles cannot move (they are confined spatially because we have a oscillator). This is a bit different from a real particle, which can move in space. How can we have a real particle? Using waves, instead of a single oscillator. Waves is nothing but couple harmonic oscillators (we put a harmonic oscillator at each position x, and couple them together). For each position x, we have one harmonic oscillator, and thus we can define a creation operator at this point a†(x) and also an annihilation operator a(x) at this point. We can also ask what is the particle number for each oscillator n (x) = a†(x) a(x) The coupling between harmonic oscillators allows the particle at x to move to y (i.e. motion of a particle). Non-harmonic parts gives interactions between this particles. This is the foundation of particle wave duality.
2.2.7. Summary The bottom line: second quantization and/or the quantum field theory are applicable to any quantum systems described by canonical coordi- nates and canonical momenta: (bosonic particles). (1) All these systems can be described as a bunch of bosonic particles. These particles are called fundamental particles in particle physics problems, and quasi-particles in condensed matter. (2) If the Hamiltonian (or Lagrangian) contains only quadratic terms, the particles are non-interacting, and the problem can be exactly solved like a quantum harmonic oscillator. (3) Higher order terms beyond quadratic describes “interactions between particles”. Interacting problem are in general NOT exactly solvable, but we can treat them perturbatively using the diagrams described before, if the interactions are weak enough. (4) The ground state has no particles (i.e. vacuum) and the low-energy excited states can be described as having a small number of particles (i.e. even if we are considering a solids with billions of atoms, the ground state can be described as the vacuum with no quasi-particle, and the low- energy excitations above the ground state can be described by the motion of a small number of particles). (5) In many case, the theory conserve particle number [H, N] = 0, although not always. (6) In the EOM, quadratic terms in H or L becomes linear terms, while higher order terms becomes non-linear terms. So interactions are also known as non-linear effects. (7) If we replace canonical communication relations with canonical anti-commutation relations, the theory is still self-consistent. These type of theory describes fermionic particles. Fermions have no classical picture, and thus we cannot see them if we start from a classical theory and try to quantize it. But they are perfectly fine theories, and thus nothing prevents their existence (and they do exist in our universe). One more comment on strongly-correlated systems Weakly-correlated systems are easy to handle (perturbation theory and diagrammatic expansions). But there exist many strongly-correlated systems, which are more challenging both theoretically and experimentally. Which systems are strongly-correlated? There are two ways to become strongly-correlated Phys540.nb 33
Weakly-correlated systems are easy to handle (perturbation theory and diagrammatic expansions). But there exist many strongly-correlated systems, which are more challenging both theoretically and experimentally. Which systems are strongly-correlated? There are two ways to become strongly-correlated Path 1, making interactions strong, i.e. the higher order terms beyond quadratic. Path 2, non-canonical systems, i.e., if a system doesn’t follow canonical commutation relations, even if we only have quadratic terms, the system can NOT be described as non-interacting particles. Instead, it will show strong interaction effects. Example: quantum spins. Spins don't follow canonical commutation relations. The reason is rotational along x, y, z axes are NOT independent. Sometimes, this problem (lack of canonical commutation relations) is not important and we can figure out the ground state straightforwardly. But sometimes, this problem becomes very important (e.g. frustrated magnets), we need to map the problem into bosons/fermions with infinite interactions, i.e. a strongly-correlated problem.