4. Chapter. Principle of Stationary Action and Lagrangian Dynamics

In Chapter 3, we learned the dynamics of quantum mechanics where the Hamiltonian (\(H\)) acts as the ‘generator’ of time evolution (\(U(t)=e^{-iHt/\hbar}\)). This reflects the perspective that “given a state, the future is determined by \(H\).”

However, classical mechanics has a dynamics system that is equivalent yet fundamentally different in philosophy. This is the Lagrangian Dynamics. This perspective states that “a particle, when going from \(t_1\) to \(t_2\), selects only one path that ‘stationary’-izes the value of ‘Action’ among all possible paths.”

This chapter provides the essential mathematical and conceptual preparation for understanding Path Integral (Path Integral) that will be learned in Chapter 5. The Path Integral states that “quantum particles pass through all paths simultaneously,” and this is why the ‘principle of stationary action’ in this Chapter 4 provides the answer to the question “why do we observe only one classical path among the countless quantum paths?”

1. Fundamental Concepts (Fundamental Concepts)

  • Functional and Action (\(S\)):

    • A function \(f(x)\) takes a number \(x\) as input and returns a number \(f(x)\).
    • A functional \(S[x(t)]\) takes a function (path) \(x(t)\) as a whole input and returns a number \(S\).
    • Action is a functional that assigns a ‘cost’ or ‘score’ to a specific path \(x(t)\). This ‘cost’ is calculated by integrating the Lagrangian \(L\) over the entire path. \[S[x(t)] = \int_{t_1}^{t_2} L(x(t), \dot{x}(t), t) dt\]
    • The Lagrangian \(L\) is defined (generally) as the difference between kinetic energy (\(T\)) and potential energy (\(V\)), \(L = T - V\).

  • Principle of Stationary Action: Nature follows a physical law that, when moving from \(x_1\) at \(t_1\) to \(x_2\) at \(t_2\), chooses a path among the infinite possible paths that makes the action \(S\) minimum (minimum), maximum (maximum), or saddle point (saddle point) in between. These three are collectively referred to as ‘stationary (stationary)’.

    • “Stationary” means that even if we slightly (\(\delta x\)) deviate from the ‘correct’ classical path \(x_{cl}(t)\), the total action \(S\) value does not change at the first approximation level (\(\delta S = 0\)).

  • Calculus of Variations: A mathematical tool used to find paths that satisfy the ‘principle of stationary action’ (i.e., paths where \(\delta S = 0\)). This extends the method of using differentiation (\(df/dx=0\)) to find extrema in general functions to the world of functionals.

  • Euler-Lagrange Equation: This is the result obtained by solving the condition \(\delta S = 0\) using the calculus of variations. The ‘solution’ of this equation is exactly that ‘correct’ classical path. In other words, this is the new equation of motion that replaces Newton’s \(F=ma\). \[\frac{\partial L}{\partial x} - \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) = 0\]

  • Stationary Phase Approximation: This concept is the most important bridge connecting Chapter 4 (classical mechanics) and Chapter 5 (quantum mechanics). > 💡 Why is the path with \(\delta S=0\) the only one observable in the classical world? > > Feynman’s path integral, which we will learn in Chapter 5, states that the total probability amplitude \(K(2, 1)\) for a particle traveling from \(x_1\) to \(x_2\) is the sum of the amplitudes for all possible paths. > > \[K(2, 1) = \sum_{\text{all paths}} e^{iS[x]/\hbar}\] > > * Non-classical path (\(x_{\text{non-cl}}\)): Paths near this one experience rapid changes in \(S\) values. Thus, the phase of \(e^{iS/\hbar}\) rotates randomly, causing the amplitudes of different paths to destructively interfere and cancel out. > * Classical path (\(x_{\text{cl}}\)): This path satisfies \(\delta S = 0\). That is, near this path, the \(S\) value changes very little. Therefore, the phase of \(e^{iS/\hbar}\) remains nearly constant, causing the amplitudes of all nearby paths to constructively interfere and strongly survive. > > In the macroscopic world where \(\hbar\) (Planck’s constant) is nearly zero, this constructive interference effect becomes extremely strong, making only the single classical path with \(\delta S=0\) visible to us. This is the intuitive idea behind the ‘stationary phase.’

2. Symbols and Key Relations

  • Path and Variation (Path and Variation):
    • Classical path: \(x_{\text{cl}}(t)\)
    • Virtual path: \(x(t) = x_{\text{cl}}(t) + \delta x(t)\) (where \(\delta x(t_1) = \delta x(t_2) = 0\), the start and end points are fixed)
    • Stationary Action Condition: The first variation of \(S\) must be zero. \(\delta S = 0\)
  • Euler-Lagrange Equation (E-L Equation):
    • One-dimensional: \(\frac{\partial L}{\partial x} - \frac{d}{dt}\left(\frac{\partial L}{\partial \dot{x}}\right) = 0\)
    • Multi-dimensional (vector notation): \(\nabla_q L - \frac{d}{dt}\left(\nabla_{\dot{q}} L\right) = 0\) (where \(\nabla_q = (\partial/\partial q_1, \dots)\), \(\nabla_{\dot{q}} = (\partial/\partial \dot{q}_1, \dots)\))
  • Noether’s Theorem (Connection to Chapter 3): The ‘symmetry \(\to\) conserved quantity’ relationship learned in Chapter 3 clearly appears in the Lagrangian formulation as well.
    • Generalized momentum (\(p_k\)): \(p_k \equiv \frac{\partial L}{\partial \dot{q}_k}\)
    • Spatial symmetry (cyclic coordinates): If \(L\) is independent of a particular coordinate \(q_k\) (\(\frac{\partial L}{\partial q_k} = 0\)), the E-L equation implies \(\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_k}) = 0\), meaning the generalized momentum \(p_k\) for that coordinate is conserved.
    • Time symmetry (energy conservation): If \(L\) is explicitly independent of time (\(\frac{\partial L}{\partial t} = 0\)), the energy \(E = \sum_k \dot{q}_k \frac{\partial L}{\partial \dot{q}_k} - L\) is conserved.
  • Constraints and Lagrange Multipliers:
    • When there is a constraint \(g(q, t) = 0\), we define the ‘extended’ Lagrangian \(L' = L + \lambda(t) g(q, t)\).
    • Treat \(\lambda\) (the Lagrange multiplier) as a new coordinate, and derive the E-L equations for \(q\) and \(\lambda\) separately.

3. Easy Examples (Examples with Deeper Insight)

  • Example 1: Free Particle (Uniform Motion)
    • Lagrangian: \(L = T - V = \frac{1}{2}m\dot{x}^2 - 0\)
    • E-L Equations: \(\frac{\partial L}{\partial x} = 0\), \(\frac{\partial L}{\partial \dot{x}} = m\dot{x}\) \(0 - \frac{d}{dt}(m\dot{x}) = 0 \implies m\ddot{x} = 0\)
    • Interpretation: The equation of motion is \(\ddot{x}=0\) (zero acceleration), i.e., uniform straight-line motion. When the potential is zero, the path that minimizes the ‘action’ (integral of kinetic energy) is the most ‘relaxed’ uniform motion path.
  • Example 2: Harmonic Oscillator
    • Lagrangian: \(L = T - V = \frac{1}{2}m\dot{x}^2 - \frac{1}{2}kx^2\)
    • E-L Equations: \(\frac{\partial L}{\partial x} = -kx\), \(\frac{\partial L}{\partial \dot{x}} = m\dot{x}\) \((-kx) - \frac{d}{dt}(m\dot{x}) = 0 \implies -kx - m\ddot{x} = 0 \implies m\ddot{x} + kx = 0\)
    • Interpretation: We obtain the same equation of motion as \(F=ma\) (here \(F=-kx\)). This demonstrates that Lagrangian mechanics is equivalent to Newtonian mechanics.
  • Example 3: Shortest Path (Straight Line)
    • Scenario: Problem of minimizing the length \(S\) of a path from \((x_1, y_1)\) to \((x_2, y_2)\).
    • ‘Action’: \(S = \int ds = \int \sqrt{dx^2 + dy^2} = \int_{x_1}^{x_2} \sqrt{1 + (y')^2} dx\) (where \(y' = dy/dx\))
    • ‘Lagrangian’: \(L(y, y') = \sqrt{1 + y'^2}\)
    • E-L Equation: (Here \(x\) is the parameter instead of \(t\)) \(\frac{\partial L}{\partial y} - \frac{d}{dx}(\frac{\partial L}{\partial y'}) = 0\) \(\frac{\partial L}{\partial y} = 0\) (cyclic coordinate \(\to\) conserved quantity!) \(0 - \frac{d}{dx}\left(\frac{y'}{\sqrt{1 + y'^2}}\right) = 0 \implies \frac{y'}{\sqrt{1 + y'^2}} = C\) (constant)
    • Interpretation: This implies \(y' = \text{constant}\), i.e., \(y(x) = ax + b\), which is a straight line. In Euclidean space, the path that minimizes the ‘action’ (length) is a straight line.
  • Example 4: Noether’s Theorem (Conservation of Angular Momentum)
    • Situation: Particle moving in polar coordinates (\(r, \theta\)) under a central force \(V(r)\).
    • Lagrangian: \(L = T - V = \frac{1}{2}m(\dot{r}^2 + r^2\dot{\theta}^2) - V(r)\)
    • E-L ( \(\theta\) coordinate): \(\frac{\partial L}{\partial \theta} = 0\) ( \(L\) is independent of \(\theta\), i.e., ‘rotational symmetry’) \(\frac{d}{dt}\left(\frac{\partial L}{\partial \dot{\theta}}\right) = 0\)
    • Conserved quantity: \(\frac{\partial L}{\partial \dot{\theta}} = m r^2 \dot{\theta} = \text{const}\).
    • Interpretation: \(mr^2\dot{\theta}\) is angular momentum (\(J_z\)). Noether’s theorem, which states that ‘rotational symmetry’ leads to ‘conservation of angular momentum’, is naturally derived via the E-L equations from the knowledge of Chapter 3.

4. Practice Problems

  1. (E-L Derivation): Using \(S[x]=\int_{t_1}^{t_2}L(x,\dot x)\,dt\) and \(x(t) = x_{cl}(t) + \epsilon \eta(t)\) (with \(\eta(t_1)=\eta(t_2)=0\)), derive the Euler-Lagrange equation directly from the condition \(\delta S = \left.\frac{dS}{d\epsilon}\right|_{\epsilon=0} = 0\). (Hint: Integration by parts)
  2. (Free Particle): Derive the E-L equation (\(m\ddot{x}=0\)) and the general solution (\(x(t)=at+b\)) for a free particle.
  3. (Harmonic Oscillator): Derive the E-L equation (\(m\ddot{x}+kx=0\)) and the general solution (\(x(t)=A\cos(\omega t)+B\sin(\omega t)\)) for a harmonic oscillator.
  4. (Energy Conservation): Prove that the energy function \(E = \dot{q}\frac{\partial L}{\partial \dot{q}} - L\) is constant (\(dE/dt=0\)) when \(L=L(q, \dot{q})\) does not explicitly depend on time (\(\partial L/\partial t = 0\)).
  5. (Cyclic Coordinate): Prove that the generalized momentum \(p_k = \partial L/\partial \dot{q}_k\) is conserved when \(L\) is independent of a coordinate \(q_k\) (\(\partial L/\partial q_k=0\)).
  6. (Multidimensional E-L): Derive the vector form of the E-L equation \(\nabla_q L - \frac{d}{dt}\nabla_{\dot{q}} L = 0\) when \(q=(q_1, \dots, q_n)\).
  7. (Natural Boundary Conditions): When \(t_1\) is fixed (\(\delta x(t_1)=0\)) but \(t_2\) is free (\(\delta x(t_2)\neq 0\)), derive what additional ‘boundary condition’ \(\delta S=0\) imposes besides the equation of motion.
  8. (Lagrange Multipliers): Derive the equation of motion for \(L=\frac{1}{2}m(\dot{x}^2+\dot{y}^2) - V(x,y)\) under the constraint \(x^2+y^2=R^2\) using a Lagrange multiplier \(\lambda\).
  9. (Classical Path Intuition): Suppose two paths \(x_A, x_B\) have \(S[x_A] = 100\hbar, S[x_B] = 100.5\hbar\). Do their quantum amplitudes \(e^{iS/\hbar}\) interfere constructively or destructively? Why?
  10. (Second Variation): For \(L=\frac{1}{2}\dot{x}^2\) (free particle), compute \(\delta^2 S\) (second variation) and show that \(\delta^2 S > 0\). This implies the classical path minimizes the action. ### 5. Explanation
  11. \(\delta S = \int \left( \frac{\partial L}{\partial x}\delta x + \frac{\partial L}{\partial \dot{x}}\delta \dot{x} \right) dt = \int \left( \frac{\partial L}{\partial x}\eta + \frac{\partial L}{\partial \dot{x}}\dot{\eta} \right) \epsilon dt\). \(\frac{dS}{d\epsilon} = \int \left( \frac{\partial L}{\partial x}\eta + \frac{\partial L}{\partial \dot{x}}\dot{\eta} \right) dt\). Perform partial integration on the second term: \(\int (\frac{\partial L}{\partial \dot{x}}\dot{\eta}) dt = \left[ \frac{\partial L}{\partial \dot{x}}\eta \right]_{t_1}^{t_2} - \int \left( \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} \right) \eta dt\). The first term is zero due to the boundary condition \(\eta(t_1)=\eta(t_2)=0\). \(\frac{dS}{d\epsilon} = \int \left( \frac{\partial L}{\partial x} - \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} \right) \eta(t) dt\). Since this must be zero for all \(\eta(t)\), the term inside the parentheses must be zero (Euler-Lagrange equation).
  12. (Refer to Example 1) \(m\ddot{x}=0 \implies \dot{x}=a \implies x(t)=at+b\).
  13. (Refer to Example 2) \(m\ddot{x}+kx=0 \implies \ddot{x} + \omega^2 x = 0\) (where \(\omega^2=k/m\)). The solution is \(x(t)=A\cos(\omega t)+B\sin(\omega t)\).
  14. \(dE/dt = \frac{d}{dt}(\dot{q}\frac{\partial L}{\partial \dot{q}} - L) = \ddot{q}\frac{\partial L}{\partial \dot{q}} + \dot{q}\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}}) - (\frac{\partial L}{\partial q}\dot{q} + \frac{\partial L}{\partial \dot{q}}\ddot{q})\).
  15. From the Euler-Lagrange equation \(\frac{\partial L}{\partial q_k} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_k}) = 0\), since \(\frac{\partial L}{\partial q_k}=0\), we have \(\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_k}) = 0\). Thus, \(p_k = \frac{\partial L}{\partial \dot{q}_k}\) is a constant (conserved quantity).
  16. In the partial integration process of the first explanation, the boundary term \(\left[ \frac{\partial L}{\partial \dot{x}}\eta \right]_{t_1}^{t_2}\) remains as \(\frac{\partial L}{\partial \dot{x}}\Big|_{t=t_2} \eta(t_2)\) because \(\eta(t_1)=0\).
  17. The two amplitudes differ in phase by \(0.5\pi\) (i.e., 90 degrees), so they are in an intermediate state neither reinforcing nor canceling each other.
  18. The free particle solution corresponds to the case where the potential energy is zero.
  19. The two amplitudes differ in phase by \(0.5\pi\) (i.e., 90 degrees), so they are in an intermediate state neither reinforcing nor canceling each other.
  20. The solution for the free particle is given by \(x(t)=at+b\).
  21. The integral \(\int (\frac{\partial L}{\partial \dot{x}}\dot{\eta}) dt\) is evaluated using partial integration, leading to the boundary term \(\left[ \frac{\partial L}{\partial \dot{x}}\eta \right]_{t_1}^{t_2}\).
  22. The term \(\left[ \frac{\partial L}{\partial \dot{x}}\eta \right]_{t_1}^{t_2}\) is zero due to the boundary condition \(\eta(t_1)=\eta(t_2)=0\).
  23. The final expression for \(\frac{dS}{d\epsilon}\) is derived by integrating the Euler-Lagrange equation.
  24. The Euler-Lagrange equation is obtained by requiring that the variation of the action is zero for all admissible variations \(\eta(t)\).
  25. The conserved quantity \(p_k = \frac{\partial L}{\partial \dot{q}_k}\) is a constant of motion under the assumption that the Lagrangian does not explicitly depend on \(q_k\).
  26. The natural boundary condition arises when the variation \(\eta(t)\) is not constrained at the endpoints of the interval.
  27. The solution to the equation \(\ddot{x} + \omega^2 x = 0\) is a combination of sine and cosine functions with frequency \(\omega\).
  28. The energy \(E\) is conserved when the system is described by a Lagrangian that does not explicitly depend on time.
  29. The partial integration process shows that the boundary term contributes to the variation of the action only if \(\eta(t)\) is non-zero at the endpoints.
  30. The variation \(\eta(t)\) must be zero at the endpoints to ensure that the boundary term vanishes, leading to the Euler-Lagrange equation.
  31. \(\delta S = \int \left( \frac{\partial L}{\partial x}\delta x + \frac{\partial L}{\partial \dot{x}}\delta \dot{x} \right) dt = \int \left( \frac{\partial L}{\partial x}\eta + \frac{\partial L}{\partial \dot{x}}\dot{\eta} \right) \epsilon dt\). \(\frac{dS}{d\epsilon} = \int \left( \frac{\partial L}{\partial x}\eta + \frac{\partial L}{\partial \dot{x}}\dot{\eta} \right) dt\). Integrate the second term by parts: \(\int (\frac{\partial L}{\partial \dot{x}}\dot{\eta}) dt = \left[ \frac{\partial L}{\partial \dot{x}}\eta \right]_{t_1}^{t_2} - \int \left( \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} \right) \eta dt\). By the boundary condition \(\eta(t_1)=\eta(t_2)=0\), the first term is zero. \(\frac{dS}{d\epsilon} = \int \left( \frac{\partial L}{\partial x} - \frac{d}{dt}\frac{\partial L}{\partial \dot{x}} \right) \eta(t) dt\). Since this must be zero for all \(\eta(t)\), the term in parentheses must be zero (Euler-Lagrange equation). 2. (See Example 1) \(m\ddot{x}=0 \implies \dot{x}=a \implies x(t)=at+b\). 3. (See Example 2) \(m\ddot{x}+kx=0 \implies \ddot{x} + \omega^2 x = 0\) (where \(\omega^2=k/m\)). The solution is \(x(t)=A\cos(\omega t)+B\sin(\omega t)\). 4. \(dE/dt = \frac{d}{dt}(\dot{q}\frac{\partial L}{\partial \dot{q}} - L) = \ddot{q}\frac{\partial L}{\partial \dot{q}} + \dot{q}\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}}) - (\frac{\partial L}{\partial q}\dot{q} + \frac{\partial L}{\partial \dot{q}}\ddot{q} + \frac{\partial L}{\partial t})\). Since \(\partial L/\partial t = 0\) and substituting the Euler-Lagrange equation \(\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}}) = \frac{\partial L}{\partial q}\), all terms cancel and the result is zero. 5. From the Euler-Lagrange equation \(\frac{\partial L}{\partial q_k} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_k}) = 0\), since \(\frac{\partial L}{\partial q_k}=0\), we have \(\frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_k}) = 0\). Therefore, \(p_k = \frac{\partial L}{\partial \dot{q}_k}\) is a constant (conserved quantity). 6. Applying the derivation from step 1 independently to each \(q_k\) component, we obtain \(\frac{\partial L}{\partial q_k} - \frac{d}{dt}(\frac{\partial L}{\partial \dot{q}_k}) = 0\) (for \(k=1 \dots n\)). Combining these into a vector form gives \(\nabla_q L - \frac{d}{dt}\nabla_{\dot{q}} L = 0\). 7. In the partial integration process of the first explanation, the boundary term \(\left[ \frac{\partial L}{\partial \dot{x}}\eta \right]_{t_1}^{t_2}\) vanishes because \(\eta(t_1)=0\), leaving only \(\frac{\partial L}{\partial \dot{x}}\Big|_{t=t_2} \eta(t_2)\). Since \(\delta S = 0\) must hold for all \(\eta(t)\) (including cases where \(\eta(t_2)\neq 0\)), the ‘natural boundary condition’ \(\frac{\partial L}{\partial \dot{x}}\Big|_{t=t_2} = 0\) must be added alongside the equation of motion. 8. \(L' = \frac{1}{2}m(\dot{x}^2+\dot{y}^2) - V(x,y) + \lambda(x^2+y^2-R^2)\). \(x\)-coordinate: \(\frac{\partial L'}{\partial x} - \frac{d}{dt}(\frac{\partial L'}{\partial \dot{x}}) = 0 \implies -\frac{\partial V}{\partial x} + 2\lambda x - m\ddot{x} = 0\). \(y\)-coordinate: \(\frac{\partial L'}{\partial y} - \frac{d}{dt}(\frac{\partial L'}{\partial \dot{y}}) = 0 \implies -\frac{\partial V}{\partial y} + 2\lambda y - m\ddot{y} = 0\). \(\lambda\)-coordinate: \(\frac{\partial L'}{\partial \lambda} - \frac{d}{dt}(\frac{\partial L'}{\partial \dot{\lambda}}) = 0 \implies x^2+y^2-R^2 = 0\) (original constraint equation). 9. \(e^{iS_A/\hbar} = e^{i 100}\) (phase 100 radians). \(e^{iS_B/\hbar} = e^{i 100.5} = e^{i 100} e^{i 0.5\pi} = e^{i 100} (i)\). The two amplitudes differ in phase by \(0.5\pi\) (i.e., 90 degrees), so they are in an intermediate state, neither constructive nor destructive interference. If \(S_B = 100.1\hbar\), it would be nearly constructive interference, and if \(S_B = 100.5\pi \hbar \approx 100.157\hbar\), it would be perfect destructive interference (\(e^{i\pi}=-1\)). Since \(\hbar\) is very small, even a slight difference in \(S\) causes a violent change in phase, resulting in cancellation. 10. \(L(x+\epsilon\eta) = \frac{1}{2}(\dot{x}+\epsilon\dot{\eta})^2 = \frac{1}{2}(\dot{x}^2 + 2\epsilon\dot{x}\dot{\eta} + \epsilon^2\dot{\eta}^2)\). \(S[x+\epsilon\eta] = \int L dt = S[x] + \epsilon \int \dot{x}\dot{\eta} dt + \frac{\epsilon^2}{2} \int \dot{\eta}^2 dt\). \(\delta S = \int \dot{x}\dot{\eta} dt = 0\) (first variation, using free particle solution \(\ddot{x}=0\)). \(\delta^2 S = \frac{1}{2} \int \dot{\eta}^2 dt\). As long as \(\eta(t)\) is not zero, \(\dot{\eta}^2 \ge 0\), so \(\delta^2 S > 0\). Therefore, the path of a free particle minimizes the action.