Feynman and Hibbs - Quantum Mechanics and Path Integrals
Chapter 2 - The quantum-mechanical law of motion
Problem 2-1 : For a free particle $L = (m/2) \dot{x}^2$. Show that the action $S_{cl}$ corresponding to the classical motion of a free particle is
\begin{equation} S_{cl} = \frac{m}{2} \frac{(x_b - x_a)^2}{t_b - t_a} \end{equation}Solution : Given the Lagrangian, the corresponding Lagrange equation is
\begin{equation} \frac{d}{dt} \frac{\partial L}{\partial \dot{x}} - \frac{\partial L}{\partial x} = 0 \end{equation}with
\begin{equation} \frac{\partial L}{\partial \dot{x}} = m\dot{x},\ \frac{\partial L}{\partial x} = 0 \end{equation}giving the equation of motion
\begin{equation} m \ddot{x}(t) = 0 \end{equation}As a simple ordinary differential equation, the solution is simply integrated to
\begin{equation} \dot{x}(t) = v_0,\ x(t) = v_0 t + x_0 \end{equation}At the two boundaries, we have that
\begin{eqnarray} x(t_a) &=& x_a &=& v_0 t_a + x_0\\ x(t_b) &=& x_b &=& v_0 t_b + x_0 \end{eqnarray}The difference of those two is then
\begin{equation} x(t_b) - x(t_a) = x_b - x_a = v_0 (t_b - t_a) \end{equation}And therefore,
\begin{equation} v_0 = \frac{x_b - x_a}{t_b - t_a} = \frac{x_b - x_a}{T} \end{equation}The on-shell action is therefore given by
\begin{eqnarray} S_{cl} &=& \int_{t_a}^{t_b} \frac{m}{2} (\frac{x_b - x_a}{T})^2 dt \\ &=& \left[ \frac{m}{2} (\frac{x_b - x_a}{T})^2 t \right]_{t_a}^{t_b}\\ &=& \frac{m}{2} (\frac{x_b - x_a}{T})^2 (t_b - t_a)\\ &=& \frac{m}{2} \frac{(x_b - x_a)^2}{T} \end{eqnarray}Problem 2-2 : For a harmonic oscillator $L = (m/2) (\dot{x}^2 - \omega^2 x^2)$. With $T$ equal to $t_b - t_a$, show that the classical action is
\begin{equation} S_{cl} = \frac{m\omega}{2\sin(\omega T)} [(x_b^2 + x_a^2) \cos(\omega T) - 2 x_b x_a] \end{equation}Solution : Given the Lagrangian, the corresponding Lagrange equation is
\begin{equation} \frac{d}{dt} \frac{\partial L}{\partial \dot{x}} - \frac{\partial L}{\partial x} = 0 \end{equation}with
\begin{equation} \frac{\partial L}{\partial \dot{x}} = m\dot{x},\ \frac{\partial L}{\partial x} = m \omega^2 x \end{equation}giving the equation of motion
\begin{equation} m \ddot{x}(t) + m \omega^2 x = 0 \end{equation}This is the classic equation of free oscillation, which can be put in several forms, but will be here put as
\begin{equation} x(t) = A \cos(\omega t) + B \sin(\omega t) \end{equation}where its derivative is then
\begin{equation} \dot{x}(t) = - A \omega \sin(\omega t) + B \omega \cos(\omega t) \end{equation}Using trigonometric identities ($\cos(a + b)= \cos(a) \cos(b) - \sin(a) \sin(b)$ and $\sin(a + b) = \sin(a) \cos(b) + \cos(a) \sin(b)$), the on-shell Lagrangian is then
\begin{eqnarray} L &=& \frac{m}{2} (\dot{x}^2 - \omega^2 x^2)\\ &=& \frac{m}{2} \left[ (- A \omega \sin(\omega t) + B \omega \cos(\omega t))^2 - \omega^2 (A \cos(\omega t) + B \sin(\omega t))^2\right] \\ &=& \frac{m}{2} [ A^2 \omega^2 \sin^2(\omega t) + B^2 \omega^2 \cos^2(\omega t) - 2AB \omega^2 \cos(\omega t) \sin(\omega t) \\ && - A^2 \omega^2 \cos^2(\omega t) - B^2 \omega^2 \sin^2(\omega t) - 2AB \omega^2 \cos(\omega t) \sin(\omega t) ] \\ &=& \frac{m}{2} [ A^2 \omega^2 (\sin^2(\omega t) - \cos^2(\omega t)) - B^2 \omega^2 (\sin^2(\omega t) - \cos^2(\omega t)) - 4AB \omega^2 \cos(\omega t) \sin(\omega t) ] \\ &=& \frac{m \omega^2}{2} [ (\sin^2(\omega t) - \cos^2(\omega t)) (A^2 - B^2) - 2AB \sin(2\omega t) ] \\ &=& \frac{m \omega^2}{2} [ - \cos(2\omega t) (A^2 - B^2) - 2AB \sin(2\omega t) ] \\ &=& \frac{m \omega^2}{2} [ \cos(2\omega t) (B^2 - A^2) - 2AB \sin(2\omega t) ] \end{eqnarray}The on-shell action is then (using the identities $\sin(2a) - \sin(2b) = 2 \cos(a+b) \sin(a-b)$ and $\cos(2a) - \cos(2b) = -2 \sin(a-b) \sin(a+b)$)
\begin{eqnarray} S_{cl} &=& \frac{m \omega^2}{2} \int_{t_a}^{t_b} [ \cos(2\omega t) (B^2 - A^2) - 2AB \sin(2\omega t) ] dt \\ &=& \frac{m \omega^2}{2} \left( \left[ \frac{\sin(2\omega t)}{2\omega} \right]_{t_a}^{t_b} (B^2 - A^2) + 2AB \left[ \frac{\cos(2\omega t)}{2\omega} \right]_{t_a}^{t_b} \right)\\ &=& \frac{m \omega}{4} \left( \left[ \sin(2\omega t_b) - \sin(2\omega t_a) \right] (B^2 - A^2) + 2AB \left[ \cos(2\omega t_b) - \cos(2\omega t_a) \right] \right)\\ &=& \frac{m \omega}{2} \left( \left[ \cos(t_a + t_b) \sin(T) \right] (B^2 - A^2) - 2AB \left[ \sin(T) \sin(t_a + t_b) \right] \right)\\ &=& \frac{m \omega}{2} \sin(T) \left( \cos(t_a + t_b) (B^2 - A^2) - 2AB \sin(t_a + t_b) \right) \end{eqnarray}Now we need to figure out what these quantities correspond to in terms of the boundary conditions, $x_a$ and $x_b$. Exploiting the identity $\cos^2 + \sin^2 = 1$ seems like a possible venue to get from $x_a$, $x_b$ to some sum of squares of the amplitudes $A$ and $B$, so let's look at the values of those squares.
\begin{eqnarray} x_i^2 &=& (A \cos(\omega t_i) + B \sin(\omega t_i))^2 \\ &=& \left[ A^2 \cos^2(\omega t_i) + B^2 \sin^2(\omega t_i) + 2AB \cos(\omega t_i) \sin(\omega t_i) \right] \end{eqnarray}We also want our results to be independent of $t_a$ and $t_b$ by themselves, since the action isn't time dependent (and therefore time translation invariant), and only on the time interval $T = t_b - t_a$. Let's see what the sum of those squares look like, as sums of products of trigonometric functions will in general lead to the sum and difference of their terms.
\begin{eqnarray} x_a^2 + x_b^2 &=& \left[ A^2 (\cos^2(\omega t_a) + \cos^2(\omega t_b)) + B^2 (\sin^2(\omega t_a) + \sin^2(\omega t_b)) + 2AB (\cos(\omega t_a) \sin(\omega t_a) + \cos(\omega t_b) \sin(\omega t_b)) \right] \end{eqnarray} \begin{eqnarray} \cos^2(\omega t_a) + \cos^2(\omega t_b) &=& \frac{1}{2} (2 + \cos(2 \omega t_a) + \cos(2\omega t_b))\\ &=& 1 + \cos(\omega (t_a + t_b)) \cos(\omega T) \end{eqnarray} \begin{eqnarray} \sin^2(\omega t_a) + \sin^2(\omega t_b) &=& \frac{1}{2} (2 - \cos(2 \omega t_a) - \cos(2\omega t_b))\\ &=& 1 - \cos(\omega (t_a + t_b)) \cos(\omega T) \end{eqnarray} \begin{eqnarray} x_a^2 + x_b^2 &=& (A^2 + B^2) + \cos(\omega T) [A^2 \cos(\omega (t_a + t_b) - B^2 \cos(\omega(t_a + t_b))\\ && + AB (\sin(2 \omega t_a) + \sin(2 \omega t_b)))]\\ &=& (A^2 + B^2) + (A^2 - B^2) \cos(\omega T) \cos(\omega (t_a + t_b)) + 2AB \sin(\omega (t_a + t_b)) \cos(\omega T) \end{eqnarray}We still need some term to absorb the unwanted terms that we have accumulated here. If we want a term of the same dimension, and to try the simplest case, the most obvious candidate is $x_a x_b$ :
\begin{eqnarray} x_a x_b &=& (A \cos(\omega t_a) + B \sin(\omega t_a)) (A \cos(\omega t_b) + B \sin(\omega t_b))\\ &=& A^2 \cos(\omega t_a) \cos(\omega t_b) + B^2 \sin(\omega t_a) \sin(\omega t_b) + A B (\cos(\omega t_a) \sin(\omega t_b) + \cos(\omega t_b) \sin(\omega t_a))\\ &=& \frac{A^2}{2} (\cos(\omega T) + \cos(\omega (t_a + t_b))) + \frac{B^2}{2} (\cos(\omega T) - \cos(\omega(t_a + t_b)))\\ && + AB \sin(\omega (t_a + t_b))\\ &=& \frac{1}{2} (A^2 + B^2) \cos(\omega T) + \frac{\cos(\omega (t_a + t_b))}{2} (A^2 - B^2) + AB \sin(\omega (t_a + t_b)) \end{eqnarray}All the important terms that we want here only differ from $x_a^2 + x_b^2$ by a factor of $\cos(\omega T) / 2$. If we consider the following quantity :
\begin{eqnarray} (x_a^2 + x_b^2) \cos(\omega T) - 2 x_a x_b &=& (A^2 + B^2)\cos(\omega T) + (A^2 - B^2) \cos^2(\omega T) \cos(\omega (t_a + t_b)) + 2AB \sin(\omega (t_a + t_b)) \cos^2(\omega T)\\ && - (A^2 + B^2) \cos(\omega T) \cos(\omega (t_a + t_b)) (A^2 - B^2) - 2 AB \sin(\omega (t_a + t_b))\\ &=& (A^2 - B^2) \cos(\omega (t_a + t_b)) (\cos^2(\omega T) - 1) + 2AB \sin(\omega (t_a + t_b)) (\cos^2(\omega T) - 1)\\ &=& (A^2 - B^2) \cos(\omega (t_a + t_b)) \sin^2(\omega T) + 2AB \sin(\omega (t_a + t_b)) \sin^2(\omega T) \end{eqnarray}This is the quantity that we found for the action, up to a factor of $\sin(\omega T)$. Therefore we can write our action as
\begin{equation} S_{cl} = \frac{m\omega}{2\sin(\omega T)} [(x_b^2 + x_a^2) \cos(\omega T) - 2 x_b x_a] \end{equation}Problem 2-3 : Find $S_{cl}$ for a particle under a constant force $f$, that is, $L = (m/2) \dot{x}^2 + fx$.
Solution : The equation of motion is trivially what the problem says it to be, a particle under a constant force :
\begin{equation} m \ddot{x}(t) = f \end{equation}As it is a constant force, it can simply be integrated :
\begin{eqnarray} \ddot{x}(t) &=& \frac{f}{m}\\ \dot{x}(t) &=& \frac{f}{m} t + v_0\\ x(t) &=& \frac{f}{2m} t^2 + v_0 t + x_0 \end{eqnarray}What is the value of $v_0$ as expressed by our boundary conditions? Unlike the free case here, this isn't gonna be trivially computed, since the velocity at $t_a$ and $t_b$ is not the same, due to the acceleration in between.
\begin{eqnarray} v_a &=& \frac{f}{m} t_a + v_0\\ v_b &=& \frac{f}{m} t_b + v_0 \end{eqnarray}The gain in speed here is linear in the acceleration,
\begin{eqnarray} \Delta v = v_b - v_a = \frac{f}{m} T \end{eqnarray}Now let's see the difference in position
\begin{eqnarray} x(t_b) - x(t_a) = \frac{f}{2m} (t_b^2 - t_a^2) + v_0 (t_b - t_a) \to v_0 = \frac{x_b - x_a}{T} - \frac{f}{2m} (t_b + t_a) \end{eqnarray} \begin{eqnarray} x(t_b) + x(t_a) = \frac{f}{2m} (t_b^2 + t_a^2) + v_0 (t_b + t_a) + 2 x_0 \end{eqnarray}