Deriving particle's wave equations

Question

My book tries to derive two equations. Unfortunately the book is in a language you wouldn't understand, so I will try to translate it as much as I can.

• $E = ℏw$
• $p = ℏk$

It starts by lagrangian:

$ S = \int_{t_0}^{t_1} L(x, y, \dot x, \dot y, t) dt$

Then the book writes down:

$ p = \nabla s$ (eq 1)
$ S(x,y, t) = S_0$ (eq 2)
$ \frac{\partial S}{\partial t} + H(r, \nabla s, t) = 0$ (eq 3)
$ H = \frac{1}{2m}(\nabla s)^2 + V(r,t)$ (when problem is stationary, then we get:)
$H = E$
(eq 4) $E = -\frac{\partial S}{\partial t}$ (eq 5)

To be honest, I didn't understand a single thing here. Let me write down my questions.

1. I don't know why eq:1 is true. I believe gradient of $S$ is partial derivatives of $S$ with respect to $\dot x$, $\dot y$. I belive we're describing the scenarios where Kinetic energy is $\frac{1}{2}m\dot x^2$, $\frac{1}{2}m\dot y^2$. In which case their partial derivatives are $m\dot x$ and $m\dot y$ which are momentums in $x$ and $y$ axis. If I'm correct till now, now question, why do we the gradient over the $S$ ? if we had written $p = \nabla L$, I would understand but doing it on $S$ must be giving us momentums summed up over the whole path ? what advantage do we get from this ? It gets used in (eq 3), but it seems wrong to put the whole momentum summed up over the whole path for calculating total energy as (eq 4) is dependent on time and plugging in time, it gives tota energy at that specific time, but by $\nabla s$, we're putting the total momentum over the whole path summed up. Doesn't make sense.
2. I couldn't fully understand what hamiltonian is, but I believe it's the $KE + U$ (total energy). but, don't see (eq 3) idea at all.

I might be asking more than one question, but I hope it's possible to explain the all the equations in a clear way.

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Update

Q1: In the following: $\frac{\partial S}{\partial t_1}$, we ask how much S changes when we change $t_1$. This basically means we're not changing the path in variation, but we're just saying what if we had started the path from $t1+\delta h$ which means the remaining path is the same. Correct ?

Q2: I wonder how we derived $\frac{\partial S}{\partial t_1} = H(t1). My idea is as follows - we know that $H = vp - L$ from which we have: $L = vp - H$.

I also understand that $\frac{\partial S}{\partial t_1} = -L(t1)$ from calculus. So we have:

$\frac{\partial S}{\partial t_1} = -(vp - H)(t1)$

It seems like $vp$ is considered 0 here. Is this because at time $t1$, the speed is 0 ? but if we're observing the motion from $t=2$ to $t=5$, at $t=2$, speed wouldn't be 0. So why and how did we make it 0 ?

Question 3: In most of the cases, Hamiltonian is total energy which is usually conserved. So at each point on the path, particle has the same total energy which is $H$ as well.

We derived:

$H = E = \frac{\partial S}{\partial t_1} = -\frac{\partial S}{\partial t_2}$.

This is clear, but my book says:

$E = -\frac{\partial S}{\partial t}$ note here that we're doing with respect to $t$ and not the initial or final point.

Where did we get this from rigorously ?

Question 4 I realize now book must be correct. I even asked this to the professor. $p$ must be the gradient of $S$, not only at initial/final points, but overally everywhere as long as potential energy is not dependent on velocity. What would be the proof though ?

Answer 1

It starts by lagrangian:

$ S = \int_{t_0}^{t_1} L(x, y, \dot x, \dot y, t) dt$

Then the book writes down:

$ p = \nabla s$ (eq 1)

This Eq. (1) must be taken to mean that the momentum at the endpoint is equal to the gradient of the classical action with respect to the endpoint position.

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Consider, for example, a free particle in one dimension. The action is: $$ S[x(t)] = \int_{t_1}^{t_2} \frac{1}{2}m\dot{x}^2 dt\;. $$

By varying the action with fixed endpoints ($\delta x(t_1)=\delta x(t_2) = 0$) we find the classical path to be a straight line: $$ x_{cl}(t) = x_1 \frac{(t-t_2)}{(t_1 - t_2)} + x_2 \frac{t-t_1}{t_2 - t_1} $$ and note that $$ \dot x_{cl}(t) = \frac{x_2 - x_1}{t_2 - t_1} $$ and further note that $$ p_2 \equiv m\dot x_{cl}(t_2) = \frac{m(x_2 - x_1)}{t_2 - t_1} $$

The classical action can be evaluated in this example to find: $$ S_{cl} = S[x_{cl}(t)] = S[x_1 \frac{(t-t_2)}{(t_1 - t_2)} + x_2 \frac{t-t_1}{t_2 - t_1}] $$ $$ =\frac{m}{2}\frac{(x_2 - x_1)^2}{t_2 - t_1}\;.\tag{A} $$ Eq. (A) above shows that the classical action (the action evaluated for a specific path, the classical path) is a function of four variables: the two time endpoints and the two space endpoints.

Using Eq. (A), we find: $$ \frac{\partial S_{cl}}{\partial x_2} = \frac{m(x_2 - x_1)}{t_2 - t_1} = p_2\;, $$ as required.

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Of course, the above example and Eq. (A) are only true in the simple free particle case. However, it turns out (but is left as an exercise to the reader to show) that in the general one-dimensional case we can still write: $$ \frac{\partial S_{cl}}{\partial x_2} = \left.\frac{\partial L}{\partial \dot x}\right|_{x_{cl}}(t_2) = p_2 $$ and also $$ \frac{\partial S_{cl}}{\partial x_1} = -\left.\frac{\partial L}{\partial \dot x}\right|_{x_{cl}}(t_1) = p_1. $$

(Hint: To prove this consider variations about the classical path where one of the endpoint variations is not zero.)

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The generalization to three dimensions is: $$ \vec \nabla_2 S_{cl}\equiv\frac{\partial S_{cl}}{\partial \vec x_2} = \left.\frac{\partial L}{\partial \dot{\vec{x}}}\right|_{\vec x_{cl}}(t_2) $$

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To be honest, I didn't understand a single thing here. Let me write down my questions.

I have answered one of the questions. It is too much to ask to answer all your questions here. But I hope that the above answer to where your Eq 1 comes from will set you on the right path.

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While I can not go into detail about your other questions, I will note that the (Hamiltonian) energy $H$ is defined (in one dimension) as: $$ H = \dot x p - L = \dot x \frac{\partial L}{\partial \dot x} - L $$ and further, by varying the initial time $t_1$ (along with a corresponding variation in path to make sure the path remains the classical path), it can be shown that: $$ \frac{\partial S_{cl}}{\partial t_1} = H(t_1)\;. $$ and by varying the final time, it can be shown that: $$ \frac{\partial S_{cl}}{\partial t_2} = -H(t_2)\;. $$

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For many systems the Hamiltonian energy $H$ is the same as the total mechanical energy $E$. This is not always true, but often true. Some other answers on this site discuss the specific conditions that must hold for $H$ to equal $E$. But, suffice it to say that whenever the potential $V$ is independent of the velocity $\dot x$ (which is often the case) and also when the kinetic energy $T$ is a homogeneous quadradic function of the velocity, we can write: $$ H(t_2) = E(t_2) = -\frac{\partial S_{cl}}{\partial t_2}\;, $$ that is, the Hamiltonian energy is equal to the total energy, which is equal to the negative of the derivative of the classical action with respect to time.

When the Hamiltonian energy and total mechanical energy are conserved (constant in time) we can drop the time argument and write: $$ H = E = -\frac{\partial S_{cl}}{\partial t_2}=\frac{\partial S_{cl}}{\partial t_1}\;, $$

For example, in our simple free partial example $$ \frac{\partial }{\partial t_1}\left(\frac{m}{2}\frac{(x_2 - x_1)^2}{t_2 - t_1}\right) = \frac{m}{2}\frac{(x_2 - x_1)^2}{(t_2 - t_1)^2} = \frac{1}{2}m\dot x_{cl}^2 = E_{free} $$

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Reference for you: Feynman and Hibbs, Quantum Mechanics and Path Integrals, Chapter 2-1.