I'm reading the McQuarrie_Physical Chemistry textbook Chapter.3 part. In here, author derives the time-independent Schrödinger equation using classical one-dimensional wave equation, which makes me feel weird! Let me first show the derivation in this textbook.
Start with the classical one-dimensional wave equation
$$ \frac{\partial^2u}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2u}{\partial t^2}$$
This equation can be solved by the method of separation of variables and we will express the temporal part as $cos(wt)$(what?) and write $u(x,t)$ as
$$u(x,t) = \psi(x)cos(wt)$$
Substitute it into the classical equation and we obtain an equation for $\psi(x)$.
$$ \frac{d^2 \psi}{d x^2} + \frac{w^2}{v^2}\psi = 0$$
Using the fact that $w=2\pi\nu$ and $\nu\lambda=v$,
$$ \tag{*} \frac{d^2 \psi}{d x^2} + \frac{4\pi^2}{\lambda^2}\psi = 0 $$
Since the total energy of a particle is $E=\frac{p^2}{2m} + V(x)$, we find $p=(2m[E-V])^{1/2}$. According to de Broglie formula, $\lambda = \frac{h}{p} = \frac{h}{(2m[E-V])^{1/2}}$ holds. Let's substitute this into (*) and we find
$$\frac{d^2 \psi}{d x^2} + \frac{2m}{\hbar^2}(E-V)\psi = 0 $$
This can be rewritten in the form, which is so called the Time Independent Schrödinger equation.
$$-\frac{\hbar^2}{2m}\frac{d^2 \psi}{d x^2} +V(x)\psi(x) = E\psi(x)$$
This derivation is something very different from what I've learned in usual undergraduate level quantum mechanics. Like the description in Griffiths, I usually set the time depenent Schrödinger equation as a postulate. Then using separation of variables, I could derive the Time independent Schrödinger equation.(Related post) But this textbook starts derivation with classical wave equation and later author explains the time dependent case.
So my question is this. What would be the proper manner to understand or encapsulate the above derivation? Is this just a piece of 'old quantum' theory? Is this derivation really invoked by someone historically in the development of QM-theoretical foundation or does this just hold for some special case?(like expressing the temporal part as $cos(wt)$)
Reference
McQuarrie, D. A.; Simon, J. D. Physical Chemistry: A Molecular Approach; University Science Books: Sausalito, California, 1997. ISBN 978-0-935702-99-6.
Best Answer
The derivation is nice, because it shows the similarity between the time-independent Schrödinger equation and a standing-wave equation. I would like to point out some aspects of this derivation that may clarify its meaning and limitations.
First of all, this derivation has not derived the Schrödinger equation from classical mechanics. It is still introducing some new postulates, just they are different from Griffiths'. Namely, the authors are postulating:
Additionally, they will have to add some postulate to justify the time dependent Schrödinger equation.
Therefore I think that this is just a matter of taste. Griffiths needs fewer postulates, he abruptly gives the correct equation. Here they work slower towards the goal, but trying to give a better physical intuition of what is going on.
That said, let's talk about the derivation and what it means. It starts from the wave equation in one dimension. When talking about the wave equation one usually means that the velocity $v$ is a constant. It may be the speed of light in vacuum, the speed of sound in a particular medium, but it is a number fixed beforehand. Yet, it may be meaningful and useful to consider also cases where the speed varies with position $v=v(x)$. For example, if a sound wave moves from a region of colder air to a region of hotter air, its speed will increase. In this case, we consider an equation where the velocity is given by the energy relation of a point particle moving in a potential:
$$E = {1 \over 2}m v^2 + V(x) \implies v(x) = \sqrt{{2 \over m}(E-V(x))}$$
Imagine you have a point particle with energy $E$ in a potential $V(x)$. The velocity of the wave in every point will be the same as the velocity of the particle. Notice that this reasoning introduced a new parameter to the equation: the energy. Now, waves that have different energy will propagate at different speeds! This means that the waves will be dispersive. They will have a different phase velocity and group velocity.
We have now our wave equation and we would like to solve it
$${\partial^2 u \over \partial x^2} = {1 \over v(x)^2}{\partial ^2 u \over \partial t^2}$$
It will have many general solutions. It is possible that a subset of all solutions can be written as $u(x,t) = \psi(x)\cos(\omega t)$. It is also possible that no solution exists in this form. Or that it exists only form some value of $\omega$. We are free to plug this into the equation and verify whether any solution like this exists. Of course, the authors already know that this is the correct substitution, but this is in general a common method to solve differential equations. You just guess a solution and plug it in to see if you were right.
In this case, we plug the solution in the equation and find a new equation for $\psi$. $\psi$ doesn't depend on time, therefore our solution (if it exists) will be a standing wave.
Now you have an equation for $\psi$ that depends on two parameters: $\omega$ and $E$. Using the de Broglie relation you find out that $E=\hbar \omega$. Also, when trying to solve the equation for $\psi$ you find that not all values of $\omega$ (i.e. of $E$) allow a solution. You have found that a wave described by this equation cannot have any frequency (energy). The energy is quantized.