# Interesting proof of the time-independent Schrödinger equation

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.

• Do you understand that the SE is referred to as a wave equation? What is it specifically that you don't understand? For example, what do you think is wrong with separating the function into a spatial and temporal part? Sep 21, 2021 at 9:52
• @josephh That's a misleading comment. The SE is a diffusion equation with a purely imaginary diffusion coefficient, not a wave equation.
– user87745
Sep 21, 2021 at 10:10
• The TISE Schrödinger equation is derivable from, and is analogous to, a wave equation, and is not a diffusion equation. The SE in its original form (the one you refer to) looks like a diffusion equation, but contains the imaginary number 𝑖 that allows oscillatory solutions. Does the diffusion equation allow oscillatory solutions? Sep 21, 2021 at 10:58
• Oops, I named the wrong author. But the comment still applies: For the excerpt shown in the question, is the author claiming to have derived quantum mechanics from classical field theory? It looks to me like the author is just making an analogy, using the mathematical form of the equation and borrowing some mathematical intuition about it from the other physical context. Sep 21, 2021 at 13:05
• @ChiralAnomaly Um.. first the author does not explain any classical field theory before the above derivation. He just explains that there is so called the classical wave equation and I could not find any contextual meaning of using such equation in the quantum mechanics. And yeah, I think showing such mathematical intuition might be the intention of author, but I'd like to know if such argument could be justified in some sense or not(i.e. making correspondence btw classical and quantum equation explicitly). Sep 21, 2021 at 13:13

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:

• the de Broglie relation.
• that a time independent quantum system can be described by a wave equation.

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.

• $\psi$ doesn't have to be a standing wave. E.g. $\exp(ikx)$ is a running wave, and it's a solution of both the hyperbolic wave equation and the Schrödinger's equation for a free particle (up to temporal part). Sep 22, 2021 at 12:47

It is not a proper derivation. It was probably meant to show that intuitively the Schrödinger equation is similar to the classical wave equation.

In Principles of Quantum Mechanics, chapter 4 -The Postulates a General Discussion, Shankar mentions several postulates of quantum mechanics and compares them with postulates of classical mechanics. The 4th postulate is

IV) The state vector $$|\psi(t)\rangle$$ obeys the Schrödinger equation $$i\hbar\frac{d}{dt}|\psi(t)\rangle=H|\psi(t)\rangle$$ where $$H(\hat{x}, \hat{p})= \mathcal{H}(x\to\hat{x}, p\to\hat{p})$$ is the quantum Hamiltonian operator and $$\mathcal{H}(x,p)$$ is the Hamiltonian for the corresponding classical problem.

They are indirectly using the above postulate and the classical wave equation to "intuitively derive" the Schrödinger equation.

A more proper derivation is done using the Dirac-von Neumann axioms.

Why? Why should I? I have no apriori reason to believe that the state vector of a quantum mechanical system ought to obey something akin to a wave equation. This author is simply one of the snake-oil salesmen who claim to derive quantum mechanics from classical mechanics as they tell you a story about how they once had a beer with a bigfoot. The issue with all such attempts is that they make assumptions that are completely unnatural and unjustified unless you already accept the results of quantum mechanics as known -- in which case, the need for a "derivation" is self-defeated.

Also, the author seems confused about how the Schrodinger equation and the classical wave equation are related. The former is a diffusion equation with a purely imaginary coefficient, not a wave equation. Both admit plane wave solutions but their dispersion relations are completely different. See What is the difference between solutions of the diffusion equation with an imaginary diffusion coefficent and the wave equation's?.

• Sep 22, 2021 at 13:02
• @Ruslan Thanks for the comment. Yeah, I don't object to @DavidZ's terminology that any equation that admits wave-solutions is a wave-equation (and then there are various wave equations with different dispersion relations) but the author cited by OP is explicitly using "the wave equation" which is not the kind of wave equation that the Schrodinger equation is.
– user87745
Sep 22, 2021 at 13:04
• Both the hyperbolic wave equation and the free-particle Schrödinger's equation have Helmholtz equation as an eigenmode equation. This makes them very closely related, and justifies the analogy. (No one is claiming that it's a rigorous derivation.) Sep 22, 2021 at 13:07
• @Ruslan My objection is not about rigor, my objection is that it is not a derivation at all. Like you say, it is a fair enough and perhaps pedagogically useful analogy to help a student familiarize themselves with the relevant mathematical objects. But the OP has claimed that it is presented as a derivation. That's my issue.
– user87745
Sep 22, 2021 at 13:09

Math
In strictly mathematical terms the proof is incorrect:
The first part of the proof uses separation of variables, by writing the solution as $$u(x,t)=\psi(x)f(t),$$ which allows us to write $$\frac{1}{\psi(x)}\frac{\partial^2\psi(x)}{\partial x^2}= \frac{1}{v^2 f(t)}\frac{\partial^2f(t)}{\partial t^2}=const$$ The constant is then replaced by an expression explicitly dependent on $$x$$... OOPS!!!

Intuition
Now, if we considered a particle moving in a constant potential (i.e., if $$V(x)$$ were independent on $$x$$), the proof would work. But in this case it is merely an elaborate mathematical restatement of the De Broglie hypothesis that a particle can be associated with a wave of certain frequency and wave-length: indeed, as soon as we are talking about a wave, it is very likely that it will be described by the wave equation (although there are exist some more complex equations resulting in wave-like solutions).

Helmholtz equation
Equation $$\nabla^2\psi(x)=-k^2\psi(x)$$ is known as Helmholtz equation. It arises whenever we are dealing with an equation, whose coordinate part is expressed by the Laplace operator $$\nabla^2$$, and apply the separation of variables. Specifically, we could get it for $$\nabla^2u=\frac{1}{v^2}\frac{\partial^2 u}{\partial t^2} \text{ (wave equation)},\\ \nabla^2\psi=i\frac{\partial \psi}{\partial t} \text{ (Schrödinger equation)},\\ \nabla^2u=D\frac{\partial u}{\partial t} \text{ (Diffusion equation)},\\ \nabla^2\zeta=i\frac{\partial^7 \zeta}{\partial t^7} + Q \frac{\partial^5 \zeta}{\partial t^5}\text{ (Vadim's equation)},$$ and so on.