# Can we derive Schrödinger equation from classical wave equation?

In classical mechanics wave equation is $$y=A\sin(kx-\omega t)$$

$$y$$=instantaneous displacement, $$A$$=maximum displacement, $$\omega$$=angular velocity, $$x$$=position of particle, $$k$$=wave number

Now in quantum mechanics, wave equation is $$\Psi=A e^{j(kx-\omega t)}$$

Also, $$\sin x= \frac{(e^{jx}- e^{-jx})}{2j}$$ j=imaginary number;

I can see some sort of resemblance in classical and quantum wave equation, can we derive quantum wave equation $$\Psi=A e^{(kx-\omega t)}$$ using $$y=A\sin(kx-\omega t)$$?

I have knowledge of classical mechanics up to high school, I have just begun studying quantum mechanics.

• Jan 29 '19 at 14:49
• That is not the wave equation--it is the equation of a wave. Google: wave equation. Jan 29 '19 at 17:30
• Apparently you are insufficiently familiar with the concept of a wave equation. What you show is the solution to a wave equation. Jan 30 '19 at 8:45
• No time to write a full answer, but you should note that the Hamilton-Jacobi equation is highly related to the Schrodinger equation.
– zh1
Jan 30 '19 at 14:26
• This question might be of interest to you physics.stackexchange.com/q/218983/94257 Jan 30 '19 at 14:28

## 3 Answers

The Schrödinger’s equation is a differential equation that does not have necessarily the form you pointed out as a solution.

Also, the expression you wrote down is the complex analog of what you called “the classical wave equation”. With this, I mean that the “classical one” is the imaginary part of the “Schrödinger one”. If you go from one to the other, you are not deriving, but rather going from a Real space (real numbers) to a Complex one (complex numbers)

• The (free-particle) Schrödinger equation is not a wave equation. It is a diffusion equation that admits plane-wave solutions. Mar 13 '19 at 0:44

As a simple matter of a general rule, you cannot derive a more fundamental theory from a less fundamental theory using any form of deductive logic. So no, you cannot possibly derive the Schrödinger equation from the classical wave equation. If you see anyone claiming to have done such witchcraft, beware of them.

So, the question remains (and always was) as to whether the resemblance between $$y=A\sin(kx-wt)$$ and $$\Psi=Ae^{i(kx-wt)}$$ is purely accidental or it can be understood more closely. Now, first of all, this question doesn't translate to the question you ask in the title of your question. Because, $$y=A\sin(kx-wt)$$ is not the classical wave equation and neither is $$\Psi=Ae^{i(kx-wt)}$$ the Schrödinger equation. They are specific solutions to the respective equations.

In particular, $$\Psi=Ae^{i(kx-wt)}$$ is a solution to the free particle Schrödinger equation. The Schrdödinger equation, in the position basis, reads $$i\hbar\frac{\partial \Psi(x,t)}{\partial t}=-\frac{\hbar^2}{2m}\frac{\partial ^2\Psi(x,t)}{\partial x^2}+V(x)\Psi(x,t)$$. I will omit the discussion about the time-independence of the potential $$V(x)$$ and the caveats of such a claim, because, the case we are interested in is that of a free particle where the potential simply vanishes for all positions and for all time. So, the free-particle Schrödinger equation reads $$\frac{\partial \Psi(x,t)}{\partial t}=i\frac{\hbar}{2m}\frac{\partial ^2\Psi(x,t)}{\partial x^2}$$. This is a diffusion equation (not a wave equation) with a purely imaginary diffusion coefficient. Such an equation admits plane-wave solutions of the form $$\Psi=Ae^{i(kx-wt)}+Be^{i(-kx-wt)}$$ where the dispersion relation reads $$w(k)=\frac{\hbar k^2}{2m}$$, Here, $$k$$ is related to energy as $$E=\frac{\hbar^2k^2}{2m}$$. The $$e^{i(kx-wt)}$$ part of the solution has a definite momentum $$p=\hbar k$$ and the $$e^{i(-kx-wt)}$$ has a definite momentum $$p=-\hbar k$$. So, a plane-wave solution with a specific energy $$E$$ and a specific momentum $$p=\hbar k$$ is simply $$\Psi=Ae^{i(kx-wt)}$$--the solution that you are interested in.

On the other hand, $$y=A\sin(kx-wt)$$ is a solution to the classical wave equation $$\frac{\partial^2y(x,t)}{\partial t^2}=c^2\frac{\partial ^2y(x,t)}{\partial x^2}$$ where $$c$$ is a positive constant. This equation admits plane-wave solutions of the form $$y(x,t)=Ae^{i(kx-wt)}+Be^{i(kx+wt)}$$ with the dispersion relation $$w(k)=|k|c$$. This is a solution with a specific wave-number $$k$$ and thus, a specific angular frequency $$w(k)$$. The $$e^{i(kx-wt)}$$ part of the solution has the velocity of propagation $$\frac{kc}{|k|}$$ and the $$e^{i(kx+wt)}$$ part has the velocity of propagation $$-\frac{kc}{|k|}$$. So, a plane-wave solution with a specifc wave-number $$k$$ and a specific direction of propagation $$\frac{k}{|k|}$$ is simply $$y(x,t)=Ae^i(kx-wt)$$--the purely imaginary part of which is $$A\sin(kx-wt)$$--the solution that you are interested in.

So, the resemblance that we set out to explore arises out of the fact that the free-particle Schrödinger equation, which is a diffusion equation with a purely imaginary diffusion constant, admits plane-wave solutions just like a classical wave equation does. There is, however, a crucial difference between the two seemingly resembling solutions: their dispersion relation. In particular, the free-particle solution of the Schrödinger equation has a quadratic dispersion relation while the solution of the classical wave-equation has a linear dispersion relation.

So, despite both of them being wave solutions, we can see that they are structurally quite different as the structure of a wave-solution is contained in its dispersion relation.

Yes, you can reformulate some classical wave equations in the form of a Schrödinger equation in the mathematical sense: to be precise, you can find a hermitian/selfadjoint operator $$H$$ so that the original wave equation is equivalent to $$\mathrm{i} \partial_t \psi(t) = H \psi(t)$$. Here, $$\psi(t)$$ is a complex wave representing the real wave $$u(t) = 2 \mathrm{Re} \, \psi(t)$$. Note that $$\psi$$ does not have anything to do with probabilities here. One of my recent papers spells this out in detail; in case you can't access Annals of Physics articles from where you are, you can also grab the arxiv version.