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There is a very old equation known as the "wave equation". It's an ordinary classical non-relativistic differential equation which applies to just about every kind of ordinary wave you can imagine (fluid dynamics, acoustics, mechanical waves, etc). It can be derived from Hooke's law simply by imagining a sequence of beads on a string, where each bead has the same mass. In the continuum limit of a string with constant tension and mass per unit length, such as that on a violin, you get this: $$ \displaystyle {\frac {\partial ^{2}u}{\partial t^{2}}}=c^{2}{\frac {\partial ^{2}u}{\partial x^{2}}} $$ where $c = \sqrt{T/\rho}$ is the velocity of traveling waves on the string, T is the tension of the string, and $\rho$ is the mass per unit length of the string.

This equation was discovered by d’Alembert in 1746 (see, eg. https://en.wikipedia.org/wiki/Wave_equation) The 3D version of this, for ordinary fluids, mechanical waves, etc. is: $$ \displaystyle {\frac {\partial ^{2}}{\partial t^{2}}}u -c^2\mathbf {\nabla } ^{2}u =0. $$

Fast forward 150+ years, and we have Einstein's theory of special relativity, which says that the energy of any body in motion is: $$\displaystyle E = \sqrt{(pc)^2 + (mc^2)^2}$$

Moving forward a few more decades, we can combine this with the assumption from quantum mechanics that momentum should not be viewed as an ordinary real number, but instead a non-commuting functional operator in a complex Hilbert space $ \displaystyle p = -i\hbar \frac{d}{dx} $. This gives us the Klein-Gordon equation, the relativistic version of the Schrodinger Equation:

$$ \displaystyle {\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}\psi -\mathbf {\nabla } ^{2}\psi +{\frac {m^{2}c^{2}}{\hbar ^{2}}}\psi =0 $$ For the ultra-relativistic case when $E >> m$, this reduces to exactly the classic wave equation we derived from Hooke's law: $$ \displaystyle {\frac {\partial ^{2}\psi}{\partial t^{2}}}=c^{2}{\frac {\nabla ^{2}\psi}{\partial x^{2}}} $$

My question is: why does the non-relativistic non-quantum wave equation for a string on a violin look more like the wave equation for a relativistic quantum particle than a non-relativistic quantum particle? Is that purely coincidence, or could it be seen as an early hint of the equations of relativity showing up in the 1700's? Without working out the math, a priori I would have guessed that the wave equation for a non-relativistic string would look more like the Schrodinger Equation than Klein-Gordon, since relativity is not involved, ie there is no reason to expect it to have a Lorentz invariant form: $$ \displaystyle i\hbar {\frac {\partial \psi }{\partial t}}=-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi. $$

Also, it seems interesting that this equation was first derived when thinking about the motion of a 1-dimensional string. This is a long shot, but... does this by any chance relate to the fact that quantum gravity turned out to have a consistent embedding in a quantum mechanics of strings, but not in a quantum mechanics of particles?

reductionista
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    Everything before the last paragraph is a good description of how one system without gravity looks a lot like another system without gravity. So the last paragraph is a long shot indeed! – Connor Behan Nov 16 '21 at 18:54
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    Related, possibly addresses the question: Why does nature favor the Laplacian? In particular, the second part of knzhou's answer there is closely related to why the same operator pops up in both places. – Michael Seifert Nov 16 '21 at 18:54
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    As a very simple note: I guess asking for the time reversibility of your wave equation is reasonable. To satisfy this, you need an even order derivative in time (if you don't go in the realm of imaginary number and complex conjugates anyway, which is plausible for classical mechanics), so something which would look like a schrodinger equation would be a diffusion equation and won't be invariant by a time reversal transformation. – Syrocco Nov 16 '21 at 18:55
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    Klein Gordon is NOT the relativistic Sch EQ , it's just the equation of motion for a (classical) massive scalar field. Remove the mass term (that is more like a quadratic self interaction of the field) and you have the same classical framework that gives rise to the wave equation... nothing exotic. – Quillo Nov 16 '21 at 19:08
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    @Quillo It most certainly is. "The Klein–Gordon equation was first considered as a quantum wave equation by Schrödinger in his search for an equation describing de Broglie waves. The equation is found in his notebooks from late 1925, and he appears to have prepared a manuscript applying it to the hydrogen atom. " https://en.wikipedia.org/wiki/Klein–Gordon_equation#History – my2cts Nov 16 '21 at 19:47
  • Klein-Gordon can be interpreted as arising directly as the Schrodinger equation for a quantized classical point particle in proper time $\tau$: $i \frac{d}{d\tau} \psi = \hat{H} \psi = 0$ where $\hat{H} = \hat{p}^2 - m^2$, from a constrained quantization approach. See section 3 here. – bolbteppa Nov 16 '21 at 20:02
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    Possible duplicate: Why is there an emergent quasi Lorentz symmetry in classical string lagrangian? – Qmechanic Nov 16 '21 at 22:59
  • @Syrocco: I'm not sure I see intuitively why it has to be time-symmetric. Why is it okay for the Schrödinger wave equation to be time-asymmetric but not the classical wave equation? – reductionista Nov 17 '21 at 01:03
  • @Quillo I guess that's mostly true... KG is an on-shell equation describing the motion of a massive scalar field. But if you're going to interpret it as "massive" that means you're assuming wave-particle duality... ie, that this is not just a field, but a particle with a mass. And more importantly... it's specifically a relativistic equation, it's not the equation for a non-relativistic massive scalar field which would look different. So why did an eq for a relativistic scalar field show up in the 1700's? – reductionista Nov 17 '21 at 01:21
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    Oh! @Quillo's answer leads me to reallize: maybe the simplest answer is that the connection has to do with Maxwell's equations being Lorentz invariant, which is actually how relativity was discovered. ie, electromagnetic waves satisfy a massless version of the Klein Gordan equation. So, perhaps the only real mystery here is why the differential equation for electromagnetic waves happened to be the same as the differential equation for non-relativistic mechanical waves. I guess it's true then, that this is not about quantum mechanics. – reductionista Nov 17 '21 at 01:30
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    Probably relevant: a discussion of what equations describe waves – Roger V. Nov 19 '21 at 09:18

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With one time dimension and one space dimension, the wave equation's most general solution is $f(t-x/c)+g(t+x/c)$ with $f,\,g$ twice-differentiable, i.e. each solution is a sum of two speed-$c$ waves, one moving right, the other left. Equivalently, the former's value at a time $t$ depends on the behaviour of the wave's source a time $x/c$ earlier, so is a "retarded" solution, while the latter is an "advanced" solution.

I've moved from the maths to the physics, but we can just as easily go the other way. If we only want the retarded them, a solution is of the form $f(t-x/c)$, and the PDE this fits is $u_t=-u_x/c$. But as soon as we work in multiple space dimensions, e.g. because we're thumping a drum skin, the requirement that both sides are scalars effects a generalization such as $u_t=-\hat{e}\cdot\nabla u/c$, with $\hat{e}$ an arbitrary unit vector. Too arbitrary. It picks out a privileged direction in space. (We can fix that with a spacetime-unifying idea like $\gamma^\mu u_\mu=0$, but let's put that aside for a moment.) If we use second-order derivatives too we avoid this problem, viz. $u_{tt}=u_{ii}$. The only downside is there are now advanced solutions too. (The general selection in multiple space dimensions looks a fair bit more complicated, but the principle is similar; we just need a $k$-space integral instead of a $2$-term sum.)

But one of the most important ideas in science is there's more than one place you can start your theory. We're not choosing axioms, then seeing which theorems they give us, like a mathematician with a preferred set theory. The universe just "is", all at once. So let's do something very ahistorical: suppose you tried to write down a relativistic quantum field theory without first knowing what classical relativity looks like. I know that sounds crazy, but bear with me.

Uniting time and space means actions become spacetime integrals of Lagrangian densities expressible in terms of fields and spacetime derivatives, rather than just time integrals of Lagrangians expressible in terms of coordinates and their time derivatives. Even before you work out ideas like Lorentz-invariance, you already realize a field ought to vibrate in a similar retarded-plus-maybe-advanced way. This quickly gives you the Klein-Gordon equation before you know Einstein's energy-momentum relation! Or if you work with a spinor with enough components, the privileged-direction idea comes off as more workable, because of a $\gamma^\mu\partial_\mu$ operator. (Hilariously, though, it actually doesn't get rid of the advanced solutions; or on a related point, it doesn't get rid of $E<0$ solutions to $E=m^2c^4+p^2c^2$.) But Lagrangians tend to give us even-order time derivatives in our EOMs, although there's a way round that.

So the real question is why does plucking a string look like Klein-Gordon rather than Dirac? (Schrödinger is out of the running because of the $f(t-x/c)$ Ansatz; we need $\nabla$ to have the same degree as $\partial_t$.) Well, the unhelpful answer is "because this derivation says so". The intuitive summary of that unhelpful answer is the string's behaviour is built on top of Newton's second law, so you need a second-order time derivative. If you think we should make our inference in the opposite direction, that won't satisfy you, in which case I'll need to motivate why the string amplitude shouldn't be a spinor with at least $4$ components. However, I suspect you don't find that negative fact surprising enough to warrant an "explanation".

J.G.
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  • Thanks... I think that mostly answers it, and covers several interesting aspects I hadn't considered. And yes, no need to explain why the Dirac equation was not needed to describe a violin string, lol! That would have been more surprising, not less ;-) I think the only two aspects of this I'm still a bit fuzzy on is 1. Why is the Schrodinger equation allowed to be time-asymmetric, but not the Klein Gordan equation or the classical wave equation? And is it just a coincidence that Maxwell's equations with no sources satisfy the same equations as those describing mechanical waves, with v -> c – reductionista Nov 17 '21 at 01:51
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    @reductionista 1. The TDSE requires you to also conjugate $\psi$ for time symmetry to work, and that's the C and T in CPT, which has a (complicated!) relativistic QFT justification. 2. I shouldn't have claimed the relativistic field theory one invents prematurely by thinking about Lagrangian densities need be quantum; the same logic applies to classical field theories, and is how you could predict the electromagnetic wave equation (although it has a constant speed due to relativity, and not a medium as per strings' sound waves). – J.G. Nov 17 '21 at 08:02
  • I think I see. The Klein Gordon equation is symmetric under both parity and time-reversal (and trivially, complex conjugation) individually. The Schrödinger Equation is symmetric under parity $P$. It's not symmetric under "naive time reversal" $T_{n} : t \rightarrow -t$ or under $C$ (complex conjugation). But since the phase of $\psi$ is not physically measurable anyway, it doesn't matter that it's technically not time-symmetric since it's the dynamics of $|psi|^2$ which matters. So we're free to just define the time reversal operator to be $T = CT_{n}$, restoring the symmetry. – reductionista Nov 19 '21 at 00:51
  • But wait... if $T = CT_n$, and by the CPT theorem $P = CT$, then wouldn't that imply $P = T_n$? That doesn't seem right... I think I'm still missing something here. Maybe it's true in the case of a simple single-particle Schrödinger Equation only because both P and CT are essentially the identity operation (when applied to both sides). But not true in general for QFT, because in addition to complex conjugation, you have to replace each field with one that transforms oppositely under the gauge symmetry? – reductionista Nov 19 '21 at 01:14
  • @reductionista If $S$ is the solution set, the correct condition is $x\in S\implies PCT_nx\in S$. – J.G. Nov 19 '21 at 05:52
  • That makes sense for the non-relativistic Schrodinger equation. What bothers me is that the abstract Schrodinger equation should also apply in a relativistic context. If the Hamiltonian is fully Lorentz invariant and Hermitian, then the CPT theorem must apply... which is almost like what you have but with $T_n$ replaced with $T$. This implies $T = CT_n = CP$. Which then implies $T_n = P$ which reduces your formula to $x \in S \rightarrow Cx \in S$ ! Which we know is not true for chiral fermions. – reductionista Nov 24 '21 at 07:38
  • I think I see where things went wrong: there are two slightly different forms of "complex conjugation" here. There's $C_{\psi}$ as in "complex conjugation of $\psi$ and a different operation $C_{\rm all}$ as in "complex conjugation of everything, including all operators and c-numbers". I think it's $C_{\psi}$ which is a more direct analog of the $C$ symmetry from QFT, where you conjugate only operators but not c-numbers. On the other hand, the definition of $T$ in QFT is $T = C_{\rm all}T_n$ which is NOT the same thing as $CT_n$ or $C_{\psi}T_n$. – reductionista Nov 24 '21 at 07:59
  • Not all operators I guess, just local field operators. (Which seems close enough to, even though here it's technically a state.) – reductionista Nov 24 '21 at 17:51
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Wave equation is the simplest equation that describes wave-like solutions in an isotropic homogeneous medium. In case of Klein-Gordon equation these correspond to the symmetries of the space-time, whereas in case of elasticity or hydrodynamics they result from making some simplifying assumptions - in fact in both latter fields one frequently comes up with equations that are different from the standard form $$ \frac{1}{c^2}\frac{\partial^2 u(\mathbf{x},t)}{\partial t^2} - \nabla^2 u(\mathbf{x},t)=0 $$ (See this answer for overview of some of the equatiions.)

E.g., in hydrodynamics one needs to do quite some work to get from the basic Navier-Stokes equation to a wave equation, and one gets somewhat different answers depending on whether one looks at the gravity waves in a shallow water, or deep water, or the pressure waves. Moreover, the Navier-Stokes equations are themselves an approximation to a more precise description in terms of Boltzmann equation (assuming local equilibrium).

In elasticity one often has to deal with anisotropy, and in both fields the standard wave equation can be obtained only under the assumption of non-linear response (aka Hooke's law in elasticity).

Finally, Schrödinger equation can be obtained from the Klein-Gordon equation as an equation for the envelope function, i.e., assuming that $$ \psi(\mathbf{x},t)=\phi(\mathbf{x},t)e^{-imc^2 t/\hbar}, $$ where $\phi(\mathbf{x},t)$ is slowly varying with time. Plugging the above into the Klein-Gordon equation and neglecting the second time derivative of the envelope function $\phi(\mathbf{x},t)$ one ends up with the Schrödinger equation for a free particle. Note that this "diffusion" approximation is not unqiue to quantum mechanics - e.g., it is used for propagation of EM waves via an inhomogeneous media.

Roger V.
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