Wave equations take the form:

$$\frac{ \partial^2 f} {\partial t^2} = c^2 \nabla ^2f$$

But the Schroedinger equation takes the form:

$$i \hbar \frac{ \partial f} {\partial t} = - \frac{\hbar ^2}{2m}\nabla ^2f + U(x) f$$

The partials with respect to time are not the same order. How can Schroedinger's equation be regarded as a wave equation? And why are interference patterns (e.g in the double-slit experiment) so similar for water waves and quantum wavefunctions?

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    $\begingroup$ For a connection between Schr. eq. and Klein-Gordon eq, see e.g. A. Zee, QFT in a Nutshell, Chap. III.5, and this Phys.SE post plus links therein. $\endgroup$
    – Qmechanic
    Commented Jul 27, 2014 at 17:47
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    $\begingroup$ "In what sense is the Schrödinger equation a wave equation?" in a loose sense. Its solutions are intuitively wave-like. From a mathematical point of view, things are not as easy. Standard classifications of PDE's dont accommodate the Schrödinger equation, which kinda looks parabolic but it is not dissipative. It shares many properties with hyperbolic equations, so we can say that it is a wave-equation -- not in the technical sense, but yes in a heuristic sense. $\endgroup$ Commented May 16, 2017 at 22:21
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    $\begingroup$ I had left a comment on one of the answers below, but then deleted it... I'll post something similar here because it's along the lines of what @AccidentalFourierTransform said. I wouldn't call this equation a wave equation. It's not hyperbolic. Wave-like? Maybe. But I don't think I would try to defend the statement that it's a wave equation. To me, hyperbolic <-> wave equation and anything else is just something else. $\endgroup$
    – tpg2114
    Commented May 16, 2017 at 22:28
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    $\begingroup$ A variant on this question - why does double-slit interference produce such similar interference patterns for water waves as for the electron wavefunction, if their underlying differential equations are so different? $\endgroup$
    – tparker
    Commented May 16, 2017 at 22:50
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    $\begingroup$ @tparker We see that all the time in, say, fluid dynamics. Linear potential equations can generate very similar solutions as the full Navier-Stokes equations under some circumstances despite the vast differences in their underlying equations. But, there are solutions that can't be produced by one or the other. I'm reluctant to say it's all just coincidental, but it's not unheard of that fundamentally different equations can produce similar solutions in a limited number of situations. $\endgroup$
    – tpg2114
    Commented May 16, 2017 at 23:15

7 Answers 7


Actually, a wave equation is any equation that admits wave-like solutions, which take the form $f(\vec{x} \pm \vec{v}t)$. The equation $\frac{\partial^2 f}{\partial t^2} = c^2\nabla^2 f$, despite being called "the wave equation," is not the only equation that does this.

If you plug the wave solution into the Schroedinger equation for constant potential, using $\xi = x - vt$

$$\begin{align} -i\hbar\frac{\partial}{\partial t}f(\xi) &= \biggl(-\frac{\hbar^2}{2m}\nabla^2 + U\biggr) f(\xi) \\ i\hbar vf'(\xi) &= -\frac{\hbar^2}{2m}f''(\xi) + Uf(\xi) \\ \end{align}$$

This clearly depends only on $\xi$, not $x$ or $t$ individually, which shows that you can find wave-like solutions. They wind up looking like $e^{ik\xi}$.

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    $\begingroup$ Doesn't any translationally-invariant PDE satisfy this criterion, even if it isn't rotationally invariant or even linear? $\partial f({\bf \xi}) / \partial x_i = \partial f({\bf \xi}) / \partial \xi_i$ and $\partial f({\bf \xi}) / \partial t = -{\bf v} \cdot {\bf \nabla}_{\bf \xi} f({\bf \xi})$, so if you take any translationally invariant PDE and replace every $\partial / \partial t$ with $-\bf{v} \cdot {\bf \nabla}_{\bf \xi}$, then can't any solution $f({\bf \xi})$ of the resulting 3D PDE be converted into a "wave-like" solution to the original 4D PDE by ... $\endgroup$
    – tparker
    Commented May 8, 2017 at 6:35
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    $\begingroup$ ... letting $f({\bf \xi}) \to f({\bf x} - {\bf v} t)$? $\endgroup$
    – tparker
    Commented May 8, 2017 at 6:35
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    $\begingroup$ I've expanded the comment above into an answer. $\endgroup$
    – tparker
    Commented May 10, 2017 at 8:28
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    $\begingroup$ I disagree. For the wave equation any function f(x-vt) (with correctly fixed v) is a solution. In you Schroedinger example only very special special functions do fulfill the equation. $\endgroup$
    – lalala
    Commented May 17, 2017 at 5:19
  • $\begingroup$ @lalala ... for non-dispersive waves. However, plenty of other phenomena that your really do want to keep calling 'waves' (like slinky waves, sound in solids, light in glass, or ripples in a pond) no longer support that property: they do have an infinite basis of solutions of the form $e^{i(kx-\omega t)}$, but they no longer sustain $f(x-vt)$ as a solution, exactly in the way that the Schrödinger equation does. "Wave" is a bit of a fluffy term, but if you use that basis to write out the Schrödinger equation, you've got to be prepared to kick out the others. $\endgroup$ Commented May 25, 2017 at 18:32

In the technical sense, the Schrödinger equation is not a wave equation (it is not a hyperbolic PDE). In a more heuristic sense, though, one may regard it as one because it exhibits some of the characteristics of typical wave-equations. In particular, the most important property shared with wave-equations is the Huygens principle. For example, this principle is behind the double slit experiment.

If you want to read about this principle and the Schrödinger equation, see Huygens' principle, the free Schrodinger particle and the quantum anti-centrifugal force and Huygens’ Principle as Universal Model of Propagation. See also this Math.OF post for more details about the HP and hyperbolic PDE's.

  • $\begingroup$ Could you elaborate on this? I am trying to understand how the DS experiment can be predicted via solutions to the Schrödinger equation, usually it seems just heuristically done with wave equation solutions but these don't typically solve Schrödinger. Maybe you have a source which explains this? Looking through the sources you posted, they don't discuss the DS experiment. $\endgroup$ Commented Sep 18, 2020 at 14:33
  • $\begingroup$ It seems like, for solutions consisting entirely of modes with the same frequency, Schrödinger/wave equations might have the same solutions—just differing by whether they include an imaginary part and possibly a factor of 2. Since they're both defined in terms of ∇², they have the same eigenspace. The differing relationships between wavelength and frequency won't be visible, qualitatively, if you're only considering one frequency/wavelength (as is mostly the case in the double-slit)? $\endgroup$
    – MRule
    Commented Oct 6, 2022 at 19:22

Both are types of wave equations because the solutions behave as you expect for "waves". However, mathematically speaking they are partial differential equations (PDE) which are not of the same type (so you expect that the class of solutions, given some boundary conditions, will present different behaviour). The constraints on the eigenvalues of the linear operator are also particular to each of the types of PDE. Generally, a second order partial differential equation in two variables can be written as

$$A \partial_x^2 u + B \partial_x \partial_y u + C \partial_y^2 u + \text{lower order terms} = 0 $$

The wave equation in one dimension you quote is a simple form for a hyperbolic PDE satisfying $B^2 - 4AC > 0$.

The Schrödinger equation is a parabolic PDE in which we have $B^2 - 4AC = 0$ since $B=C=0$. It can be mapped to the heat equation.

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    $\begingroup$ Shouldn't it be $B^2 - 4AC$ or maybe $2B$ in the PDE? $\endgroup$
    – Luzanne
    Commented May 18, 2017 at 15:21
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    $\begingroup$ Is there an insightful explanation why Schroedinger equation and heat equation are so similar (actually identical with $U=0$ except for being complex-valued) yet result in such different behavior? $\endgroup$
    – divB
    Commented Apr 4, 2020 at 22:40
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    $\begingroup$ @divB I think it's similar to how the ODE f'(x) = -k f(x) gives exponential decay (for real k > 0), but add an i before the k and you instead get oscillating solutions like you get for f''(x) = -k f(x) $\endgroup$ Commented Jun 6, 2020 at 2:20
  • $\begingroup$ see also: physics.stackexchange.com/q/62135/226902 $\endgroup$
    – Quillo
    Commented Jul 4, 2022 at 17:53
  • $\begingroup$ @divB Yes, google Wick rotation. $\endgroup$
    – Bcpicao
    Commented Apr 2 at 16:47

As Joe points out in his answer to a duplicate, the Schrodinger equation for a free particle is a variant on the slowly-varying envelope approximation of the wave equation, but I think his answer misses some subtleties.

Take a general solution $f(x)$ to the wave equation $\partial^2 f = 0$ (we use Lorentz-covariant notation and the -+++ sign convention). Imagine decomposing $f$ into a single plane wave modulated by an envelope function $\psi(x)$: $f(x) = \psi(x)\, e^{i k \cdot x}$, where the four-vector $k$ is null. The wave equation then becomes $$(\partial^\mu + 2 i k^\mu) \partial_\mu \psi = ({\bf \nabla} + 2 i\, {\bf k}) \cdot {\bf \nabla} \psi + \frac{1}{c^2} (-\partial_t + 2 i \omega) \partial_t \psi= 0,$$ where $c$ is the wave velocity.

If there exists a Lorentz frame in which $|{\bf k} \cdot {\bf \nabla} \psi| \ll |{\bf \nabla} \cdot {\bf \nabla} \psi|$ and $|\partial_t \dot{\psi}| \ll \omega |\dot{\psi}|$, then in that frame the middle two terms can be neglected, and we are left with $$i \partial_t \psi = -\frac{c^2}{2 \omega} \nabla^2 \psi,$$ which is the Schrodinger equation for a free particle of mass $m = \hbar \omega / c^2$.

$|\partial_t \dot{\psi}| \ll \omega |\dot{\psi}|$ means that the envelope function's time derivative $\dot{\psi}$ is changing much more slowly than the plane wave is oscillating (i.e. many plane wave oscillations occur in the time $|\dot{\psi} / \partial_t \dot{\psi}|$ that it takes for $\dot{\psi}$ to change significantly) - hence the name "slowly-varying envelope approximation." The physical interpretation of $|{\bf k} \cdot {\bf \nabla} \psi| \ll |{\bf \nabla} \cdot {\bf \nabla} \psi|$ is much less clear and I don't have a great intuition for it, but it seems to basically imply that if we take the direction of wave propagation to be $\hat{{\bf z}}$, then $\partial_z \psi$ changes very quickly in space along the direction of wave propagation (i.e. you only need to travel a small fraction of a wavelength $\lambda$ before $\partial_z \psi$ changes significantly). This is a rather strange limit, because clearly it doesn't really make sense to think of $\psi$ as an "envelope" if it changes over a length scale much shorter than the wavelength of the wave that it's supposed to be enveloping. Frankly, I'm not even sure if this limit is compatible with the other limit $|\partial_t \dot{\psi}| \ll \omega |\dot{\psi}|$. I would welcome anyone's thoughts on how to interpret this limit.


As stressed in other answers and comments, the common point between these equations is that their solutions are "waves". It is the reason why the physics they describe (eg interference patterns) is similar.

Tentatively, I would define a "wavelike" equation as

  1. a linear PDE

  2. whose space of spacially bounded* solutions admits a (pseudo-)basis of the form $$e^{i \vec{k}.\vec{x} - i \omega_{\alpha}(\vec{k})t}, \vec{k} \in \mathbb{R}^n, \alpha \in \left\{1,\dots,r\right\}$$ with $\omega_1(\vec{k}),\dots,\omega_r(\vec{k})$ real-valued (aka. dispersion relation).

For example, in 1+1 dimension, these are going to be the PDE of the form $$\sum_{p,q} A_{p,q} \partial_x^p \partial_t^q \psi = 0$$ such that, for all $k \in \mathbb{R}$ the polynomial $$Q_k(\omega) := \sum_{p,q} (i)^{p+q} A_{p,q} k^p \omega^q$$ only admits real roots. In this sense this is reminiscent of the hyperbolic vs parabolic classification detailed in @DaniH answer, but without giving a special role to 2nd order derivatives.

Note that with such a definition the free Schrödinger equation would qualify as wavelike, but not the one with a potential (and rightly so I think, as the physics of, say, the quantum harmonic oscillator is quite different, with bound states etc). Nor would the heat equation $\partial_t \psi -c \partial_x^2 \psi = 0$: the '$i$' in the Schrödinger equation matters!

* Such equations will also often admit evanescent waves solutions corresponding to imaginary $\vec{k}$.


This answer elaborates on my comment to David Z's answer. I think his definition of a wave equation is excessively broad, because it includes every translationally invariant PDE and every value of $v$. For simplicity, let's specialize to linear PDE's in one spatial dimension. A general order-$N$ such equation takes the form

$$\sum_{n=0}^N \sum_{\mu_1, \dots, \mu_n \in \{t, x\}} c_{\mu_1, \dots, \mu_n} \partial_{\mu_1} \dots \partial_{\mu_n}\, f(x, t) = 0.$$

To simplify the notation, we'll let $\{ \mu \}$ denote $\mu_1, \dots, \mu_n$, so that

$$\sum_{n=0}^N \sum_{\{\mu\}} c_{\{\mu\}} \partial_{\mu_1} \dots \partial_{\mu_n}\, f(x, t) = 0.$$

Let's make the ansatz that $f$ only depends on $\xi := x - v t$. Then $\partial_x f(\xi) = f'(\xi)$ and $\partial_t f(\xi) = -v\, f'(\xi)$. If we define $a_{\{\mu\}} \in \mathbb{N}$ to simply count the number of indices $\mu_i \in \{\mu\}$ that equal $t$, then the PDE becomes

$$\sum_{n=0}^N f^{(n)}(\xi) \sum_{\{\mu\}} c_{\{\mu\}} (-v)^{a_{\{\mu\}}} = 0.$$

Defining $c'_n := \sum \limits_{\{\mu\}} c_{\{\mu\}} (-v)^{a_{\{\mu\}}}$, we get the ordinary differential equation with constant coefficients

$$\sum_{n=0}^N c'_n\ f^{(n)}(\xi) = 0.$$

Now as usual, we can make the ansatz $f(\xi) = e^{i z \xi}$ and find that the differential equation is satisfied as long as $z$ is a root of the characteristic polynomial $\sum \limits_{n=0} c_n' (iz)^n$. So our completely arbitrary translationally invariant linear PDE will have "wave-like solutions" traveling at every possibly velocity!

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    $\begingroup$ Interesting point. I'm not ready to admit that the definition is overly broad; maybe all translationally invariant linear PDEs are "wave equations". But I am wondering whether there's more to the story. For example, do some of these PDEs admit other solutions which cannot be expressed as a linear combination of waves $f(x \pm vt)$? $\endgroup$
    – David Z
    Commented May 10, 2017 at 19:07
  • $\begingroup$ @DavidZ The PDE's don't even have to be linear, as mentioned in my original comment (I just considered the linear case for simplicity). If you allow the phrase "wave equation" to cover general TI nonlinear PDE's, in my opinion it becomes so broad that you might as well just say "translationally invariant PDE." $\endgroup$
    – tparker
    Commented May 10, 2017 at 23:02

While it is not very technical in nature it is worth going back to the first definition of what a wave is (that is, the one you use before learning that "a wave is a solution to a wave-equation"). The wording I use in the introductory classes is

a moving disturbance

where the 'disturbance' is allowed to be in any measurable quantity, and simple means that the quantity is seen to vary from its equilibrium value and then return to that value.

The surprising thing is not how general that expression is, but that it is necessary to use something that general to cover all the basic cases: waves on strings, surface waves on liquids: sound and light.

And by that definition Schrödinger's equation is used to describe the moving variation of various observables, so arguably qualifies.

There is room to quibble—the wave-function itself is not an observable, and even the distributions of values that can be observed are often statistical in nature—but I've always been comfortable with this approach.

  • $\begingroup$ Yes, I'm starting to regret placing the bounty on this question instead of creating my own. The thing I really want to understand is the much more concrete question of why slit interference patterns look so similar (both qualitatively and quantitatively) for the free-particle Schrödinger equation and for "the" wave equation, even the the differential equations are so mathematically different. $\endgroup$
    – tparker
    Commented May 24, 2017 at 2:15
  • $\begingroup$ I see. That is an interesting question, but not one I've given a lot of thought to before. A line of inquiry which presents itself is considering the TDSE as the Newtonian approximation to the underlying relativistic, quantum, wave-equations, which have the symmetry between time and space that we see in the "the" wave-equation. Certainly that fits with the usual heuristic picture in which $H = p^2/2m + V(x)$ plus the time derivative of $\Psi_0\exp(kx - \omega t)$ resulting in energy while spacial derivatives result in momentum (to within appropriate constants, of course). $\endgroup$ Commented May 24, 2017 at 2:40
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    $\begingroup$ @tparker just to let you know, reading your comments prompted me to ask this related question. physics.stackexchange.com/questions/335225/… $\endgroup$
    – CDCM
    Commented May 25, 2017 at 1:06

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