If we do the wick rotation such that τ = it, then Schrödinger equation, say of a free particle, does have the same form of heat equation. However, it is clear that it admits the wave solution so it is sensible to call it a wave equation.

  1. Whether we should treat it as a wave equation or a heat equation with imaginary time, or both?
  2. If it is a wave equation, how do we express it in the form of a wave equation?
  3. Is there any physical significance that Schrödinger equation has the same form of a heat equation with imaginary time? For example, what is diffusing?
  • $\begingroup$ I don't understand what physics principle you're asking about. There are many linear equations. So what? There are many exponential decay equations. It merely gives us analogues. $\endgroup$
    – Bill N
    Commented Sep 19, 2016 at 17:15
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    $\begingroup$ Possible duplicate of How is the Schroedinger equation a wave equation? $\endgroup$ Commented Sep 19, 2016 at 17:35
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    $\begingroup$ Is there any reason why Schrödinger equation can be written in the form of a heat equation? Also, Schrödinger has first order derivative with respect to time (instead of second order), why is it a wave equation? $\endgroup$
    – Kevin Kwok
    Commented Sep 19, 2016 at 17:46
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    $\begingroup$ The part with the heat equation is closely related to physics.stackexchange.com/q/80131/50583 and physics.stackexchange.com/q/144832/50583, the part about it being a wave equation is a duplicate of the already linked question. Please ask a single, focused question per post, or at least questions so closely related they can't meaningfully be split up. $\endgroup$
    – ACuriousMind
    Commented Sep 20, 2016 at 13:25

1 Answer 1


1) Both: it is apparently a heat equation in imaginary time and it is a wave equation because its solutions are waves.

2) Nonstationary Schrodinger equation (let us assume free particle) $$ i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^2\nabla^2}{2m}\psi $$ is essentially complex: it can never be satisfied by a real function, only by a complex one.

Nevertheless, its solutions are waves because the complex $\psi$ means it is actually a system of two real equations of the first order in time. Assuming $\psi=u+iv$ we have: $$ \hbar\frac{\partial u}{\partial t}=-\frac{\hbar^2\nabla^2}{2m}v,\qquad \hbar\frac{\partial v}{\partial t}=\frac{\hbar^2\nabla^2}{2m}u. $$ Eliminating, say, $v$, we get: $$ \hbar^2\frac{\partial^2 u}{\partial t^2}=-\frac{\hbar^4\nabla^4}{4m^2}u. $$ In two dimensions, this equation has the same form as a wave equation for bending (flexural) waves on a thin rigid plate. It is also of the 2-nd order in time and 4-th order in coordinates. The analogy extends also to wave dispersions: the bending waves have a quadratic dispersion $\omega\sim q^2$ similarly to free particle obeying Schrodinger equation $E=p^2/2m$.

3) This analogy is widely used in the diffusion Monte-Carlo method, where the Schrodinger equation is solved in imaginary time. In this case, its solution is decaying instead of being oscillatory and, if we normalize it properly, it will converge to the ground state wave function:



What is diffusing here? Taking imaginary time $\tau=it$, we have the following imaginary time Schrodinger equation for a particle in a potential $V$: $$ \hbar\frac{\partial\psi}{\partial t}=\frac{\hbar^2\nabla^2}{2m}-V\psi. $$ The first term in right hand side is usual diffusion. The second is something like heat production or burning, and its "minus" sign means this heat production is more intense in the minima of $V$.

Thus, the picture of diffusion in imaginary time is the following: the first term ("diffusion") tries to delocalize $\psi$, while the second term tries to lure $\psi$ to the minima of the potential $V$. Their interplay is the same as that between kinetic and potential energies in quantum mechanics, and its result is a ground state wave function - exactly what is used in diffusion Monte-Carlo calculations.

  • $\begingroup$ As a reference, the governing equation of the dynamics of a rigid plate can be found at en.wikipedia.org/wiki/… $\endgroup$
    – Kevin Kwok
    Commented Sep 20, 2016 at 7:36
  • $\begingroup$ In the heat equation formulation Why does the $V$ take on a minus sign? It’s clear that $\frac{d}{dt} = \frac{1}{i} \frac{d}{d\tau} = -i \frac{d}{d\tau}$ so your LHS will necessarily be $h \frac{d\psi}{d\tau}$ after cancelling $i$ and negative sign. On your RHS I see no reason for the sign to flip. Unless you assumed that minus sign in the original schrodinger equation (which is not standard I believe) $\endgroup$ Commented Dec 5, 2023 at 17:10

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