I noticed a long time ago the similarity between Schrodinger equation and Euler-Bernoulli beam equation. Namely, Euler-Bernoulli equation is equivalent to the system of Schrodinger equation for a free particle and its complex conjugate.

Euler-Bernoulli equation for vibrations of beams ($\Psi$ here - amplitude of bending, Y - Young modulus, etc):

$$ \frac{YI}{\rho A} \frac{\partial ^4 \Psi}{\partial x^4}+\frac{\partial ^2 \Psi}{\partial t^2}=0 $$

If we transform the beam parameters to a particle parameters appropriately:

$$ \frac{\hbar^2}{4m^2} \frac{\partial ^4 \Psi}{\partial x^4} + \frac{\partial ^2 \Psi}{\partial t^2}=0 $$

$$ \left( \frac{\hbar}{2m} \frac{\partial ^2 }{\partial x^2}+i\frac{\partial }{\partial t} \right) \left( \frac{\hbar}{2m} \frac{\partial ^2 }{\partial x^2}-i\frac{\partial}{\partial t} \right) \Psi =0 $$

Which is obviously a system of two Schrodinger equations ($\Psi$ becomes a wavefunction):

$$ \frac{\hbar}{2m} \frac{\partial ^2 \Psi}{\partial x^2}+i\frac{\partial \Psi}{\partial t}=0 $$ $$ \frac{\hbar}{2m} \frac{\partial ^2 \Psi}{\partial x^2}-i\frac{\partial \Psi}{\partial t}=0 $$

The main (mathematical) difference between the solution of this system and a single Schrodinger equation is of course that here


Which is not true in quantum mechanics, since complex conjugation is equivalent to time-reversal (and momentum-reversal for a free particle).

Also, the general solution of B-E contains $\cosh$ and $\sinh$ as well as $\cos$ and $\sin$ and thus requires four boundary conditions instead of two.

But the $\omega(k)$ spectrum is the same for both equations.

So, do you think there is any physical meaning for this mathematical similarity, or can it prove to be in any way useful, like in modeling the behavior of particles with acoustic experiments?

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    $\begingroup$ I think it is false that Euler-Bernoulli equation is equivalent to a system of two Schrodinger equations. Instead it is equivalent to a Schrodinger equation with source which, in turn, is given by a solution of another (time reversed) Schrodinger equation. $\endgroup$ Sep 1, 2016 at 20:08
  • $\begingroup$ $$ \left( \frac{\hbar}{2m} \frac{\partial ^2 }{\partial x^2}+i\frac{\partial }{\partial t} \right) \left( \frac{\hbar}{2m} \frac{\partial ^2 }{\partial x^2}-i\frac{\partial}{\partial t} \right) \Psi =0 $$ does not imply $$ \frac{\hbar}{2m} \frac{\partial ^2 \Psi}{\partial x^2}+i\frac{\partial \Psi}{\partial t}=0 $$ $$ \frac{\hbar}{2m} \frac{\partial ^2 \Psi}{\partial x^2}-i\frac{\partial \Psi}{\partial t}=0 $$ $\endgroup$ Sep 1, 2016 at 20:11
  • $\begingroup$ The first equation implies that $$\Phi= \left( \frac{\hbar}{2m} \frac{\partial ^2 }{\partial x^2}-i\frac{\partial}{\partial t} \right) \Psi$$ is a solution of $$ \left( \frac{\hbar}{2m} \frac{\partial ^2 }{\partial x^2}-i\frac{\partial}{\partial t} \right) \Phi =0 \:.$$ So that, as I wrote, $\Psi$ is the solution of a S. equation with source and this source is a solution of another time-reversed S. equation. This is the right viewpoint to adopt to write the general solution of your equation. $\endgroup$ Sep 1, 2016 at 20:16
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    $\begingroup$ However your question is very interesting, I did not noticed this relation between S. equation and the equation describing vibrations of a rigid plate... $\endgroup$ Sep 1, 2016 at 20:20
  • $\begingroup$ Schroedinger equation is non-causal exactly as the heat equation, it propagates signals with infinite speed. Instead Euler-Bernoulli must be causal is it describes vibrations of, for instance, the body of a violin. This fact is intriguing to me! $\endgroup$ Sep 1, 2016 at 20:24

3 Answers 3


This correspondence is a special case of the Madelung equations - a formulation of quantum mechanics that is equivalent to the Schrodinger equation. The Madelung equations are more mathematically complicated than the Schrodinger equation, but some people find them more physically intuitive because they are formally almost identical to the Euler equation for a fluid. They are useful in the de Broglie-Bohm formulation of quantum mechanics.


The free 1D TDSE

$$ i\frac{\partial\Psi}{\partial t}~=~-\frac{\hbar}{2m} \frac{\partial^2\Psi}{\partial x^2}, \qquad \Psi~\equiv~u+iv,$$ $$\Updownarrow$$ $$ \frac{\partial u}{\partial t}~=~- \frac{\hbar}{2m} \frac{\partial^2v}{\partial x^2}, \qquad \frac{\partial v}{\partial t}~=~ \frac{\hbar}{2m} \frac{\partial^2 u}{\partial x^2},\tag{1}$$ is an auto-Bäcklund transformation for the free EBE

$$ \frac{\partial^2\Psi}{\partial t^2}+\frac{\hbar^2}{4m^2} \frac{\partial^4\Psi}{\partial x^4}~=~0, \qquad \Psi~\equiv~u+iv, $$ $$\Updownarrow$$ $$ \frac{\partial^2u}{\partial t^2}+\frac{\hbar^2}{4m^2} \frac{\partial^4 u}{\partial x^4}~=~0, \qquad \frac{\partial^2 v}{\partial t^2}+\frac{\hbar^2}{4m^2} \frac{\partial^4v}{\partial x^4}~=~0.\tag{2}$$

This means in particular that if $u$ is a solution to the EBE (2), then we can solve the TDSE (1) wrt. the imaginary part $v$ [i.e. the EBE (2) for $u$ serves as the integrability condition for the existence of the $v$-solution to the TDSE (1)]. The $v$-solution is unique modulo an affine function $a_2x+b_2$, where $a_2,b_2\in \mathbb{R}$. Then we have generated a partner solution $v(x,t)+a_2x+b_2$ to the EBE (2).

Similarly, note that the TDSE (1) is unaffected if we replace $u(x,t)$ with $u(x,t)+a_1x+b_1$, where $a_1,b_1\in \mathbb{R}$. The two solutions $u(x,t)+a_1x+b_1$ and $v(x,t)+a_2x+b_2$ to the EBE (2) then constitute a "Schrödinger pair".

  • $\begingroup$ Thank you. I did not consider this way of relating the two equations. This might lead to the better understanding of the more general problem I'm trying to solve. $\endgroup$
    – Yuriy S
    Sep 2, 2016 at 13:58

Schrödinger himself noted the similarity in Schrödinger (1926): Quantisation as a problem of proper values (Part IV), where he is deriving the time-dependent Schrödinger equation:

It is evidently no longer of the simple type arising for vibrating membranes, but it is of fourth order, and of a type similar to that occurring in many problems in the theory of elasticity (Footnote: E.g. for a vibrating plate $\nabla^2\nabla^2 u + \frac{\partial^2 u }{\partial^2 t} = 0$, Cf. Courant-Hilbert chap. v. §8, p. 256)

I have problems following his arguments in the following paragraphs. However later he then writes

I have taken a somewhat different route, which is much easier for calculations, and which I consider is justified in principle.

We need not to raise the order of the wave equation to four, in order to get rid of the energy parameter ... We thus arrive at one of the two equations $$ \nabla^2 \psi - \frac{8\pi^2}{h^2}V\psi\mp\frac{4\pi i}{h}\frac{\partial \psi}{\partial t} = 0$$ We will require the complex wavefunction $\psi$ to satisfy one of these two equations.

It seems to me Schrödinger himself couldn't really justify his step in going from the beam equation to two complex equations, except that this was easier to calculate and seemed to give physically reasonable results.

  • $\begingroup$ Thank you so much for the reference! I haven't read Schrödinger's work to my shame. This is very interesting. Especially, since there's also a (less obvious) connection between the more general Timoshenko beam equation and Klein-Gordon equation. I don't have the necessary backround to explore it further $\endgroup$
    – Yuriy S
    Aug 24, 2017 at 15:25

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