In the book "Group theory and it's Applications to the Quantum Mechanics of atomic spectra " by Eugene P. Wigner

in chapter 4 The elements of quantum mechanics it is written

Consider a many dimensional space with as many coordinates as the system considered as position coordinates. Every arrangement of the positions of the particles of the system corresponds to a point in this multidimensional configuration space. This point will move in the course of time tracing out a curve by which the motion of the system can be completely described classically. There exists a fundamental correspondence between the classical motion of this point, the system point in configuration space, and the motion of a wave packet also considered in configuration space, if only we assume that the index of refraction for these waves is $\frac{\sqrt{2m(E-V)}}{E}$, where $E$ is the total energy of the system; $V$ is the potential energy as a function in the configuration space.

What does the wave-packet and the refractive index implies here. How to interpret this?

The author is trying to derive Schrodinger equation by using configuration space.

This is related to the development of Schrodinger equation using the action, in configuration space where the waves are shells traveling in configuration space just like huygene's principle in the optics, where the phase is action, and phase velocity is the resiprocal of the above mentioned refractive index.

This is the same process used by Schrodinger in his 1926 paper where he first introduced the Schrodinger wave equation,

And these waves are in configuration space, which will lead to quantum mechanics, he is trying to say that as ray optics won't work for phenomenon like diffraction or interference, similarly mechanics as they knew it won't work for micro-things as the wavelength of these waves is roughly of the order of radius of curvature of path configuration space

  • $\begingroup$ Your question isn't clear. Are you asking how the expression for the refractive index was obtained? $\endgroup$
    – Miyase
    Commented Apr 4, 2023 at 18:40
  • $\begingroup$ yes, I am asking how to interpret this and how this was obtained. $\endgroup$
    – Pradyuman
    Commented Apr 4, 2023 at 18:42

2 Answers 2


The Schrodinger in his works uses the analogy of ray-optics being useless when the wavelength of light is comparable to the things it encounter as it starts interference or diffraction which can't be understood using ray-optics, similarly he pointed out that macro-mechanics (being equivalent to ray optics) can't work when things become small.

Consider a configuration space and lagrangian $L$. The Hamiltonian function of action is

$$ W=\int_{t_0}^t L\ dt \ \tag{1}$$

and we know,

$$ \frac{\partial W}{\partial t} + H = 0 \tag{2}$$

Which can be written as

$$ \frac{\partial W}{\partial t} + \frac{1}{2m}\left(\frac{\partial W}{\partial x_i}\right)^2 + V(x_i) = 0 \tag{3}$$

Where $x_i$ are generalized coordinates and $p_i = \frac{\partial W}{\partial x_i}$

We solve it by considering $$W=-Et +S(x_i) \tag{4}$$

Equation 3 becomes

$$ \vert \text{grad}W\vert = \sqrt{2m(E-V)} \tag{5}$$

Its geometric interpretation is as follows

Assume $t$ is constant, any function $W$ of space alone can be described by system of surfaces over which the $W$ is constant, choose an arbitrary surface with value say $W_0$ and the solution of equation 5 can be constructed out of initial arbitrarily chosen condition, by simply extending normal at every point say

$$ dn= \frac{dW_0}{\sqrt{2m(E-V)}} \tag{6}$$

The points on first surface will lead us to second surface with value $W= W_0 +dW_0$. Continuing it will lead to the whole system.

Now if we vary time the constant value of the surface will vary with velocity

$$ u=\frac{E}{\sqrt{2m(E-V)}} \tag{7} $$

Now instead of thinking the value is varying from surface to surface, one can imagine that the surface with constant value is varying in space, Which look like surface traveling like stationary waves in a medium (Huygen's principle), and the velocity given above will become the normal velocity of surface,

So now let's consider this wave system described by a wave-function

$$\psi =A(x_i)\sin{\left(\frac{W}{K}\right)}\tag{8}$$

$$\psi =A(x_i)\sin{\left(\frac{-Et+S(x_i)}{K}\right)} \tag{9}$$

Which leads to

$$\nu = \frac{E}{2πK} \tag{10}$$

Now $K$ is considered as universal constant as it does not depends on nature of mechanical systems and set it as $\hbar$ giving us

$$ h\nu=E \tag{11}$$

Which is an established relationship,

The wave-function should also satisfy wave equation

$$u^2\nabla_{x_i}^2\psi - \ddot\psi=0\tag{12}$$

Since frequency is known the wave-function depends on time only through factor $e^{\pm itE/\hbar}$

And $$\ddot\psi=-\left(\frac{E}{\hbar}\right)^2\psi \tag{13}$$

From equation 7,12 and 13

$$\nabla_{x_i}^2\psi + \frac{2m(E-V)\psi}{\hbar^2}=0\tag{14}$$

Which is the famous Schrodinger equation,

The phase of these waves that are in configuration space is given by the action.


If you develop the wave packet on a stationary-state basis, each component satisfies the time-independent Schrödinger equation: $$-\frac{\hbar^2}{2m}\varphi''(x)+V\varphi(x)=E\varphi(x)$$ If the potential is constant and $E>V$, this can be written as a harmonic oscillator equation: $$\varphi''(x)+k^2\varphi(x)=0 \quad\text{with}\quad k=\sqrt{\frac{2m(E-V)}{\hbar^2}}$$ The full stationary wavefunction is of the form: $$\psi(x,t)=\varphi(x)e^{-i\omega t} \quad\text{with}\quad E=\hbar\omega$$ Now, about the refractive index, using phase velocity: $$v_\varphi =\frac{\omega}{k} =\frac{E}{\sqrt{2m(E-V)}}$$ The author is slightly wrong calling the inverse of this expression a "refractive index" since it's not even dimensionless, but it is proportional to it, so it's harmless.

  • $\begingroup$ Well you got it the other way, the author is using these arguments to prove Schrodinger equation $\endgroup$
    – Pradyuman
    Commented Apr 4, 2023 at 19:03
  • 3
    $\begingroup$ @Pradyuman If that's the case, your question is missing a lot of context. As it stands, it seems that the author pulls this refractive index out of thin air. Moreover, how does he define those waves if they aren't solutions to a given equation? Please edit your question to add more detail. $\endgroup$
    – Miyase
    Commented Apr 4, 2023 at 19:08
  • 1
    $\begingroup$ @Pradyuman The harmonic oscillator naturally emerges from the Schrödinger equations when you're choosing a stationary basis. If the author assumes that a given wave has a defined energy E, then it has to be this basis. $\endgroup$
    – Miyase
    Commented Apr 4, 2023 at 19:09
  • $\begingroup$ The author is taking a configuration space of particles and deriving Schrodinger equation using action, it has nothing to do with harmonic oscillator, sorry for not adding much details. $\endgroup$
    – Pradyuman
    Commented Apr 4, 2023 at 19:17
  • $\begingroup$ @Pradyuman I know, you said so above. But you don't provide any detail about the author's process. If possible edit your question with additional details. Or try to narrow down your question to one specific aspect. $\endgroup$
    – Miyase
    Commented Apr 4, 2023 at 21:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.