# Intuitively Re-Deriving Equations of Mathematical Physics

Using the intuitive interpretation of the Laplacian $\vec{\nabla}^2$ as the difference between the average value of a field in the neighbourhood of a point & the value of the field at that point, one can pretty easily & quickly derive the form of the heat equation, Poisson's equation & the wave equation (as is done in that link if anyone's interested: Davis - Fourier Series & Orthogonal Functions P196). I quite honestly cannot remember those equations, I re-derive them using the intuition that the Laplacian affords me mixed with physical reasoning drawn from the situation (drawn from the field we're using whether it's temperature, concentration, electric potential or displacement).

I'm wondering if similar intuition can be used to derive the form of the Helmholtz equation, the Schrodinger equation, the Dirac equation, & really any other nice equation from mathematical physics that people have nice intuition for in their head & wouldn't mind sharing. It doesn't have to be in any way rigorous or even necessarily completely logical so long as you get the right result, though it should really be quick & to the point, thanks.

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My mnemonic is to remember a simple special case, and heuristically rederive the equations of motion from that. In the case of wave-related equations, you can obtain most of them by just inventing differential equations that describe different properties of plane waves.

## Properties of a plane wave

The simplest solution to a wave equation is the plane wave $\Psi(x,t) = \exp(i\mathbf{k}\cdot\mathbf{x}-i\omega t)$. The time-derivative of this is: $$\frac{\partial\Psi}{\partial t} = -i\omega\Psi$$ Insert the de Broglie relation $E = \hbar \omega$, and solve for $E$: $$E = i\hbar\frac{\partial}{\partial t}$$

Then try differentiating the plane wave $\Psi$ with respect to position $\mathbf{x}$: $$\nabla \Psi = i\mathbf{k}\Psi$$ Now insert the de Broglie relation $\mathbf{p} = \hbar\mathbf{k}$ and solve for $\mathbf{p}$: $$\mathbf{p} = -i\hbar\nabla$$ We now have two equations $E = i\hbar\partial_t$ and $\mathbf{p} = -i\hbar\nabla$ relating spacetime derivatives of a wave function to physical observables.

## Schroedinger equation (energy)

Remember that the Schroedinger equation connects the time-derivative of a wave function to it's energy. Just right-multiply our equation for $E$ above with a wave function $\Psi$, and we get the Schroedinger equation:$^\dagger$ $$E\Psi = i\hbar \frac{\partial\Psi}{\partial t}$$

To get the familiar form of the equation for a single particle in a potential $V(\mathbf{x})$, just remember that classically we have $E = \mathbf{p}^2/2m + V(\mathbf{x})$, use $\mathbf{p} = -i\hbar\nabla$ from above, and insert it into the Schroedinger equation: $$\left[-\frac{\hbar^2}{2m}\nabla^2 + V(\mathbf{x})\right]\Psi = i\hbar \frac{\partial\Psi}{\partial t}$$

## Helmholtz equation (momentum)

The Helmholtz equation relates the momentum of a wave to its spatial derivative. To obtain it, just square the relation $\mathbf{p} = -i\hbar\nabla$, and right-multiply the result by a wave $\Psi$: $$\mathbf{p}^2\Psi = -\hbar^2\nabla^2\Psi$$ To obtain the conventional form, divide the equation by $\hbar^2$, and use the de Broglie relation $\mathbf{p} = \hbar\mathbf{k}$: $$\left[\nabla^2 + \mathbf{k}^2\right]\Psi = 0$$

## Wave equation (energy-momentum)

The energy and momentum of a free massless particle is related by $E^2 = \mathbf{p}^2c^2$. Insert $E = i\hbar\partial_t$ and $\mathbf{p} = -i\hbar\nabla$, and we get: $$-\hbar^2 \frac{\partial^2}{\partial t^2} = -c^2\hbar^2\nabla^2$$ Divide this by $\hbar^2c^2$ and right-multiply by a wave function, and you've got the wave equation: $$\nabla^2\Psi = \frac{1}{c^2}\frac{\partial^2\Psi}{\partial t^2}$$ Using the relativistic notation $\partial^2 = \partial_t^2/c^2-\nabla^2$, we may write the equation more compactly as: $$\partial^2\Psi = 0$$

## Klein-Gordon equation (energy-momentum)

To get a wave equation for a free massive particle, we just start from the relativistic expression $E^2 = (\mathbf{p}c)^2 + (mc^2)^2$, and insert the relations $\mathbf{p} = -i\hbar\nabla$ and $E = i\hbar\partial_t$ that we got above: $$-\hbar^2\frac{\partial^2}{\partial t^2} = -\hbar^2c^2\nabla^2 + m^2c^4$$ Then you right-multiply this by a wave $\Psi$, and you've got your equation of motion. To get the conventional form, insert $\partial^2 = \partial_t^2/c^2 - \nabla^2$ and rewrite it: $$\partial^2\Psi = -\left(\frac{mc}{\hbar}\right)^2\Psi$$

## Where does it all come from?

Although the heuristic derivations above consider only the special case of a plane wave, the relations $E \sim \partial/\partial t$ and $\mathbf{p} \sim \nabla$ have quite deep origins. Technically, we say that the Hamiltonian is the generator of time translations, and momentum is the generator of space translations. If you have studied classical mechanics, this is related to Noether's theorem, which connects the time-translation invariance of a theory to conservation of energy, and space-translation invariance to conservation of momentum. (Another important quantity is the angular momentum, the generator of rotations, which is related to rotational invariance of the theory.)

$^\dagger$ We usually refer to the operator as $H$ and the eigenvalues as $E$ in quantum mechanics, so just rename $E$ to $H$ to get the standard form of the Schroedinger equation.

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