Schrödinger's Equation is one of the most important equations in modern physics. It's also fairly easy to derive (in one dimension) with some basic Calculus and a few fundamental concepts from Newtonian Mechanics and modern particle physics. So that's what we'll do here; derive the time-independent version of Schrödinger Equation in one dimension.

The approach I take in this post mimics the one taken here with additional details filled in since some observations made are not immediately apparent to non-physicists.

## Preliminaries

There are several axioms and basic results we need to know in order for the rest of the derivation to make sense. We'll start with three basic relationships;$E=KE+PE\\

E=hf\\

p=\frac{h}{\lambda}$

$h=6.62607004 × 10^{-34} m^2 kg / s$ is Planck's constant.

$f$ is the frequency of the wave, $\lambda$ is its wavelength.

$\omega=2\pi f$ is the wave's angular frequency.

We define the angular wavenumber as $k=\frac{2\pi}{\lambda}$.

We will also need the reduced Planck's constant $\hbar=\frac{h}{2\pi}$.

From Newtonian Mechanics, we remember that $KE=\frac{1}{2}mv^2$ where $m$ is the mass of the object and $v$ is its velocity. We also recall that we can write potential energy as $U$, thus transforming the energy equation above into;

$$E=\frac{1}{2}mv^{2}+U$$

We can write momentum a couple different ways. What we see above, $p=\frac{h}{\lambda}$ is de Broglie’s hypothesis. From Newtonian Mechanics, we recall that $p=mv$. It's worth noting that $E=hf$ is Planck’s hypothesis for the energy of a photon.

It will be useful later to have the energy equation written as $E=\frac{p^2}{2m}+U$. We can do this by starting from the classical equation for momentum as such;

$$p=mv\\

p^2=m^2v^2\\

\frac{p^2}{2m}=\frac{m^2v^2}{2m}\\

\frac{p^2}{2m}=\frac{mv^2}{2}$$

## Choosing a Wave Function

The final preliminary piece we need is a wave function written in this form;

$$\Psi=e^{i(kx-\omega t)}$$

To derive this function we're going to remember Euler's formula

$$e^{ix}=\cos{x} + i\sin{x}$$.

If we just look at $kx$, then we see that $\cos{kx}=\cos\big(\frac{2\pi}{\lambda}x\big)$. If we take $\lambda=1$ we see this function looks like figure 1.

Figure 1 - $g(x)=\cos\big(\frac{2\pi}{\lambda}x\big)$ |

This is a standing wave and we need a wave function that takes time into account as well. But why do we choose to use $kx-\omega t$ in Euler's formula?

Let's start with the velocity of the wave. Like a lot of things in physics, being able to write velocity in a couple different ways turns out to be immensely useful. In this case, we know we can write the velocity as the distance $x$ over the time $t$ to get $v=\frac{x}{t}$. In terms of waves, we can also write the velocity as the frequency $f$ times the wavelength $\lambda$ to get $v=f\lambda$.

We need to be clever, so we're going to multiply this second version of the velocity equation by $\frac{2\pi}{2\pi}$ to get

$$v=\frac{2\pi f\lambda}{2\pi}$$

Then if we bring $\lambda$ to the denominator, we can rewrite $v$ as

$$v=\frac{2\pi f}{\frac{2\pi}{\lambda}}=\frac{\omega}{k}$$

Now, we get

$$v=\frac{\omega}{k}=\frac{x}{t}$$

So that

$$\omega t=kx\implies kx-\omega t=0$$

Thus $\cos(kx-\omega t)$ gives us a moving form of the wave we found in Figure 1, which is, obviously, $1$ whenever $\omega t=kx$.

Heading back to Euler's formula, and using $kx-\omega t$ we get

$$\cos(kx-\omega t)+i\sin(kx-\omega t)=e^{i(kx-\omega t)}$$

We call this $\Psi$ and thus arrive at

$$\Psi=e^{i(kx-\omega t)}$$

## The Time-Independent Schrödinger's Equation

Starting with $\Psi=e^{i(kx-\omega t)}$, and since we want the time-independent version, we take the partial with respect to $x$ of $\Psi$ to get$$\frac{\partial\Psi}{\partial x}=\frac{\partial}{\partial x}e^{i(kx-\omega t)}=ike^{i(kx-\omega t)}=ik\Psi$$

Taking the second partial with respect to $x$ we get

$$\frac{\partial^2\Psi}{\partial x^2}=(ik)^2\Psi$$

Now, remembering de Broglie’s hypothesis we have $p=\frac{h}{\lambda}$, but $\lambda=\frac{2\pi}{k}$. Thus we can rewrite

$$p=\frac{h}{\lambda}=\frac{h}{\frac{2\pi}{k}}=\frac{hk}{2\pi}$$

We remember that $\frac{h}{2\pi}=\hbar$ (reduced Planck's constant) and we have $p=\hbar k$. Thus $k=\frac{p}{\hbar}$.

Now, going back to our second partial of $\Psi$ we have

$$\frac{\partial^2\Psi}{\partial x^2}=(ik)^2\Psi=-\big(\frac{p^2}{\hbar^2}\big)\Psi$$

Now we can multiply both sides by $-\hbar^2$ to get

$$-\hbar\frac{\partial^2\Psi}{\partial x^2}=p^2\Psi$$

But, $E=\frac{p^2}{2m}+U$. If we multiply through by $\Psi$ we get

$$E\Psi=\frac{p^2\Psi}{2m}+U\Psi$$

But, $p^2\Psi=-\hbar\frac{\partial^2\Psi}{\partial x^2}$ so

$$E\Psi=\frac{p^2\Psi}{2m}+U\Psi=\frac{-\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2}+U\Psi$$

And thus we arrive at the time-independent Schrödinger Equation in one dimension.

$$E\Psi=-\frac{\hbar}{2m}\frac{\partial^2\Psi}{\partial x^2}+U\Psi$$

Next time we'll look at the time-dependent Schrödinger Equation, which will be easier since we've already covered most of the physics requirements already.