Wednesday, October 12, 2016

Time-Independent Schrödinger Equation in One Dimension

Time-independent Schrodinger Equation in one dimension

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.


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;


$h=6.62607004 × 10^{-34} m^2 kg / s$ 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;


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;


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


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


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


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.