Suppose a particle travelled one dimensionally. Classically, the particle’s behavior should be in line with Newton’s laws of motion. Quantum mechanically, however, one cannot determine the exact position and momentum of the particle and as such is reduced to using a probabilistic interpretation. Here we will explore the notion of a probability density and how one could derive such an artifact from a simple Taylor expansion argument.

Let $I$ be a one dimensional interval of points taken from the real line, i.e., $\mathbb{R}$. Now suppose that $P(x)$ is the probability of finding a particle at $x\in I$ and, for simplicity, let’s keep the probabilities positive.

**Example**: Suppose that the probability function looks like $$P(x)=\frac{x^3}{3}$$ and $x$ ranges from $[0,3^{1/3}]$. So when $x=0$, $P(x)=0$ and when $x=3^{1/3}$, $P(x)=1$.

Now, how does a probability density arise, and what is a probability density anyway? Recall that a density is simply a quantity divided by its volume. More explicitly, a density has the form $$\rho = \frac{\text{mass}}{\text{volume}}$$ where $\text{mass}$ generally refers to the amount of stuff and volume generally refers to the space the stuff exists in. If we are dealing with probabilities, the mass is then the amount of probability, so $m=P(x)$, and the volume in one dimension is simply a length, $V=L$. We will soon see how the probability density naturally arises.

Now that all the pieces are set into place, suppose that we have a particle that is traveling one dimensionally. It zips by some point $x_1$ and is making its way over to $x_2$. If we want to find the probability that the particle is somewhere within the region $\Delta x\equiv x_2 – x_1$, then we want to consider $$\frac{P(x_2)-P(x_1)}{\Delta x}.$$

How do we come about this quantity? Well, we use a simple trick, we can always write

\begin{align}P(x_2)&=P(x_2-x_1+x_1)\\&=P((x_2-x_1)+x_1)\\ &=P(\Delta x+x_1)\\&=P(x_1+\Delta x).\end{align}

Then by Taylor expanding, we find

\begin{align}P(x_1+\Delta x) &= P(x_1) + \frac{\Delta x}{1!}\frac{\text{d}P(x)}{\text{d}x}\Bigg|_{x=x_1}+\mathcal{O}((\Delta x)^2)\\P(x_1+\Delta x)-P(x_1)&=\frac{\Delta x}{1!}\frac{\text{d}P(x)}{\text{d}x}\Bigg|_{x=x_1}+\mathcal{O}((\Delta x)^2)\\\frac{P(x_1+\Delta x)-P(x_1)}{\Delta x}&=\frac{\text{d}P(x)}{\text{d}x}\Bigg|_{x=x_1}+\frac{1}{\Delta x}\mathcal{O}((\Delta x)^2).\end{align}

Now making $\Delta x$ very small, we see that $\mathcal{O}((\Delta x)^2)/(\Delta x)$ is small enough to neglect, leaving us with $$\frac{P(x_1+\Delta x)-P(x_1)}{\Delta x}=\frac{P(x_2)-P(x_1)}{\Delta x}=\frac{\text{d}P(x)}{\text{d}x}\Bigg|_{x=x_1}.$$

So how do we connect this quantity to the probability density? Well, it **is** the probability density, with some small tweaking. We call the “mass” as $$m = |P(x_2)-P(x_1)|$$ and the “volume” as $$V=L=|\Delta x|.$$ Doing the absolute value gives us a positive mass which is nice, **although not necessary**. Realizing this we find

\begin{align}\rho(x_1)&=\Bigg|\frac{P(x_2)-P(x_1)}{\Delta x}\Bigg|=\Bigg|\frac{\text{d}P(x)}{\text{d}x}\Bigg|_{x=x_1}\Bigg|\end{align}

So this is the probability density of finding a particle in a region between $x_2$ and $x_1$. We could also make this more general and make $x_1$ arbitrary, then we would find our final function

\begin{align}\rho(x)&=\Bigg|\frac{\text{d}P(x)}{\text{d}x}\Bigg|\end{align}

**Example:** Returning to our example, if $P(x)=x^3/3$ and $\Delta x=3^{1/3}$, then we can find the density easily; $$\rho(x)=\Bigg|\frac{\text{d}P(x)}{\text{d}x}\Bigg| = |x^2|=x^2$$ since $x^2\ge0$ for all $x$ values. Now, to check that the particle is * somewhere* in between $[0,3^{1/3}]$, we simply need to integrate the density over the full volume, i.e.,

\begin{equation}\int_{0}^{3^{1/3}}\rho(x)\text{d}x =\int_{0}^{3^{1/3}}x^2\text{d}x = \frac{(3^{1/3})^3}{3}=1.\end{equation}

Since the answer is $1$, we know that the particle will be found somewhere in that interval and we are done. Quantum mechanics offers a (largely correct) way of understanding the behavior of particles at the fundamental level. Although there are always philosophical issues with such a probabilistic interpretation of mechanics, the underlying ideas are quite simple. Anyone can do quantum mechanics, to understand quantum mechanics, however, that’s another story.