### Integrating the Gaussian Function

As I take my first quantum mechanics class, I’ve come to find that integrating probability density functions are a pain. From only a few problems I’ve worked on, the integrals are long and tedious to do. If there’s a single theme present in these integrals, it’s the strategy of integrating by parts. However, I wanted to show a specific integration today, because it’s quite ingenious and allows one to integrate a function that would otherwise be very difficult.

But first, a little context. The integrals that one sees in quantum mechanics are essentially known as probability density functions. The idea is that, at each value of the function, one has an associated probability density. Multiplying this by a small interval $dx$ gives one a probability. This also means that any one point doesn’t have a probability. Instead, one requires an interval in order to get a probability. Formally, we get the following:

Additionally, we have some properties that we want for our probability. First, we want any probability to be greater or equal to zero, and second, we want the probability over the whole space that we are interested in to add up to one. These are reasonable requirements if we want to capture what we mean by probability when talk about it in everyday life.

What this second condition usually means though is that we have to normalize a function. We have to integrate over the whole space, and introduce a constant such that the integral then gives one.

This is where my quantum mechanics assignment comes in. I had to normalize what is known as the Gaussian distribution, and the integral looks like this:

My job was to solve for the constant A. At first glance, this integral doesn’t look too bad. However, after a few moments, you’ll realize that this isn’t as easy as it looks. Indeed, the only way I know of to solve this is to use neat trick. To start, we’ll square both sides of the equation. Since the one won’t change, let’s just look at how the left-hand side changes.

This might not seem any better, but what if we change one of the variables of integration on the right-hand side? If we go from $x$ to $y$, we get the following:

Now everything is coming together. If you’ve taken a calculus course and seen other coordinate systems, you should immediately recognize this to be a good candidate for switching into polar coordinates. Doing so gives us the following:

This is something that is much easier to integrate, since we have the factor of $r$ in the denominator. Performing the integration gives us:

This is quite a nice expression, because we know that this is equal to one (from the fact that we need the equation to be normalized to one). Therefore, we get $A = \sqrt{\frac{\lambda}{\pi}}$.

Without squaring our original integral, it would have been very difficult to evaluate. However, by seeing this clever workaround, we were able to turn a difficult integral into one we could evaluate without too much trouble.

### The Sweet Spot

Despite what many schools actually do, I think most of us can agree that learning is highly personal. What works for you might not work for me, and there’s nothing wrong with that. Thankfully, there’s more than one way to learn a subject.

However, what I fear happens early on in school is that, if a student struggles with the explanations and teaching style of a particular teacher, the student writes off the subject they are learning as unattainable. This can only be made worse when one sees the other students around them succeeding, while the ideas in the subject can’t seem to sink in to the student. It wouldn’t be surprising to me if this is a frustrating experience that also makes one have bad memories of a particular subject.

I’m pretty sure I’ve mentioned this before, but I’m eternally thankful to my parents for helping me learn arithmetic early on (as well as to my own, perhaps slightly better than average ability), which meant that the early subjects in mathematics during elementary school was easy for me. Without this boost, I feel like I would have been like many of my friends, where mathematics and science were subjects that one “survived”, but never really liked. From this small advantage, I was able to turn it into a whole education trajectory, where I’m now studying mathematics and physics in university.

The tragedy here is twofold. First, there’s the simple fact that mathematics is not linear. Despite what one may think after looking at a school curriculum, there’s not a required progression. Of course, it’s always nice to know arithmetic to speed up calculations within other fields of mathematics, and learning algebra is definitely a prerequisite to calculus, but should you take calculus first, or a class on proofs? Should you study graph theory before probability, or linear algebra before number theory? The answers to these questions depend heavily on the subjects, but the point I want to illustrate is that you don’t necessarily have a set progression in which you have to learn one subject before the next. Instead, you have a range of fields, and many can be studied independently of the others (though of course, one can always find interesting connections between fields).

Coming back to arithmetic, the problem with this is that arithmetic is now seen as the first “hurdle” to get over while learning mathematics at school. If you can’t get a grasp of arithmetic, then it’s as if you’ll be incapable of being good at other branches of mathematics (or, at least, you’ll have a lot of difficulty). But as I mentioned in the last paragraph, there are other fields of mathematics where the reliance on arithmetic may not be as great, so the student would be fine. However, the way mathematics progression is set up now, these small discrepancies between students now grow very large after only a few years, which begs the question, “Why arithmetic first?” If we started with a different subject, would that create a different set of students who seem to “excel” at mathematics? My feeling is that it would to some extent, which means we need to be careful in treating the beginning of mathematics education as a hurdle.

The second part of the tragedy is that, as I mentioned in the beginning, learning is personal. As such, there are different ways to teach subjects to a student. Different ways will work for different students, so the most critical thing to do when you don’t understand a subject is to try and learn it in a different way. If your class has a large emphasis on equations and solving abstract systems of equations, perhaps you need to try and transform this into a visual. It’s incredible how helpful it can be to have a visual when solving abstract equations. It might not make the manipulations easier, but it gives clarification as to why a particular strategy is used. And often, that can be the big difference.

I know that I’ve rambled on about both “higher” mathematics and more elementary mathematics, but my message is the same. If you’re having difficulty with a subject (which happens at every level), look for a different way to understand the material. You might look for a visual, or perhaps you rather work with only the abstract equations. Maybe diagrams help. The possibilities are numerous, so don’t feel like you have to learn things a certain way.

What I’m referring to here is your sweet spot, which is the amount of abstraction that you feel comfortable with while learning. I feel like I have a good grasp on calculus, but if you make it too mathematical and abstract, I get lost. It’s the same story with physics. I like to have some mathematical sophistication, but I’m not able to go off the deep end with it. Therefore, when learning a new subject, I try to find explanations and material that’s at my level. I know there’s no point trying to learn physics in the most mathematical and abstract way for now, because I’ll get lost. Instead, I start at my level, and I let myself become familiar with the subject. Then, when I start feeling more comfortable, I can extend my studies to the more sophisticated concepts. There’s no use to metaphorically bang your head against a brick wall when you simply don’t understand a concept. Use different or easier material as a stepping stone for what you want to learn.

This is exactly what I had to do when I worked in the physics department of my university this summer. I had to learn general relativity, so did I take the most mathematical and sophisticated book on general relativity? Of course not! I used the much more introductory book by Sean Carroll (which is great, mind you) to help me get through the initial hurdle of learning the subject. Without this book, it would have been extremely difficult to understand what was going on in my work. However, since this book was made for people like myself (who were just starting to learn), it made my journey into general relativity much easier.

Therefore, my advice is simple. Figure out what your sweet spot is when learning, and don’t feel afraid to seek out alternative explanations to what you get in class. Almost always, there’s another way to understand a concept, so use that alternative perspective to help you learn. Remember, you’re not necessarily like every other student in your class, so chances are high that you’ll get a class where the explanations the professor uses doesn’t mesh well with you. When this happens, seek the alternatives, and this will help you a lot as you advance your studies.

### Linear versus Cyclic Permutations

One aspect of probability I’ve always found to be a little tricky is the part where you need to count things. In theory, this sounds easy enough. After all, it’s just looking at the complete list of things you’re studying, and enumerating them, right?

Well, we know that things become much more subtle when you have a big number to count and you have to be careful when using the “tricks” of multiplication in order to avoid duplication. This is the part that has sometimes seemed straightforward, while at other times being totally mystifying.

If you’ve taken an introductory probability class, you’ve undoubtedly come across the notion of permutations. This concept simply deals with answering the question, “How many ways can I order these items I have?” Just to briefly review, let’s look at an example. Suppose I want to know how many different ways I can order four items: a, b, c, and d. The answer is given by $4!$, which can be seen below.

This makes sense, and it’s usually the image we have in mind when thinking about permutations. This readily generalizes to $n$ items, where the number or permutations is $n!$.

But here’s a question. What are the possible orderings of the four items from above if I arrange them in a circle like this?

Suddenly, the idea has shifted. Now, the absolute placement of the object doesn’t matter (eg. Is it first, second, third, or fourth?), but it’s placement relative to the other objects matters. If we look at object a, the only thing that matters is that it is in between objects $d$ and $b$. How the circle is oriented doesn’t matter. As such, these scenarios are now equivalent.

The question now becomes, “How do we address this change in possibilities within our expression?”

If we consider the circles I drew above, you may notice that there are four of them. This is not a coincidence. The reason is that if we have a certain configuration of the circle with $n$ objects comprising the circle, we can rotate the circle $n$ times without changing the relative positions of the objects. This is because the circle possesses rotational symmetry. As such, the number of ways you can permute the four objects is the the number of ways we can permute them when they are in a line, divided by $n$. Mathematically, this looks like: $P_{cyclic} = \frac{n!}{n} = (n-1)!$ Here, I’ve used the symbol $P_{cyclic}$ to represent the permutations that can be made in a cycle.

This wasn’t meant to be a long post, but it’s something that I thought was interesting, since (without thinking about) we usually only consider the linear case when looking at permutations. I wanted to show here that circular permutations are just as easy as regular permutations.

### Period of a Pendulum

The pendulum is a classic physical object that is modeled in many introductory physics courses. It’s a nice system to study because it is so simple, yet still allows students to see how to study the motion of a system. Here, I want to do through the steps of deriving what is usually seen as an elementary result, the period of a pendulum, and show how it is actually more complicated than what most students see.

To begin with, what exactly is a pendulum? This may seem like an easy question, but it’s a good idea to have a well-defined system. So, the pendulum we will be looking at today is called a simple pendulum. Surprising no one, a simple pendulum is the most idealized pendulum, consisting of a point mass attached by rod of fixed length. This means we aren’t dealing with a pendulum that has a flexible rope that changes length, nor do we have something like a boat, which doesn’t quite act like a point mass since the mass is distributed throughout the object and isn’t localized. In other words, our situation should look something like this:

You may be wondering why we aren’t using Cartesian coordinates, and the reason is quite simple. In Cartesian coordinates, we would need to specify both the $x$ and $y$ coordinates, which requires two degrees of freedom and is also a pain in this particular setup. By contrast, using polar coordinates is more compact since the radius $r$ is fixed (in this case, $r=l$), which means we only have one degree of freedom, the angle $\theta$.

To begin our analysis, we will start with our generic equation for conservation of energy, which looks like this:

Here, the kinetic energy is $T$ and the potential energy is $U$. To know the kinetic energy, we need to know the magnitude of the velocity of the object, which we don’t know at the moment (and which changes depending on the angle $\theta$). We do know though that the kinetic energy is given by $T = \frac{1}{2} m v^2$, where $v$ is the magnitude of the velocity (the speed), so we will keep that to the side.

We also know that the potential energy is given by the familiar equation $U = mgh$ on Earth, where $h$ is the height of the object from the ground. To find this height $h$, we need to draw some clever lines and invoke some geometry:

From the diagram above, we can see that the height is given by $h = l \left( 1 - \cos \theta \right)$. Therefore, the potential energy is:

With this, we almost have everything we need in our equation. The goal is to isolate for our speed $v$, so we can then integrate it over a whole cycle to find the period. To do this, let’s remember our conservation of energy equation: $E = T + U$. This equation states that the total energy $E$ is always a constant in time. In other words, $\frac{dE}{dt} = 0$, and so we can simply find the total energy at one particular instant, and then substitute it for $E$.

What we will do is consider the energy that the pendulum initially has, just before it is allowed to fall. At that moment, it has an initial angle which we will call $\theta_0$, and since it isn’t moving, the pendulum has no kinetic energy. Therefore, the energy of the pendulum is:

We can now make this equal to the sum of the kinetic and potential energy at any time to get:

Since each term in this equation has the mass $m$ in it, we can see that our result will be independent of the mass. If we then isolate for $v^2$, we get:

At this point, we need to think about what the speed $v$ is. The definition of speed is $v = \frac{ds}{dt}$, where $s$ is the path length. Fortunately, the path length of a pendulum is very easy to find, since it’s simply the arc length of a circle!

From the above diagram, we can see that the path length is given by $s = l \theta$. Therefore, the speed is:

We can now substitute this into Equation \ref{vSquared}, and solve for $\frac{d\theta}{dt}$:

Note here that I’m only considering the positive solution for $\frac{d\theta}{dt}$, since we will be solving for the period, which is a positive quantity. What we will now do is employ the method of separation of variables to integrate this quantity. If you aren’t familiar with this method, I suggest taking a look at a resource on differential equations such as here. Separating our variables gives us:

This is good. We now have an expression for $dt$, which means we can integrate it for the angle between $0$ and $\theta_0$, and this will be one quarter of the period. To see why it’s only a quarter of the period, look at the following sketch (each arrow is a quarter period):

Integrating gives us:

And solving for the period $T$ gives:

This is the full expression for the period of a pendulum at any initial angle $\theta_0$. The only slight issue is that, while correct, this expression is not easily integrated. In fact, I don’t know how to integrate it at all. What we would like the period to be is of the form:

The expression above would be what is called a Taylor expansion, with the first term being what you might have already seen to be the period of a pendulum, plus some correction factors that are contained in the ellipsis. To get it into this form, we want to be able to use the binomial expansion, which is given by:

To do this, we need to transform Equation \ref{fullPeriod}. First, we will perform what may seem like a totally random substitution, but bear with me. We will change coordinates and go from $\theta \rightarrow \psi$. This mapping will be done using the following relation:

Looking at this relation, we can see that when $\theta$ ranges from 0 to $\theta_0$, the corresponding variable $\psi$ varies from $0$ to $\pi / 2$.

Implicitly differentiating each side gives us:

We can then pull out a handy trigonometric identity called the double angle identity, which is given by:

Using this identity, we can rewrite the expression inside the square root of Equation \ref{fullPeriod} as:

From here, we can insert our original substitution from Equation \ref{transform} into the second term above, giving us:

Just to note, from the second to third line, I simply used the Pythagorean theorem. Now, since we wanted the square root of $\cos \theta - \cos \theta_0$, we can take the square root of the above expression. Furthermore, we can use Equation \ref{dTheta} in order to find an expression for $d \theta$:

From this, we can insert everything into the integral of Equation \ref{fullPeriod} and simplify. Note here that I’ve omitted the prefactor in the front of the integral just to get things a little cleaner, but we won’t forget about it.

We’re almost there. Now, we can simply used a rearranged version of the Pythagorean theorem to write:

Here, I’ve made use of equation (13) again in order to write this expression in terms of $\psi$. Throwing this all together and reintroducing the prefactor in front for the period gives us the following result for the period:

I don’t know about you, but that was a lot of work. This integral is actually a special kind of integral. It’s called a complete elliptic integral of the first kind, and is defined by:

In our case, $m = \sin^2 \left( \frac{\theta_0}{2} \right)$. What’s nice about this form of the integral is that it is indeed in binomial form, so we can expand it. We therefore have:

This looks like quite the jumbled expression, but we can can write it quite succinctly in the following form:

Here, the double factorial sign (!!) means that we skip a number each time we do the multiplication. Therefore, $5!! = 5 \cdot 3 \cdot 1$ and $6!! = 6 \cdot 4 \cdot 2$. You can verify that this does represent the above expression of Equation \ref{long}. We are now in a better position to evaluate the integral. It looks like this:

This last integral is a bit of tricky one, but we will show that the integral is given by:

To get this result, we will use recursion. First, we note that the values of $n$ are all positive, which is clear from Equation \ref{sum}. This means our lowest value of $n$ will be one. If we label the integral in Equation \ref{In} as $I(n)$, then we can evaulate this function to get:

With the base case out of the way, we now tackle the whole integral. Let’s start by splitting up the integral as such:

We can now use integration by parts to partially evaluate this integral. If we use $u = \sin^{2n-1} \psi$ and $dv = \sin \psi$, we get:

The first term evaluates to zero, and so we are only left with the integral. We can then change the cosine into a sine and rearrange things to give:

If you look at this and compare it to our definition of $I(n)$ from Equation \ref{In}, you’ll notice that we can write the above equation as:

Solving for $I(n)$ gives:

This is a recurrence relation, which means it tells us how to construct the next term from the previous one, as long as we have a beginning “seed”. Thankfully, we do have one, which is $I(1) = \pi/4$.

What we want to do at this point is to keep on applying the recurrence relation to the term $I(n-1)$, until we get all the way to $I(1)$, where we stop. I’ll illustrate a few of these for you, and hopefully it becomes clear what the pattern is.

I could continue, but this is a good representation of what happens. In summary, the numerators of the fractions are odd numbers (since they are in the form $2k+1$), and the denominators are even numbers (since they are in the form $2k$). Furthermore, as you go down the fraction, you go from an odd number to the next closest odd number, and the argument is the same for the even numbers. Therefore, what we are really doing is another factorial all the way until we get to $I(1)$, which we can evaluate since it is our starting seed. Therefore, we get:

Now that we have this result, we can put it all together to give us:

Expanding this gives us the following infinite series:

If we recall that $m = \sin^2\left( \frac{\theta_0}{2} \right)$ and we insert the prefactors for the period from Equation \ref{Period} in, we get the following result for the period of the pendulum:

This is the full expression for the period of the pendulum with any starting angle $\theta_0$. What’s quite nice about this expression is that we can immediately see that if $\theta_0 \approx 0$, then all of the sine functions become very close to zero and so the only important term in the square brackets is the one. At this point, the period becomes what one usually learns (for small angles): $T = 2\pi \sqrt{\frac{l}{g}}$.

Furthermore, we can see that when our initial angle gets bigger, it becomes more important to take on successive terms within the brackets of Equation \ref{Final}.

Hopefully, this wasn’t too bad. I wanted to go through the calculation as explicitly as possible, since I remember being a bit confused when I saw it for the first time. As such, I want to make sure things are illustrated nice and slow so everyone can follow.

What I love the most about these long analytical expressions is how you can recover the simpler result you had from simplifying the problem. We can easily see that our “usual” period is nestled within the long infinite expression. Lastly, I just wanted to make clear that one assumption we did make was that we were dealing with a point mass pendulum. In other words, we still weren’t quite modelling a physical pendulum, which requires taking into account the centre of mass of the bob and the rod of the pendulum together. Still, this is enough precision for today, so we will leave it at that.