The Schrödinger Equation — 1, 2
If we wanted to answer the question of what’s truly fundamental in this Universe, we’d need to investigate matter and energy on the smallest possible scales. On such scales, reality starts behaving in strange, counterintuitive ways. We can no longer describe reality as being made of individual particles with well-defined properties like position and momentum. Instead, we enter the realm of the quantum: where fundamental indeterminism rules, and we need an entirely new description of how nature works. But even quantum mechanics itself has its failures here. They doomed Einstein’s greatest dream — of a complete, deterministic description of reality — right from the start. If we lived in an entirely classical, non-quantum Universe, making sense of things would be easy. As we divided matter into smaller and smaller chunks, we would never reach a limit. There would be no fundamental, indivisible building blocks of the Universe. Instead, our cosmos would be made of continuous material, where if we build a proverbial sharper knife, we’d always be able to cut something into smaller and smaller chunks.
With quantum physics, new rules are needed, and to describe them, new counterintuitive equations. The idea of an objective reality goes out the window, replaced with notions like probability distributions rather than predictable outcomes, wavefunctions rather than positions and momenta, and Heisenberg uncertainty relations rather than individual properties.
These paradigm shifts necessitated something completely new. Something that described the quantum world. Enter: the Schrödinger equation. Postulated by Erwin Schrödinger in 1926, it is undoubtedly one of the most brilliant equations known to humanity.
It would be prudent to kick this off with some groundwork. Below, is a brief explanation of what the Schrödinger equation really is but if you’d like a deeper understanding, I’d recommend this:
Quoting Wikipedia:
The Schrödinger equation is a linear partial differential equation that governs the wave function of a quantum-mechanical system.
Though that probably made little to no sense. What does it mean to say that a function is linear? Or that it’s partially differentiated? Well answering that would mean delving deeper into the mathematics of it all. But as far as we’re concerned today:
The Schrödinger equation provides a way to calculate the wave function of a system and how it changes dynamically in time.
Conceptually, the Schrödinger equation is the quantum counterpart of Newton’s second law in classical mechanics. Given a set of known initial conditions, Newton’s second law makes a mathematical prediction as to what path a given physical system will take over time. The Schrödinger equation does the same but instead gives the evolution of a wave function over time.
The Schrödinger equation can be seen written in many different forms. Each of which corresponds to a different physical or mathematical scenario.
The simplest form is naturally the first one. It just has four terms and seems easy to deal with. When expanded, however, it’s a lot less friendly.
This is my attempt at explaining how the Schrödinger equation is solved. The logic discussed in this article is exactly what is used to solve the equation in more complicated systems as well. For example, the hydrogen atom’s Schrödinger equation takes the form:
The logic I will follow here is what would be required to solve this monstrous equation too. We don’t need to necessarily bother ourselves with what any of this means but what I’d like for you to take away from this is the level of complexity that goes behind computing such equations.
To proceed further, we’ll break the Schrödinger equation into smaller parts. First, we see that the Greek letter, Psi, Ψ, appears often.
If you’re familiar with what Ψ is, you’d understand that the Schrödinger equation deals with a quantity known as the wave function.
Ψ → the wave function is a mathematical expression for everything we know about a particular quantum system.
So with our newly established understanding, we can also conclude that all the other variables must affect the wave function in some form. Or in other words, they must describe the wave function’s behavior.
Well so far, this works. We know that the equation describes wave functions. But how exactly does it achieve that? To do this, we explore each term individually.
The first term represents the kinetic energy of the system.
The second term happens to represent the potential energy due to the environment.
Finally, the letter E can be thought of as the total energy but this is a bit hand-wavy. Essentially, what we’re looking at is this:
Kinetic energy + potential energy = total energy…ish
It’s important to note that this is a very simplified example. Most systems include other terms for energy too. Terms that are a lot more complex than just kinetic and potential energy. Here, we are only working with the time-independent Schrödinger equation. So to simplify things further, the total energy remains constant through time. This equation tells us about the behavior of the wave function depending on its energy and solving it tells us what the wave function could be.
I’ll spare the excessive mathematics here but to further simplify our equation, even more, we can consider the movement of a particle in a single dimension. Confining it to one arbitrary axis would allow us to simplify the math a great deal. What this essentially means is that our system now consists of one particle that can travel left or right but not up or down. This isn’t exactly easy to recreate in reality but remember, reality doesn’t matter here.
We’re now going to place two barriers at r = 5 and r = 9. These two “barriers” are infinitely thick, impenetrable plates. Again, another non-physical assumption but we do this to eliminate quantum entanglement. What we’re now saying is that the particle cannot be found at any position on the r-axis that isn’t between 5 and 9. It must always be in that region of space.
This setup is known as a one-dimensional particle in a box. In this case, we’re saying that the potential in between the walls is 0. In the sense that there’s nothing influencing the particle in any way between the barriers. We can show this on an energy diagram.
From here, we can take the convenient V=0 region and substitute that into our original Schrödinger equation. So now our equation would look like this:
Now, we could also solve the equation beyond the barriers — specifically, in the regions where the potential is infinite. This would, however, be balanced out by the fact that the wave function would be 0. 0 times infinity would yield 0 again. The wave function, as we’ve already established, corresponds to the probability distribution of a particle. And if that particle has a zero probability of being found in the infinite potential region, then its probability distribution, i.e., the wave function, must also be 0. This infinite potential is immediately offset by the zero wave function. What we understand is that the wave function can’t exist in those regions and consequently we don’t have a need to solve for it.
The main question, really, is what does the wave function look like in the non-trivial region where the potential is 0. Well, at this point, we’ve got a fairly straightforward equation to solve.
Those of you that are familiar with calculus will recognize that this is a second-order differential equation. The left side of this equation refers to the second derivative of the wave function, Ψ, with respect to its position.
Essentially, if we find an equation that describes the gradient of our wave function, then we’ve solved for the first derivative of Ψ; dΨ/dx. This is essentially the equivalent of finding the rate at which the wave function changes with position.
Now if we take our dΨ/dx and find an expression for the gradient at each point of that, then we’ve solved for the derivative of the derivative of Ψ. Or in other words, the second derivative of Ψ. This is a very brief explanation of second derivatives. For more, check this out.
But here’s where things get a little weird. We know how to find the second derivative of Ψ if we know what Ψ is. We can simply find Ψ’s derivative and then solve for its derivative again. d²Ψ/dx² can be calculated given Ψ. Here, however, we have no idea what Ψ is. We’re working backward.
Our question is now this: Find Ψ from
Even with our massive simplifications, we’re left with a very daunting equation. After eliminating the derivative, we still have the EΨ term to deal with on the right side. Fortunately, all of these terms are constants:
hbar is the reduced Planck’s constant, 2m is, well, the mass of the particle multiplied by 2, and E, as we established at the very beginning, does not change with time. Now, we can rearrange our equation so that what we’re left with, is this:
And since we’ve got a bunch of constants on the right, we can combine all of them into one larger constant -k²:
At this point, with some fairly simple mathematical manipulation, what we’ve found is that the second derivative of Ψ is equal to some constant, -k², times Ψ itself. We’ve gone through all this trouble to try and find a wave function, Ψ, that fits our equation. One such equation that obeys this, is a sinusoid.
Here is a sine wave, y=sin(x):
Here is its derivative, y=cos(x):
Here is the derivative of the derivative, y=-sin(x):
What you see is that we’ve got the original function, sin(x), multiplied by a factor of -1. In other words, the second derivative of a sine curve spits the sine curve back out but with a -1 by its side.
So, if we carefully account for the constant in our equation, -k², our solution is going to look like a sinusoid:
Where the term in red, if you notice, is essentially the constant k:
This may seem like we’ve got what we wanted but we’re not fully done here. This is only half the picture. What we’re yet to consider is the behavior of the wave function at the barriers. If you recall, we said that the wave function must be 0 within the walls but this also means that the wave function must be zero at the walls.
There are a ton of different reasons for this but here’s a particular one for why Ψ must be 0. We’ll proceed with contradiction. Let’s assume that we have a non-zero wave function at the walls. In other words, there’s some probability of finding our electron at r = 5 and 9. But at the same time, before and beyond these values, the wave function drops to 0 and as a consequence, so does the probability. This means that our particle, at 5, must have a non-zero wave function value, and our system, at 4, must have a zero wave function value. The only way this would be physically possible is if the wavefunction witnessed a sudden jump. So, what we’d get, as an extension of the existing wave function, is this:
Notice that at the walls, the wave function goes from 0 to a non-zero value. At 5 and 9, our gradient is infinite. And that means the derivative of our wave function at those points must be undefined and as a consequence, the second derivative must be undefined too.
Here, the Schrödinger equation begins to break down. At the walls, since the second derivative of Ψ is undefined and the potential is infinite, we get something that looks like:
“undefined” + ∞ = total energy.
This is even worse than within the walls where we at least had Ψ = 0 to offset the infinities. So, to summarize, the value of the wave function at both the walls must be zero.
Solving the Schrödinger Equation — 2
Boundary conditions
Here we take inspiration from mathematicians and like they always do, we generalize our system to death. First, we throw the r-axis out the window. Instead, we now replace our barriers at positions on the x-axis where x = 0 and x = a (a is just some arbitrary number, doesn’t matter). The electron is discarded and now we just have some particle. Our system is now “some particle” confined within the boundaries of x = 0 and x = a whose wave function takes the form of sin(kx)…ish
Well, the logical way to proceed forward is to find a way to encode the boundary conditions into our wavefunction. In other words, we must have a wave function that yields 0 at x = 0 and x = a.
We know that when x = 0,Ψ = 0 as well. This works out quite nicely since sin(k(0)) = 0. Zero is equal to zero, the math checks out.
At x = a, however, things are different. We know that Ψ = 0 and as a consequence, have sin(ka)= 0. What we find is a restriction to the kind of sine wave our system takes. For example, this is one possible solution:
It’s half a sine wave. The value of Ψ at each wall is zero, as we need it to be. Looking at a, we know that we’ve gone through half a sine wave and this must mean that the argument of our original function, sin(ka) must equal 180° because that’s the period of half a sine wave.
sin(180) = 0 and this checks out.
If we switch over to radians, the more natural unit of angle measure, then we know ka = 𝜋 = 180°.
Resubstituting k, we know that if our solution is half a sine wave then this equation must hold:
With some fairly trivial mathematical manipulation, we now have an equation that tells us the energy, E, of our system.
If our wave function looks like half a sine wave, then the energy of our particle is given by the equation shown above.
Another possible solution is a full sine wave fitting into this region 0 < x < a:
At the wall on the right, the argument must now equal 360°. Or in other words, 2𝜋.
Manipulating our equation again, we have:
But this doesn’t just stop here. We could have three sine waves, four sine waves, and so on.
The sine wave is periodic — it repeats every 2𝜋. So, in principle, the solution to our equation could take any form of n𝜋 where n is an integer. And since the solution can take any form of n𝜋, the solution can also take any form of half n𝜋. So again, in principle, we could have three half-sine waves, four half-sine waves, and so on.
Each scenario that we have corresponds to a different energy. What we witness here is a quantum phenomenon known as quantization. We find that since we have specific wave functions, we must have specific energies. Or in other words, our particle must have specific energies. It cannot have an energy that is in between two levels. And considering the most rudimentary setup, our particle must have minimum energy of
This is because we can fit a half-sine wave in there, not a fourth or a third of a sine wave. The half sine wave solution corresponds to the lowest energy and by virtue of that, it also means that a particle cannot have energy lesser than it.
So with different solutions, we can find different energies. But this doesn’t end here. We have one last thing to worry about: normalization. This is a really interesting idea and I will probably write about it in the future but I don’t want to saturate you with more. The crux of it is that normalization adds a factor to our solution.
Well, we know that the wave function has something to do with the probability of finding a particle at different positions in space. So this added factor must affect that in some way. With the lowest energy level we first explored — the half-sine wave — we know that the probability of finding our electron is highest in the middle of our region of space. That’s a fairly straightforward conclusion. But when we switch energy levels and consider the full sine wave solution, the highest probability isn’t the middle anymore. This normalized factor corrects that and scales the wave functions so that all the probabilities add to 1.
With that, I’ll end this here! Thank you for reading me.
