Schrödinger's Cormorant

6.2 Bringing it All Back Home

Combining the wave equation, the de Broglie equation and the principle of conservation of energy leads us to the Schrödinger equation.

Let’s set about finding the Schrödinger equation. This is where everything starts to come together: we are going to need to combine a whole lot of the ideas we’ve developed in these first five chapters.

The wave function of an electron in an atom should be a standing wave.

We can begin by remembering our proposition from chapter one: if an electron is in a stable state in an atom, its wavefunction should represent a standing wave, meaning that the peaks oscillate in sign and intensity but do not shift in space. In chapter four, we explored the idea that a standing wave can be understood as a superposition of travelling waves, identical except that they are travelling in opposite directions. This led us to an equation for a simple sinusoidal standing wave:

Ψ = 2Asin(2πx/λ)cos(2πvt/λ) (VI-1)

The crucially helpful thing about this expression is that the space and time-dependent terms are separate – just two terms multiplied together – which makes it different from equations describing travelling waves. This is going to help us immensely in working out the maths to come.

Now we know from the previous chapter that a particle waveform won’t be a simple sinusoid like this – although, as Prof. Fourier was once again about to remind us, it should be possible to represent it as a sum of sinusoids - because that would not localise the particle in space. Moreover, particles are three dimensional, whereas all our theory has been confined to 1-dimension. But don't worry: the principles are exactly the same, the maths just gets more complicated and challenging once we introduce extra dimensions. So we'll keep thinking 1-dimensionally and trust that we can generalise down the line.

If we can describe our particle in terms of a standing wave, it seems reasonable to suggest that, since the space and time-dependent parts of the function are separate, only the spatial part needs to be different in order to get a standing wave that's appropriate to a localised particle. Therefore, by direct analogy with equation IV-2, we can propose that our wavefunction is of the form:

Ψ(x,t) = ψ(x).cos(2πvt/λ) (VI-2)

where ψ(x) is the spatial part. This massively simplifies our problem: now we only have to try to find ψ(x). How are we going to do that?

The wave function must obey the wave equation

In Chapter 4, we developed the wave equation (equation IV-5). This should be obeyed by any wave, including our proposed wave function for a particle. It provides a connection between the time and space parts of the wave function, at the level of second derivatives. Now that we have introduced Ψ as the wave function, we can write the wave equation as:

(VI-3)

Let’s see where this takes us.

Question 6-1

This process gets us to something very interesting – an equation which relates the spatial component of the wave function to its second derivative: a second order differential equation. This looks like something that might be soluble:

The de Broglie equation lets us bring in momentum.

So this is good. It means we have a way to find possible waveforms for our particle: they will be functions of x which are solutions to this differential equation. Cormorant, however, however spots the fly in the ointment: “How the hell do we know what the wavelength is, though?”

What we are after in our mission to find the wave function is an equation built around fundamental constants and particle properties that we can readily define and measure, rather than this mysterious wavelength.

Of course – Cormorant gets there even before Prince Louis has sprung to his ghostly feet: λ is not entirely mysterious. We can relate it to momentum, using the de Broglie relationship (equation V-1): λ = h/p. Let's do that:

Question 6-2

This gives us a more mechanical-looking equation, featuring momentum rather than wavelength.

But, points out Nefertiti, this doesn’t seem all that helpful either. How should we know what the momentum of a stationary wave should be? Remember that a stationary wave can be seen as a superposition of travelling waves which are identical except that they are travelling in opposite directions. The momentum can in turn be understood as being associated with the individual component waves which are moving, even though their superposition is not. These component waves will have equal and opposite momenta, p and -p, but the key thing to notice in equation VI-5 is that the quantity we need is actually not the momentum but its square. Since p2 = (-p)2 the value of p2 will be non-zero for the stationary wave. But that still leaves us at square one: to find these individual component momenta, we would seem to need to know what the wave function is, and that is what we are trying to find. Cormorant feels like he’s flying round in circles.

“Well OK”, says Cormorant – impatiently now – “what can we relate momentum to, which would give us a properly useful equation? How about kinetic energy”? Nefertiti still looks unmoved: the kinetic energy of a stationary wave is still a bit strange because the wave isn’t going anywhere: like the magnitude of the momentum, it’s associated with the individual components that are superposed to construct the stationary wave. That doesn’t, at first sight, seem any more helpful than momentum in terms of avoiding quantities that we can’t easily define. Except .....

The Principle of Conservation of Energy to turn the equation into a more useful form

There is a way forward. The kinetic energy of a stationary wave might seem mysterious but it still has to obey the fundamental law of conservation of energy: if the total energy of the particle is E and the potential energy of the particle at any point in space is V, we can state that:

E = V + Ek

The total energy is something that in principle we can measure, independently of our quantum mechanical model, and the potential energy is something we can calculate, based on the external forces acting on it. So we can rewrite kinetic energy in terms of these two, more straightforward, terms: Ek = E-V. If we can then transform our wave equation to switch the parameter from wavelength to momentum to kinetic energy, we’ll be in a position to get something really useful.

Question 6-3

This is a big moment in our journey. What you have now constructed is a differential equation, the solution(s) to which are wave functions. Provided we know the mass of a particle and how its potential energy depends on position in space, the equation generates a function which (when squared) describes the probability distribution of the particle’s position, together with its total energy. So here it is, in bold, in a box, for you to celebrate. Ladies and Gentlemen, meet the Schrödinger equation:

Nefertiti feels we should celebrate our achievement. Cormorant senses a big night at the pub. But Nefertiti has other ideas: “Let’s construct a summary picture of all the ideas that contributed to our arrival here". Cormorant is thrilled. Here’s the result:

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