This is the next step in working through Susskind's quantum mechanics book, Quantum Mechanics: The Theoretical Minimum^[1]. I'm a visual thinker, and I find that a geometric understanding of maths is essential to truly understand it, alongside the symbols that represent these concepts. In reading Chapter 2, I realised that each time I have previously studied quantum mechanics, I have mixed up various concepts and not truly understood how they fit together. Part of the challenge, in my opinion, is the lack of resources to get hands-on: to actually see what experiments show on a quantum scale for yourself, and to pull real levers and see how the underlying maths responds. This post is my own exploration as I followed Susskind through this chapter. I may not yet fully grasp it, but I'm certainly getting closer. There may also be some mistakes in these widgets, but if you can spot them it will certainly be a sign you are thinking about the underlying concepts in sufficient detail to gain a real understanding.

The last post built up vector spaces; this one puts them to work on the state of the simplest quantum system there is. The chapter is only a few pages long, but it packs in ideas that took me a while to see, so this post builds pictures, and widgets you can play with, until they feel (hopefully) obvious.

The objective:

1. See why two measurement results that sound like opposites, such as up and down, are perpendicular axes in the state space.
2. Find out what changes when you can look from more than one direction, and why that forces the state-vector components to be complex numbers if the world is actually 3D.
3. Show that a quantum state is not just a coin flip hidden in a box.
4. Bring it all together in one picture: a single arrow on a sphere (i.e. an interactive Bloch sphere).

Chapter 2 turns the spin experiments of chapter 1 into vectors. The simplest quantum object there is has just two possible results when you measure it, and its state lives in $\mathbb{C}^2$, the two-dimensional complex vector space from the last post. For an electron's spin, Susskind calls the two basis states "up" and "down", and writes any state as

There's one thing here that kept on confusing me with the electron's quantum spin states: |u⟩ and |d⟩ are orthogonal: perpendicular axes, with ⟨u|d⟩ = 0. But the words "up" and "down", and the values these measurements yield $+1$ and $-1$, make them sound like opposites, as if |d⟩ were minus |u⟩. It isn't.

I therefore wanted to represent the quantum state as something visual and not numerical. This is grounded in the fact that a quantum state is already a confusing thing, and here we are mixing the mathematics of probability and experimentally determined outcomes with trying to understand what exactly the quantum state itself is. My way around that is to park the spin state and think of it as a physical object that is totally relatable. It is just a thing here. This allows us to focus on what is strange about how that thing appears when we change our perspective (how we look at it), as well as the fact that we always see one of two things when we look. You can ignore "Spin" and being confused about what it is for now. There are some other concepts here to understand that don't rely on understanding what spin is. They just require observing the experimentally discovered facts and logically deducing what mathematical framework must produce those results. Ultimately that's what I guess the forefathers of quantum mechanics did when they invented this stuff: Heisenberg, Schrödinger, Dirac and the rest.

A box with one window

Imagine a box you can't see into, except through one small window in its lid. Open the window and you see one of exactly two things:

• a hat (a magician's pointy one), or
• a tulip.

A hat and a tulip have nothing in common, so nobody will mistake a tulip for "minus a hat", and nobody will think they're numbers. A tulip is not the opposite of a hat. It's just something else.

We are using the hat and the tulip to represent the basis vectors of our state vector. The basis states are chosen relative to an actual observation direction in real space (i.e. looking from above), and they are chosen because every time we look we find that in reality we see either the tulip or the hat, never something in between. Quantum. Discrete. A hat or a tulip. These basis vectors are orthogonal, which means they are perpendicular to each other. There is no 'tulipness' to the hat and no 'hatness' in the tulip.

While the window is shut, what's in the box is neither: call it the wisp. The wisp is the quantum state. Every time you look, it turns into a hat or a tulip. The wisp is a symbolic representation of the state vector. A linear combination of the two basis vectors.

Every widget in this post has the same Picture switch. Flip it to Spin and the hat becomes spin up, the tulip spin down, the wisp a spin state |A⟩, and the notation in the text switches to Susskind's letters. The maths is identical either way; only the pictures change.

Looking in

Here's the box that contains our illustrative quantum mechanical object. This widget allows you to conduct experiments that simulate what we actually find when we measure quantum states. Work through the three experiments in order.

• 1. Look again. Whatever you saw the first time, you see again, every time. Looking doesn't just reveal what's inside; it sets it. The box now holds what you saw.
• 2. Fresh boxes. Each look is at a new box holding the same wisp: hat or tulip, with no way to predict which. Over 100 boxes the tally settles near 50:50, but no single look ever shows half a hat.
• 3. A certain box. A wisp that is all hat, |A⟩ = |u⟩, shows a hat every time.

Two states, two perpendicular axes

Give each state its own axis in state-space. A box that is prepared in a state certain to show a hat can be represented by a state vector |u⟩; one certain to show a tulip is |d⟩. Neither has any of the other in it, so they're orthogonal, and each has length one:

In columns, |u⟩ is $(1, 0)^T$ and |d⟩ is $(0, 1)^T$. The axes are labelled by states, not numbers: the first slot of the column means hat, not 1, and the component α_u is "how much hat" the wisp has in it.

Before we look, we suppose the state of the box is described by the wisp. Every wisp is a combination of the two. So far there is no need for any complex numbers: the components are real.

To take the inner product of two states, write both as columns, take the complex conjugate of each entry in the first one (that's what turning a ket into a bra does), then multiply matching entries and add them up, as in the last post. For the wisp |A⟩ and another state |B⟩ = β_u|u⟩ + β_d|d⟩:

While the components are real, the conjugate changes nothing. The basis states check out: ⟨u|d⟩ = 1 · 0 + 0 · 1 = 0 and ⟨u|u⟩ = 1 · 1 + 0 · 0 = 1. And the inner product with |u⟩ picks out the first entry, ⟨u|A⟩ = 1 · α_u + 0 · α_d = α_u, which is why the components are inner products.

The chance of seeing each state when we look is the squared size of its component:

which is why a state must be normalised, ⟨A|A⟩ = |α_u|² + |α_d|² = 1: the box always shows something with probability 1. The probability of each of the two basis states appearing when we look must therefore sum to 1, which is geometrically the same as saying the length of the state vector (wisp) must be 1.

Now let's take a look at this state vector in state space. For now, in this 2D example, the components α_u and α_d are real numbers, so we can plot the wisp as an arrow on an ordinary pair of axes. The arrow's shadow on each axis is that component.

Notice −|u⟩: the arrow points the other way, but the chances are exactly those of |u⟩, a hat every time. A minus sign isn't a tulip.

So far so good, but it seems like we are adding a lot of complexity and confusing symbols without actually explaining anything new. That is because we need to move on to a more realistic number of spatial dimensions. So far we have only looked into the box from the top, and we just see a randomly appearing hat or tulip. This is strange, but just looks random in a classically random kind of way: heads or tails. We look, and we learn whether it was heads or tails. Things get much stranger when we start looking from another direction after already looking from the top.

Looking from other directions

Before we get to why these components need to be, in reality, complex numbers, we need a more realistic set of experiments: looking into the box from different directions in space.

The first thing we find is that, whichever direction we look from, we always see a hat or a tulip pointing directly towards us. This is extremely bizarre, but it is how quantum states work. This is another way of saying that the physical direction from which we observe the state (i.e. the measurement axis) results in the object manifesting a state that is aligned with that direction.

The second finding is stranger still. In the first experiment, once we had looked and seen a hat, every further look into the same box (without resetting it) showed the hat again. It never became a tulip. So suppose we look from above and see a hat pointing up, and then look into that same box from the side. We never find the hat still pointing up. Instead, half the time we see a hat pointing towards us, and half the time a tulip pointing towards us.

Read these two paragraphs again, as many times as it takes to let that sink in.

So seeing a hat from above didn't turn the box into a hat for good. It only made the box certain to show a hat from above. The box still held a quantum state, a wisp, and looking from the side reveals it. The kicker comes when we look at an angle. We still only ever see hats and tulips, but the closer the window is to the top, the direction we prepared the state in, the more often we see a hat.

Try it, with windows on the top, the side and the front. These are the experiments from Susskind's chapter 1.

The tilted window gives a formula worth keeping: from a box certain to show a hat through the top, the chance of a hat through a window tilted $\theta$ from the top is $\cos^2(\theta/2)$. Certain at $0°$, 50:50 at $90°$, and never at $180°$: through a window underneath, a box sure to show a hat through the top is sure to show a tulip.

The side window

The side window gives two new states: |r⟩, certain to show a hat through the side, and |l⟩, certain to show a tulip there. They're not new dimensions. $\mathbb{C}^2$ is still two-dimensional, so they must be combinations of the top-window states. We'll derive them properly in the next section; the answer is

Now let's put state space next to real space. On the left, the wisp shows as an arrow in state space; on the right, the direction you'd have to look from to be certain of seeing the hat (with the Picture set to Spin, the direction the spin points). This is also an interesting result. For a given state vector, there exists a direction in physical space from which we can be confident we will always see a hat pointing towards us. In a sense, that polarisation vector (a vector in physical space) also characterises the quantum state. You can see from this widget that there is a mapping between the two, but it's two-to-one: |A⟩ and −|A⟩ point the same way in real space.

• Drag from |u⟩ to |d⟩. A quarter turn on the left is a half turn on the right. Every angle in state space is doubled in real space. That's how "up" and "down", opposite in space, are perpendicular in state space.
• Stop at 45° for |r⟩, or −45° for |l⟩.

Everything so far is real. The sign of the components decides whether the certain direction is towards $+x$ or $-x$, and every real wisp is certain through some window in the plane of the top and the side.

Why we cannot avoid complex components for the state vector

This section is algebra, so it uses Susskind's letters whatever the Picture switch says. Each window has its own pair: |u⟩ and |d⟩ are a hat and a tulip through the top, |r⟩ and |l⟩ a hat and a tulip through the side, and |i⟩ and |o⟩ a hat and a tulip through the front.

The front window (along $y$) is at right angles to both the top and the side. To find its states, we'll go in three steps: first work out the side window's states from nothing but the experiments, then try exactly the same thing for the front window, see where it breaks, and then fix it.

Step 1: the side window

Take a box that is certain to show a hat through the side. Its state is a vector in the same two-dimensional space as everything else, so it's some combination of the top-window states:

Here |u⟩ and |d⟩ are the basis vectors, the two perpendicular axes of state space, and $a$ and $b$ are the components of |r⟩ along them: how much of |u⟩ and how much of |d⟩ it takes to build |r⟩. They're its coordinates in the top-window basis, just as $x$ and $y$ are the coordinates of an arrow on a page. Each one is an inner product with its basis vector, a = ⟨u|r⟩ and b = ⟨d|r⟩.

We don't know $a$ and $b$ yet, but they tell us the chances through the top. The chance of each result is the squared size of its component: $|a|^2$ for a hat and $|b|^2$ for a tulip.

The side window is at $90°$ to the top, so by the tilted-window formula, $\cos^2(90°/2) = \tfrac12$: this box shows a hat through the top half the time. So

$$|a|^2 = |b|^2 = \tfrac12.$$

With real components, that leaves $a = \pm 1/\sqrt2$ and $b = \pm 1/\sqrt2$. Multiplying a whole state by $-1$ changes no chance, so we can take $a = +1/\sqrt2$. That leaves just two candidates:

Their inner product is $(1 \cdot 1 + 1 \cdot (-1)) / 2 = 0$, so they're orthogonal, which is exactly what a hat and a tulip through the same window need. So one is |r⟩ and the other |l⟩; which is which is a convention (which way we call $+x$):

So with real components, being 50:50 through the top allows exactly two states, and the side window uses both of them.

Step 2: the front window

This is great, but we're not done. In three dimensions there's another window at right angles to the top: the front window. By the symmetry of space you might expect the same answer, something like this:

It passes the top-window test: 50:50, just like the side. But look at what we've written. That |i⟩ is exactly |r⟩, and that |o⟩ is exactly |l⟩. A box certain to show a hat through the front would then be certain to show a hat through the side too:

But the front window is at $90°$ to the side window as well, so through the side it must be 50:50, not certain. And with real components there's nothing else to try: being 50:50 through the top allows only those two states, and the side window has already used both. No real components can do it.

Step 3: complex components

From Step 1 we have |r⟩ and |l⟩ in terms of |u⟩ and |d⟩. Now we want the front-window hat in terms of |u⟩ and |d⟩ too, this time letting the components be complex:

Constraint 1: the top window. The front window is at $90°$ to the top, so through the top it's 50:50. The chances are the squared sizes of the components, so

$$|\alpha|^2 = |\beta|^2 = \tfrac12.$$

Constraint 2: the side window. The front window is at $90°$ to the side as well. So if we wrote |i⟩ in the side-window basis instead,

its components there (ρ for r, λ for l) would have to pass the same test:

$$|\rho|^2 = |\lambda|^2 = \tfrac12.$$

Linking the two. Substitute Step 1's |r⟩ and |l⟩:

Both lines describe the same |i⟩, so the components along |u⟩ and |d⟩ must match:

$$\alpha = \frac{\rho + \lambda}{\sqrt2}, \qquad \beta = \frac{\rho - \lambda}{\sqrt2}.$$

We want this the other way round, with the side-window components in terms of the top-window ones, because $\alpha$ and $\beta$ are what we're solving for. Adding the two equations gives $\alpha + \beta = 2\rho/\sqrt2 = \sqrt2\,\rho$, and subtracting them gives $\alpha - \beta = \sqrt2\,\lambda$. So

$$\rho = \frac{\alpha + \beta}{\sqrt2}, \qquad \lambda = \frac{\alpha - \beta}{\sqrt2}.$$

Now Constraint 2 is a condition on $\alpha$ and $\beta$ too. (This is the same as taking inner products: ρ = ⟨r|i⟩ and λ = ⟨l|i⟩.)

Solving for $\alpha$ and $\beta$. Multiplying a whole state by any complex number of size 1 (like $-1$, or $i$) changes no chance, so we can take $\alpha = 1/\sqrt2$. Constraint 1 then says only that $\beta$ has size $1/\sqrt2$. Write it as $\beta = x + iy$, with $x^2 + y^2 = \tfrac12$. Then $\alpha + \beta = \left(\tfrac{1}{\sqrt2} + x\right) + iy$. The squared size of a complex number is its real part squared plus its imaginary part squared, $|x + iy|^2 = x^2 + y^2$, and dividing by $\sqrt2$ halves the squared size, so

$$|\rho|^2 = \frac{\left(\tfrac{1}{\sqrt2} + x\right)^2 + y^2}{2}.$$

Expanding the bracket gives $\tfrac12 + \sqrt2\,x + x^2$, and $x^2 + y^2 = \tfrac12$, so the top is $1 + \sqrt2\,x$:

$$|\rho|^2 = \frac{1 + \sqrt2\,x}{2}.$$

Constraint 2 needs this to be $\tfrac12$, so $x = 0$. ($|\lambda|^2$ works out as $(1 - \sqrt2\,x)/2$, which gives the same.) With no real part, $y^2 = \tfrac12$, so $y = \pm 1/\sqrt2$ and

$$\beta = \pm \frac{i}{\sqrt2}.$$

Compare real components: $y = 0$ and $x = \pm 1/\sqrt2$, which makes $|\rho|^2$ either 1 or 0, certain through the side. That's Step 2's dead end again. The only way through is for $\beta$ to be purely imaginary.

That gives two states, and they're the front window's hat and tulip. Which is which is a convention again; Susskind calls them "in" and "out":

They're orthogonal, as a hat and a tulip through the same window should be. The bra ⟨o| conjugates the $-i$ to $+i$, so

Two windows could be described with real numbers alone. The third window is what forces in $i$.

The wisp is not a coin flip

Is the wisp anything more than a coin-flip box? In a coin-flip box, someone tossed a coin, put in a box certain to show a hat or a tulip through the top accordingly, and closed the lid. It really does hold one or the other all along; you just don't know which.

The difference shows in the order you look.

• Through the top, then the side. The two boxes behave identically. The top window shows a hat or a tulip, and then the side is 50:50, for the wisp and the coin-flip box alike. Looking through the top turns the wisp into a definite hat or tulip, and after that it is no different.
• Through the side only, with the top window left shut. The coin-flip box is still 50:50: whichever box the coin chose, the side shows a hat half the time. But the wisp |r⟩ shows a hat every time.

Look at |r⟩, |l⟩, |i⟩ and |o⟩. All four have components of the same size, so through the top window all four are 50:50, indistinguishable from each other and from a fair coin-flip box. They differ only in the phase of α_d relative to α_u: $1$, $-1$, $i$ and $-i$, the four quarter turns.

So the phase the top window couldn't see is exactly what the side and front windows measure. Write a general wisp as

The mix θ sets what you see through the top. The relative phase φ is invisible there, but decides what you see through the side and the front.

Things to try:

• Press Play. The arrow of α_d goes round its dial. Through the top window, nothing changes. Through the side and front, the wisp swings between always-a-hat and never-a-hat, while the coin-flip box sits at 50%.
• Watch the compass. The direction you'd look from to be certain of a hat goes round the equator as φ turns. The phase is a direction in space that the top window can't see.
• Change the mix, θ. It tips that direction up or down.
• Press × e^(iπ/3) to multiply the whole state by a phase. Both arrows turn together, the relative phase is unchanged, and nothing moves at all, through any window. That's the phase that really is unobservable.

That last point is behind a counting argument Susskind makes in this chapter. Two complex numbers are four real numbers. Normalisation, ⟨A|A⟩ = 1, fixes one. The overall phase can't be seen through any window, so it doesn't count either. That leaves two, θ and φ: exactly the two angles you need to point in a direction in space. So every wisp, however it was prepared, is certain to show a hat through some window. A box that is 50:50 through every window isn't a wisp at all; it's a fair coin-flip box.

Why not just use the two angles?

If θ and φ say everything about the state, why keep the complex components? Because the angles list the states, but they don't say how states combine. Directions can't be added: "up" plus "down" as arrows is zero, while |u⟩ plus |d⟩ as kets is |r⟩, a perfectly good state. The chance of any result is always an inner product, squared. And the angles stop working with two boxes: their joint states need four complex components, and some of them, the entangled ones, point in no direction at all. The complex vectors carry on unchanged.

The two pictures, side by side

• The hat and the tulip are spin up and spin down along $z$: |u⟩ and |d⟩.
• The wisp is the state |A⟩, a vector in $\mathbb{C}^2$.
• Hat and tulip have nothing in common: ⟨u|d⟩ = 0. Physically opposite states are orthogonal.
• The chance of seeing a state is the squared size of the matching component: P_u = |⟨u|A⟩|².
• "Pointing straight at you through any window" is the rule that a spin measured along any axis gives $+1$ or $-1$.
• The top window sees only the sizes of the components. The side window sees their relative sign; the front window needs, and sees, a complex relative phase. No window sees the overall phase.

The whole picture: one arrow

Put everything together and the physical state is just an arrow in ordinary 3D space: the one direction the wisp is certain to show a hat from. Once you know that arrow, the uncertainty is completely known too. Every other direction you could look from has a definite chance of a hat, $\cos^2(\gamma/2)$, where $\gamma$ is the angle between that direction and the arrow.

The sphere of all the directions that arrow can point in is called the Bloch sphere, after the physicist Felix Bloch.

To calculate those chances, though, we built something else: the two-dimensional complex state space. That space is always written relative to a choice of window, which is a choice of basis. A window is a single axis, so its basis is a pair of states pointing in opposite directions along that axis: a hat through the window and a tulip through it. Two possible results along one axis is why the space has two dimensions.

Choosing a different window rotates the basis, and the state gets new components in it. The rule that ties it all together is that every choice of window has to reproduce the chances we actually see. For windows in the plane of the top and side, real components would do. For windows anywhere in 3D, including the front, the components have to be complex.

The widget starts in a 2D world, with only the top and side windows, where real components are enough. Switch the World to 3D to add the front window, and watch what state space needs.

Things to try:

• In 2D, move the window round the circle. In state space, its two states are a pair of axes at right angles, and they turn half as fast as the window: a quarter turn of the window (top to side) is an eighth of a turn in state space (from the |u⟩, |d⟩ axes to the |r⟩, |l⟩ axes). The state's shadows on the axes are its components, all real.
• Point the window along the arrow. The state lies along one axis: a single component, and the window is certain to show a hat.
• Move the state. The coloured dots follow the arrow. The dots at right angles to it are always 50:50.
• Run experiments. Look into 100 fresh boxes through a few different windows, then move the state and do it again. Every result lands on the same curve: the chance depends only on the angle between the window and the arrow.
• Switch to Observe and press Look once. The arrow jumps to point straight at the window, or straight away from it, and the state in state space lands on one of the window's two axes. Reset and look again starts each look from the state you set, so you can watch the hats and tulips come up at the predicted rate.
• Switch to 3D and choose the front window. A real plane has no room for its pair of axes, so state space becomes one complex plane per component, and the components turn complex.

---

1. Susskind, Leonard, and Art Friedman. Quantum Mechanics: The Theoretical Minimum. Basic Books, 2014. ↩︎

I'm working through Leonard Susskind and Art Friedman's Quantum Mechanics: The Theoretical Minimum^[1]. Chapter 1 is about spin and simple experiments, but before the physics can go anywhere it pauses for a mathematical interlude on complex numbers and vector spaces, because the state of a quantum system is a vector in a vector space. Susskind's definition is short and abstract. This post is me making it concrete.

The objective is simple:

1. Understand what a vector space is, starting from an ordinary arrow in three dimensions.
2. See how the same idea extends to as many dimensions as you like, even when you can't draw them.
3. See a vector made of three complex numbers, the kind quantum mechanics actually uses.

Every picture below can be dragged, nudged or rebuilt. The notation matters, but the aim is that each symbol has a picture behind it.

A vector is an arrow, and a list of numbers

Start with one dimension. A vector on a line is a single number, $a_1$: how far to walk, and in which direction (the sign). If $\hat{e}_1$ is a step of length one along the line, the vector is

$$\vec{A} = a_1 \hat{e}_1.$$

Now add a second direction, at right angles to the first. To describe a vector in the plane, you need two numbers: walk $a_1$ along the first direction, then $a_2$ along the second.

$$\vec{A} = a_1 \hat{e}_1 + a_2 \hat{e}_2.$$

Add a third direction, at right angles to both, and you need a third number:

$$\vec{A} = a_1 \hat{e}_1 + a_2 \hat{e}_2 + a_3 \hat{e}_3 = \begin{pmatrix} a_1 \\ a_2 \\ a_3 \end{pmatrix},$$

where

$$\hat{e}_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix},\; \hat{e}_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix},\; \hat{e}_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix}.$$

That's the whole pattern: each new dimension is one more independent direction, and one more number in the list. The column on the right is just the list of how far to walk in each direction. The unit steps $\hat{e}_1, \hat{e}_2, \hat{e}_3$ are called basis vectors, and the numbers $a_i$ are the vector's components in that basis.

Build one up below. Start in 1D, then switch to 2D and 3D and watch the picture gain a direction each time. The coloured arrows are the components, laid head-to-tail; the plain arrow is the vector they add up to.

The length of the vector comes from Pythagoras, applied twice: once across the floor ($a_1$ and $a_2$), and once up the wall ($a_3$):

$$|\vec{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2}.$$

What makes it a space

A single arrow isn't the interesting part. What makes a collection of vectors a vector space is what you can do with them. There are exactly two operations.

Addition, component by component. Geometrically, put the arrows head-to-tail:

$$\vec{A} + \vec{B} = \begin{pmatrix} a_1 + b_1 \\ a_2 + b_2 \\ a_3 + b_3 \end{pmatrix}.$$

Multiplication by a number (a scalar), again component by component. Geometrically, stretch the arrow, shrink it, or flip it round if the number is negative:

$$\lambda \vec{A} = \begin{pmatrix} \lambda a_1 \\ \lambda a_2 \\ \lambda a_3 \end{pmatrix}.$$

A vector space is any collection of objects where these two operations always give you back something in the collection, and behave the way you'd hope. Written out, for any vectors $\vec{A}, \vec{B}, \vec{C}$ in the space and any scalars $\lambda, \mu$:

• Closure: $\vec{A} + \vec{B}$ and $\lambda\vec{A}$ are in the space too.
• The order of adding doesn't matter: $\vec{A} + \vec{B} = \vec{B} + \vec{A}$.
• Nor does the grouping: $(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C})$.
• There's a zero vector, $\vec{0}$, with $\vec{A} + \vec{0} = \vec{A}$.
• Every step can be undone: each $\vec{A}$ has a $-\vec{A}$ with $\vec{A} + (-\vec{A}) = \vec{0}$.
• Scaling distributes: $\lambda(\vec{A} + \vec{B}) = \lambda\vec{A} + \lambda\vec{B}$ and $(\lambda + \mu)\vec{A} = \lambda\vec{A} + \mu\vec{A}$.
• Scaling is consistent: $\lambda(\mu\vec{A}) = (\lambda\mu)\vec{A}$, and $1\vec{A} = \vec{A}$.

None of this mentions arrows or columns of numbers. That's the point Susskind is making: anything that obeys these rules is a vector space, and everything we prove using only these rules applies to all of them. Arrows are just the example we can draw.

Put the two operations together and you get a linear combination, $\alpha\vec{A} + \beta\vec{B}$. Try it: scale $\vec{A}$ and $\vec{B}$ with the sliders, and watch the sum move. Then switch on the span.

Some things to try:

• Set $\beta$ to 0 and sweep $\alpha$ from $-2$ to $2$. The tip of $\alpha\vec{A}$ traces out a line through the origin.
• Show the span. Every combination $\alpha\vec{A} + \beta\vec{B}$ lies on one flat plane through the origin. However you set the sliders, two vectors can't get you off it.
• Type B = 4, 1, 0 (twice $\vec{A}$). Now $\vec{B}$ adds no new direction, and the span collapses to a line.
• Turn the picture until you're looking along the plane edge-on, and it becomes a line. That's the third dimension the plane is missing.

This is what "three-dimensional" really means. Two vectors pointing in different directions reach a plane. To reach every point in space you need a third that isn't in that plane. The dimension of a vector space is the number of independent directions you need to reach every vector in it, and any set of that many independent vectors (such as $\hat{e}_1, \hat{e}_2, \hat{e}_3$) is a basis.

Past three dimensions

We can't draw four mutually perpendicular axes, but notice that nothing in the algebra above cared about the number three. A vector in $n$ dimensions is a list of $n$ real numbers, and the set of all of them is called $\mathbb{R}^n$:

$$\vec{A} = \begin{pmatrix} a_1 \\ a_2 \\ \vdots \\ a_n \end{pmatrix} = \sum_{i=1}^{n} a_i \hat{e}_i.$$

Addition and scaling are still component by component, every axiom still holds, and the length formula just grows more terms:

$$|\vec{A}| = \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2} = \sqrt{\sum_{i=1}^{n} a_i^2}.$$

The trick for building intuition is to stop picturing the arrow and picture the list. Draw each component as a bar. For $n \le 3$ the bars and the arrow are two pictures of the same thing; past that, the arrow runs out but the bars carry on happily. Add components below and watch the arrow go first from a line to a plane to space, then disappear, while the bars and the length don't care.

High-dimensional vectors are everywhere once you look. A yield curve quoted at ten maturities is a single point in $\mathbb{R}^{10}$; a greyscale photo with a million pixels is a point in $\mathbb{R}^{1{,}000{,}000}$. You can still add them, scale them, and measure how far apart they are.

One more tool before we leave real vectors. The inner product (or dot product) of two vectors multiplies matching components and adds them up:

$$\vec{A} \cdot \vec{B} = \sum_{i=1}^{n} a_i b_i.$$

It measures how much two vectors point the same way: it's zero when they're at right angles (orthogonal), and the inner product of a vector with itself is its length squared, $\vec{A} \cdot \vec{A} = |\vec{A}|^2$. The basis vectors are each of length one and orthogonal to each other, which makes them an orthonormal basis, and the components of any vector are just its inner products with them: $a_i = \hat{e}_i \cdot \vec{A}$.

Complex numbers, briefly

A complex number $z = x + iy$, with $i^2 = -1$, is a pair of real numbers, so it's natural to draw it as a point (or an arrow) in a plane: the real part $x$ across, the imaginary part $y$ up. The same point can be described by its distance from the origin, $r$, and its angle, $\theta$, called its phase:

$$z = x + iy = r(\cos\theta + i\sin\theta) = r e^{i\theta}.$$

These are two ways of describing the same point, not two different numbers. The Cartesian form, $x + iy$, says how far across and how far up. The polar form, $r e^{i\theta}$, says how far out and in which direction. Each can be turned into the other with a little trigonometry:

$$x = r\cos\theta, \quad y = r\sin\theta \qquad\Longleftrightarrow\qquad r = \sqrt{x^2 + y^2}, \quad \tan\theta = \frac{y}{x}.$$

(For $\theta$, take the angle that puts the point in the right quarter of the plane: $\tan\theta$ alone can't tell $1 + i$ from $-1 - i$.) Drag $z$ in either picture below, or move either pair of sliders, and watch the other description follow.

Each form is better at something. Adding complex numbers is easiest in Cartesian form: add the $x$s and add the $y$s. Multiplying is easiest in polar form, as we're about to see. Which is why, from here on, you'll see both.

Two facts do almost all the work in what follows:

• The complex conjugate $z^* = x - iy = r e^{-i\theta}$ flips the sign of the imaginary part: a reflection in the real axis. Multiplying a number by its conjugate gives a real, non-negative number, the length squared:

$$z^* z = (x - iy)(x + iy) = x^2 + y^2 = r^2 = |z|^2.$$

• Multiplying complex numbers multiplies their lengths and adds their angles: $r_1 e^{i\theta_1} \cdot r_2 e^{i\theta_2} = r_1 r_2 \, e^{i(\theta_1 + \theta_2)}$. So multiplying by $i = e^{i\pi/2}$ is a quarter turn, multiplying by $-1 = e^{i\pi}$ is a half turn, and multiplying by $e^{i\varphi}$ turns by $\varphi$ without changing the length.

A complex 3-vector

Now the step Susskind takes. Keep everything about a vector space exactly as it was, but let the components, and the scalars you multiply by, be complex numbers. Following Susskind (and Dirac), we write the vector as a ket, $|A\rangle$:

$$|A\rangle = \begin{pmatrix} \alpha_1 \\ \alpha_2 \\ \alpha_3 \end{pmatrix} = \alpha_1 |1\rangle + \alpha_2 |2\rangle + \alpha_3 |3\rangle, \qquad \alpha_i \in \mathbb{C}.$$

Here $|1\rangle, |2\rangle, |3\rangle$ play the role $\hat{e}_1, \hat{e}_2, \hat{e}_3$ did before: a column with a 1 in one place and 0 elsewhere. Adding kets is still component by component, scaling $|A\rangle$ by a complex number $z$ multiplies every component by $z$, and every axiom from earlier holds word for word. This space is called $\mathbb{C}^3$.

So why can't we just draw it as an arrow? Count the real numbers. Each component $\alpha_i = x_i + i y_i$ is two real numbers, so $|A\rangle$ holds six. As far as the number of dials to turn goes, $\mathbb{C}^3$ is as big as $\mathbb{R}^6$, and we've just seen that six dimensions won't fit in one picture.

But the bar-chart trick still works, upgraded. Instead of a bar for each component, give each component its own complex plane, and draw it there as an arrow with a length and a phase. Three components, three little planes. Drag the arrows around, and try the buttons, which multiply the whole vector by a complex number.

Things to try:

• Press × i. All three arrows turn a quarter turn together. Each component's phase changes, but the angles between them (shown under the picture) don't, and neither does any length.
• Press × −1. A half turn, all together. In a real vector space, multiplying by $-1$ flips an arrow; here you can see it's really a rotation through $180°$ that happens to land on the opposite side.
• Press × 2, then × ½. Scaling by a real number stretches every component equally, leaving all the phases alone.
• Press Make real. Every arrow drops onto its real axis. What's left is three real numbers: an ordinary vector in $\mathbb{R}^3$, like the one at the top of this post. The real vectors sit inside the complex ones.
• Show the bra. More on this next.

Bras, and the length of a complex vector

How long is $|A\rangle$? The obvious guess is to reuse the real formula, $\sqrt{\alpha_1^2 + \alpha_2^2 + \alpha_3^2}$. But try it on the one-component vector $(i)$: we get $i^2 = -1$, a negative length squared, which is nonsense.

The fix is the conjugate. Every ket $|A\rangle$ has a partner, the bra $\langle A|$, which is a row whose entries are the conjugates of the ket's components:

$$\langle A| = \begin{pmatrix} \alpha_1^* & \alpha_2^* & \alpha_3^* \end{pmatrix}.$$

Switch on the bra in the widget above and you'll see each $\alpha_i^*$ as the reflection of $\alpha_i$ in its real axis. The inner product of a bra with a ket (a "bra-ket", or bracket) multiplies matching entries and adds them up:

$$\langle B|A\rangle = \beta_1^* \alpha_1 + \beta_2^* \alpha_2 + \beta_3^* \alpha_3 = \sum_{i} \beta_i^* \alpha_i.$$

With a vector and itself, each term is $\alpha_i^* \alpha_i = |\alpha_i|^2$, which is real and never negative:

$$\langle A|A\rangle = |\alpha_1|^2 + |\alpha_2|^2 + |\alpha_3|^2 \;\ge\; 0.$$

That's the length squared of $|A\rangle$, and it's the number shown under the widget. Each component contributes its own length squared, the strip at the bottom shows how the total is shared out between them, and the Normalise button divides $|A\rangle$ by $\sqrt{\langle A|A\rangle}$ so that the total is exactly 1.

A small worked example. Take

$$|A\rangle = \begin{pmatrix} 1 \\ i \\ 0 \end{pmatrix}, \qquad |B\rangle = \begin{pmatrix} i \\ 1 \\ 1 \end{pmatrix}.$$

Then $\langle A|A\rangle = 1 \cdot 1 + (-i)(i) + 0 = 1 + 1 = 2$, and

$$\langle B|A\rangle = (i)^* \cdot 1 + (1)^* \cdot i + (1)^* \cdot 0 = -i + i + 0 = 0.$$

Neither vector looks perpendicular to the other at a glance, but their inner product is zero, so they are orthogonal. Without the conjugate, we'd have got $i + i = 2i$ instead. One consequence of the conjugate is that the order now matters, a little: swapping bra and ket conjugates the answer, $\langle A|B\rangle = \langle B|A\rangle^*$.

Everything we built for real vectors carries over. The basis kets are orthonormal, $\langle i|j\rangle = 1$ if $i = j$ and $0$ otherwise, and the components of any ket are its inner products with them: $\alpha_i = \langle i|A\rangle$.

Two real arrows inside one complex vector

The dials group the six real numbers in $|A\rangle$ one component at a time. They can be grouped the other way too. Write each component as its real and imaginary parts, $\alpha_k = x_k + i y_k$, and lay them out in a table:

	Real part	Imaginary part
$\alpha_1$	$x_1$	$y_1$
$\alpha_2$	$x_2$	$y_2$
$\alpha_3$	$x_3$	$y_3$

Each row is one dial. Each column is a list of three real numbers, which is an ordinary real 3-vector:

$$\mathbf{x} = \begin{pmatrix} x_1 \\ x_2 \\ x_3 \end{pmatrix}, \qquad \mathbf{y} = \begin{pmatrix} y_1 \\ y_2 \\ y_3 \end{pmatrix}, \qquad |A\rangle = \mathbf{x} + i\,\mathbf{y}.$$

Both $\mathbf{x}$ and $\mathbf{y}$ are real: the $i$ sits outside $\mathbf{y}$, not inside it. So both can be drawn as arrows in one 3D picture, just like the vectors at the start of this post, with the three axes now the directions $|1\rangle, |2\rangle, |3\rangle$. Reading across dial $k$ gives the $k$-th component of $\mathbf{x}$; reading up it gives the $k$-th component of $\mathbf{y}$. For example, $(1, i, 0)$ is $\mathbf{x} = (1, 0, 0)$ plus $i$ times $\mathbf{y} = (0, 1, 0)$: two arrows at right angles.

The length splits neatly between the two arrows. Each $|\alpha_k|^2 = x_k^2 + y_k^2$, so

$$\langle A|A\rangle = \sum_k |\alpha_k|^2 = \sum_k x_k^2 + \sum_k y_k^2 = |\mathbf{x}|^2 + |\mathbf{y}|^2.$$

A phase mixes the two arrows

Now multiply $|A\rangle$ by a phase, $e^{i\varphi} = \cos\varphi + i\sin\varphi$, and gather up the real and imaginary parts:

$$\begin{aligned} e^{i\varphi}|A\rangle &= (\cos\varphi + i\sin\varphi)(\mathbf{x} + i\,\mathbf{y}) \\ &= (\mathbf{x}\cos\varphi - \mathbf{y}\sin\varphi) + i\,(\mathbf{x}\sin\varphi + \mathbf{y}\cos\varphi). \end{aligned}$$

So after the phase, the real and imaginary arrows are

$$\mathbf{x}' = \mathbf{x}\cos\varphi - \mathbf{y}\sin\varphi, \qquad \mathbf{y}' = \mathbf{x}\sin\varphi + \mathbf{y}\cos\varphi.$$

Each new arrow is a blend of the two old ones. As $\varphi$ turns, they swing round together, always staying in the plane the original two share. At $\varphi = 90°$, multiplying by $i$, the blend is complete: $\mathbf{x}' = -\mathbf{y}$ and $\mathbf{y}' = \mathbf{x}$. The real arrow has become the imaginary one, and the imaginary arrow has become (minus) the real one.

In general, neither arrow keeps its length. Square them, using $|\mathbf{u} + \mathbf{v}|^2 = |\mathbf{u}|^2 + 2\,\mathbf{u}\cdot\mathbf{v} + |\mathbf{v}|^2$:

$$\begin{aligned} |\mathbf{x}'|^2 &= |\mathbf{x}|^2\cos^2\varphi - 2\,(\mathbf{x}\cdot\mathbf{y})\sin\varphi\cos\varphi + |\mathbf{y}|^2\sin^2\varphi, \\ |\mathbf{y}'|^2 &= |\mathbf{x}|^2\sin^2\varphi + 2\,(\mathbf{x}\cdot\mathbf{y})\sin\varphi\cos\varphi + |\mathbf{y}|^2\cos^2\varphi. \end{aligned}$$

But add them, and the middle terms cancel, while $\cos^2\varphi + \sin^2\varphi = 1$ collects the rest:

$$|\mathbf{x}'|^2 + |\mathbf{y}'|^2 = |\mathbf{x}|^2 + |\mathbf{y}|^2 = \langle A|A\rangle.$$

Length flows back and forth between the real arrow and the imaginary arrow, but the total never changes. That's the length of $|A\rangle$, and a phase can't touch it.

See it happen below. The dials at the top are the components of $e^{i\varphi}|A\rangle$; the 3D picture shows the real arrow $\mathbf{x}$ (solid) and the imaginary arrow $\mathbf{y}$ (dashed) built from them; and the bar at the bottom splits $\langle A|A\rangle$ into $|\mathbf{x}|^2$ and $|\mathbf{y}|^2$.

Things to try:

• Press Play, or drag the phase slider. All three dials turn together. In 3D, $\mathbf{x}$ and $\mathbf{y}$ swing round, growing and shrinking. In the bar, the dividing line between $|\mathbf{x}|^2$ and $|\mathbf{y}|^2$ slides back and forth, but the total, marked in pink, stays exactly where it is.
• Break down into components shows $\mathbf{x}$ (or $\mathbf{y}$) laid head-to-tail along $|1\rangle, |2\rangle, |3\rangle$, as in the first widget. Each step is one dial's reading: across for $\mathbf{x}$, up for $\mathbf{y}$.
• Press × i and check the algebra: the solid arrow moves to where the dashed one was, and the dashed one moves to minus where the solid one was.
• Start from $(1, 1, 0)/\sqrt{2}$, a real vector. At $\varphi = 0$, $\mathbf{y}$ is zero and all the length is in $\mathbf{x}$. Turn to $90°$ and it's all in $\mathbf{y}$. Both arrows stay on one line, and the ellipse is squashed flat.
• Start from $(1, i, 0)/\sqrt{2}$. Now $\mathbf{x}$ and $\mathbf{y}$ are the same length and at right angles, so $\mathbf{x}\cdot\mathbf{y} = 0$, the middle terms above vanish, and the split stays at exactly half and half. The arrows just rotate, and the ellipse is a circle.
• Watch the ellipse. As the phase turns, the tip of $\mathbf{x}$ runs round it, and so does the tip of $\mathbf{y}$, a quarter turn behind. The ellipse itself doesn't move, whatever the phase: it's a picture of $|A\rangle$ that ignores the overall phase. Its size is set by $\langle A|A\rangle$, while its shape (flat, oval or round) comes from the relative phases between the components.
• Drag a dial. That changes $|A\rangle$ itself, not just its phase, so now the total does change.

The inner product can be written in the same terms. Multiplying out $\sum_k (x_{B,k} - i y_{B,k})(x_{A,k} + i y_{A,k})$ gives

$$\langle B|A\rangle = (\mathbf{x}_B\cdot\mathbf{x}_A + \mathbf{y}_B\cdot\mathbf{y}_A) + i\,(\mathbf{x}_B\cdot\mathbf{y}_A - \mathbf{y}_B\cdot\mathbf{x}_A),$$

a complex inner product built from four ordinary dot products. For the worked example above, $|A\rangle = (1, i, 0)$ has $\mathbf{x}_A = (1, 0, 0)$ and $\mathbf{y}_A = (0, 1, 0)$, and $|B\rangle = (i, 1, 1)$ has $\mathbf{x}_B = (0, 1, 1)$ and $\mathbf{y}_B = (1, 0, 0)$. The real part is $0 + 0 = 0$, the imaginary part is $1 - 1 = 0$, and $\langle B|A\rangle = 0$, as before.

Why physics bothers

Susskind uses all of this straight away. The spin of an electron, the system at the heart of his first chapters, is described by a vector in a two-dimensional complex vector space, $\mathbb{C}^2$, with a basis of two kets he calls $|u\rangle$ (up) and $|d\rangle$ (down). A general spin state is a linear combination

$$|A\rangle = \alpha_u |u\rangle + \alpha_d |d\rangle,$$

which is the widget above with one fewer plane. The lengths $|\alpha_u|^2$ and $|\alpha_d|^2$ turn out to be probabilities, which is why normalising (making them add up to 1) matters. And the phases, the part with no counterpart in real vectors, turn out to be where interference, and much of what makes quantum mechanics strange, lives. Notice what the × i button showed: turning every component by the same angle changes none of the lengths and none of the angles between components. That's a hint of something Susskind comes back to: multiplying a state by an overall phase $e^{i\varphi}$ changes nothing you can measure.

Real and complex, side by side

	Real vectors, $\mathbb{R}^n$	Complex vectors, $\mathbb{C}^n$
Components	$n$ real numbers	$n$ complex numbers ($2n$ real numbers)
Scalars	real	complex
Picture	one arrow (if $n \le 3$), or a bar per component	an arrow in its own complex plane per component
Basis	$\hat{e}_1, \dots, \hat{e}_n$	$\vert 1\rangle, \dots, \vert n\rangle$
As real arrows	one arrow	two: $\vert A\rangle = \mathbf{x} + i\,\mathbf{y}$
Inner product	$\vec{A} \cdot \vec{B} = \sum_i a_i b_i$	$\langle B \vert A\rangle = \sum_i \beta_i^* \alpha_i$
Length squared	$\sum_i a_i^2$	$\langle A \vert A\rangle = \sum_i \lvert\alpha_i\rvert^2 = \lvert\mathbf{x}\rvert^2 + \lvert\mathbf{y}\rvert^2$
Swapping order	$\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$	$\langle A \vert B\rangle = \langle B \vert A\rangle^*$

Exercises

A few to check your understanding. Answers are hidden; try them first.

1. Is the set of all real 3-vectors whose components add up to 1 a vector space? What about those whose components add up to 0?

Answer

The first is not. $(1, 0, 0)$ and $(0, 1, 0)$ are both in it, but their sum $(1, 1, 0)$ has components adding up to 2, so it isn't closed under addition (and it doesn't contain $\vec{0}$ either). The second is: if two vectors' components each add up to 0, so do those of their sum and of any multiple. Geometrically, it's a plane through the origin, just like the span in the second widget.

2. How many real numbers does it take to specify a vector in $\mathbb{C}^4$? In $\mathbb{C}^2$, the space of electron spin states?

Answer

Eight, and four: two real numbers (a real and an imaginary part, or a length and a phase) for each complex component.

3. Find $\langle A|A\rangle$ for $|A\rangle = (1 + i,\; 2,\; -i)^T$, then normalise $|A\rangle$.

Answer

$|1 + i|^2 = 1^2 + 1^2 = 2$, $|2|^2 = 4$ and $|-i|^2 = 1$, so $\langle A|A\rangle = 7$. Dividing by $\sqrt{7}$ gives the normalised ket $\tfrac{1}{\sqrt{7}}(1 + i,\; 2,\; -i)^T$.

4. Show that multiplying $|A\rangle$ by $e^{i\varphi}$ leaves $\langle A|A\rangle$ unchanged, as the × i button suggested.

Answer

Every component becomes $e^{i\varphi}\alpha_i$, and its conjugate becomes $e^{-i\varphi}\alpha_i^*$. So each term of the inner product becomes $e^{-i\varphi}\alpha_i^* \, e^{i\varphi}\alpha_i = \alpha_i^* \alpha_i = |\alpha_i|^2$, exactly as before.

5. Using the widget, make $|A\rangle$ real (press Make real), then press × i. Is the result still a real vector? Is it still in $\mathbb{C}^3$?

Answer

No, and yes. Every component is now purely imaginary, so it's no longer in $\mathbb{R}^3$: multiplying a real vector by a complex scalar can take you out of the real vectors. But it's still a perfectly good vector in $\mathbb{C}^3$. That's why the choice of scalars is part of the definition of a vector space: $\mathbb{R}^3$ is closed under multiplication by real numbers, not complex ones.

---

1. Susskind, Leonard, and Art Friedman. Quantum Mechanics: The Theoretical Minimum. Basic Books, 2014. ↩︎

This is the second post in my series on how derivatives are priced. In the first post we built up the discount factor in an imaginary world with one bank and one interest rate that never changed. I promised to start relaxing those assumptions, and the obvious one to go first is the rate that never changes.

In the real world, the interest rate that sits underneath almost every sterling derivative and bank loan is quoted fresh every single day. It's called SONIA, and you can see where it is now on my markets page.

As before, I'm writing for someone with high-school maths and no finance background.

What SONIA is

SONIA stands for the Sterling Overnight Index Average. Every business day, banks and other large financial institutions lend each other money in pounds overnight: borrow today, pay it back tomorrow. SONIA is a measure of the interest rate paid on that money.

A quick note on words. Lending money to a bank overnight is the same thing as putting it on deposit there: the same transaction, seen from the lender's side or the bank's. The Bank of England calls them deposits, so I will too.

A few details make it trustworthy:

• It's based on real deposits, not opinions. The Bank of England collects the actual overnight, unsecured deposits of £25 million or more that took place that day.^[1]
• It's a trimmed average. The Bank sorts the deposits by rate, throws away the top and bottom quarters (weighted by size), and averages the middle half. A handful of odd deposits can't move it much.
• It's published the next morning. Monday's SONIA is published at 9am on Tuesday. You only find out what yesterday's rate was today.
• It's quoted as an annual rate, with the "actual days ÷ 365" day count from the last post.

Nobody chooses SONIA's level, but it is steered. It tracks the Bank of England's Bank Rate very closely, usually sitting a few hundredths of a percent below it. When the Monetary Policy Committee changes Bank Rate, SONIA moves with it from the day the new rate takes effect; because of the publication lag, you see the change in the figure published the following morning. The charts on the markets page show the two lines moving almost on top of each other, stepping up and down together.

Trying out the trimmed average

The widget below shows a day's deposits. Each bar is one deposit: its width is the amount lent and its height is the rate. The bars are lined up from the lowest rate to the highest, so trimming simply cuts the same width of volume off each end. Solid bars are kept; faint bars are trimmed away. The trimmed mean is the average height of what's left.

Try the examples and move the slider. At 0% the trimmed average is just the ordinary volume-weighted mean. As the trim gets close to 50%, only the deposit in the very middle is left, which is the median. SONIA uses 25%, somewhere in between: it ignores odd deposits at either extreme, like the median, but still takes an average of the many deposits in the middle, like the mean.

Why it replaced LIBOR

For decades, the reference rate in sterling markets was LIBOR, the London Inter-Bank Offered Rate, which was set each morning by asking a panel of banks what they thought they could borrow at. After it emerged that some of those answers had been manipulated, regulators pushed markets to move to rates based on real transactions. Sterling markets chose SONIA, and LIBOR has now stopped being published altogether.^[2]

There's a second difference which matters much more for the maths. LIBOR was a term rate: on day one, you knew the rate for the next three months. SONIA is an overnight rate: on day one, you only know the rate for one night. That changes how interest is worked out, and generally adds complexity to interest calculations. Having said that, the market has converged on a standard approach to these calculations meaning that there is rarely any confusion.

Why it matters

Any contract where one side pays a floating rate of interest needs a rate that both sides can look up, that nobody involved can nudge in their own favour, and that will still be published in thirty years' time. That's what a benchmark is for. SONIA itself has been around since 1997,^[3] but since LIBOR's departure it has become the benchmark for sterling: the official "risk-free rate" that the Bank of England and the industry chose for sterling markets to build on. It's considered "risk-free" because you can't easily lend money for a period of time of less than one day, and the chances of the bank you are very happy to lend to today going under by tomorrow are very slim. It also gives the Bank a clean read on how changes in Bank Rate pass through to the rates banks actually lend at.

The most familiar place to see it is a sterling bank loan. A company borrowing in pounds typically pays compounded SONIA plus a margin. The two parts do different jobs:

• The SONIA part is floating: it goes up and down with the market, and it's the same for every borrower. It's the price of money itself, and has nothing to do with who is borrowing.
• The margin is a fixed extra amount of interest, set for that particular loan. It reflects the risk of lending to that borrower: a riskier company pays a bigger margin.

One thing it isn't usually linked to is your mortgage. A UK tracker mortgage typically follows Bank Rate, not SONIA, although the two move together so closely that the difference rarely matters.

SONIA matters for this series for one more reason. When banks trade derivatives with each other, they usually post cash as collateral, and that cash typically earns SONIA. That makes SONIA the natural rate for discounting, so the discount curve at the end of this post is built from it.

Earning SONIA over more than one night

Suppose you have £1,000,000 and you want to earn SONIA on it for a week. You can't lock in a week's rate; you can only lend overnight. So you do what we did in stage 2 of the last post: a chain of one-period deposits.

1. Lend £1,000,000 overnight on Monday at Monday's SONIA.
2. On Tuesday you get it back with a day's interest. Lend all of it again overnight at Tuesday's SONIA.
3. Repeat every day until the following Monday.

Each night you earn interest on the interest from the nights before, so this is daily compounding, exactly as before. The only change is that each link in the chain now has its own rate.

Weekends

Overnight lending only happens on business days. Money lent on a Friday comes back on Monday, three nights later, at Friday's rate. So Friday's rate counts three times. Bank holidays work the same way.

A worked example

Here's a made-up week. To make it interesting, the Monetary Policy Committee cuts Bank Rate by a quarter of a percentage point on Thursday, and SONIA drops with it. Each row is one overnight deposit of everything we've built up so far:

Day	SONIA	Nights	Growth factor that night	Balance afterwards
Mon	3.9700%	1	1.00010877	£1,000,108.77
Tue	3.9700%	1	1.00010877	£1,000,217.55
Wed	3.9600%	1	1.00010849	£1,000,326.06
Thu	3.7200%	1	1.00010192	£1,000,428.01
Fri	3.7200%	3	1.00030575	£1,000,733.90

Each night's growth is just the one-period formula from the last post, $1 + r\,\tfrac{n}{365}$, with that day's rate $r$ and that day's number of nights $n$. Chain them together by multiplying:

$$\text{Growth factor} = \prod_{i} \left(1 + r_i \, \frac{n_i}{365}\right)$$

The $\prod$ symbol just means "multiply all of these together", the way $\sum$ means "add them all up". Here it comes to 1.00073390, so the week earns £733.90.

It's common to express this as a single annual rate for the whole period, by asking what simple rate over the whole period would have given the same growth:

$$\begin{aligned} 1 + \text{Compounded SONIA} \times \frac{D}{365} &= \text{Growth factor} \\[4pt] \text{Compounded SONIA} &= \left(\text{Growth factor} - 1\right) \times \frac{365}{D} \\[4pt] &= \left(1.00073390 - 1\right) \times \frac{365}{7} = 3.8268\% \end{aligned}$$

where $D = \sum_i n_i$ is the total number of days in the period (7 in our example).

This is called compounding in arrears: the rate for the period is built up a day at a time, and you only know it once the period is over.

"Compounded SONIA" for a period is therefore a simple interest rate, which when applied for that period, gives a final interest cash flow of the exact same amount as if the daily compounding calculation using each day's SONIA fixing was done properly.

For comparison, a plain weighted average of the five rates (counting Friday three times) is 3.8257%, which would earn £733.70. Compounding adds 20p. Over a week the effect is tiny, but on large sums over long periods it adds up, and pricing models have to get it exactly right.

Try it yourself below. Change the daily rates, or mark a day as a bank holiday and see the previous day's rate stretch to cover it.

The problem: you don't know the rate in advance

In the last post, a deposit of £1,000 at 5% for a year turned into £1,050, and we knew that on day one. That's what let us say the discount factor was $\tfrac{1}{1.05}$ and value a future payment today.

With SONIA, if I deposit £1 today and roll it overnight for three months, I'll end up with something, but I don't know what until the three months are up. If the Bank of England cuts rates next month, I'll earn less; if it raises them, I'll earn more.

So where does a discount factor come from now?

Locking in a rate: the overnight index swap

The answer is that the market lets you swap an unknown rate for a known one. The contract that does this is called an overnight index swap, or OIS. In its simplest form:

• You and a counterparty agree a notional amount, say £1,000,000, and a period, say three months (91 days).
• You agree a fixed rate $K$ today.
• At the end of the period, you pay the fixed interest, $N \times K \times \tfrac{91}{365}$.
• In return, you receive the SONIA interest, compounded in arrears exactly as in the worked example above.

The £1,000,000 itself never changes hands. It only exists to scale the payments. In practice only the difference is paid: whoever owes more pays the other.

If you have money rolling overnight at SONIA, take the other side of this swap: receive the fixed rate and pay SONIA. That turns your unknown SONIA interest into a known fixed amount. Whatever happens to rates, the SONIA you earn on your deposit cancels the SONIA you hand over in the swap, and you're left with the fixed rate $K$. You've locked in a three-month rate. And because the fixed rate is agreed in a busy market, $K$ is a price that everyone can see.

The widget below puts £1,000,000 on deposit at SONIA for three months, alongside a swap where you receive $K$ and pay SONIA. Each simulation draws a random path for SONIA, with the Bank of England cutting, holding or raising rates at two meetings along the way. Run it a few times: the deposit on its own earns a different amount every time, but the deposit plus the swap always earns exactly the fixed interest.

From the swap rate to a discount factor

This is where it gets neat. Let's work out what the swap is worth today, using the same "I'm indifferent" argument as the last post.

The SONIA side. Lend £1 overnight and keep rolling it. It costs £1 today, and at the end of the period it pays back $\text{Growth factor}$, whatever that turns out to be. So receiving "$\text{Growth factor}$" at the end is worth exactly £1 today, even though we don't know what it will be. The SONIA interest is $\text{Growth factor} - 1$, and the "1" is just a fixed payment at the end, worth $DF$ today. So, per £1 of notional:

$$\text{Value of SONIA side} = 1 - DF$$

The fixed side. This is a known amount, $K\tau$, paid at the end, where $\tau = \tfrac{91}{365}$ is the day count. A known payment at the end is worth that amount times the discount factor:

$$\text{Value of fixed side} = K\,\tau \times DF$$

The fair swap rate is the one that makes the deal fair for both sides, so that neither side is worth more than the other on the day it's agreed:

$$K\,\tau \times DF = 1 - DF$$

Rearranging:

$$DF = \frac{1}{1 + K\,\tau}$$

This is exactly the one-period discount factor from the first post, with the swap rate $K$ playing the part of the bank's rate. The difference is that in the first post we made the rate up. Here the market tells us what it is.

For example, if the three-month SONIA swap rate is 3.90%:

$$DF = \frac{1}{1 + 0.039 \times \tfrac{91}{365}} = 0.990370$$

So £1,000,000 promised in three months' time is worth £990,370.34 today.

What I've assumed, and what comes next

We've dropped the idea of a single rate that never changes, but a few simplifications remain:

• Still no risk. SONIA loans are unsecured, so there is a small chance a borrower doesn't pay back. It's overnight and between very large institutions, so the risk is small, and in practice SONIA is treated as close to "risk-free". That's why it's called a risk-free rate.
• Payment timing. In practice SONIA is published the next morning, so the final day's rate isn't known until after the period has ended. Real contracts allow for this by looking back a few days (usually five business days), or paying a couple of days late. I've ignored this here.
• One period. Real swaps can run for 30 years, with payments every year. The maths is the same idea, repeated.

That last point is where the next post goes. One swap rate gives us one discount factor. The market quotes SONIA swap rates for many lengths, from one week to fifty years. Line them up, and we can work out a discount factor for every date, one after another, in a process called bootstrapping. The result is the discount curve: the single most important input to pricing almost any sterling derivative.

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1. Bank of England, SONIA key features and policies. It sets out which deposits count (unsecured, overnight, £25 million or more), the trimmed mean of the central 50% of volume, rounding to four decimal places, and publication at 9am on the following London business day. ↩︎

2. Bank of England, Transition to sterling risk-free rates from LIBOR. Most LIBOR settings ended on 31 December 2021; the last "synthetic" sterling LIBOR setting ended in March 2024, and all 35 LIBOR settings have now ceased. ↩︎

3. Bank of England, SONIA interest rate benchmark. SONIA was introduced in March 1997; the Bank of England took over running it in 2016 and reformed it in 2018. ↩︎