In 1949, Marcel Golay was thinking about spectrometry.

As he described it some time later the situation was as follows.

You have a spectrometer. The point of spectrometry is to find the frequency of light (or electromagnetic radiation more generally — but for convenience I’ll just say “light” from now on). Given a light source, spectrometry aims to find which frequencies (or colours) of light occur in it, and how they are distributed across the optical spectrum.

The spectrometer Golay had in mind was a cleverly designed “multislit” one. As the name suggests, it had many slits. Each slit could be open or closed. Light would come in on one side, pass through the contraption, and then exit on the other side, where detectors would be placed to record the output.

Both the entrance side and the exit side had many slits — the same number \(4N\) on either side. (Why a multiple of 4? It’s all part of the clever design, read on…)

Moreover, each entrance slit had a natural pathway through to an exit slit. The slits were designed so that light entering a particular entrance slit would pass through to a specific exit slit. The entrance and exit slits were thus matched up in a one-to-one fashion. This “matching up” in fact “inverted” the light: light coming in through the top slit on the left, would exit through the bottom slit on the right; light entering through the second-from-top slit on the left would exist through the second-from-bottom slit on the right; and so on.

At least, that’s what would happen for *one* particular colour of light, i.e. one particular frequency — let’s say pure crimson red. The point of the spectrometer is to pick out distinct frequencies, and so this contraption is “tuned” to perfectly align the slits for crimson red light.

What about other frequencies? They get *shifted*. When light of another frequency, let’s say green, passed through an entrance slit, it *did not* end up in the same place as crimson red light, opposite to where it came in; rather, it ended up shifted across by some number \(j\) of slits.

In other words, if red light and green light enter through the same lit, they exit through slits which are \(j\) spots apart from each other.

Golay’s idea was to arrange the slits and detectors in a clever way, so as to eliminate all the light of other freuqencies, and isolate the preferred (red) light. By an ingenious arrangement of detectors and open and closed slits, the red light would be greatly enhanced, with other colours (frequencies) completely filtered out.

How did this arrangement go? In a slightly complicated way. The entrance slits would be split into four equal length sections, each of length \(N\), as would the exit slits. Light entering through a slit in a particular section would go out a slit in the corresponding (opposite) section of exit slits.

These sections were separated from each other. In particular, non-red light could be shifted across slits within one section, but it could not cross over to another section.

Golay imagined there to be two detectors. The first detector \(D_1 \) would cover the bottom two exit sections, measuring the total amount of light exiting the top half of the slits, i.e. the bottom \( 2N \) exit slits. The other detector \(D_2\) would cover the top half of the exit slits, i.e. the bottom two sections, the bottom \( 2N \) exit slits. The detectors \( D_1 \) and \( D_2 \) simply capture the amount of light coming out of the bottom and top \( 2N \) slits respectively, or equivalently for our purposes, the number of those slits through which light emerges.

So in effect the whole contraption is in four separated parts, and there are two detectors, each detecting the output from two of the parts.

Now, how to arrange the open and closed slits? Let’s denote open slits by a \( +1 \) (or just \( + \) for short), and closed slits by a \( -1 \) (or just \( – \) for short). So a sequence of open and closed slits can be denoted by a sequence of \( + \) and \( – \) symbols.

(You might think \( 1 \)s and \( 0 \)s are more appropriate for open and closed slits then \( +1\)s and \( -1 \)s. You could indeed use \(1\)s and \(0\)s; in that case I’ll leave it up to you to adjust the mathematics below.)

Now Golay suggested taking *two* sequences \(a\) and \(b\) of \(+\)s and \(–\)s, each of length \(N\) . They would be used to configure the slits. Let’s write \( a = (a_1, \ldots, a_N) \) and \(b = (b_1, \ldots, b_N)\) , where every \(a_i \) or \( b_i \) is either a \( +1 \) or a \( -1 \) .

Now, sequences \(a \) and \( b \) each have length \( N\), but there are \( 4N \) entrance slits and \( 4N \) exit slits.

What to do? Golay said what to do. Golay said also to take the *negatives* of \( a \) and \( b\). The negative of a sequence is given by multiplying all its terms by \(-1 \) (just like how you take the negative of a number). In other words, to take the negative of the sequence \( a\), you replace each \( + \) with a \( – \), and each \(– \) with \(+\). We can write \(-a\) for the negative of \(a\), and \(-b\) for the negative of \(b\).

Golay suggested, very cleverly, that the \(4N \) entrance slits, from top to bottom, should be should be arranged using \(a \) (for the top \( N \) slits), then \( -a \) (for the next \( N\)), then \( b \) (for the next \(N\)), and finally \(-b\) (for the bottom \(N\) slits). So as we read down the slits we read the sequences \( a,-a,b,-b\).

On the exit side, because the light is “inverted”, we now read bottom to top. Golay suggested that, as we read up the slits, we use the sequences \( a,-a,-b,b\). That’s not *quite* the same as what we did on the entrance side. The top \(N\) entrance slits, set according to the sequence \(a\), correspond to the bottom \(N \) exit slits, also set according to the sequence \(a\). The next \(N\) entrance slits are set according to \(-a\), as are the next \( N \) exit slits. But after that, the entrance slits set according to \( b \) correspond to the exit slits set to \( -b \) ; and the final \( N \) entrance slits are set to \( -b\), with corresponding exit slits set to \( b\). So the \(a \) and \( -a \) slits “match”, but the \( b \) and \( -b \) “anti-match”.

We can now see what the a’s and b’s mean in Golay’s diagram. (Golay writes \( a’ \) and \( b’ \) rather than \(-a\) and \(-b\).)

One final twist: the output of the contraption is measured by the two detectors \( D_1 \) and \( D_2\). But Golay proposed not to *add* their results, but to *subtract* them. So the final number we want to look at is not \(D_1 + D_2\), but \( D_1 – D_2\).

Anyway, that was Golay’s prescription.

So what happens to light going through this spectroscopic contraption, now with its numerous slits configured in this intricate way?

First let’s consider red light — which, recall, means the light goes straight from entrance slit to opposite exit slit. We’ll take the four sections separately, which, we recall, are labelled \(a,-a,b,-b \) at the entrance, and \( a,-a,-b,b \) at the exit.

- For light hitting one of the top \( N \) entrance slits, one encoded by \( a_i\), it is blocked if \( a_i = -1\). But if \(a_i = 1 \) then the light sails through the open slit, out to the corresponding exit slit, which is also labelled \( a_i = 1\), and through to the detector \( D_1\).
- Similarly, consider one of the entrance slits in the next section, encoded by some \( -a_i\). Light is blocked if \(-a_i = -1 \) but if \( -a_i = 1 \) then the light sails over the the corresponding exit slit, also labelled \( -a_i = 1 \) , through to the detector \( D_1\).
- Now consider the third section, where entrance slits are encoded by \( b \) but exit slits are encoded by \( -b\). Light hits a slit encoded by some \(b_i\). If \(b_i = -1\), the entrance slit is closed, and the light is blocked there. If \( b_i = 1\), the entrance slit is open, and the light enters, but then the exit slit is encoded by \( -b_i = -1\), so is closed, and the light is blocked here. Either way, the light is blocked.
- The final section is similar. The entrance slit is labelled by some \( -b_i\), and the exit slit by \(b_i\). If \( -b_i = -1\), the entrance slit is closed and light is blocked; if \( -b_i = 1\), then the entrance slit is open, but as \( b_i = -1\), the exit slit is blocked. Either way the light is blocked.

Now detector \(D_1 \) counts the number of slits in the first two sections from which light emerges. In the first section, those slits are the ones encoded by \( a_i \) such that \( a_i = 1\). In the second section, those slits are the ones encoded by \( -a_i \) such that \( -a_i = 1 \) , i.e. \( a_i = -1\). On the other hand, \(D_2 \) detects nothing, as everything is blocked. So we have

\[

D_1 = ( \# i \text{ such that } a_i = 1) + ( \#i \text{ such that } a_i = -1), \\

D_2 = 0.

\]

The expression for \( D_1 \) simplifies rather dramatically, because *every* \( a_i \) is either \( +1 \) or \( -1\). If you add up the number of \(+1\)s and the number of \(-1\)s, you simply get the number of terms in the sequence, which is \(N\). Thus in fact

\[

D_1 = N, \quad D_2 = 0,

\]

and the final result (remember we *subtract* the results of the two detectors) is

\[

D_1 – D_2 = N.

\]

So, we end up with a nice result, when we feed Golay’s spectroscope light of the colour it’s designed to detect (i.e. red).

Now, what happens with other colours? Let’s now feed Golay’s spectroscope some other colour (i.e. frequency, i.e. wavelength) of light, which means that the light gets shifted across \( j \) slots. Let’s say the light is green.

- Consider green light hitting one of the top \( N \) entrance slits, encoded by \( a_i\). The light is blocked if \( a_i = -1\). But if \(a_i = 1 \) then the light sails through the open slit, over to the corresponding exit slits, which are also encoded by the sequence \( a\). The light is shifted across \(j \) slots in the process, and so arrives at the exit slit encoded by \( a_{i+j}\). If \(a_{i+j} = 1\), the light proceeds to detector \( D_1\); otherwise, the light is blocked. In other words, the green light gets to the detector if and only if \( a_i = a_{i+j} = 1\).
- (Note also that if \( i+j > N \) or \( i+j < 1\), then the light beam gets shifted so far across that it hits the end of the section of the machine; and the sections are separated from each other. So we only need to consider those \( i \) (which are between \( 1 \) and \( N \) ) such that \( i+j \) is also between \( 1 \) and \( N\). In other words, (assuming \( j \) is positive) \( i \) only goes from \( 1 \) up to \( N-j\).
- Now consider green light hitting the second section, where entrance and exit slits are labelled by \( -a_i\). If \( a_i = -1\), then light is blocked at the entrance. If \( -a_i = 1\), light enters, and proceeds with a shift over to an exit slit encoded by \( -a_{i+j}\). If \( -a_{i+j} = -1\), light is blocked at the exit, but if \( -a_{i+j} = 1\), then the light proceeds to detector \( D_2\). In other words, light gets to the detector \( D_1 \) if and only if \( -a_i = -a_{i+j} = 1\), or equivalently, \( a_i = a_{i+j} = -1\).
- In the third section, entrance slits encoded by \( b \) and exit slits by \( -b\). For light to get through, we must have \( b_i = 1 \) and \( -b_{i+j} = 1\).
- Finally, in the fourth section, entrance slits are encoded by \( -b \) and exit slits by \( b\). Light gets through when \( -b_i = 1 \) and \( b_{i+j} = 1 \).

Putting these together, we have

\[

D_1 = ( \# i \text{ such that } a_i = 1 \text{ and } a_{i+j} = 1) + ( \# i \text{ such that } a_i = -1 \text{ and } a_{i+j} = -1), \\

D_2 = ( \# i \text{ such that } b_i = 1 \text{ and } b_{i+j} = -1) + ( \# i \text{ such that } b_i = -1 \text{ and } b_{i+j} = 1).

\]

Now let’s manipulate these sums a little. Note that, for any \(i\), \( a_i = \pm 1 \) and \( a_{i+j} = \pm 1\). Thus the product \( a_i a_{i+j} = \pm 1\). But note that \( a_i a_{i+j} = 1 \) precisely when \( a_i = a_{i+j} = 1\), or \( a_i = a_{i+j} = -1\), i.e. when \( a_i \) and \( a_{i+j} \) are equal. These are precisely the cases counted in the sum for \( D_1 \) above. When \( a_i \) and \( a_{i+j} \) are not equal, they multiply to \( -1 \) instead.

Similarly, consider \( b_i \) and \( b_{i+j}\). The product \( b_i b_{i+j} \) is equal to \( -1 \) precisely when \( b_i = 1 \) and \( b_{i+j} = -1 \) , or when \( b_i = -1 \) and \( b_{i+j} = 1 \) . And these are precisely the cases counted above for \( D_2 \) .

So we have

\[

D_1 = ( \# i \text{ such that } a_i a_{i+j} = 1 ),\\

D_2 = ( \# i \text{ such that } b_i b_{i+j} = -1).

\]

Now, as we’ve said, for each \(i\), \( a_i a_{i+j} \) is \( 1 \) or \( -1\). For how many \( i \) do we get \( +1\)? Precisely \(D_1 \) times! Because that’s exactly what the equation above for \( D_1 \) says. All the other terms must be \( -1\). And we said above that \( i \) goes from \( 1 \) up to \( N-j\). So there are \( N-j-D_1 \) times that \( a_i a_{i+j} = -1\).

Let’s now just add up all the terms \( a_i a_{i+j}\), all the way from \( i=1\), i.e. the term \( _1 a_{1+j}\), to \( i=N-j\), i.e. the term \(a_{N-j} a_{N-j+j}\). We get \(+1 \) sometimes — precisely \( D_1 \) times — and \( -1 \) sometimes — precisely \( N-j-D_1 \) times. It follows that

\[

a_1 a_{1+j} + \cdots + a_{N-j} a_{N-j+j} = 1 \cdot D_1 + (-1) \cdot (N-j-D_1)

\]

or if we tidy up,

\[

\sum_{i=1}^{N-j} a_i a_{i+j} = 2D_1 – N + j.

\]

We can do the same for the terms \( b_i b_{i+j}\). We get \( -1 \) precisely \( D_2 \) times, as the equation for \( D_2 \) says above. And we get \( +1 \) all the other times, but there are \( N-j \) times overall, so we get \( +1 \) precisely \( N-j-D_2 \) times. Hence

\[

b_1 b_{1+j} + \cdots + b_{N-j} b_{N-j+j} = 1 \cdot (N-j-D_2) + (-1) \cdot D_2,

\]

or equivalently,

\[

\sum_{i=1}^{N-j} b_i b_{i+j} = -2D_2 + N – j.

\]

We want to get the final result of the detectors, which is \( D_1 – D_2\) . So let’s rearrange the equations above to obtain \( D_1 \) and \( D_2\),

\[

D_1 = N – j + \frac{1}{2} \sum_{i=1}^{N-j} a_i a_{i+j}, \\

D_2 = N – j – \frac{1}{2} \sum_{i=1}^{N-j} b_i b_{i+j},

\]

and subtract. When we do so, things simplify considerably!

\[

D_1 – D_2 = \frac{1}{2} \sum_{i=1}^{N-j} a_i a_{i+j} + b_i b_{i+j}

\]

This is a very nice result. And it reduces what Golay wanted to a very interesting maths problem. Two sequences \( a = (a_1, \ldots, a_N) \) and \( b = (b_1, \ldots, b_N) \) of \( \pm 1\)s are called a *complementary pair* or a *Golay pair* if, for all \( j \neq 0\), this sum is zero:

\[

\sum_{i=1}^{N-j} a_i a_{i+j} + b_i b_{i+j} = 0.

\]

Sums like these are often called *autocorrelations*. So the property we are looking for is a property of autocorrelations. Golay pairs are all about autocorrelations. Hence the title of this post.

If you can find a pair of Golay complementary sequences, then you can configure all the slits in the multislit spectrometer according to the sequence, and for any colour except the one you are looking for (red), the detectors will perfectly cancel out that colour! So your spectrometry will be greatly enhanced.

Now you might wonder, do any such pairs exist?

Yes, that do. Oh yes, they do. And that is also a very interesting question — not yet completely solved, with lots of ongoing research.

Stay tuned for more.

P.S. Yes, the title of this blog post is based on a song by Chumbawumba. It’s a very excellent song.