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dimension, which means it is located somewhere on a line. The more realistic three-dimensional case is not fundamentally different – it’s just harder to draw. In Figure 4.3 we’ve sketched this situation, representing the particle by a line of three clocks. We should imagine that there are many more than this – one at every possible point that the particle could be – but this would be very hard to draw. Clock 3 sits at the left side of the initial clock cluster and clock 1 is at the right side. To reiterate, this represents a situation in which we know that the particle starts out somewhere between clocks 1 and 3. Newton would say that the particle stays between clocks 1 and 3 if we do nothing to it, but what does the quantum rule say? This is where the fun starts – we are going to play with the clock rules to answer this question.

Figure 4.3. A line of three clocks all reading the same time: this describes a particle initially located in the region of the clocks. We are interested in figuring out what the chances are of finding the particle at the point X at some later time.

Let’s allow time to tick forward and work out what happens to this line of clocks. We’ll start off by thinking about one particular point a large distance away from the initial cluster, marked X in the figure. We’ll be more quantitative about what a ‘large distance’ means later on, but for now it simply means that we need to do a lot of clock winding.

Applying the rules of the game, we should take each clock in the initial cluster and transport it to point X, winding the hand around and shrinking it accordingly. Physically, this corresponds to the particle hopping from that point in the cluster to point X. There will be many clocks arriving at X, one from each initial clock in the line, and we should add them all up. When all this is done, the square of the length of the resulting clock hand at X will give the probability that we will find the particle at X.

Now let’s see how this all pans out and put some numbers in. Let’s say that the point X is a distance of ‘10 units’ away from clock 1, and that the initial cluster is ‘0.2 units’ wide. Answering the obvious question: ‘How far is 10 units?’ is where Planck’s constant enters our story, but for now we shall deftly side-step that issue and simply specify that 1 unit of distance corresponds to 1 complete (twelve-hour) wind of the clock. This means that the point X is approximately 102 = 100 complete windings away from the initial cluster (remember the winding rule). We shall also assume that the clocks in the initial cluster started out of equal size, and that they all point to 12 o’clock. Assuming they are of equal size is simply the assumption that the particle is equally likely to be anywhere in between points 1 and 3 in the figure and the significance of them all reading the same time will emerge in due course.

To transport a clock from point 1 to point X, we have to rotate the clock hand anti-clockwise 100 complete times, as per our rule. Now let’s move across to point 3, which is a further 0.2 units away, and transport that clock to X. This clock has to travel 10.22 units, so we have to wind its hand back a little more than before, i.e. by 10.22, which is very close to 104, complete winds.

We now have two clocks landing at X, corresponding to the particle hopping from 1 to X and from 3 to X, and we must add them together in order to start the task of computing the final clock. Because they both got wound around by very close to a whole number of winds, they will both end up pointing roughly to 12 o’clock, and they will add up to form a clock with a bigger hand also pointing to 12 o’clock. Notice that it is only the final direction of the clock hands that matters. We do not need to keep track of how often they wind around. So far so good, but we haven’t finished because there are many other little clocks in between the right- and left-hand edges of the cluster.

And so our attention now turns to the clock midway between the two edges, i.e. at point 2. That clock is 10.1 units away from X,