Demystifying Quantum Mechanics - Part III  In Part I, I tried to show how one of the main assumptions of Quantum Mechanics (QM) arises, namely that measurements are the diagonal values of certain blocks of numbers. These blocks of numbers are called "linear operators." Part II showed how modelling a random variable as columns of numbers, called "kets", leads naturally to taking the overlap between kets to compare their similarity. I then showed how squaring this gives a probability. This is one of the other main assumptions of QM.I'd now like to look at how we can model mutually incompatible measurements. Our example of the die is a bit too complicated for this. Instead, I'm going to switch to a coin. When a coin gets tossed in the air it tumbles unpredictably until it lands either heads up or tails up. So there are two possible measured values, heads or tails. Just like with the die, we model these two states as columns of numbers.`|h> = 1 |t> = 0 0 1`The measurement operator will be represented by a 2x2 block of numbers this time. I'll call it S (for "side" of the coin).`S = 1 0 0 -1`The numbers 1 and -1 in the block represent heads and tails respectively. They are an arbitrary choice, we could have chosen any two numbers. In fact, to show that the choice is arbitrary, I'll mostly represent S as:`S = h 0 0 t`Just remember that "h" and "t" are any two distinct numbers. With these definitions in place we get our typical, QM-like measurement equations.`S|h> = h|h>S|t> = t|t>`Just like the case of the die, we can create a ket of complete uncertainty, which this time I'm going to give the symbol |c>.`|c> = 0.71 0.71`As before, we ensure that the sum of the squared values equals one, this ensures that probabilities come out correctly. e.g. The chance of getting heads is 2 = 0.712 = 1/2.But I can also define a second ket, which this time I'll give the name |a>.`|a> = 0.71 -0.71`This gives the same probabilities for heads and tails as |c> does, but it's clearly different from |c>. Both |c> and |a> represent states of equal uncertainty. I'm going to choose to interpret |c> as the state where the coin is tumbling clockwise and |a> as the state where it is tumbling anti-clockwise. (You may be a little worried about the use of the word "choose" in that sentence - I'll come back to that.)We can certainly observe a coin tumbling clockwise or anti-clockwise, so according to our assumption that observables can be modelled as linear operators, there ought to be a linear operator, call it T (for "tumbling") that satisfies:`T|c> = c|c>T|a> = a|a>`Where c and a are any two, different, numbers that we choose (there's that word again) to represent tumbling clockwise and tumbling anti-clockwise. There is.`T = 0.5c+0.5a -0.5c+0.5a -0.5c+0.5a 0.5c+0.5a`If you apply the "multiply row into column and add" rule to T|c> and T|a> you'll find that it works.There are a number of interesting things to point out here. The first is that the |c> state is a combination of the |h> and |t> states. When it's tumbling clockwise the coin is neither definitely heads nor definitely tails. The same is true for the |a> state. But you can also reverse this logic. When the coin is definitely heads it is not in a state of tumbling clockwise or anti-clockwise. In fact, it can be regarded as in a combination of the tumbling states.`|h> = 0.71 |c> + 0.71 |a>|t> = 0.71 |c> - 0.71 |a>`This is an example of two mutually incompatible observations. If the system is in a state of definite "heads or tails" then it cannot be in a definite tumbling state, and vice versa. Something very similar happens in QM with position and momentum. It turns out that the position and momentum operators have different sets of states. A particle in a definite state of position is always in a combination of momentum states, i.e. its momentum is completely uncertain. Conversely, when a particle is in a definite momentum state, its position is completely uncertain - in fact, the particle is everywhere, filling the entire universe at the same time.In reality, particles never find themselves in states of definite position or momentum. These states are mathematical idealisations, a bit like a mathematical point or a perfect circle - useful abstractions but never found in the real world. Real world particles are always found in combinations of position and momentum states. If the actual state is close to being a position state then it behaves like a particle. If the state is close to being like a momentum state then it looks more like a wave. In all cases there is some uncertainty in both the position and the momentum. This is the origin of Heisenberg's famous uncertainty principle.There's something else that's interesting about the T operator. Our previous examples of measurement operators, D and S, were both diagonal blocks of numbers. T clearly has off diagonal entries. This is because we chose S to be diagonal which gives its basis kets the very simple form that |h> and |t> have. Once we choose S to be diagonal, any mutually incompatible measurement, such as T, will not be diagonal and T's basis kets will not consist of a simple 1 with the rest zeros. However there is nothing special about S. We could have chosen T to be diagonal and then its basis kets would look simple. In that case S's form and its basis kets would look more complicated. QM often switches between different sets of basis kets like this.And now a final word about choices in QM. I defined the operator S as:`S = h 0 0 t`I could have defined it as:`S = t 0 0 h`The |h> and |t> basis kets would then have swapped their zeros and ones. As long as a consistent model is adopted, everything still works. In many cases there is a natural order to the values in an operator. E.g. the die operator, D.`1 0 0 0 0 00 2 0 0 0 00 0 3 0 0 00 0 0 4 0 00 0 0 0 5 00 0 0 0 0 6`Just as with S, I could have jumbled up the numbers along the diagonal. It would be a confusing, and probably pointless thing to do but, provided I did this consistently, everything would still work. In QM too there are often natural orderings, such as in position or momentum representations. In fact they're so natural that I can't think of any good reason for choosing alternatives, but you could choose alternative orderings if you want to.In Part IV I'll show how to combine operators and, in particular, how to construct the Schrodinger Equation. Links Archives View Archives 2018 Search Search  Counter Totals Total: 44,721Today: 26Yesterday: 53 - Page Generated in 0.0246 seconds