The Quantum Thomist

Order matters

Last modified on Sat Jul 13 18:39:23 2019

I have (sporadically) been writing a series of posts outlining the basis of quantum field theory. So far I have discussed how uncertainty is parametrised in quantum physics (including a notion that the same being can exist in numerous different states or potentia, with change being movement from one potentia to another); that particles are created and destroyed; that physics is ultimately indeterminate (meaning that even given complete knowledge of the universe at one moment in time, and complete knowledge of the laws of physics, it is impossible to predict what the universe will be like at a future moment in time -- the best we can do is calculate probabilities or likelihoods for different possible outcomes); and the crucial role that symmetry plays in contemporary physics.

In my previous post, I discussed how we might start to put these ideas into practice. My goal is to answer a single question, which is worth re-iterating:

Given that the universe has a certain likelihood of being in a state A at one moment of time, and given an understanding of physics consistent with the ideas listed above, what is the likelihood that the universe will be in a state Z at a future moment in time?

In that post, I focussed on definitions. In particular I sought to introduce a notation which would allow us to represent the physical situation abstractly. Firstly, I looked at the idea of a state, and introduced a notation to represent states (the notation is intended, eventually, to each representation in a one to one relationship with the physical states). Secondly, I introduced a notation which represents the process of comparing one state with another. Thirdly, I introduced a way in which we can assign likelihoods to states. This allows us to represent our knowledge of the situation. In practice, we can never say that "reality is in this state." Rather, we can only say, "reality is either in this state, or that state, or that state," and if we are smart about it we can assign a particular likelihood to it being in each particular state. Fourthly, I introduced the concept of operators, which act on state vectors. These operators represent the process of change from one state to another. In particular, I introduced the creation operator and an annihilation operator, which represent creation and destruction of particles in particular states. This allows us to represent a change of state as the simultaneous and localised destruction of a particle in one state and the creation of that particle in another state.

My goal in this post is to discuss two things. Firstly, how to manipulate creation and annihilation operators, and secondly how to convert from one representation of reality to another. These two things are an important step in being able to extract quantities from the abstract notation which we can compare against the real physical world.

Descartes famously hoped to be able to get from one induitable axiom, cogito ergo sum to construct a complete understanding of reality. He equally famously failed, and his failure has led many people to say that such a process is doomed to failure. Such people are cowards. Even if we fail, we learn more in the failed attempt than we would in any successful inaction. I would like to do something similar; to start from a few basic axioms, and construct a theory of fundamental physics (and, ultimately, from physics, all the rest of the sciences). Obviously, the bulk of this work is not my own -- I am relying most of all on the physicists, mostly from the 1930s to 1950s, who first developed this theory (but also the later physicists who brought it to maturity). I differ from Descartes' project in three main ways. Firstly, I don't begin with a method of doubt, entirely distrusting the senses. I assume that our senses are reliable in most circumstances, but rarely perfectly precise. We can therefore not say that "this is," but only "it is either this, or this, or this," or that it lies between certain well defined bounds. Thus our knowledge is limited by uncertainty, but that uncertainty can be parametrised by likelihoods. Since it can be systematically treated, we can still make predictions despite our uncertainty. Secondly, though Descartes was a far more intelligent mathematician than I am (his contributions were far more important to the history of mathematics than mine will ever be), I nonetheless know a lot more mathematics than was available in his time. Thirdly, I proceed from very different axioms than Descartes used.

So what are these axioms? In addition to the above, so far I have assumed that

Some things are in motion, in the classical rather than modern meaning of motion. The modern meaning is restricted to what used to be referred to as locomotion, movement from one place to another. The assumption (without proof) of this restriction is itself one of the most unfortunate fruits of the enlightenment. Make no assumptions except those which are neccessary for any reasoned empirical investigation, or which can be confirmed by unambiguous observation. I am speaking of motion as any type of change. For example, when a banana goes from green to yellow to brown, it is changing and thus (by the definition of "motion") in motion. I do not assume that there are more types of motion than just locomotion, but neither do I assume (as Descartes and the mechanists did) that there aren't.
That implies that the same being can exist in a multitude of states.
We are capable of comparing one state against another.
The universe is intelligible. We can understand the universe, and in particular how it evolves from one moment to another.
This evolution need not be deterministic (i.e. predictable even with complete knowledge and infinite computing resources). It might be, but we have no justification in assuming before we set out that it is. But, we can make predictions for the likelihoods of various outcomes given an initial state.
There are fundamental building blocks of matter, with complex objects built from simpler parts (which might not necessarily be the electrons, quarks and so on which are the most fundamental particles we know of today, or they could be).
That we can represent those states abstractly.
These states of matter are not immutable, and can matter can move from one state to another, and, indeed, fundamental particles might be able to come into and out of existence in certain circumstances.

I will need to add more axioms later. In particular, my method is only going to give us a general framework of thinking about physics. To convert that framework into a precise theory is going to need direct experimental input in various places.

My first five of these axioms are very much Aristotlean in spirit. What have have described there is a slightly eccentric and over-simplified presentation of Aristotle's central idea of potentiality and actuality, from which the bulk of Aristotle's metaphysics can be derived (in a full introduction, I would be more precise and express them in a more standard way, but since this is intended to be a simplified introduction to a different topic, I am not going to go into that detail here). Many of those assumptions were explicitly denied by the early modern scientists and philosophers of science; or rather, where I say that we have no justification of assuming one way or the other, they explicitly made that assumption. And most of the time got it wrong (as we now know). Where I differ from Aristotle is in the two penultimate of my axioms. Firstly, I believe that the universe can be represented in an abstract, mathematical langauge. Secondly, I have assumed the existence of fundamental particles, which makes me an atomist (in the classical sense of the word). So while my acceptance of the first premises puts me in the Aristotlean family, I am that weird cousin who doesn't quite fit in and nobody likes to talk to at family reunions.

So my desire is to proceed from simple axioms, which are either necessary to make any progress at all or which come from basic observation, and to rigorously develop a framework for physical theory based on those axioms. That framework is then tested and fine-tuned against experiment. Once we have the physics in place, we can look at and develop the metaphysics that inspired it. From that metaphysics, we can start to construct a theology, which will allow us to address the discussion between atheism and theism.

That's the plan, which I carried out to the best of my ability in my book. But here I cannot go into the required detail. So my goal is instead to show how it might be done, and convince the reader that the goal is at least possible.

So let us start with a state vector for an electron at a definite location. I will call that location $x$ . $^ayx$ represents the creation operator for that state, $^ax$ represents the annihilation operator for that state, and $j1xi$ (meaning that there is one particle at location $x$ represents the state itself.

But hang on. We have already made an assumption we shouldn't. How do we know that the electron states, corresponding to the particles observed in experiment, are localised at a single point? We assume (it comes directly from the axioms) that there is some way of distinguishing between states, but that way need not be the location. So let us invent another index, $p$ , that distinguishes between electron states. This might ultimately turn out to be the same as location, or it might not. We are in no position right now to judge.

But if we attempt to make a measurement of location, we will get a definite value. But perhaps this is just an average value, or perhaps it is indeterminate. So for an electron in a state $p$ , there is a certain likelihood that our instruments will record its location as $x$ , and a certain likelihood that our instruments will record its location as $y$ , and so on. The particle of a determinate state might be in an indeterminate location. We haven't proved that it is so, but neither have we proved that it isn't, so we have to keep our notation general enough to cope with both possibilities.

We already know, however, how to deal with indeterminate states; the situation is anologous to the knowledge states I constructed in the last post to cope with our lack of certainty about the state something is in. This time, however, the necessity doesn't come from our own uncertainty, but is a bit more fundamental.

So we use the same approach, and write,

j1p1i = d3xcx1 j1x1i+ d3xcx2 j1x2i +d3xcx3 j1x3i + ::::

We can't define states to be at a particular point, because a point is infinitely small volume. The probability of finding a particle in that volume is zero. You can never hit it exactly. For that reason, we instead consider states defined within a tiny volume, which I denote as $3 dx$ , surrounding that point. You can hit a small volume. By being careful with the mathematics, we can then gradually reduce the volume and smoothly approach a representation of the real-world continuum.

What I have done here is what is known as change the basis. The set of states $fj1x1i;j1x2i;:::g$ are mathematically a perfectly good representation. They are orthogonal (you can't be in two difierent states at the same time); they give the correct normalisation (the likelihood of you being in a particular state given that you are in that state is one). We just don't know that this is the representation that best reflects reality. There are plenty of other options which have the same mathematical properties; and are self-consistent possibilities.

An analogy of this is the rotatation of a coordinate system. Draw two lines (coordinate axes) at right angles to each other, mark off equidistant points along each line, and number those points. Any point on the plane can then be associated with two numbers. Each number is a measurement of how far the point is from each of the lines. But the numbers are not intrinsic to the point. If we rotate the coordinate axes, while keeping the point in the same place, then it will be described by a different set of numbers. Each of these different sets of coordinate axes is a valid way of describing the plane, as good as any other. But the system is also objective; we always cite which coordinate system we use when we specify the two numbers which specify the point. Saying that a point is at position $(1;2)$ is meaningless. Saying that it is at position $(1;2)$ in a particular coordinate system tells us everything we need to know; and if it is not a coordinate system we find useful we can always convert the numbers from that system and work out what they would be in our favoured one.

The situation of the states, whether the states should represent particles at a particular location, or with a particular value of $p$ , is similar. It is similar because the system used to represent states and the system used to represent a coordinate space both obey the rules of a mathematical vector space. The state system is more complicated (there are an uncountable infinite number of dimensions), but the mathematician knows how to deal with that. Each of these bases can completely describe any physical situation, so in some sense each of them are equally valid. Now ultimately we will be interested in states that remain meta-stable in time (since these will describe the actual material particles we see around us), which means that one basis will be more convinient for us. Just like when mapping out the dynamics of the solar system, a basis with a stationary sun is more convinient than one in which the sun moves around. But being less convinient is not the same as being invalid. We could use a different basis if we wanted to; it would just make our life and calculations much harder. While constructing the formulism (which is what we are doing right now), this freedom to rotate the basis is something we want to keep, since it has important consequencies.

In each basis, we can identify creation operators associated with each state. So there is one operator $^ayx$ which creates a particle in a state at a definite location, and another operator $^ayp$ which creates a particle in a state with a definite value of $p$ .

For example, consider the case when we create two different particles. Does it matter which order we create them in? The classical answer would be \no". What is the quantum answer? Formally, what we are asking is whether $^ayx ^ayx j0i 1 2$ , which means that we start with an empty state, then create a particle at location $x2$ , then a particle at location $x1$ , is the same as $^ayx ^ayx j0i 2 1$ . Since we can take any state as input and ask the same question, what we want to know is whether

In classical physics, this would be true. And we would naturally think $7 5 = 5 7$ , and want to apply the same logic here. But $y ^a$ is an operator, and operators don't always follow the same commutation relations as numbers. Usually they don't. Thus we can't assume that the two sides of the equation are equal. And in practice, they are not.

Now I am going to give you a choice. The next part of this post is full of technical detail. If you want to read it, then read on. If not, then click here, and you can skip to the end.

If we apply the operator $^ayx ^ayx j0i 1 2$ to the empty state, what we get is a state representing the two particles existing at the two specified locations. And $^ayx ^ayx j0i 2 1$ also represents a state with the same two particles existing at the same locations. But the outcome of these two processes need not be the same; the final result of each is not a state by itself, but a state vector multiplied by a likelihood, and the two likelihoods could be different. Any likelihood on the circumfrence of the circle represents a particle being present. We don't know where on that circumfrence we are after each procedure. It might be the same likelihood in each case, but we have no right to assume that. What we measure experimentally are frequencies, but our uncertainty is parametrised as a likelihood. A probability (which can be mapped to the frequency) is the radius squared (or complex modulus squared) of a likelihood. This means that in principle the two quantities need not be the same; the only experimental demand we have is that the modulus square of the two quantities is the same. We can thus write

We don't yet know what the number $12$ is, except that $j 12j2 = 1$ . I have written this with indices $1$ and $2$ because what this number is might depend on the two states involved. Now, we can also see that

y y y y y y
a^x1^ax2 = 12^ax2^ax1 = 12 21^ax1^ax2;

But now we use our freedom to rotate to another basis. We write that $^ayp = R1j^ayx 1 j$ , with an implicit sum over all $j$ states. If we then suppose that

We should not assume that we need the same number when comparing $p$ states as when comparing states based on location, so I will use a different symbol. Then we have

X R R ^ay ^ay = X R R ^ay ^ay :
kj 1j 2k xj xk 12 kj 1j 2k xk xj

$Pkj$ means that we are summing over every value of $k$ and $j$ . Or,

Now this is true no matter which state we apply it too, and no matter which state we compare it against. So we will apply it to the empty state $j0i$ , and compare against the state in which we create the particle at location $x1$ and then the particle at location $x2$ . This will filter out all the operators except those involving the creation operators for those states. There are two members of the sum remaining. Firstly when $j = 1$ and $k = 2$ , and secondly when $j = 2$ and $k = 1$ (we have to swap the order of the operators in one of them) and we have

But the $p$ basis is arbitrary, which means that the numbers $R11$ , $R22$ , $R12$ and $R21$ could take a huge range of possible values (restricted only by the normalisation conditions). This equation would still have to be satisfied whatever numbers we substitute in for the Rs. The only way this could happen is if $12 = 12$ and $21 = 12$ , which implies that $12 = 21$ . This doesn't just apply for the states $1$ and $2$ . It applies for any pair of states.

Thus every $\text{[math]}$ , and $\text{[math]}$ for that matter, has the same value. But we also saw that $12 21 = 1$ . This means that $212 = 1$ , or $12$ is either $1$ or $1$ .

If it is $1$ , we find ourselves with the equation we naively expect,

i.e. it doesn't matter in which order you create the states, you end up in the same place. $^ 0$ is an operator that takes in any state as input, and returns a state with likelihood zero. Particles which obey this rule are known as Bosons. The other option gives

which is a bit weirder. If we consider the case when $x1 = x2$ , we see that we can't have two particles occupying the same state. These types of particles are known as Fermions, and I will concentrate my discussion on Fermions from this point. (Why? Why not!)

The same considerations apply when we have two annihilation operators.

Now let's consider what happens when we destroy a particle and then create another particle. The argument is very similar, so I won't go over the details again. Once again, if we take a state and destroy a particle $A$ and create a particle $B$ , then we must get something proportional to the case if we create $B$ or destroy $A$ . The only thing that can be different is the numerical factor that multiplies the state. Thus,

for some number $12$ . And we can go through the same argument as before, and show that $12$ must be either one or minus one.

But that is not true in every case. You cannot destroy something that isn't there. This means that applying the annihilation operator to a state with no particles in that state shouldn't give us anything. For example, the empty state by definition doesn't have any particles at location $x1$ . So we write $^ax1 j0i = 0$ . Thus $^ayx ^ax1 j0i = 0 1$ . On the other hand, given the empty state, we can create a particle and then annihilate it. So $^ax1^ayx j0i 6= 0 1$ .

In general, $^ax1^ayx 1$ acting on a state will give a result proportional to that state. If you start with thirteen green marbles, add a purple marble, and then take away the purple marble, you are left with thirteen green marbles. Similarly, if you start with thirteen green marbles, take away a green marble and add a green marble, you are left with thirteen green marbles. The same principle applies here. So both $^ax1^ayx 1$ and $^ayx ^ax1 1$ acting on a state must be proportional to that state (even if the constant of proportionality is zero. Thus, for example, we can write that

where $X$ is some number, which might be a function of $x1$ , and $jXi$ an arbitrary state.

So let us try applying this operator to the empty state, and compare it against the empty state. We get

h0j^a ^ay + ^ay ^a j0i = h0j^a ^ay j0i
x1 x1 x1 x1 x1 x1

But $^ayx1 j0i = j1x1i$ , the state with one particle at that location (this is the definition of the one particle state). So we have

h0j^ax1^ayx1 + ^ayx1^ax1 j0i = h1x1j1x1i = 1

Which means, since we know that $y ^ax1^ax1$ acting on an empty state gives the empty state multiplied by some (until now unknown) likelihood, that $y ^ax1^ax1 j0i = j0i$ .

Now let us look at the state with one particle in it, and again apply it against the operator and compare against the original state.

h0j^ax(^ax ^ay + ^ay ^ax)^ay j0i = h0j^ax ^ax ^ay ^ay j0i+ h0j^ax ^ay^ax ^ay j0i:
1 1 x1 x1 1 x1 1 1 x1 x1 1 x1 1 x1

But, since these are fermions, we know that we can't have two particles of the same location in the same place, so $^ayx1^ayx1 j0i = 0$ , and the first of these terms disappears. But we also know that $^ax1^ayx1 j0i = j0i$ , so we can simplify the second expression. In fact we can simplify this twice, and we are left with just $h0j0i = 1$ .

Whatever state, $jXi$ we apply the operator to, we will find that

which means that $^axa^y + ^ay^ax 1 x1 x1 1$ is just the identity operator, which we can write as $^1$ . The identity operator is an operator that takes an input state, and spews out the same state as output, no matter what that state is.

This means that we have the following rules for fermions:

We combine the first two equations by inventing an operator $^ ij$ , which is $^0$ if $i$ and $j$ are different and $^1$ if they are the same.

For Boson creation and annihilation operators, $^by$ and $^b$ , the rules are similar (I am not going to prove these expressions here, but it requires similar reasoning to the previous case):

Boson operators just commute straight-forwardly with fermion operators.

Well, if you are still with me so far, then well done. If you haven't seen this before, then you are probably asking "What on earth was the point of all that?", or somewhat less polite words to that effect.

Well, at the end of the previous post, I expressed the likelihood of a system evolving from one state to another as

h0j^a ^a ^a ::: ^S:::^ay^ay ^ay j0i;
x1 x3 x5 x6x4 x2

where $^S$ represents the operator which describes the time evolution of the system. Now that operator is also going to be made from creation and annihilation operators; the operator describes the change of the system, and every instance of change involves destroying one or more particle-states and creating a set of particle states. We also know that $^a j0i = 0$ and $h0j^ay = 0$ . All we need to do to evaluate this likelihood is move all the annihilation operators to the right of the expression and all the creation operators to the left. The creation and annihilation operators will disappear. All that remains will be a simple numerical expression which can be evaluated. We move the operators around by swapping their positions. So we can replace $^axi^ayx j$ (which is in the undesired order) with $^ y i;j ^axj^axi$ (where the operators are in the order we want).

So this allows us to simplify the expression by removing all the creation and annihilation operators. What we will be left is a series of $^$ objects multiplied by numerical factors. In short, a number. We can take the modulus square of that number to get a probability; and we can compare probabilities against experimental results.

We still don't yet have an answer, of course. We still need an exact form for the operator $^ S$ which describes the time evolution of the system. I'll discuss that next time.

On the time evolution of fields.

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