2

Sorry if the title sounds meta-sciency, allow me to clarify.

In physics, our goal is to understand how the universe works. To this end, we construct a theory, which hopefully makes falsifiable predictions, and then carry out experiments to test the theory.

In classical mechanics, for example, our theory allows us to determine the position and momentum of the particles in a system given some set of initial conditions. That is to say, in this theory we care about position and momentum, which we can measure.

In quantum mechanics, the situation is slightly more complicated, in that the whole theory revolves around the wave function, something which we can't measure directly. However, knowing the state of a system (its wave function), we can extract measurable information, like the probability of a particle having some position.

However, I'm starting to learn some classical field theory (to go on to QFT) and I'm completely lost from the very beginning. Here, the focus seems to be on some field $\phi(x)$. But, what does this field represent? I'm familiar with the mathematical notion of field, and could for example understand a scalar field like the temperature as a function of position. But what in the world does $\phi(x)$ represent? And how do we turn it into something we can measure?

Thanks in advance.

Physics Llama
  • 1,593
  • 13
  • 24
  • Related: http://physics.stackexchange.com/q/13157/2451 and links therein. – Qmechanic Feb 13 '16 at 21:29
  • My question was more along the lines of the following: if in Maxwell's equations, the field that the equations refer to is the electromagnetic field, what field does the Klein-Gordon equation refer to? – Physics Llama Feb 14 '16 at 00:49

1 Answers1

1

You'll want to start with the electric field, or better, the electromagnetic field: this is the abstraction of a force, which according to a force law originates at a source, and is applied to an object. With a field we simply ignore the object, removing it from the equation, and consider the situation for a hypothetical test object dropped into the field at any point in time or space. The resultant field theory permits use to work with differential equations due to the continuity of the field. In the case of EM it is a vector field.

What are the fields of interest in a typical QFT? Anything that results in an action will do, and we typically want to remove all of the classical potentials so that we have a completely quantum formulation - unlike the Shroedinger equation, with its classical potential! The goal is to obtain manifestly relativistic equations so that they work for all cases, e.g., particle physics.

Since we always start with an action to be explained, we already know of some experimental result that applies. If, however, your work is completely theoretical, looking at abstract QFts, just proving theorems - well, then the experimentalist may not know how to obtain information, and you may not know what the operators mean. After all, when looking at an axiomatic statement of geometry, there are undefined terms like point and line - only the relationships are specified.

Peter Diehr
  • 7,165
  • Hi, thanks for your answer! However, the question, more than directed along the philosohpical meaning of a general field in physics, was asking what specific field is referred to in the Klein-Gordon equation. I probably should have been clearer – Physics Llama Feb 15 '16 at 23:39
  • 1
    The Klein-Gordon equation is a relativistically covariant "equivalent" of the Schroedinger equation. Thus it refers to the same basic fields. – Peter Diehr Feb 15 '16 at 23:46