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Is it correct to visualize operators existing as matrices parameterized by spacetime coordinates in the context of QFT?

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up vote 7 down vote accepted

If you mean that QFT operators are matrices $M_{ij}$ whose indices $i,j$ are really points in space (or spacetime), so that the operator is really represented by the function $M(x,x')$, then the answer is No. The actual matrices corresponding to QFT operators are much much larger than that.

An object expressed by $M(x,x')$ is pretty much equivalent to operators in ordinary quantum mechanics of a single particle: $$M(x,x') = \langle x| \hat{M} |x'\rangle$$ The function $M(x,x')$ knows everything about the operator $\hat{M}$ because we have evaluated all matrix elements of this operator with respect to a basis - in this case, the position eigenvector basis. (It is just space, not spacetime.)

However, quantum field theory has a much larger Hilbert space. Instead of the simple functions above, you may imagine that an operator is expressed by the following functional: $$M[\phi(x,y,z),\phi'(x,y,z)] = \langle \phi(x,y,z)|\hat{M}| \phi'(x,y,z)\rangle$$ Note that the object on the left hand side is a functional - it is a function whose two arguments are functions of 3 variables themselves - field configurations of a Klein-Gordon field, in this case. Again, the left hand side knows everything about the operator $\hat{M}$ that acts on the Hilbert space of the Klein-Gordon quantum field theory. The formula above chose a specific basis - a "truly" continuous basis of field configurations of $\phi(x,y,z)$.

However, this prescription may make the operators of QFT look more complicated than they really are. Any operator in any quantum mechanical theory may be specified by its matrix elements with respect to any basis. And there are usually more intuitive bases. In particular, in free quantum field theories, a simple-to-imagine basis is the Fock space basis. $$|0\rangle, a^\dagger_{i}|0\rangle, a^\dagger_{i} a^\dagger_{j}|0\rangle, \dots$$ It's the vacuum and all of its excitations by an arbitrary number of creation operators that add particles into any state. In this way, the Hilbert space is represented as an infinite-dimensional harmonic oscillator. The transformation "matrix" from the continuous basis of the field configurations to the Fock space basis may be found by copying the same transformation for a normal harmonic oscillator infinitely many times and taking the tensor product. (You don't have to mechanically repeat the same work infinitely many times, so this prescription may sound more intimidating than it is.)

If you meant that the quantum fields $\hat{F}(x,y,z)$ are operators and for each $x,y,z$, then the answer is almost Yes. They're operator distributions - that generalize operators much like the distributions such as the delta-function generalize the concept of a function. However, they're damn large matrices. They're matrices expressed with respect to a basis - for example one of the two bases above. But if this paragraph captures what you meant, then you were just asking whether operators are matrices. Indeed, at least morally, they are. But this fact is true in any quantum mechanical theory and isn't specific to QFT.

An operator contains a lot of numbers and may always be represented as a matrix. However, the values of the matrix elements always depend on the basis you chose - sometimes in ways that can't be recognized quickly. However, there is something physical about each operator that doesn't depend on any basis: you may "feel" an operator. You may learn many of its properties. Some of them are more easily understood in one basis; other properties naturally lead to another basis. So it's useful to think about operators in a more general way than matrices in a particular fixed basis - because the latter viewpoint usually discourages one from gaining the insights that are more obvious in another basis. One shouldn't forget that the transition from one basis to another is just a "trivial" linear algebra that doesn't contain any special "physics" knowledge.

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@lumo: that rocks – Humble Jan 16 '11 at 23:50

Given that a quantum field is a field of operators and given that you can think of operators as (possibly infinite) matrices acting on a Hilbert (or, better, Fock) space, your assumption is correct. However, I wouldn't call it "visualizing" as I find operators more tangible than Matrices.

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I'm not quite sure if you're asking the same question in the title of your post as you are in the body of your post. But the answer to the question in your title is yes, you can think of an operator field as having a matrix associated with each point in space. That's what a field is, after all: a mapping that associates some value (in this case, an operator, which can be represented by a matrix) with each point in space.

However, keep in mind that there's a subtle difference between an operator and a matrix. An operator is an abstract entity that has some defined action on quantum states. There are various ways to describe the action of that operator, and giving the matrix representation in a certain basis is just one possible way. For example, in the basis $\{|1\rangle, |2\rangle, |3\rangle, \cdots\}$, the particle creation operator can be represented as

$$a^\dagger = \begin{pmatrix}0 \\ 1 & 0 \\ & \sqrt{2} & 0 \\ & & \sqrt{3} & \ddots\end{pmatrix}$$

but the same operator can also be represented by its action on a generic basis state,

$$a^\dagger|n\rangle = \sqrt{n+1}|n+1\rangle$$

among other ways.

The point is, the operator itself is something more than any of these representations. So don't get stuck thinking that one operator corresponds to just one particular matrix. If you have an operator field, then sure, you can choose a particular basis and use that basis to create a matrix out of the operator at each spacetime point, and then you'll have a matrix at each point, like you were talking about. But you could choose a different basis, and have a different matrix representation at each point, but it'd still be the same field. You could even do certain things without picking a basis at all, and in that case you wouldn't have any matrices at all. You'd be working directly with the operators themselves.

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Thanks, abstraction is admittedly not my strong suit, I tend to think in terms of absolutes. I can think in terms of matrices better simply because I can think in terms of spreadsheets and different heirarchies in a filing system. I am a sad creature of excel. – Humble Jan 17 '11 at 0:09
@Humble: sure, that's reasonable. I'm a little like that myself, so I think of different matrices which are related by a similarity transform as being essentially the same thing. – David Z Jan 17 '11 at 0:35

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