This questions stems from Axiomatic Quantum Field Theory and is mathematical in nature. However, I feel that an answer from physicists is more in line with what I will be asking.

Let $\phi$ be a real quantum field, namely $\phi$ is an operator-valued distribution. One of the requirements of $\phi$ is that it is local.

In case $f\in C{^\infty _0 }_{real}$, then the assumption of locality requires that $$\phi(f)\phi(g)=\phi(g)\phi(f)$$ when the supports of f and g cannot be connected by a light ray. One says that such supports are space-like separated.

Reference: Page 7 http://www.arthurjaffe.com/Assets/pdf/Quantum-Theory_Relativity.pdf

I'd like to gather further insight into this statement. Namely, how does one picture the supports? Should I have light cones in mind?


2 Answers 2


I think this is more of a comment because I don't think I really understand your question, but it is long so I will post it as an answer. I will delete if it looks like I was completely on the wrong track.

The way you are explaining the question doesn't make sense with how I think about QFT. $\phi$ is an operator valued field. The space that $\phi$ operates on is the hilbert space, and we don't really say that space-time points in the hilbert space are space-like separated or not, because they are not space time points; they are points in the hilbert space.

Now let me make a few points about notation. Points in the hilbert space are usually represented by kets. Let's write your $f$ and $g$ as $|f\rangle$ and $|g \rangle$. This is more standard notation. To write the operation of $\phi$ on a ket we usually use juxtaposition, so we would write $\phi |f\rangle$. To express the dependence of $\phi$ on spacetime coordinate, we use parentheses. So the operator $\phi$ at the spacetime point $x$ would be written $\phi(x)$.

The statement "locality" that I am used to (I would have called it causality) is actually the operator equation $\phi(x) \phi(y) = \phi(y) \phi(x)$ whenever $x$ is space-like separated from $y$. Another way of writing the equation is $[\phi(x), \phi(y)]=0$.

Maybe you are dealing a more complicated notion of a QFT where $\phi$ is a field on $C{^\infty _0 }_{real}$. Is this the case? This is the main point I was curious about.

Anway, your question becomes how do I make sense of what it means for two space time points to be space-like separated. Here is a related question this website, and I can also refer you to an article on wikipedia. For two points to be space-like separated, they have to be outside of each other's light cone. See the links for further details.

If the fields are really supposed to be defined on $C{^\infty _0 }_{real}$, then I don't know what to expect the supports to look like because I don't have any experience with this. I suppose saying the supports are space-like separated just means there is all pairs of points with one in each support that are space-like separated.


Ok I read the pdf and I saw what he did. He is using $\phi(f)$ as a shorthand for $\int \phi(x) f(x) dx$. I think he prefers dealing with $\phi(f)$ over $\phi(x)$ because he is trying to be mathematically rigorous, and this $\phi(f)$ notation will be convenient for that purpose.

Anyway his $[\phi(f),\phi(g)]=0$ condition is equivalent to my $[\phi(x),\phi(y)]=0$ condition.

To see the forward implication take $f(x')=\delta(x'-x)$ and $g=\delta(y'-y)$. Then $f$ and $g$ are spacelike separated if $x$ and $y$ are. Thus if $x$ and $y$ are space-like separated then $[\phi(f),\phi(g)]=0$, on the other hand, $\phi(f)=\int \phi(x') f(x') dx'=\int \phi(x') \delta(x'-x) dx' = \phi(x)$, and similarly $\phi(g)=\phi(y)$. Thus we have that $[\phi(x),\phi(y)]=0$.

To get the reverse direction, see that $$[\phi(f),\phi(g)]=[\int \phi(x') f(x') dx',\int \phi(y') g(y') dy']=\int \int f(x') g(y')[\phi(x'),\phi(y')]dxdy$$. Now if $f$ and $g$ are spacelike separated, then the only times that $f(x')$ and $g(y')$ are both nonzero are when $x'$ and $y'$ are spacelike separated, but then $[\phi(x'),\phi(y')]=0$, so we conclude that $\int \int f(x') g(y')[\phi(x'),\phi(y')]dxdy=0$. This proves the other direction.

Thus to get an intuition for what he means, it is sufficient to just think about the condition at single points. You can read the part of the answer above the edits to see what it means for two points to be spacelike separated. I am guessing that to be mathematically rigorous, he needs to state the conditionally formally in terms of this $\phi(f)$.

  • $\begingroup$ I edited my original question to include a reference for the question $\endgroup$
    – sunspots
    Commented Mar 18, 2014 at 3:22
  • $\begingroup$ I don't want to discourage you, but you have misunderstood what $f$ and $g$ mean in this context. The OP is asking about the smeared operator $\phi(f) = \int f(x) \phi(x) dx$. $\endgroup$
    – user1504
    Commented Mar 18, 2014 at 4:02
  • $\begingroup$ @user1504 Ok thanks, I see that now after reading the pdf, but I couldn't discern it from the original question statement. I edited my answer. $\endgroup$ Commented Mar 18, 2014 at 4:10

No, don't picture light rays. The test function is meant to be a little bump supported near some event.

For free scalar fields, it's particularly helpful to think of a test function $f(x,t) = \delta_{t_0}(t) \psi(x)$, where $\delta_{t_0}$ is some smooth approximation to a delta function and $\psi$ is a wave function. In this special case, the operator $\phi(f)$ acts on the Hilbert space by creating a particle whose wave function at time $t_0$ is $\psi$.

More generally, the test function is a source for your fields. This principle gives some help when you're dealing with gauge fields (where you have to couple to conserved currents, which can't be completely localized).


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