Understanding Local and Nonlocal Operators in Quantum Field Theory

[martinbn]

WHAT is also true for the local ones?

What you wrote in your post.

 

[Demystifier]

What you wrote in your post.

So you are saying that $O(f)$ is a local operator?

 

[vanhees71]

Since this thread turns out to be about nitpicking on sloppy (physicist's?) and precise (mathematician's?) definitions, it should be mentioned that in mathematics direct product and tensor product are different things, but physicists often say "direct" product when in fact they mean tensor product. In the example above, I think it was really the direct (not the tensor) product.

Do we have to define all well-known terminology on PF? It all started with "reference frame". Now we get lost in discussions about standard terminology in quantum (field) theory. It's getting somewhat annoying!

A direct, Kronecker, or tensor product in quantum theory is a construct like $|x \rangle \otimes |y \rangle \otimes |z \rangle=|\vec{x} \rangle$ for vectors or, in this case, generalized eigenvectors and the corresponding definitions for operators $\hat{O}_1 \otimes \hat{O}_2 \otimes \hat{O}_3$ etc.

Of course you can define all kinds of "direct products" between all kinds of mathematical objects:

https://en.wikipedia.org/wiki/Direct_product

 

[Demystifier]

That is not how i read it. Where is that definition? To me this looked like an example of a local operator. Similar to saying that a polynomial is a continuous function. That is not what the definition of a continuous function.

The paper says: "... for any local quantum field operator ${\cal O}(\vec{x}) = e^{-i{\vec P}\cdot{\vec x}}{\cal O}e^{i{\vec P}\cdot{\vec x}}$ ..." How could the word "any" be interpreted as an example?

 

[Demystifier]

Do we have to define all well-known terminology on PF.

For @martinbn , yes we do.

 

[martinbn]

So you are saying that $O(f)$ is a local operator?

No, I said that local operators are also of that type, they evaluate on test function. Are you saying they are not?

The paper says: "... for any local quantum field operator ${\cal O}(\vec{x}) = e^{-i{\vec P}\cdot{\vec x}}{\cal O}e^{i{\vec P}\cdot{\vec x}}$ ..." How could the word "any" be interpreted as an example?

Because it means any of these. And they have a definition for the general case early on, which I already quoted. How would you understand the statement "For any polynomial $a_nx^n+\cdots+a_1x+a_0$, we have..."? As a definition of polynomial? That all polynomials are of that form? Then is $x^2+y^2$ a polynomial?

 

[martinbn]

Do we have to define all well-known terminology on PF?

You don't have to do anything, but if asked, why would you argue about nonrelevant things for two pages instead of give the definition? Have you never asked about a definition here? Is it not allowed to ask about definitions here?

 

[vanhees71]

The definition was given in #1. I don't undestand, why we have to discuss standard definitions. I stop from watching this thread now, because obviously I've nothing to add to clarify the confusion. I don't understand, how you can come to the conclusion that @Demystifier 's $O[f]$ might be a local operator. It hasn't even a space-time argument and is an integral of a local field operator over all space!

 

[Demystifier]

Because it means any of these. And they have a definition for the general case early on, which I already quoted. How would you understand the statement "For any polynomial $a_nx^n+\cdots+a_1x+a_0$, we have..."? As a definition of polynomial? That all polynomials are of that form? Then is $x^2+y^2$ a polynomial?

The latter is a polynomial in $x$ with $a_0=y^2$.

 

[Demystifier]

No, I said that local operators are also of that type, they evaluate on test function. Are you saying they are not?

Local operator $O(x)$ in QFT does not evaluate on test function. For instance, for the free Klein-Gordon field operator $\phi(x)$ you can calculate $\phi(x)|0\rangle$ without using any test function. If you smear the local operator $O(x)$ with a test function $f(x)$, the resulting operator $O(f)=\int dx \, O(x)f(x)$ is no longer a local operator.

 

[Nullstein]

What if I define $O(0) := \int \phi(x) f(x) dx$ and $O(x) := e^{-i P x} O(0) e^{i P x}$? Is $O(x)$ a local operator then, because it agrees with the definition?

 

[Demystifier]

What if I define $O(0) := \int \phi(x) f(x) dx$ and $O(x) := e^{-i P x} O(0) e^{i P x}$? Is $O(x)$ a local operator then, because it agrees with the definition?

Good example! Yes, it is a local operator.

 

[Nullstein]

But if we expand it, we get $O(x) = e^{-i P x} \int \phi(y) f(y) dy e^{i P x} = \int e^{-i P x} \phi(y) e^{i P x} f(y) dy = \int \phi(y + x) f(y) dy = \int \phi(y) f(y - x) dy$, so $O(x) = \int \phi(y) g_x(y) dy$ with the new function $g_x(y) := f(y-x)$ and so $O(x)$ is again just $\phi(x)$ smeared with some non-local function $g_x$ and should really be a non-local operator according to your earlier post? I don't see what I'm missing.

 

[MathematicalPhysicist]

But if we expand it, we get $O(x) = e^{-i P x} \int \phi(y) f(y) dy e^{i P x} = \int e^{-i P x} \phi(y) e^{i P x} f(y) dy = \int \phi(y + x) f(y) dy = \int \phi(y) f(y - x) dy$, so $O(x) = \int \phi(y) g_x(y) dy$ with the new function $g_x(y) := f(y-x)$ and so $O(x)$ is again just $\phi(x)$ smeared with some non-local function $g_x$ and should really be a non-local operator according to your earlier post? I don't see what I'm missing.

I think in the integral it should be $e^{-iPx}\phi(y)f(y)e^{iPx}=\phi(x+y)f(x+y)$, am I wrong here?

 

[Nullstein]

For the translation operator, $f(x)$ is just a constant real number and can be pulled out. But even if it were true, we would just get $O(x) = \phi[f]$ instead of $O(x) = \phi[g_x]$, which would still not solve the problem.

 

[MathematicalPhysicist]

For the translation operator, $f(x)$ is just a constant real number and can be pulled out. But even if it were true, we would just get $O(x) = \phi[f]$ instead of $O(x) = \phi[g_x]$, which would still not solve the problem.

Notice that $O$ should satisfy also: $\langle \Omega , O\Omega \rangle =0$ for $\Omega$ a vacuum vector.
Does you example satisfy this condition?

 

[Nullstein]

In fact, it seems like we can make any operator $O$ a local operator according to the definition. Just define $O' := e^{i P x} O e^{-i P x}$. Then $O$ can be written as $O = e^{-i P x}O' e^{i P x}$ and satisfies the definition. Something must be wrong.

 

[Nullstein]

Notice that $O$ should satisfy also: $\langle \Omega , O\Omega \rangle =0$ for $\Omega$ a vacuum vector.
Does you example satisfy this condition?

Yes, because $<\Omega, \phi[f]\Omega> = 0$. But even if not, we can just shift $\phi[f]$ by the constant value $<\Omega, \phi[f]\Omega>$ to make its expectation value 0, so that doesn't help a lot.

 

[Demystifier]

Notice that $O$ should satisfy also: $\langle \Omega , O\Omega \rangle =0$ for $\Omega$ a vacuum vector.

That's not a part of the definition of local operator. The paper discusses local operators that satisfy this condition as an additional condition.

 

[Demystifier]

But if we expand it, we get $O(x) = e^{-i P x} \int \phi(y) f(y) dy e^{i P x} = \int e^{-i P x} \phi(y) e^{i P x} f(y) dy = \int \phi(y + x) f(y) dy = \int \phi(y) f(y - x) dy$, so $O(x) = \int \phi(y) g_x(y) dy$ with the new function $g_x(y) := f(y-x)$ and so $O(x)$ is again just $\phi(x)$ smeared with some non-local function $g_x$ and should really be a non-local operator according to your earlier post? I don't see what I'm missing.

Smearing with $f(x)$ is not the same as smearing with $f(x,y)$. For the latter kind of smearing, see post #4.

 

[Demystifier]

In fact, it seems like we can make any operator $O$ a local operator according to the definition. Just define $O' := e^{i P x} O e^{-i P x}$. Then $O$ can be written as $O = e^{-i P x}O' e^{i P x}$ and satisfies the definition. Something must be wrong.

It really means that you can construct a local operator (local in $x$) from any constant ($x$-independent) operator. But this, of course, does not mean that the initial constant operator was local. So there is nothing wrong with that.

 

[Nullstein]

Smearing with $f(x)$ is not the same as smearing with $f(x,y)$. For the latter kind of smearing, see post #4.

But that seems very strange. The operator $\phi[g_x]$ depends on values of $\phi(x)$ at many points, much like $\phi(x)\phi(7)$, yet it is supposed to be local. What about $\phi[g_7]$ then? Now it's just $\phi(x)$ smeared with a shifted version of $f$.

 

[vanhees71]

I think in the integral it should be $e^{-iPx}\phi(y)f(y)e^{iPx}=\phi(x+y)f(x+y)$, am I wrong here?

No, it's confusing, because nobody writes the hats on the operators, but here you have in the integrand $\hat{\phi}(y)f(y)$, i.e., $f(y) \in \mathbb{C}$ and thus $\hat{T}(x) \hat{\phi}(y) f(y) \hat{T}(x) = f(y) \hat{\phi}(y+x)$ with $\hat{T}(x)=\exp(\mathrm{i} \hat{P} \cdot x)$.

 

[Nullstein]

It really means that you can construct a local operator (local in $x$) from any constant ($x$-independent) operator. But this, of course, does not mean that the initial constant operator was local. So there is nothing wrong with that.

But what's wrong with the following argument then: Any constant operator $O$ is local (according to the definition), because it takes the form $O=e^{-i P x} O' e^{i P x}$ for some operator $O'$. $O=e^{-i P x} (e^{i P x} O e^{-i P x}) e^{i P x}$.

 

[vanhees71]

But then $O$ depends on $x$ and is no longer constant and thus not the same operator (except $O'$ commutes with $P$ ;-)).

 

[Nullstein]

$O=e^{-i P x} (e^{i P x} O e^{-i P x}) e^{i P x}$ only seemingly depends on $x$. The $x$-dependence just cancels out, independent of whether $O'$ commutes with $P$ or not.

 

[vanhees71]

Of course, applying the identity operator finally doesn't do anything, which is why it's called identity operator.

A constant operator can only be a local operator according to the above definition when it is translation invariant, i.e., a function of $\hat{P}$.

 

[Nullstein]

I don't see where a translational invariance condition suddenly comes in. The definition just says that an operator $O$ is local if and only if there is a second operator $O'$ such that $O=U(t) O' U(-t)$. I showed that there is such an $O'$ for any operator $O$, so any operator satisfies the definition and hence is local. The definition doesn't require me to check for translational invariance.

 

[vanhees71]

Again: By definition a local operator obeys $\hat{O}(x)=\exp(\mathrm{i} \hat{P}) \hat{O}(0) \exp(-\mathrm{i} \hat{P}).$ If $\hat{O}(x)=\text{const}$, then $\hat{O}(x)=\hat{O}(0)$, from which $[\hat{O}(0),\hat{P}]=0$, i.e., a constant operator can only be local if it commutes with the four-momentum, i.e., it's a function of four-momentum.

 

[martinbn]

I don't see where a translational invariance condition suddenly comes in. The definition just says that an operator $O$ is local if and only if there is a second operator $O'$ such that $O=U(t) O' U(-t)$. I showed that there is such an $O'$ for any operator $O$, so any operator satisfies the definition and hence is local. The definition doesn't require me to check for translational invariance.

The primed operator has to be constant.

 

[Nullstein]

Now that's a different criterion than the one in the first paper. This one applies to a set of operators now, rather than a single operator. Now a set $\{O(x)\}$ is said to be local if the individual $O(x)$ are related via translations. That's an important distinction. A single operator cannot be local or non-local then.

Take any operator $O$ and define the two sets $X = \{U(x) O U(-x)\}$ and $Y = \{U(2x) O U(-x)\}$. Then $O$ is local if viewed as a member of $X$, but not as a member of $Y$.

 

[Demystifier]

But that seems very strange. The operator $\phi[g_x]$ depends on values of $\phi(x)$ at many points, much like $\phi(x)\phi(7)$, yet it is supposed to be local. What about $\phi[g_7]$ then? Now it's just $\phi(x)$ smeared with a shifted version of $f$.

I think the problem is only in the sloppy language. To understand it, let me further simplify the problem. Let $\phi(x)$ be a local operator in one dimension, $x\in\mathbb{R}$. I claim that then $\phi(7)$ is not a local operator. But $\phi(7)$ is just a special case of $\phi(x)$. So how can a special case of local operator fail to be a local operator?

The solution of the paradox consists in using a more precise language. When we say "local operator", we don't really mean one operator. Instead, we mean the whole set of operators $\{ \phi(x)|x\in\mathbb{R}\}$.

This is very much analogous to the vector language. When we say that $V^{\mu}$ is a 4-vector, we really mean that the whole set $\{V^{\mu} |\mu\in\{0,1,2,3\}\}$ is a 4-vector. A component $V^3$, for instance, is not a 4-vector. So $\phi(7)$ is not a local operator in the same sense in which $V^3$ is not a 4-vector. And $\phi(x)$ is a local operator in the same sense in which $V^{\mu}$ is a 4-vector.

 

[martinbn]

This is not precise enough. How do you phrase all that in terms of distributions?

 

[Nullstein]

Okay, I agree that the locality criterion only makes sense if applied to sets of operators, rather than a single operator. However, the question remains why this is a meaningful definition. Apparently the set $\{\phi(x)\}$ is just as local as the set $\{\eta(x)\}$ where $\eta(x) = U(x)\phi[f]U(-x)$. But we would like to think of a local (set of) operator(s) as something that captures only local physics, e.g. like $\phi(x)$. However, $\eta(x)$ can easily depend on the physics of the whole galaxy if we choose $f$ accordingly and yet, it still qualifies as a local (set of) operator(s).

 

[Demystifier]

Okay, I agree that the locality criterion only makes sense if applied to sets of operators, rather than a single operator. However, the question remains why this is a meaningful definition. Apparently the set $\{\phi(x)\}$ is just as local as the set $\{\eta(x)\}$ where $\eta(x) = U(x)\phi[f]U(-x)$. But we would like to think of a local (set of) operator(s) as something that captures only local physics, e.g. like $\phi(x)$. However, $\eta(x)$ can easily depend on the physics of the whole galaxy if we choose $f$ accordingly and yet, it still qualifies as a local (set of) operator(s).

I think it's in fact quite physical that a field at one point depends on other fields at other points. For instance, static electric field at one point depends on static charge distribution at all other points.