# Non-local differential operator

1. Sep 15, 2013

### WannabeNewton

Heyo. On page 4 of Srednicki's QFT text, the following equation is given (in an attempt to make the Schrodinger equation relativistic): $i\hbar \partial_t \psi(x,t) = \sqrt{-\hbar^2c^2 \nabla^2 + m^2 c^4}\psi(x,t)$ where $\psi(x,t) = \left \langle x|\psi,t \right \rangle$ is the position representation of the state vector, as usual.

He then states that if we expand the square root in the RHS in powers of $\nabla^2$, then we get an infinite number of spatial derivatives acting on $\psi(x,t)$ which implies that the above equation is non-local in space. I don't get why this implication follows (i.e. why an infinite number of spatial derivatives acting on the position-space wavefunction implies that the equation is non-local in space), could someone explain it? Thanks!

2. Sep 15, 2013

### jostpuur

The explanation repeated by phycisists makes no sense.

The series $\sqrt{1+a}=1+\frac{1}{2}a+\cdots$ diverges if you substitute an unbounded operator in place of $a$, so you cannot define the mentioned operator with such series.

The operator $\sqrt{1-\nabla^2}$ is defined with Fourier transforms. The non-locality property can be proven from the Fourier transform definition.

I think the idea of the phycisists' explanation comes from the fact that $p(\nabla)$ is local for any polynomial $p$, and $e^{u\cdot\nabla}$ is a translation operator, which is non-local in such sense. So it seems that non-local operators can be written with infinite series of differential operators. But these are only two examples. Surely some infinite series produce local operators too? Although I haven't checked any examples.

Last edited: Sep 15, 2013
3. Sep 16, 2013

### WannabeNewton

Ah sweet, thanks! That cleared it up for me, much appreciated mate :)

4. Sep 16, 2013

### jostpuur

In fact I think it could be reasonable to search for some example of a local operator with infinite series of higher order differential operators. For example, suppose I define an operator as

$$A = \sum_{n=0}^{\infty}\frac{1}{n^n} D_x^n$$

Will $(Af)(x)$ only depend on $f$ in some infinitesimal environment of $x$, or will the value be affected by the values of $f$ far away from $x$?

I've never thought about that, except now...

5. Sep 16, 2013

### jostpuur

I just realized my question is not well posed, because we don't have a proper definition for localness or non-localness! :surprised

The entire point of the Taylor series is that local information in infinitesimal environment contains non-local information (for some functions)!

Well anyway, it is true that $\sqrt{1-D_x^2}$ has some obvious serious non-local properties. If $f$ has a bounded support, then $\sqrt{1 - D_x^2}f$ will not have a bounded support.

Apart from that special result, it would be nice to have some definition for what this "local stuff" is supposed to mean in general.

6. Sep 16, 2013

### jostpuur

Oh dear, I think I'm making mistakes, because I'm only relying on my distant memories. It probably won't hurt if these Fourier transform things are gone through again here?

I'm sure this topic has been covered on PF before (perhaps multiple times)...

I tried to check how to prove the non-locality property, and I was only able to convince myself of this: If $f$ has a bounded support and $|f|$ too is bounded, then it cannot hold that $\sqrt{1-D_x^2}f$ has bounded support and $\big|\sqrt{1-D_x^2}f\big|$ would be bounded too. In other words, the implication would only be that either $\sqrt{1-D_x^2}f$ does not have a bounded support, or then $\big|\sqrt{1-D_x^2}f\big|$ is not bounded. But this implication looks like weaker than the result I remember.

Those guys who feel confident with these mathematical quantum physics things, please, share your knowledge on what the strongest well-known result actually is.

7. Sep 17, 2013

### jostpuur

The result my memories are telling me about could be this: Assume that $f$ belongs to the Schwartz space, and also that $f$ has a bounded support. Then also $\sqrt{1-D_x^2}f$ belongs to the Schwartz space, but does not have a bounded support.

This result shows that there is something non-local about the operator.

But should this be considered a "strong result"? The assumption concerning the Schwartz space is a strong assumption, which makes the result weak in the same sense.

8. Sep 17, 2013

### vanhees71

Well for quantum physics it's a pretty strong result since it tells us that there is particle creation and absorption in relativistic quantum theory and that you thus need a many-body theory if you want to keep it local and thus relativistic QFT, upon which the Standard Model of elementary particle physics is based :-)).

9. Sep 17, 2013

### jostpuur

It is truly amazing what things phycisists can deduce from some mathematical results considering that they cannot even separate diverging series from convergent ones.

And how could the ordinary relativistic QFT be "local" while nobody knows what the theory is supposed to look like in spatial representation? And while according to some, its spatial representation doesn't even exist?

In fact also the non-relativistic Schrödinger equation has a significant non-locality property (although of a different nature than the relativistic one). Strangely we don't need to deduce the many particle necessity there.

10. Sep 17, 2013

### vanhees71

Well, I guess here we have a typical "clash of civilizations" between mathematicians and physicists :-)).

Strictly speaking you are right: So far there is no example for a real-world relativistic QFT of interacting particles that could be strictly formulated in terms of mathematical rigor. Nevertheless relativistic QFT, as used in its perturbative setup to calculate S-matrix elements (cross sections) is a pretty successful theory.

Also I don't see, what divergent series have to do with the original question. As was already pointed out in this thread the definition of the operator is best established in momentum representation
$$\tilde{\psi}(\vec{p}) \mapsto \sqrt{1+\vec{p}^2} \tilde{\psi}(\vec{p}).$$
There is no need to expand this in a power series of $\vec{p}^2$.

The issue in relativistic field theory is that causality has a more restrictive meaning than in non-relativistic field theory: In the former case there must not be "faster-than-light propagation", i.e., a signal must always propagate with a speed less than the speed of light, i.e., the time evolution of a wave packet with compact support in space at an initial time must always give a wave packet with compact spatial support at any later time.

At the same time, the energy should be bounded from below. From that idea, one comes to the idea to investigate the non-local wave equation
$$\mathrm{i} \partial_t \phi(t,\vec{x}) = \sqrt{1-\vec{\nabla}^2} \phi(t,\vec{x}),$$
interpreting the operator on the right-hand side as the Hamilton operator for a non-interacting spin-less particle of mass $m=1$.

As emphasized above the operator is better defined in momentum space, and in momentum space the equation reads
$$\mathrm{i} \partial_t \tilde{\phi}(t,\vec{p})=\sqrt{1+\vec{p}^2} \tilde{\phi}(t,\vec{p}).$$
This implies
$$\tilde{\phi}(t,\vec{p})=\exp(-\mathrm{i} \omega_p t) \tilde{\phi}_0(\vec{p}).$$
In spatial representation this means
$$\phi(t,\vec{x})=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x}' U(t,\vec{x}-\vec{x}') \phi_0(\vec{x}')$$
with
$$U(t,\vec{y})=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \exp(-\mathrm{i} \sqrt{1+\vec{p}^2}) \exp(\mathrm{i} \vec{p} \cdot \vec{y}).$$
It can be easily shown that this does not vanish for space-like arguments. For $\vec{y}^2 \gg t^2$ we can estimate the integral with the stationary-phase method giving
$$U(t,\vec{y}) \propto \exp[-\sqrt{\vec{y}^2-t^2}],$$
which is small but nowhere 0.

This means that parts of the wave propagate faster than the speed of light, and this violates causality in a relativistic framework. The only known solution within local relativistic QFT is to introduce creation operators in addition to annihilation in the mode decomposition of the Klein-Gordon field, i.e., to introduce anti-particles into the game to fulfill the boundedness of energy and causality at the same time. See Weinberg, The Quantum Theory of Fields, Vol. I for further details.

11. Sep 17, 2013

### jostpuur

IMO the causality argument has always appeared too philosophical to be reliable.

One point of view is that the solutions of relativistic Schrödinger equation are also solutions of the Klein-Gordon equation, and the Klein-Gordon equation does not violate causality.

Another point of view is that the relativistic Schrödinger equation is Lorentz invariant, and therefore it cannot violate causality.

So IMO it would be more reasonable to accept the fact that the wave packet solutions are always non-local, and that's the way they are. It is not surprising at this point, that sometimes nature is strange.

12. Sep 17, 2013

### vanhees71

The KG equation itself does not violate causality, but for the causal solutions you need modes with both signs of frequencies. The attempt of a one-particle interpretation of the solutions of the KG equation together with the assumption of boundedness of energy from below violates causality.

Both physical requirements can, within a local theory, only be fulfilled by a many-body interpretation, which best is formulated as a quantum-field theory. It's known as the "Feynman-Stückelberg trick". In the canonical quantized theory you simply write down a creation instead of a annihilation operator for the negative-frequency contributions (and flip the sign of momentum), so that
$$\hat{\phi}(x)=\int_{\mathbb{R}^3} \frac{\mathrm{d}^3 \vec{p}}{\sqrt{(2 \pi)^3 2 \omega(\vec{p}}} \left [\hat{a}(\vec{p}) \exp(-\mathrm{i} p \cdot x) + \hat{b}^{\dagger}(\vec{p}) \exp(+\mathrm{i} p \cdot x) \right]_{p^0=E(\vec{p})}.$$
Here
$$E(\vec{p})=+\sqrt{\vec{p}^2+m^2}.$$
This model for (free) scalar particles has the following desired properties:

(a) The (normal ordered) Hamiltonian is a positive semidefinite operator:
$$\hat{H}=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{p} [\hat{a}^{\dagger}(\vec{p}) \hat{a}(\vec{p}) + \hat{b}^{\dagger}(\vec{p}) \hat{b}(\vec{p})],$$

(b) Then theory is micro-causal, i.e.,
$$[\hat{\phi}(x),\hat{\phi(y)}]=0 \quad \text{if} \quad (x-y)^2<0.$$

(c) The Poincare group is locally represented, i.e., there exists a unitary operator such that
$$\hat{U}(\Lambda,a) \hat{\phi}(x) \hat{U}^{\dagger}(\Lambda,a)=\hat{\phi}(\Lambda^{-1} x+a).$$

13. Sep 18, 2013

### jostpuur

To be more precise, I would say that for local (non-zero only locally) solutions you need modes with both signs of frequencies.

If a solution is non-zero (almost) everywhere, I wouldn't interpret it as a paradox directly.

This is an interesting result, but its meaning is mysterious.