Don't understand equation with overleftrightarrow symbol

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Equation in Srednicki QFT book
I've started reading Srednicki's book on QFT, which was starting well. Then I hit on an equation which I just don't understand at all. Since I don't know what the symbol is called, I can only refer to it by its latex name.
Here's the bit. Srednicki defines the following object:
$$f \overleftrightarrow{\partial_{\mu}}g := f (\partial_{\mu}{g}) - (\partial_{\mu}f ) g $$.
Already, it is not clear to me if the second term is a function of a derivative or a product.
He goes on by deriving
$$i \overleftrightarrow{\partial_{0}} \phi(x) =i \partial_0 \phi(x) + \omega \phi(x) $$
clearly using
$$ \partial_0 \phi = i \omega \phi $$.

I will probably feel like an idiot when someone explains this to me, but I just can't get it. How are those two equations compatible? Does this double-arrow beast have a name?
 
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Page number or other pointer?
 
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The second term is a product. You first apply the differential operator to the left member of the original product of functions. You obtain a new function which you then multiply with the function which was/is at the right of the original product. Pay attention, the functions of spacetime may not commute in a product, so keep the exact order.
 
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Nugatory said:
Page number or other pointer?
Srednicki, QFT. page 26, just after equation 3.21. Sorry I didn't mention that.
 
  • #5
It looks a bit like differentiation by parts, but with a minus sign.
 
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dextercioby said:
The second term is a product. You first apply the differential operator to the left member of the original product of functions. You obtain a new function which you then multiply with the function which was/is at the right of the original product. Pay attention, the functions of spacetime may not commute in a product, so keep the exact order.
Uh, sorry. I didn't get that. I'm wondering what the "beast" with the double arrow over it is. And what f is.
 
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I think here is the answer.
 
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Aha, yes indeed. So it is sortuv like differentiation by parts of a product of functions, only with a minus sign. I'm guessing this has to do with the metric signature, which would explain why it could be the inverse.
Why oh why did people start using opposite metric signatures?
In any case, thanks to all.
 
  • #9
What Sredinicki evaluates in (3.21) using this symbol is
$$\begin{split}
\exp(-\mathrm{i} k x) \overleftrightarrow{\partial_0} \varphi(x)&=\exp(-\mathrm{i} k x) \partial_0 \varphi(x) - [\partial_0 \exp(-\mathrm{i} k x)] \varphi(x) \\
&= \exp(-\mathrm{i} k x) [\partial_0 \varphi(x) +\mathrm{i} k_0 \varphi(x)]\\
&=\exp(-\mathrm{i} k x) [\partial_0 \varphi(x) - \mathrm{i} \omega \varphi(x)].
\end{split}$$
In the last step I used that the four-momentum is on-shell, ##\omega=\sqrt{\vec{k}^2+m^2}##, and that Srednicky uses the (-+++) convention of the Lorentz fundamental form.
 
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joneall said:
So it is sortuv like differentiation by parts of a product of functions, only with a minus sign. I'm guessing this has to do with the metric signature, which would explain why it could be the inverse.
Why oh why did people start using opposite metric signatures?
The minus sign has nothing to do with any particular convention for the signature of the 4D metric. Even in ordinary quantum mechanics, conservation of probability requires that the wavefunction ##\psi## of a particle of mass ##m## should satisfy:$$\frac{\partial}{\partial t}\left(\psi^{*}\psi\right)+\mathbf{\nabla}\cdot\mathbf{j}=0$$where the probability current density ##\mathbf{j}## is given by:$$\mathbf{j}\equiv-\frac{i\hslash}{2m}\left(\psi^{*}\mathbf{\nabla}\psi-\psi\mathbf{\nabla}\psi^{*}\right)=-\frac{i\hslash}{2m}\psi^{*}\mathbf{\overleftrightarrow{\nabla}}\psi$$So the double-arrow differential operator is useful even in a non-relativistic setting.
 
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  • #11
That ##k_0=-k^0=-\omega## depends on the sign convention of the Minkowski fundamental form. The physics is of course completely independent of the choice of convention.
 

FAQ: Don't understand equation with overleftrightarrow symbol

What does the overleftrightarrow symbol mean in an equation?

The overleftrightarrow symbol, typically written as \( \overleftrightarrow{AB} \), represents a line that passes through points A and B and extends infinitely in both directions. It is often used in geometry to denote a line segment that has no endpoints.

How is the overleftrightarrow symbol different from other arrow symbols?

The overleftrightarrow symbol differs from other arrow symbols like \(\overrightarrow{AB}\), which denotes a directed line segment or vector from point A to point B, and \(\overleftarrow{AB}\), which represents a directed line segment or vector from point B to point A. The overleftrightarrow symbol specifically indicates a line extending infinitely in both directions through two points.

In what types of mathematical problems is the overleftrightarrow symbol commonly used?

The overleftrightarrow symbol is commonly used in geometry and vector mathematics. It is often seen in problems involving lines, line segments, and their properties, such as intersection points, parallelism, and collinearity.

How do I type the overleftrightarrow symbol in LaTeX?

To type the overleftrightarrow symbol in LaTeX, you use the command \(\overleftrightarrow{AB}\). This command will produce a line with arrows on both ends, indicating that it extends infinitely in both directions through points A and B.

Can the overleftrightarrow symbol be used in other fields outside of mathematics?

While the overleftrightarrow symbol is primarily used in mathematics, particularly in geometry and vector analysis, it can also be found in physics, engineering, and computer science to represent concepts involving lines and directions. However, its usage outside of mathematics is less common and often context-dependent.

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