Equivalence of differential operator terms in action

In summary, the conversation is about different ways to describe the action of differential operators on fields, specifically in relation to a real field. The three expressions discussed are equivalent and can be written as (\nabla\phi) \cdot (\nabla\phi), \nabla^2 |\phi|^2, and (\nabla^2\phi) \phi. The conversation also includes a clarification about the notation and an example for better understanding.
  • #1
Mithra
16
0
Hi guys, I'm sure I'm being very stupid here but I'm reading through notes which contain various actions for fields, most of which are very similar, however there is some discrepancies with the way differential operators are shown acting on the fields and I can't for the life of me work out which are equal to which so any advice would be great.

The three forms I can see are, for a real field [itex]\phi[/itex]

[itex]
1) (∂^\mu \phi)(∂_\mu \phi)
[/itex]

[itex]
2) ∂^\mu ∂_\mu |\phi|^2
[/itex]

[itex]
3) \phi (∂^\mu ∂_\mu) \phi
[/itex]

These seem to often be used in very similar places but I can't really see how they relate? This is likely painfully simple and I'm just overthinking it but even just pointing in the right direction would be great, thanks!
 
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  • #2
Perhaps it is clearer if we just write [itex]\partial_\mu \phi[/itex] as the gradient [itex]\nabla\phi[/itex]. In that case, the three expressions you have written are:
1) [itex](\nabla\phi) \cdot (\nabla\phi)[/itex]
2) [itex]\nabla^2 |\phi|^2[/itex]
3) [itex](\nabla^2\phi) \phi[/itex]
where
[tex]\nabla \equiv \left(\frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} \right)[/tex]
and
[tex]\nabla^2 \equiv \nabla \cdot \nabla = \frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial z^2}[/tex]

It may also help you if you write out these expressions for a simple example, like [itex]\phi(x, y, z) = xyz[/itex].
 
  • #3
I could be wrong, but I think (2) as you've written it doesn't describe a charged scalar field. I think you may have meant [itex]|\partial \phi|^{2}=(\partial^{\mu}\phi^{*})(\partial_{\mu}\phi)[/itex]. Anyways, (1) and (3) are equivalent as long as you remember that these quantities appear in an integral, so that [itex]\int d^{4}x (\partial^{\mu}\phi)(\partial_{\mu}\phi)=-\int d^{4}x\, \phi \partial^{\mu}\partial_{\mu}\phi[/itex]. This means that the action for a real scalar field can be written as [itex]S=\frac{1}{2}\int d^{4}x\, [(\partial \phi)^{2}-m^{2}\phi^{2}]=-\frac{1}{2}\int d^{4}x \, \phi(\partial^{2}+m^{2})\phi[/itex]. (Note [itex]\partial^{2}\equiv\partial^{\mu}\partial_{\mu}[/itex])
 

Related to Equivalence of differential operator terms in action

1. What does "equivalence of differential operator terms in action" mean?

The equivalence of differential operator terms in action refers to the concept that certain differential operators can be represented in different forms, but still produce the same result when applied to a mathematical function or equation.

2. How do you determine if two differential operator terms are equivalent?

To determine if two differential operator terms are equivalent, you can compare their actions on a given function. If both operators produce the same result when applied to the function, then they are considered equivalent.

3. Can any differential operator term be rewritten into an equivalent form?

Yes, any differential operator term can be rewritten into an equivalent form. This is because the underlying mathematical principles and rules remain the same, regardless of the form in which they are represented.

4. Why is it important to understand the equivalence of differential operator terms in action?

Understanding the equivalence of differential operator terms is important because it allows for more flexibility and efficiency in solving mathematical problems. By recognizing equivalent forms, a scientist can choose the most suitable form for a particular problem and simplify the solution process.

5. Are there any exceptions to the equivalence of differential operator terms in action?

Yes, there are some exceptions to the equivalence of differential operator terms. For example, certain differential operators may have restrictions on the types of functions they can act on, which can affect their equivalence. Additionally, in some cases, a specific form of a differential operator may be more suitable for solving a particular problem, even if it is not equivalent to another form.

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