# Quantum theory, show variation of S zero, integrate by parts

• binbagsss
In summary: May I ask, in general how should you approach integration by parts when four -derivaitves are involved.Should I choose a different index like ##\partial_a## than u and v which are already used in the problemAm I okay to use ##\partial_v ##, but then I must be consistent throughout?You can use the usual integration by parts, or you can do partial integration with respect to each direction separately. In reality, it really is just the divergence theorem.
binbagsss

## Homework Statement

Hi,

I am trying to show the second equality , expressing all as a total derivative (I can then show that ##\delta S = ##)

See above

## The Attempt at a Solution

So the ## m ## term is pretty obvious, simply using the chain rule.

It is the first term I am stuck on. So looking by the sign, it looks like we have done integration by parts twice.

My working so far is to go by parts initially as:

##w=\partial^{u}\phi ##
##\partial w = \partial_{v}\partial^{u} \phi ##
## \partial z = \partial_{u}\partial_{v} \phi ##
## z= \partial_{u} \phi ##

to get, since we are allowed to assume vanishing of the field ##\phi ## at inifnity:

## - \int \partial_{u} \phi ( \partial_{v} \partial^{u} \phi ) ##

I am now stuck of what to do, I can't see a move that will get the desired expression for a choice of integration by parts, which makes me question whether this was the correct first move to make.

#### Attachments

• qft variaton of s.png
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Is ##a## constant? It is just the same type of relation as the mass term. Just switch the order of the derivatives.

Orodruin said:
Is ##a## constant? It is just the same type of relation as the mass term. Just switch the order of the derivatives.

The ##\partial_{u} ## is not hitting the ## \partial^{u} \phi ## though?

I have the first line as ## a^{v}\partial_{u}\partial_{v}\phi \partial^{u} \phi = a^{v}(\partial^2 _{uv} \phi)(\partial^{u} \phi )##,
whereas I have the bottom line as ## (a^{v}\partial^{2}_{uv}\phi)(\partial_{v}\partial^{u} \phi) ##..

thanks

binbagsss said:
The ##\partial_{u} ## is not hitting the ## \partial^{u} \phi ## though?

I have the first line as ## a^{v}\partial_{u}\partial_{v}\phi \partial^{u} \phi = a^{v}(\partial^2 _{uv} \phi)(\partial^{u} \phi )##,
whereas I have the bottom line as ## (a^{v}\partial^{2}_{uv}\phi)(\partial_{v}\partial^{u} \phi) ##..

thanks
No this is not the bottom line - it does not contain four derivatives ... Just use the same argument as for the mass term, or more generally ##f\, df/dx = 0.5 df^2/dx##.

Orodruin said:
No this is not the bottom line - it does not contain four derivatives ... Just use the same argument as for the mass term, or more generally ##f\, df/dx = 0.5 df^2/dx##.

erm, so I should have used the product rule on the bottom line?
Which would give arise to two terms, whereas the top line has one term..?

binbagsss said:
erm, so I should have used the product rule on the bottom line?
Which would give arise to two terms, whereas the top line has one term..?
I don't get your meaning. Both the bottom and top lines have two terms. One from the mass and one from the kinetic term.

Orodruin said:
I don't get your meaning. Both the bottom and top lines have two terms. One from the mass and one from the kinetic term.
apologies, corresponding to the kinetic term, I am only referring to here.

Is your problem that you do not see that
$$(\partial_a \partial_b \phi) (\partial^a \phi) = \frac{1}{2} \partial_b [(\partial_a \phi)(\partial^a\phi)] ?$$

binbagsss said:
apologies, corresponding to the kinetic term, I am only referring to here.
The kinetic term has one term only in both lines in your OP.

Orodruin said:
The kinetic term has one term only in both lines in your OP.

oh right got it !
by two terms I was reffering to the product rule, but then you sea-saw giving the factor of ##1/2##, thank you !

[QUOTE="binbagsss, post: 5663557, member: 252335"

My working so far is to go by parts initially as:

##w=\partial^{u}\phi ##
##\partial w = \partial_{v}\partial^{u} \phi ##
## \partial z = \partial_{u}\partial_{v} \phi ##
## z= \partial_{u} \phi ##

[/QUOTE]

May I ask, in general how should you approach integration by parts when four -derivaitves are involved.
Should I choose a different index like ##\partial_a## than u and v which are already used in the problem
Am I okay to use ##\partial_v ##, but then I must be consistent throughout? i.e. not change to ##\partial_u##
I'm a bit confused..
Many thanks !

There is nothing different from the usual integration by parts, you just do it in the direction of the given index. In other words, ##\partial_\mu## is really a lot of different terms. You can do partial integration with respect to each direction separately. In reality, it really is just the divergence theorem.

## 1. What is quantum theory?

Quantum theory is a branch of physics that explains the behavior of particles on a very small scale, such as atoms and subatomic particles. It describes how these particles interact with each other and with energy.

## 2. What does the term "S zero" refer to in quantum theory?

In quantum theory, "S zero" refers to the spin quantum number, which describes the intrinsic angular momentum of a particle. It is represented by the symbol S and has a value of either 0 or 1/2.

## 3. Can you explain the variation of S zero in quantum theory?

The variation of S zero in quantum theory refers to the change in the spin of a particle when it interacts with another particle or energy. This variation is described by the spin quantum number and can have a range of values depending on the specific interaction.

## 4. How do you integrate by parts in quantum theory?

In quantum theory, integration by parts is a mathematical technique used to simplify the calculation of integrals involving complex functions. It involves breaking down the function into smaller parts and using the properties of integration to solve for the final result.

## 5. Why is integration by parts important in quantum theory?

Integration by parts is important in quantum theory because it allows for the calculation of complex integrals that are necessary for understanding the behavior of particles on a small scale. It is also used in many theoretical and computational methods in quantum mechanics to solve equations and make predictions about the behavior of particles.

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