An Introduction to Quantum Field Theory (Peskin and Schröder) - Page 22

[Breo]

Hello,

Can someone tell me how to derive:

$$ grad\hat{\phi} $$

from:

$$ \hat{P}= -:\int \mathrm{d³}x [\pi (x) grad\hat{\phi}(x)]: = \int \mathrm{d³}p [p a⁺(p)a(p)] $$

Are all vectors. Note normal ordering ":" is used.

I want to understand well QFT and want to learn to do this calcs.

Thank you in before hand.

 

[strangerep]

Try doing it first without the normal ordering -- it's similar to their calculation of $H$ in eq(2.31). Actually, you should probably start by reproducing eq(2.31) in more detail than P+S provide. If you can do that, then eq(2.33) will become easier.

(If you need more help than this, you'll have to "show your attempt" -- similarly to the HW forums.)

BTW, the latex macro for the "grad" is \nabla, i.e., $\nabla$ .

 

[Breo]

Try doing it first without the normal ordering -- it's similar to their calculation of HH in eq(2.31).

I already did it but I have a doubt. Why I must use: $$\mathcal{H} \ \alpha \ \hat{\pi}(x)\hat{\pi}(x') $$ instead $$ \hat{\pi}^2(x) $$ due to the hamiltonians form which is written (in my course notes) as follows:

$$ \mathcal{\hat{H}}=\frac{1}{2}\int d³x \hat{\pi}^2(x) + \phi(x)(\Delta + m²)\phi(x) $$ \phi is defined in two states and \pi in one to square power.

And why they made this change from (2.25) to (2.27):

$$ a⁺_p e^{-ipx} = a⁺_{-p}e^{ipx} $$

 

[strangerep]

Why I must use: $$\mathcal{H} \ \alpha \ \hat{\pi}(x)\hat{\pi}(x') $$ [...]

I don't know what you're referring to.

And why they made this change from (2.25) to (2.27):
$$ a⁺_p e^{-ipx} = a⁺_{-p}e^{ipx} $$

So they could factor out $e^{ipx}$ -- which makes it easier to evaluate the commutator (2.30).

 

[Breo]

I don't know what you're referring to.

I mean he multiplies a conjugate momentum term of (p) and (x) times another with (p') and (x') instead the first square.

Because we are taking care of hermitians quantities?

 

[strangerep]

I mean he multiplies a conjugate momentum term of (p) and (x) times another with (p') and (x') instead the first square.
[...]

If you're talking about something in Peskin & Schroeder, you must give a precise reference, otherwise I cannot help you.

If you're talking about some other notes, then I definitely cannot help you since I'm not familiar with those.

 

[VanamaliShastry]

And why they made this change from (2.25) to (2.27):

$$ a⁺\_p e^{-ipx} = a⁺_{-p}e^{ipx} $$

In addition to this answer

I don't know what you're referring to.

So they could factor out $e^{ipx}$ -- which makes it easier to evaluate the commutator (2.30).

you should note that the operator a^† _p is the creation operator, i.e, it creates new energy levels (as said in non-relativistic quantum mechanics) or new particles (as said in quantum field theory). The change that you mentioned physically means that creation of a particle moving in the negative x direction can be considered as creation of a particle with negative momentum.

Considering the four-vector notation, the negative sign for the momentum p^μ indicates negative energy and negative sign for x^μ indicates negative direction in time. So, in the later stages (once Dirac's equations are discussed), you'll probably understand that the said substitution also signifies the inclusion of antiparticles into the Hamiltonian (because, antiparticles are considered to move backwards in time (Feynman's approach) OR possesses negative energy (Dirac's approach))

 

[ChrisVer]

I don't understand what is your question. The physical momentum operator is defined by:
P^i = \int d^3x T^{0i}
Where T^{\mu \nu} is the energy momentum tensor. The energy momentum tensor is obtained by applying the Noether's theorem on your lagrangian. That is already derived in P+S and used...
So P^i = - \int d^3 x ( \pi \partial_i \phi)

You cannot derive the \partial_i \phi from this for all I can see... the \partial_i \phi is found by applying the derivative onto the solutions \phi you found for the free scalar field after quantization:

\phi (x) =\int \frac{d^3 k}{(2 \pi)^3 \sqrt{2 \omega_\textbf{k}}} \big ( a_\textbf{k} + a_\textbf{-k}^{\dagger} \big ) e^{i \textbf{k} \cdot \textbf{x}}

 

[ChrisVer]

Also for your other question, aparart from the physical answer given by Vanamali, you can do it by hand:

\int \frac{d^3 k}{ (2 \pi )^3 \sqrt{\omega_k}} (a_k e^{ikx} + a_{k}^{\dagger} e^{-ikx})

this is two integrals, I take only the 2nd and write it:

\int \frac{d^3 k}{ (2 \pi )^3 \sqrt{\omega_k}} a_{k}^{\dagger} e^{-ikx}

Do a change k \rightarrow -k. The \omega_k = \sqrt{m^2 + |k|^2} = \omega_{-k} (doesn't change. The d^3k \rightarrow -d^3 k and also the integration limits ar going to be interchanged...to bring them back to what they were before, they wll take the minus of the differential .. a_k^\dagger \rightarrow a_{-k}^\dagger and the exp(-ikx) \rightarrow exp(ikx). The result will then be the one you ask for...
I think the best reason to explain why you do that is bcause it makes the whole algebras easier when multiplying fields...a field multiplication like: \phi \phi would else have 4 exponentials, while now it will have only 2...

 

[pongwit2534]

Hello, I also try to prove eq. 2.33 in page 22
by replacing phi(x) as you mentioned in
Pi=−∫d3x(π∂iϕ)

but i got struck with 4 term with a†a†, aa ,a†a, and aa†
I don't know how to go next to get rid off first 2 terms (because the last 2 terms can use the commutation relation to combine it as number operator)

How do i get rid of the "a†a† and aa" terms?

 

[ChrisVer]

can you give the expression?

 

[pongwit2534]

Here is the picture of my note (I wrote it in Thai but just see the equations)

you can see that the last equation has 4 terms with creation and annihilation operator

I have no idea how to eliminate a†a† and aa term I don't want to just ignore it.

sorry for adding picture instead of typing them. I am quite new for writing LATEX. Thank you

 

[pongwit2534]

in case the pic is not clear

 

[ChrisVer]

Yup the expression seems correct...
Try to think about the last (your problematic) terms, what would happen if p \leftrightarrow -p
Obviously the integral of those terms should vanish (if Peskin is right)