# P&S Chapter 2: Derivation of the KG Hamiltonian

1. Feb 8, 2005

### hellfire

I am not familiar with calculations such as the following one, and I want to be sure I do the right steps before going on with Peskin & Schröder. I want to derive step by step the Hamiltonian of the real scalar field. I have no problem to arrive at the first part of (2.31) in page 21:

$$H = \int dx^3 \int \frac {d^3p d^3p^{\prime}}{(2 \pi)^6} e^{i (p + p \prime) x} \left( - \frac{\sqrt{\omega_p \omega_{p \prime}}}{4} (a_{p} - a^{\dagger}_{-p}) (a_{p \prime} - a^{\dagger}_{-p \prime}) + \frac{-p p^{\prime} + m^2}{4 \sqrt{\omega_p \omega_{p \prime} }} (a_{p} + a^{\dagger}_{-p}) (a_{p \prime} + a^{\dagger}_{-p \prime}) \right)$$

This shall be equal to:

$$H = \int \frac{d^3p}{(2 \pi)^3} \omega_p \left(a^{\dagger}_{p} a_p + \frac{1}{2}[a_p, a^{\dagger}_{p}] \right)$$

So, I figured out I can make use of:

$$\int dx^3 e^{i (p + p \prime) x} = (2 \pi)^3 \delta^{(3)} (p + p^{\prime})$$

and

$$\int \delta(p + p^{\prime}) F(p^{\prime}) dp^{\prime} = F(- p)$$

and I can set $p\prime = - p$ and $\omega_p = \omega_p^{\prime}$ considering that $p^2 + m^2 = \omega^2_p$, remove dp' and I arrive at:

$$H = \int \frac {d^3p}{(2 \pi)^3} \frac{\omega_p}{4}} \left(- a_p a_{-p} + a_p a^{\dagger}_{p} + a^{\dagger}_{-p} a_{-p} - a^{\dagger}_{-p} a^{\dagger}_{p} + a_p a_{-p} + a_p a^{\dagger}_{p} + a^{\dagger}_{-p} a_{-p} + a^{\dagger}_{-p} a^{\dagger}_{p} \right)$$

which simplifies to

$$H = \int \frac {d^3p}{(2 \pi)^3} \frac{\omega_p}{2}} \left(a_p a^{\dagger}_{p} + a^{\dagger}_{-p} a_{-p} \right)$$

This seams to lead to the required result if $a_p = a_{-p}$, but this is not the case, isn't it? So, I may have made some error. Comments are appreciated... (I hope I made no typo...)

Last edited: Feb 8, 2005
2. Feb 9, 2005

### vanesch

Staff Emeritus
I didn't follow your calculation, but assuming you correctly got up here, I don't see the difficulty: split the integral of the sum into a sum of 2 integrals, and then change variables p => -p in the second term. Now put them together again and you're done, no ?

cheers,
Patrick.

3. Feb 9, 2005

### Norman

This is the Klein-Gordon Hamiltonian. The second term is the infinite background energy that you discard by convincing yourself that you can only measure energy differences and that this term doesn't contribute to that difference since it is present in both cases. Read the paragraph after this equation on page 21 continueing onto page 22.
Cheers

4. Feb 9, 2005

### dextercioby

$$e^{i(p+p')x}$$

is a tensor contraction of 2 3vectors or 2 4vectors.
What's the functional dependence of the operators a and a dagger...??That's the most annoying part.

Daniel.

P.S.What's the commutation relation between the operators (why is that $(2pi)^{3}$ left in the final integral)...?There must be a weird scaling of the amplitudes.

5. Feb 9, 2005

### hellfire

Yes, this would lead to the required result. Can I do that for only one of the integrals and then sum both?

(To clarify the notation: the p and p' are 3-vectors here).

6. Feb 9, 2005

### dextercioby

ADVICE:Do both integrals and try to "catch" the trick.It's an useful exercise.Many more such integrals will be present...

Daniel.

7. Feb 10, 2005

### vanesch

Staff Emeritus
Sure, why not ?

Try to calculate integral dx (f(x) + f(-x)) OVER ALL X.

You can write the first term as integral dx f(x) (duh!)
and the second one as integral dx f(-x) (duh2!)

In this second one, you do u = -x.
Because the integral is still over all x, this is: integral du f(u)
(you switched sign in du, and then inverted the integration limits again)

So now we have: integral dx f(x) + integral du f(u)

dummy change of variable in the last one u -> x:

integral dx 2 f(x)

Note that in YOUR problem there is something non-trivial:
that is that omega_p must be equal to omega_(-p)

But that's so because omega(p) = sqrt( p^2 + m^2)

cheers,
Patrick.

8. Feb 10, 2005

### vanesch

Staff Emeritus
If I remember well from P&S, all p-s here are 3-vectors, because we are still in the Schroedinger picture, so our hamiltonian is time-less.
The subtle part is indeed that you can keep exactly the same notation, but consider this time, p and x to be 4-vectors, and suddenly you're in the Heisenberg picture !

A long time ago I tried to capture this in a summary of chapter 2 of P&S:

cheers,
Patrick

9. Feb 10, 2005

### hellfire

It is clear now, thank you!

10. Sep 16, 2007

### Tipi

This is a old thread, but I think there is a better answer. P&S use the notation p, -p because it is usefull. If you make many changes of variables because of that, it become more complicated than the notation with only p, and you surely solve the problem in a way P&S would'nt.
In the following, I try another way to solve the problem, but it seems to give the bad result. I would like if you can help me with that.

Take $$\phi$$ given by eq. (2.27), impose $$\phi^*(x)=\phi(x)$$ (reality condition), and then we get easily the conditions $$a_{-p}=a_{p}^\dag$$ and $$a_{-p}^\dag=a_{p}$$. Reporting this result in the last equation of the first post of hellfire, one find
$$H=\int \frac{d^{3}p}{\left( 2\pi \right) ^{3}}\omega _{\bf{p}}a_{\bf{p}}a_{\bf{p}}^{\dag }$$​

But this is not the good result, since we have
$$a_{\bf{p}}a_{\bf{p}}^\dag = [a_{\bf{p}},a_{\bf{p}}^\dag]+a_{\bf{p}}^\dag a_{\bf{p}}$$​
which give
$$H=\int \frac{d^{3}p}{\left( 2\pi \right) ^{3}}\omega _{\bf{p}}\left([a_{\bf{p}},a_{\bf{p}}^\dag]+a_{\bf{p}}^\dag a_{\bf{p}}\right)$$​

So, the result is similar, but without the 1/2 factor which should appear in the good result :

$$H=\int \frac{d^{3}p}{\left( 2\pi \right) ^{3}}\omega _{\bf{p}}\left(\frac{1}{2}[a_{\bf{p}},a_{\bf{p}}^\dag]+a_{\bf{p}}^\dag a_{\bf{p}}\right)$$​

Could someone help me to fix the problem in my calculation?

Thanks a lot,

Tipi

11. Sep 16, 2007

### Tipi

Here is the calculation, since I found the problem in it.

We have (eq (2.27-28), p.21 of P&S)

$$\phi \left( \bf{x}\right) =\int \frac{d^{3}p}{\left( 2\pi \right) ^{3}}e^{i\bf{p\cdot \bf{x}}}\phi \left( \bf{p}\right)$$

$$\pi \left( \bf{x}\right) =\int \frac{d^{3}p}{\left( 2\pi \right) ^{3}}e^{i\bf{p\cdot \bf{x}}}\pi \left( \bf{p}\right)$$​

where

$$\phi \left( \bf{p}\right) =\frac{1}{\sqrt{2\omega _{\bf{p}}}}\left( a_{\bf{p}}+a_{-\bf{p}}^{\dag }\right) \,\,\,\, (1)$$

$$\pi \left( \bf{p}\right) =-i\sqrt{\frac{\omega _{\bf{p}}}{2}}\left( a_{\bf{p}}-a_{-\bf{p}}^{\dag }\right)\,\,\,\,(2)$$​

From (1), we have
$$a_{\bf{p}}=\sqrt{2\omega _{\bf{p}}}\phi \left( \bf{p}\right) -a_{-\bf{p}}^{\dag }\,\,\,\,(1')$$ ​

and from (2), we have
$$a_{-\bf{p}}^{\dag }=a_{\bf{p}}-i\sqrt{\frac{2}{\omega _{\bf{p}}}}\pi \left( \bf{p}\right)\,\,\,\,(2')$$​

The substitution of (2') in (1') gives
$$a_{\bf{p}}=\sqrt{\frac{\omega _{\bf{p}}}{2}}\phi \left( \bf{p}\right) +i\sqrt{\frac{1}{2\omega _{\bf{p}}}}\pi \left( \bf{p}\right)$$​

If $$\phi(x)$$ and $$\pi(x)$$ are real, then, we must have $$\phi^\dag(p)=\phi(-p)$$ and $$\pi^\dag(p)=\pi(-p)$$. So, we have

$$a_{\bf{p}}^{\dag }=\sqrt{\frac{\omega _{\bf{p}}}{2}}\phi \left( -\bf{p}\right) -i\sqrt{\frac{1}{2\omega _{\bf{p}}}}\pi \left( -\bf{p}\right)\,\,\,\,(*)$$​

and we DON'T HAVE $$a_{\bf{p}}^{\dag }=a_{-\bf{p}}$$. That was the mistakes. So it seems that the method proposed by Vanesh is better...

Sorry for that. It should be fine to find such an identity between the "a" operators... but so far I didn't find one. Damn ;)

Tipi

Last edited: Sep 16, 2007