0
$\begingroup$

From David Tong's notes: How do we derive $$ H = \int\frac{d^3p}{(2\pi)^3}\omega_p[ a_p^\dagger a_p + \frac{1}2(2\pi)^3\delta^{(3)}\ (0) ] $$ from $$H = \frac{1}2\int d^3x\ [\ \pi^2\ +\ (\nabla\phi)^2 \ +m^2\phi^2 \ ] ~? $$

I've been stuck in these calculations for days, any help would be greatly appreciated!

The notes show that $\pi$ and $\phi$ are used as shown below :

$$ \phi(x)\ = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_p}} [a_p e^{ipx} - a_p^\dagger e^{-ipx}] $$

$$ \pi(x)\ = \int \frac{d^3p}{(2\pi)^3}\ (-i)\sqrt{\frac{\omega_p}2}\ [a_p e^{ipx} - a_p^\dagger e^{-ipx}] $$

Where p and x are 3D spacial vectors.

Also, what does this teach us about the free scalar field? Whats the underlying message other than the maths? I can't seem to understand it. I've also watched the first 5 lectures of this course (understood a good 20% of what was said).

$\endgroup$
0
$\begingroup$

From your second equation, the formulae for $\phi,\,\pi$ obtain a triple-integral expression for $H$ (where I define $n$-tuple integration as over $n$ $3$-vectors). In computing the squared terms, make repeated use of $\int d^3xe^{ik\cdot x}=(2\pi)^3\delta^3(k)$ to remove the $x$ integration. These Dirac delta factors then allow you to remove one momentum integration too. I'll leave you to try it out.

As for what the maths means:

  • the formulae for $\phi,\,\pi$ tell us the field is harmonically oscillating (the plane-wave dependence is analogous to that of a classical oscillator's position and momentum);
  • your second formula for $H$ says again in analogy with classical physics that the energy has squared-momentum and squared-oscillation contributions (the $\nabla\phi$ part plays a role analogous to $\pi=\dot{\phi}$);
  • and your first formula for $H$ says that each momentum contributes to the energy a per-quantum energy $\hbar\omega_p$.

In other words, these results from QFT generalise both the classical $\frac{p^2}{2m}+\frac{k}{2}x^2$ intuition and its nonrelativistic quantum extension, in which one oscillator's eigenenergies come in discrete steps.

Following ohneVal's suggestion, I'll mention also that the quadratic formula for $H$ shows its energy is bounded below and it has a ground state, while the $a^\dagger a$-based formula adds up excitations from such number operators' eigenstates. (I alluded above to the quantum counting seen therein.)

$\endgroup$
  • $\begingroup$ You might want to add comments the number operator and on boundedness from below, which will be relevant when moving on to fermions... $\endgroup$ – ohneVal Jun 29 '18 at 14:51
  • $\begingroup$ Thanks for the answer, I'm trying this solution right now. I understand the solution, though the math is not straightforward (at least for me right now) (Hope I get it right). Your explanations of the Hamiltonian and the functions really helped thanks! $\endgroup$ – user193115 Jun 29 '18 at 15:24
  • $\begingroup$ I'm having a bit of trouble after integrating over space. Is $ \int\delta^{(3)}(\pm p\pm q)\ dq = 1 $ ?? Also, which combinations of the creation/annihilation operators cancel each other out ? $\endgroup$ – user193115 Jun 29 '18 at 16:16
  • $\begingroup$ @grand_unifier More generally, $\int\delta^{(3)}(p-q)f(p,\,q)dq=f(p,\,p)$ and similarly with other signs. It's the $f$ factor in each integrand that causes the cancellation. Don't forget that $[a_p,\,a^\dagger_q]=(2\pi)^3 2\omega_p\delta^{(3}}(p-q)$ (at least, I think that's the normalisation consistent with what we're doing). $\endgroup$ – J.G. Jun 29 '18 at 16:54
  • 1
    $\begingroup$ @grand_unifier You can work it out; use the fact that $\delta$ is even $\endgroup$ – J.G. Jun 29 '18 at 19:52
0
$\begingroup$

The $\pi$ is the field theory is the promotion of Heisnberg's canonical operator commutation relation in the 'ordinary' quantum mechanics.

$$[p,q]=-i$$

Define $$a \equiv \frac{1}{\sqrt{2 \omega}}(\omega q + ip)$$

And you get, using the cononical relation: $$[a, a^{\dagger}]=1$$

And the qunatum harmonic oscilator Hamiltonian is just: $$H=\omega(a^{\dagger}a+\frac{1}{2})$$

See the similiarty?

$\endgroup$
  • $\begingroup$ Oh, thanks I understood that part. I feel like as I progress I will find more meaning? I guess? I understand that there is much more to learn till I get to QED and Weak and Strong interactions. $\endgroup$ – user193115 Jun 29 '18 at 15:26
  • $\begingroup$ This is an important template for much to come. Furthermore the equation that pops out is the Klein-Gordon. Furthermore There is the highly popular QFT Nut book which begins with Derivation of the path integral formualtion and it is likely to be enlightning as well. If you 'feel' you understand all of this as well than continue up the moutain :) $\endgroup$ – user76568 Jun 29 '18 at 15:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy