Free Scalar Field Hamiltonian derivation

Question

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

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 :

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

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

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).

J.G. · Answer

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 $int d^3xe^{ikcdot x}=(2pi)^3delta^3(k)$ 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 $nablaphi$ 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 $hbaromega_p$.

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.)

user76568 · Answer

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?

Free Scalar Field Hamiltonian derivation

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