User:Phinneganthephysicist/sandbox/creationandannihilation

--Testbed for my updates to the article--

We seek to show how starting from the commutators between field operators, indexed by position, emergent excitations (particles) arise.

We treat the Klein-Gordon field, which is a stepping stone to more advanced cases such as Dirac fields.

The Lagrangian density is:

$$ \mathcal{L} = \frac{1}{2}\partial_\mu\partial^\mu\phi - \frac{1}{2}m^2\phi^2 = \frac{1}{2}(\partial_t \phi)^2 - \frac{1}{2}(\nabla \phi)^2 - \frac{1}{2}m^2 \phi^2 $$

Although the Lagrangian is manifestly Lorentz invariant, we must single out the time dimension to quantize. The canonical momentum for a field is defined as $$\pi = \frac{\partial \mathcal{L}}{\partial\dot{\phi}} = \dot{\phi}$$. We write the Lagrangian density in terms only the field, its spatial derivatives, and its canonical momenta as:

$$ \mathcal{L} = \frac{1}{2}\pi^2 - \frac{1}{2} (\nabla\phi)^2 - \frac{1}{2}m^2 \phi^2 $$

The Hamiltonian density is defined as the Legendre transform of the Lagrangian density:

$$ \mathcal{H} = \pi \frac{\partial\mathcal{L}}{\partial \pi} - \mathcal{L} = \frac{1}{2}\pi^2 + \frac{1}{2} (\nabla\phi)^2 + \frac{1}{2}m^2 \phi^2 $$

To quantize, we promote fields to operators and apply the canonical commutation relations. To ensure that the fields' expectation values evolve classically, we have $$ [\phi(\vec{x}),\phi(\vec{y})] = [\pi(\vec{x}),\pi(\vec{y})] = 0 $$, and $$ [\phi(\vec{x}),\pi(\vec{y})] = i\delta^{(3)}(\vec{x}-\vec{y}) $$. Because the field and its conjugate momenta are observables, the field operators are Hermitian operators. Our total Lagrangian is

$$ L = \int d^3x\bigg[\frac{1}{2}\pi(x)^2 - \frac{1}{2} (\nabla\phi(x))^2 - \frac{1}{2}m^2 \phi(x)^2\bigg]. $$

Define the Fourier transform operators as $$ \phi(k) = \int d^3x\bigg[e^{-\vec{k}\cdot\vec{x}}\phi(\vec{x})\bigg] $$, and $$ \pi(k) = \int d^3x\bigg[e^{-\vec{k}\cdot\vec{x}}\pi(\vec{x})\bigg] $$. Because the real space field operators are Hermitian, we have $$\phi^\dagger(k)=\phi(-k)$$ and $$\pi^{\dagger}(k)=\pi(-k)$$. Because the real space field operators commute with each other, and the real space momentum operators commute with each other, the Fourier field operators commute with each other, and the Fourier momentum operators commute with each other. However, the Fourier field operator will not necessarily commute with the Fourier momentum operator.

$$ [\phi(k),\pi(k')] = \int d^3x \int d^3x' \bigg[e^{-ikx-ik'x'}[\phi(x),\pi(x')]\bigg]=\int d^3x\int d^3x'\bigg[e^{-ikx-ik'x'}\delta(x-x')\bigg]=\int d^3x[(2\pi)^3 e^{-(k+k')x}] = (2\pi)^3\delta(k+k') $$

Using Plancherel's theorem, we can write the Hamiltonian as: $$ H = \int d^3x \bigg[\frac{1}{2}\pi^2 + \frac{1}{2} (\nabla\phi)^2 + \frac{1}{2}m^2 \phi^2\bigg] = \int \frac{d^3k}{(2\pi)^2} \bigg[\pi(k)^2 + (k^2+m^2)\phi(k)^2\bigg] = \int \frac{d^3k}{(2\pi)^2} \bigg[\pi(k)^2 + \omega_k^2\phi(k)^2\bigg], $$ where we have the frequency $$\omega_k = \sqrt{m^2+k^2}$$. This looks similar to the Hamiltonian for a quantum harmonic oscillator, so we can attempt a solution by ladder operators. However, the Fourier field and momentum operators are non-Hermitian, so we need to slightly modify the ladder operators. Define ladder operators $$a_k = \sqrt{\frac{\omega_k}{2}}\phi(k)+\frac{i}{\sqrt{2\omega_k}}\pi(k)$$ and $$a_k^\dagger = \sqrt{\frac{\omega_k}{2}}\phi^\dagger(k)-\frac{i}{\sqrt{2\omega_k}}\pi^\dagger(k)$$ Need to insert sources - David Tong's QFT Notes, Sidney Coleman's lectures