Second Quantization#

Let the set of annihilation and creator operators in both position and momentum be related by,

\[\begin{split} \hat a_x = \frac{1}{\sqrt{V}} \sum_{p}\hat a(p) e^{i p \cdot x}\\ \hat a^\dagger_x = \frac{1}{\sqrt{V}} \sum_{p}\hat a(p)^\dagger e^{i p \cdot x} \end{split}\]

The number operator for the corresponding number of particles in some position and momentum be,

\[\begin{split} \hat n_x = a^\dagger_xa_x\\ \hat n(p) = a^\dagger(p)a(p) \end{split}\]

The basic usage of these operators are simply by the examples below:

Create a particle at position \(x\) starting at a field with no particles \(\ket 0\),

\[\begin{split} \ket 1 = a_x^\dagger \ket 0\\ \end{split}\]
Find the number of particles at position \(x\) for a field with two particles

\[\begin{split} \ket 2 = (a_x^\dagger)^2 \ket 0 \\ n_x \ket 2 = 2 \ket 2 \end{split}\]
Find the position of a particle at position \(x\) by determining the position PDF with some general position state \(\ket{x'}\),

\[\begin{split} \ket 1 = a^\dagger\\ \braket{x'}{1} = \delta(x-x') \end{split}\]

Thus the particle only exist at position \(x'\).

Lagrangian and Hamiltonian#

Let’s begin by writing the Lagrangian for two particles in a continuous field. It’s a lot to write so let’s drop the hat on every variable (if you like, you can justify this by noting that a constant is just an operator with itself as the eigenvalue and every vector is an eigenvector).

\[ L = \int \left[i\hbar a^\dagger_x a_x - \frac{\hbar^2}{2m} \nabla a_x^\dagger \nabla a_x - \lambda a^\dagger_x a^\dagger_x a_x a_x\right] \d^3 x \]

\[ H = \int{\left[\frac{\hbar^2}{2m} \nabla a_x^\dagger \nabla a_x + \lambda a^\dagger_x a^\dagger_x a_x a_x\right]\d^3 x} \]

Wavefunction#

The wavefunction is a state generated by applying the annihilation and creation operators thus it follows in continuous space with \(N\) particles is generated by,

\[ \ket{\Psi} = \int \prod_{i=1}^N a^\dagger_x \ket{0} \d^3 x_i \]

Schrodinger Equation#

The Schrodinger equation is still valid with this second quanitzation, recall

\[ i\hbar \dot \Psi = H \Psi \]

Proof#

The proof is not very pretty and requires induction to prove for any number of particles \(N \in \mathbb C\). Let’s begin with two particle state at some arbritrary position \(x_1\) and \(x_2\).

\[\begin{split} \ket{2} := \ket{\Psi(x_1, x_2, t)}\\ \ket{\Psi(x_1, x_2, t)} = \int a_{x_1}^\dagger a^\dagger_{x_2} \ket{0} \Psi(x_1, x_2, t) \d^3 x_1\,\d^3 x_2 \end{split}\]

It’s tedious to write the index of position \(x\) so let,

\[\begin{split} x := x_1\\ y := x_2\\ \end{split}\]

\[ \ket{2} = \int a_{x}^\dagger a^\dagger_{y} \ket{0} \Psi\, \d^3 x\,\d^3 y \]

We proceed the proof by showing that \(i\hbar \dot{\ket{2}} = H\ket{2} \). The LHS is,

\[ i \hbar \int a_{x}^\dagger a^\dagger_{y} \ket{0} \dot \Psi\; \d^3 x\,\d^3 y \]

To apply the RHS, we rely heavily on the commutation relationship in order to move operators around,

\[ a_i^\dagger a_j - a_ja_i^\dagger = \delta^3(i-j) \]

The goal is to move all creation operators to the left and anihilation operators to the right

TODO: Continue proof