Fock Space and Operators

Any CFT (classical field theory), upon quantisation, gives us the Hamiltonian and Fock Space. We use these to lay out the expectation for any given observable $\mathcal{O}$ (Hermitian operator).

• Schrödinger

$⟨s,t|\mathcal{O}(\vec{x})|s,t⟩i\hslash \frac{\partial }{\partial t}|s,t⟩=H|s,t⟩$.

• Heisenberg

$⟨s|\mathcal{O}(t,\vec{x})|s⟩i\hslash \frac{\partial }{\partial t}\mathcal{O}=[\mathcal{O},H]$.

Now, these two pictures are equivalent since the position is a label.

We define the evolution operator $|s,t⟩=U_0(t,0)|s,0⟩$ and $\mathcal{O}(t,\vec{x})=U_0^{†}(t,0)\mathcal{O}(0,\vec{x})U_0(t,0)$ so both pictures give $⟨s,0|U_0^{†}(t,0)\mathcal{O}(0,\vec{x})U_0(t,0)|s,0⟩$.

For quadratic harmonic-oscillator-like theories, evolution is simple:

$i\hslash {\partial }_t{U}_0=H{U}_0{U}_0(t)=-e^{iHt/\hslash }$.

Then, $a_k(t)=e^{iHt/\hslash }{a}_k{e}^{-iHt/\hslash }$

$=a_k+\frac{it}{\hslash }\int [d{k}^{\prime }]E_{k^{\prime }}[a_{k^{\prime }}^{†}{a}_{k^{\prime }},a_k]+\mathcal{O}(H^2)=a_k+\frac{it}{\hslash }\int [d{k}^{\prime }]E_{k^{\prime }}[a_{k^{\prime }}^{†},a_k]a_{k^{\prime }}+\mathcal{O}(H^2)$

$=a_k-\frac{i{E}_{k}t}{\hslash }{a}_k+\mathcal{O}(H^2)$.

For higher terms, we use the Baker-Campbell-Housdorff formula $e^{B}A{e}^{-B}={\sum }_n\frac{1}{n!}\stackrel{\text{n times}}{\overbrace{[B[B[\dots [B,}}A]]]]$

and show that $[H,[H,a_k]]=[H,-i{E}_k{a}_k]=-E_k^2{a}_k$ where $a_k(t)=e^{-i{E}_{k}t/\hslash }{a}_k$

which carries over to ${\Phi }_I(t,\vec{x})=\hslash \int [d\vec{k}]\sqrt{\frac{N_k}{2E_k}}(e^{-i{\omega }_{k}t+i\vec{k}.\vec{x}}{a}_k+\text{h.c.})$.

The label $I$ will be useful when talking about Interactions.

We can also use the Euler-Lagrange equations for the operator to obtain this solution,

$({\partial }_t^2-c^2{\nabla }^2+\alpha ){\Phi }_i(t,\vec{x})=0$.

Now, $⟨0|a_{k}H{a}_k^{†}|0⟩=E_k⟨0|a_k{a}_k^{†}|0⟩=E_k(2\pi {)}^3{N}_k{\delta }^3(\vec{k}-\overrightarrow{k^{\prime }})$.

Setting aside the infinity (resolved by renormalisation), the Hamiltonian expectation value is the energy, its squared will be $E_k^2$, etc. But what about the field itself?

$⟨0|a_k{\Phi }_I(t,\vec{x})a_k^{†}|0⟩=\int [d\vec{k}]A\left(⟨0|a_k\left(a_{k^{\prime }}{e}^{-i{k}^{\prime }.x}+a_k^{†}{e}^{i{k}^{\prime }.x}\right)a_k^{†}|0⟩\right)$

$=\int [d\vec{k}]A\left(e^{-i{k}^{\prime }x}⟨0|a_k\left(a_k^{†}{a}_{k^{\prime }}+(2\pi {)}^3{N}_k{\delta }^3(\vec{k}-\overrightarrow{k^{\prime }})\right)|0⟩+e^{ikx}⟨0|\left(a_{k^{\prime }}^{†}{a}_k+(2\pi {)}^3{N}_k{\delta }^3(\vec{k}-\overrightarrow{k^{\prime }})\right)a_k^{†}|0⟩\right)=0$,

where we have used $a_k|0⟩=0$ and $⟨0|a_k^{†}=0$. So the expectation value of a field in a one-particle state is zero, i.e., if we measure the field value at any point at any time, we will get a $0$ on average. This however does not mean that the field is not moving since for ${\Phi }_I(t,\vec{x}{)}^2$, the expectation value is non-zero. From this, we can conclude that the field is moving (oscillating) around $0$. We can picture our states as “ripples” on the field, around an average level:

Note: $⟨0|(a_k{)}^n\varphi (a_k^{†}{)}^n|0⟩=0$ for any $n$. We can build $⟨\varphi ⟩\ne 0$ with a superposition of different $n$ states. This allows us to qualify the statement “light is made up of photons” to “light is a superposition of multiphoton states”.

Number and Momentum Operators

$\mathcal{N}=\int [d\vec{k}]a_k^{†}{a}_k|n⟩=(a_k^{†}{)}^n|0⟩$.

Thus, $\mathcal{N}|n⟩=n|n⟩$.

An eigenstate of $\mathcal{N}$ then has a well-defined number of particles and since $[\mathcal{N},\Phi ]=0$, they cannot be simultaneously $\Phi$ eigenstates.

This counting can be used to define total momentum $P^i$ as $P^i\equiv \int [d\vec{k}]\hslash k^i{a}_k^{†}{a}_k$. Then,

$P^i{a}_k^{†}|0⟩=\hslash k^i{a}_k^{†}|0⟩$ and $P^i{a}_{k_1}^{†}{a}_{k_2}^{†}|0⟩=2\hslash k_1^i{a}_{k_1}^{†}{a}_{k_2}^{†}|0⟩$.