# Quantum field theory: field operators in terms of creation/annihilation operators

I am learning Quantum Field Theory and there is a step in my notes that I do not really understand.

It starts with the classical definitions of position $q$ and momentum $p$: $$q = \frac{1}{\sqrt{2\omega}}(a+a^{\dagger})$$ and $$p = -i\sqrt{\frac{\omega}{2}}(a-a^{\dagger}).$$

$a$ and $a^{\dagger}$ being the annihilation and creation operators.

Then, it defines the field operators $\phi(\mathbf{x})$ and $\pi(\mathbf{x})$, equivalent to $q$ and $p$, in the following way:

$$\phi(\mathbf{x}) = \int \frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2\omega_{\mathbf{p}}}}[a_{\mathbf{p}}e^{i\mathbf{p}\cdot{\mathbf{x}}} + a_{\mathbf{p}}^{\dagger}e^{-i\mathbf{p}\cdot{\mathbf{x}}}]$$

and

$$\pi(\mathbf{x}) = \int \frac{d^3p}{(2\pi)^3}(-i)\sqrt{\frac{\omega_{\mathbf{p}}}{2}}[a_{\mathbf{p}}e^{i\mathbf{p}\cdot{\mathbf{x}}} - a_{\mathbf{p}}^{\dagger}e^{-i\mathbf{p}\cdot{\mathbf{x}}}].$$

Is there an obvious relation between the two expressions? What mathematical and physical assumptions have been made?

It seems there is a mistake in your $\pi (x)$ expression: there must be one minus sign near $\hat{a}^{\dagger}$.

The relation between classical and field is obvious since lagrangian (hamiltonian) of free $\varphi$ (it's not hard to see that $\varphi$ satisfies Klein-Gordon equation) field may be rewritten as lagrangian of free ossilator in terms of $\hat{\varphi} , \hat{\pi} = \dot{\hat{\varphi}}$ which have been introduced in your question. The differences between "classical" and field expressions for coordinate and momentum is caused that field in every point can be represented as set of oscillators. This follows from hamiltonian in terms of $\hat{\varphi} , \hat{\pi}$): $$L =\frac{1}{2}\left( (\partial_{\mu}\varphi )^{2} - m^{2}\varphi^{2}\right) \Rightarrow \hat{H} = \int T^{00}d^{3}\mathbf r = \int \left(\frac{\partial L}{\partial (\partial_{0}\varphi)}\partial_{0}\varphi - L \right)d^{3}\mathbf r =$$ $$= \frac{1}{2}\int d^{3}\mathbf r \left( m^{2}\varphi^{2} + \pi^{2} + (\nabla \varphi )^{2}\right).$$

By introducing canonical relations $$[\hat{a}(\mathbf p), \hat{a}^{\dagger}(\mathbf p')] = \delta (\mathbf p - \mathbf p '), \quad [\hat{a}(\mathbf p), \hat{a}(\mathbf p')] = [\hat{a}^{\dagger}(\mathbf p), \hat{a}^{\dagger}(\mathbf p')] = 0$$ you get full correspondence between "classical" and "field" operators $\hat {x}, \hat{p}$, $\hat{\varphi}, \hat{\pi}$ (up to delta-function $\delta (\mathbf x - \mathbf x ')$):

$$[\hat{x}_{i}, \hat{p}_{j}] = i\delta_{ij}, \quad [\hat{\varphi} (\mathbf x , t), \hat{\pi}(\mathbf y, t)] = i\delta (\mathbf x - \mathbf y).$$

• I corrected the typo. Could you try and rephrase your answer, maybe going into a little bit more detail, please? For example, what do you mean mathematically by "the field in every point can be represented as set of oscillators"? Aug 31, 2014 at 20:17
• @Harold : I have added some explanation. Aug 31, 2014 at 20:31
• Sorry, I don't think this answers my question. I just want to know where the expressions for $\phi$ and $\pi$ come from. I can see a similarity with the ones for $q$ and $p$, but I don't understand the minus sign in the complex exponential next to $a^{\dagger}$ Aug 31, 2014 at 20:56
• @Harold : the minus sign is needed for satisfying canonical relation $[\hat{x}, \hat{p}] = i$. Aug 31, 2014 at 20:57
• @Harold : If there were not a minus sign, $\phi$ and $\pi$ would be proportionnal, so commute... Sep 1, 2014 at 8:19