# Canonical commutation relations and a scalar interacting field

Let $\hat\phi(x)$ be a quantum scalar hermitian field and $\hat \pi(x)$ its conjugated field.

IF the scalar field evolves according to a FREE theory it is possible to write down the following normal mode expansions:

$$\hat \phi(x) = \int \frac{d\vec k}{(2\pi)^{3/2} 2 \omega_k} \left \{ \hat a(k) e^{-ikx} + \hat a^\dagger(k) e^{ikx} \right \}_{\big|_{k^0=\omega_k}}$$

$$\hat \pi(x) = \int \frac{d\vec k}{(2\pi)^{3/2} 2 \omega_k} (-i\omega_k) \left \{ \hat a(k) e^{-ikx} - \hat a^\dagger(k) e^{ikx} \right \}_{\big|_{k^0=\omega_k}}$$

and the equal time canonical commutation relations:

$$[\hat \phi(t, \vec x) , \hat \phi(t, \vec y)]=0$$ $$[\hat \pi(t, \vec x) , \hat \pi(t, \vec y)]=0$$ $$[\hat \phi(t, \vec x) , \hat \pi(t, \vec y)]=i \delta (\vec x - \vec y)$$

are equivalent to the following commutation relations between the ladder operators:

$$[\hat a(k) , \hat a(k') ]=0$$ $$[\hat a^\dagger(k) , \hat a^\dagger(k') ]=0$$ $$[\hat a(k) , \hat a^\dagger(k') ]= 2\omega_k \delta (\vec k - \vec k')$$

(by "equivalent" I mean that one set of relations can be derived from the other using the normal mode expansion)

IF the scalar field evolves according to an INTERACTING theory I can't write down a general normal mode expansion, but at a given time $t$ it should be possible to write

$$\hat \phi_t(\vec x) = \int \frac{d\vec k}{(2\pi)^{3/2} 2 \omega_k} \left \{ \hat \alpha_t (\vec k) e^{i {\vec {k}} \cdot {\vec {x}}} + \hat \alpha_t^\dagger(\vec k) e^{-i {\vec {k}} \cdot {\vec {x}}} \right \}$$

$$\hat \pi_t(\vec x) = \int \frac{d\vec k}{(2\pi)^{3/2} 2 \omega_k} \left \{ \hat \beta_t (\vec k) e^{i {\vec {k}} \cdot {\vec {x}}} + \hat \beta_t^\dagger(\vec k) e^{-i {\vec {k}} \cdot {\vec {x}}} \right \}$$

for some unknown, up to now, $\hat \alpha_t (\vec k)$, $\hat \beta_t (\vec k)$ operators.

Using the equal time canonical commutation relations, which should hold true even in an interacting theory, is it possible to show that the following relations hold true?

$$\hat \beta_t (\vec k) = (-i\omega_k) \hat \alpha_t (\vec k)$$ $$[\hat \alpha_t (\vec k) , \hat \alpha_t(\vec k') ]=0$$ $$[\hat \alpha_t^\dagger(\vec k) , \hat \alpha_t^\dagger(\vec k') ]=0$$ $$[\hat \alpha_t(\vec k) , \hat \alpha_t^\dagger(\vec k') ]= 2\omega_k \delta (\vec k - \vec k')$$

in such a way to recover, at time $t$, the usual commutation relations among the operator coefficients in the Fourier expansion of $\hat \phi_t(\vec x)$ and $\hat \pi_t(\vec x)$.

OR I can simply reformulate the question in the following way: given the operators

$$\hat \phi_t(\vec x) = \int \frac{d\vec k}{(2\pi)^{3/2} 2 \omega_k} \left \{ \hat \alpha_t (\vec k) e^{i {\vec {k}} \cdot {\vec {x}}} + \hat \alpha_t^\dagger(\vec k) e^{-i {\vec {k}} \cdot {\vec {x}}} \right \}$$

$$\hat \pi_t(\vec x) = \int \frac{d\vec k}{(2\pi)^{3/2} 2 \omega_k} \left \{ \hat \beta_t (\vec k) e^{i {\vec {k}} \cdot {\vec {x}}} + \hat \beta_t^\dagger(\vec k) e^{-i {\vec {k}} \cdot {\vec {x}}} \right \}$$

does the following statement hold true?

$$\begin{cases} [\hat \phi_t( \vec x) , \hat \phi_t( \vec y)]=0 & & \hat \beta_t (\vec k) = (-i\omega_k) \hat \alpha_t (\vec k) \\ [\hat \pi_t( \vec x) , \hat \pi_t( \vec y)]=0 & \Leftrightarrow & [\hat \alpha_t (\vec k) , \hat \alpha_t(\vec k') ]=0 \\ [\hat \phi_t( \vec x) , \hat \pi_t( \vec y)]=i \delta (\vec x - \vec y) & & [\hat \alpha_t^\dagger(\vec k) , \hat \alpha_t^\dagger(\vec k') ]=0 \\ & & [\hat \alpha_t(\vec k) , \hat \alpha_t^\dagger(\vec k') ]= 2\omega_k \delta (\vec k - \vec k')\\ \end{cases}$$

• Note that both $\alpha$ and $\beta$ depend on $t$. – AccidentalFourierTransform Oct 23 '17 at 18:14
• This will only hold at equal times, so it's not very useful. (Recall that for free fields a whole lot of calculations are done by commuting creation and annihilation operators evaluated at different times.) – knzhou Oct 23 '17 at 18:38
• @AccidentalFourierTransform yes, you are right, but since I'm interested only in relations at a fixed time $t$ I didn't write explicitly the time dependence; anyway to be more correct I have just added it as a subscript – L.R. Oct 23 '17 at 21:35
• @knzhou but are you sure that the relations above hold true at equal times? for me it would be enough – L.R. Oct 23 '17 at 22:12
• @L.R. I'm not 100% sure, but I think you need nothing but the mode expansion and the canonical commutators to prove the relations you want. Since they both hold just fine at a fixed time, the derivation goes through just like in the free case. – knzhou Oct 23 '17 at 22:17