# Do commutation relations stay the same when we use different definitions to quantize a free scalar field?

I saw different ways to write the scalar field. For example (Tong p.23):

$$\phi(x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}} \left[a_pe^{ipx}+a_p^\dagger e^{-ipx}\right].\tag{2.18}$$

And we can write the commutators as: $$[a_p, a_q] = [a_p^\dagger, a_q^\dagger] = 0, [a_p, a_q^\dagger] = (2\pi)^3\delta^3(p-q).\tag{2.20}$$

However, on my lecture note, $$\phi(x)$$ is defined as

$$\phi(x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}} \left[b_{-p}^\dagger+a_p\right]e^{ipx}$$

My question is do commutation relations stay the same when we use different definitions to quantize a free scalar field? If so, should I have $$[a_p, a_q^\dagger] = [b_p, b_q^\dagger] = (2\pi)^3\delta^3(p-q),$$ and all other commutators, including $$[a_p, b_q^\dagger]$$, are equal to $$0$$?

The first definition pertains to a real scalar field. The second definition to a complex scalar field. Moreover, the second one can be made much more alike the first as follows. Start from the second definition and distribute the exponential

$$\phi(x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}} \left[b_{-p}^\dagger+a_p\right]e^{ipx}=\int \dfrac{d^3p}{(2\pi)^3}\dfrac{1}{\sqrt{2E_p}}\left(a_p e^{ipx}+b^\dagger_{-p}e^{ipx}\right),$$

then on the second term containing $$b^\dagger_{-p}$$ change variables on the integral using $$p\to -p$$. The measure is invariant and so is the function $$E_p$$. We are left with

$$\phi(x) = \int\frac{d^3p}{(2\pi)^3}\frac{1}{\sqrt{2E_p}} \left[b_{-p}^\dagger+a_p\right]e^{ipx}=\int \dfrac{d^3p}{(2\pi)^3}\dfrac{1}{\sqrt{2E_p}}\left(a_p e^{ipx}+b^\dagger_{p}e^{-ipx}\right),$$

which is exactly like your first definition, except that we have both $$a_p$$ and $$b^\dagger_p$$ instead of $$a_p$$ and $$a^\dagger_p$$. The reason for that is the difference between a real scalar field and a complex one. The real one encodes just a spin zero particle with creation and annihilation operators $$a_p$$ and $$a^\dagger_p$$, while a complex one encodes a spin zero particle with creation and annihilation operators $$a_p$$ and $$a^\dagger_p$$, together with its spin zero antiparticle with creation and annihilation operators $$b_p$$ and $$b^\dagger_p$$.

So in the second case you have two copies of the oscillator algebra which commute with each other. In other words, the only non-vanishing commutators are: $$[a_p,a^\dagger_q]=(2\pi)^3 \delta^{(3)}(\vec{p}-\vec{q}),\quad [b_p,b_q^\dagger]=(2\pi)^3 \delta^{(3)}(\vec{p}-\vec{q})$$

• Thanks so much for the answer!! Where does the $2p^0$ come from? Sometimes I see it's not included.
– IGY
Dec 28, 2022 at 17:43
• @IGY, sorry that was my mistake. When the fields are defined as in your question, with the factor $\frac{1}{\sqrt{2E_p}}$ in the measure, we really do not have the $2p^0$ in the commutators. It appears only when we write the fields with the factor $\frac{1}{2E_p}$ in the measure. These are just two different normalizations. The one in which you get $2p^0$ in the commutators is called covariant normalization.
– Gold
Dec 28, 2022 at 17:46

Tong is considering 1 real scalar field, while OP's lecture notes are considering 1 complex scalar field $$\phi~=~\frac{\phi_1+i\phi_2}{\sqrt{2}},$$ which is equivalent to 2 real scalar fields. It is straightforward to check via canonical quantization that the corresponding commutator relations are compatible.