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 $0$?


2 Answers 2


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: $$[a_p,a^\dagger_q]=(2\pi)^3 \delta^{(3)}(\vec{p}-\vec{q}),\quad [b_p,b_q^\dagger]=(2\pi)^3 \delta^{(3)}(\vec{p}-\vec{q})$$

  • $\begingroup$ Thanks so much for the answer!! Where does the $2p^0$ come from? Sometimes I see it's not included. $\endgroup$
    – IGY
    Commented Dec 28, 2022 at 17:43
  • 1
    $\begingroup$ @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. $\endgroup$
    – Gold
    Commented 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.