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Quantization of complex scalar field

Physics Asked on December 22, 2021

I’m learning Peskin’s qft now and I’m a little confused about problem 2.2 .

Suppose I write the field $phi(x)$ as:

$phi(x) =int frac{d^3p}{(2pi)^3}frac{1}{sqrt{2E_{p}}} (a_{p}e^{-ipx}+b_{p}e^{ipx})$

I know that $b_p$ should be written as $b_p^dagger$ because it annihilate antiparticle, otherwise $b_p$ creates particle with negative energy.

However, when I calculate the Hamiltonian, all I got is:

$int frac{d^3p}{(2pi)^3}E_{p}(a^dagger_{p}a_{p}+b^dagger_{p}b_{p})$

In my result, the $b$ particles create positive energy as $a$ particle did. I’m not sure if I did something wrong in calculation or there are some other explanation in the result.enter image description here

2 Answers

To begin with, Peskin gives the following action: $$ mathcal{S} = int d^4 left[ partial_{mu} phi^{ast} partial^{mu} phi - m^2 phi^{ast} phi right]. $$ Let's begin by considering the classical field theory. The equation of motion is $$ left( partial^2 + m^2 right) phi = 0 $$ (and its complex conjugate). A general solution to this equation is given by $$ phi(mathbf{x},t) = int frac{d^3 p}{(2 pi)^3} frac{1}{sqrt{2 E_mathbf{p}}} left( a_{mathbf{p}} e^{i mathbf{p} cdot mathbf{x} + i E_{mathbf{p}} t} + b_{mathbf{p}} e^{i mathbf{p} cdot mathbf{x} - i E_{mathbf{p}} t} right), $$ where $E_{mathbf{p}} = sqrt{|mathbf{p}|^2 + m^2}$. At this point, the quantities $a_{mathbf{p}}$ and $b_{mathbf{p}}$ are just some complex functions of momentum. We can also write down the classical Hamiltonian of this system, where the conjugate momentum is $$ pi(mathbf{x},t) = partial_t phi^{ast}(x,t) = - i int frac{d^3 p}{(2 pi)^3} sqrt{frac{E_mathbf{p}}{2}} left( a_{mathbf{p}}^{ast} e^{-i mathbf{p} cdot mathbf{x} - i E_{mathbf{p}} t} - b_{mathbf{p}}^{ast} e^{-i mathbf{p} cdot mathbf{x} + i E_{mathbf{p}} t} right) $$ (and its complex conjugate).

Now we go to the quantum theory, where the fields $phi$ and $pi$ (and therefore the coefficients $a_{mathbf{p}}$ and $b_{mathbf{p}}$) become operators. The starting point of quantization is the equal-time canonical commutation relations: $$ [phi(mathbf{x},t),pi(mathbf{y},t)] = [phi^{ast}(mathbf{x},t),pi^{ast}(mathbf{y},t)] = i delta^3(mathbf{x} - mathbf{y}) $$ We cannot assume at this point that $a_{mathbf{p}}$ and $b_{mathbf{p}}$ are bosonic annihilation operators, we need to actually calculate their commutation relations from the above equations. Inverting the above formulae, we have $$ a_{mathbf{p}} = e^{- i E_{mathbf{p}} t} int d^3 x , e^{- i mathbf{p} cdot mathbf{x}} left( sqrt{frac{E_{mathbf{p}}}{2}} , phi(mathbf{x},t) - i frac{1}{sqrt{2E_{mathbf{p}}}} , pi^{ast}(mathbf{x},t) right), $$ $$ b_{mathbf{p}} = e^{- i E_{mathbf{p}} t} int d^3 x , e^{- i mathbf{p} cdot mathbf{x}} left( sqrt{frac{E_{mathbf{p}}}{2}} , phi(mathbf{x},t) + i frac{1}{sqrt{2E_{mathbf{p}}}} , pi^{ast}(mathbf{x},t) right). $$ Now we can compute the commutators directly. They are $$ [a_{mathbf{p}},a^{dagger}_{mathbf{p}'}] = - (2 pi)^3 delta^3(mathbf{p} + mathbf{p}'), qquad [b_{mathbf{p}},b^{dagger}_{mathbf{p}'}] = (2 pi)^3 delta^3(mathbf{p} + mathbf{p}'). $$ As you can see, one of these commutators is the wrong sign compared to the usual bosonic commutation relations, and as a result the construction of the usual bosonic Fock space proceeds differently. (In particular, $[a^{dagger} a, a^{dagger}] propto - a^{dagger} $ so $a^{dagger}$ is actually a lowering operator.) To get the usual commutation relations, we should define $$ tilde{a}_{mathbf{p}} = a^{dagger}_{mathbf{p}}, quad tilde{a}^{dagger}_{mathbf{p}} = a_{-mathbf{p}}, quad tilde{b}_{mathbf{p}} = b_{mathbf{p}}, quad tilde{b}^{dagger}_{mathbf{p}} = b^{dagger}_{-mathbf{p}}. $$ This gives us the usual result. In textbooks they often write down the initial expansion in such a way so that the answer comes out correctly, but in practice one needs to go through the above procedure (until you've done it enough times that it has become automatic).

Answered by Seth Whitsitt on December 22, 2021

I am afraid what is wrong is in your initial writing.

A complex scalar field should be written as
$phi(x) = int frac{d^3p}{(2 pi)^3} frac{1}{sqrt{2 omega_p}} (a_p e^{-ipx} + b_p^dagger e^{ipx})$
And, by complex conjugation,
$phi^*(x) = int frac{d^3p}{(2 pi)^3} frac{1}{sqrt{2 omega_p}} (a_p^dagger e^{ipx} + b_p e^{-ipx})$
Clearly $a_p^dagger neq b_p^dagger$ as these operators create particles of opposite charge. However in both cases $omega_p = sqrt{vec p^2 +m^2} gt 0$.

To get the normal ordering of the $b's$ operators, at the end of the demonstration you have to apply the equal-time commutation relations
$[b_p, b_{p'}^dagger] = (2 pi)^3 delta^3 (vec p - vec p')$

Note: Your notation of the $b$ operator as creator is confusing. The notation used in literature for a complex scalar field and reported here is a generalization of the notation applied to a real scalar field.

Answered by Michele Grosso on December 22, 2021

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