0
$\begingroup$

I am reading Peskin & Schroeder and in chapter 2 (p.21) he quantizes the real K-G field such as: $$\phi=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{\sqrt{2E_p}}\left(a_pe^{ip·x}+ a^{\dagger}_pe^{-ip·x}\right)$$ in analogy with the harmonic oscillator.

I understand that this allows $\phi=\phi^{\dagger}$ as it should.

The problem arises in exercise 2.2.b) when you are asked to follow the same procedure but for a Complex K-G field and "show that the theory contains two sets of particles of mass m". In this case I don't think the first expression holds true since $\phi=\phi^{\dagger}$ is not always the case. I've seen that the solution is to express the field in terms of two different creation and annihilation operators $a$ and $b^{\dagger}$ in the following way:

$$\phi=\int\dfrac{d^3p}{(2\pi)^3}\dfrac{1}{\sqrt{2E_p}}\left(a_pe^{-ip·x}+ b^{\dagger}_pe^{ip·x}\right)$$

But I don't undersand why. What is the condition on the complex field that forces you to introduce $b^{\dagger}$ ? Is it just to ensure that $\phi$ is different than $\phi^{\dagger}$ ? Later on this b's turn out to be the ladder operators for antifermions, that come as a negative-frequency solution of the Dirac equation, but I don't see the relation between this and my K-G complex field. Can anyone shed some light in this issue?

$\endgroup$
2
$\begingroup$

Maybe it would make more sense to turn the question around.

Remember that the operators are essentially the coefficients in the Fourier expansion of the field. You could just as easily have written

$$\phi = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2E_p}} \left(c_1(p)\ e^{ip\cdot x} + c_2(p)\ e^{-ip\cdot x}\right)$$

The question now becomes what, if anything, do these coefficients have to do with one another? In the absence of additional information, the answer is "nothing," so you need to use different symbols ($a$ and $b$) for them to reflect this.

On the other hand, if you make the demand that $\phi = \phi^\dagger$, it becomes immediately clear that $c_1 = c_2^\dagger$ so we can just stick with $a$.

From this perspective, the question isn't why you need to add a $b$ for the second coefficient in the complex case; it's why you don't need to add a $b$ for the second coefficient in the real case.

$\endgroup$
  • $\begingroup$ @BobBowie Please note that the etiquette here is that, when a post fully answers your question, you should accept it by clicking on the "accept" button, that is, the check mark at the top-left of the post (below the voting buttons). It is also encouraged to wait at least 24h before accepting any answer, so that you give others the chance to post alternative answers that may be even better. BTW, welcome to physics.SE. Cheers! $\endgroup$ – AccidentalFourierTransform Aug 14 '18 at 16:10
  • $\begingroup$ Thank you :) I didn't know that, I just accepted this answer because I don't think there is a better explanation. $\endgroup$ – Bob Bowie Aug 14 '18 at 16:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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