0
$\begingroup$

I have some problems about basic concepts of quantum field theory.

First let's look at Klein-Gordon field.

Klein-Gordon equation has two branches of solutions, one of which is positive frequency and the other is negative frequency. The solution can be written as $$\psi(x)=\int \frac{d^{3}p}{(2\pi)^{3}}\frac{1}{\sqrt{2E_{p}}}(a_{p}e^{-ipx+i\omega t}+b_{p}e^{-ipx-i\omega t}).$$

Then we need to quantize above solution. In the exercise part of both books, Peskin and Srednicki, the complex scalar field operator is written as

$$\hat{\psi}(x)=\int \frac{d^{3}p}{(2\pi)^{3}}\frac{1}{\sqrt{2E_{p}}}(\hat{a}_{p}e^{-ipx}+\hat{b}_{p}^{+}e^{ipx})\tag{1}$$

The question is:

Why equ(1) is not written as $$\hat{\psi}(x)=\int \frac{d^{3}p}{(2\pi)^{3}}\frac{1}{\sqrt{2E_{p}}}(\hat{a}_{p}e^{-ipx}+\hat{b}_{p}e^{ipx})~?$$ Is this because when quantizing negative frequency, annihilation operator must be replaced by creation operator and $-p$ by $p$ in exponent?

$\endgroup$

1 Answer 1

0
$\begingroup$

Of course, you can give any name for coefficients in field expansion. But when you will analyze commutation relations:

$$ [\psi(x), \psi^{\dagger}(y)] \sim \hbar \delta(x-y) $$

For your choice of "names" you will obtain the another commutation relations:

$$ [b_p, b^\dagger_k] = \delta(p-k) $$

So you see, that $b_p$ is creation operator, $b^\dagger_k$ is annihilation operator. But usually physics choose reverse conventions. Due to these, in all books, peoples use (1).

Please, ask for clarification, if you need.

$\endgroup$
2
  • $\begingroup$ So in quantum field, commutation relaship of field operator and its cojugate momenta comes before that of annihilation operator and creation operator, though they're in fact equivalent? $\endgroup$
    – xiang sun
    Jul 27, 2020 at 15:38
  • $\begingroup$ @xiangsun, no, they are not equivalent. There are creation and annihilation operators. And for this operators there is canonical symbols: $b^\dagger$ and $b$. In your question, you suggest noncanonical symbols for these operators. $\endgroup$
    – Nikita
    Jul 27, 2020 at 15:51

Your Answer

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

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