When solving for the Klein-Gordon field $\phi$, most texts and online resources that I look at say that: $$\phi(x) = \int \frac{ d^{3} p }{ ( 2 \pi )^{3} } \frac{1}{\sqrt{ 2 E_{\mathbf{p}} }} \left[ a_{\mathbf{p}} e^{-ip\cdot x} + a_{\mathbf{p}}^{\dagger} e^{ip\cdot x} \right]\tag{1}$$

In this case, we'd have $$[a_{\mathbf{k}}, a_{\mathbf{p}}^{\dagger}]=(2\pi)^{3}\delta^{(3)}(\mathbf{k} - \mathbf{p}).$$

$\ $

However, my teacher right now has given me a solution $\phi$ such that:

$$\phi(x) = \int \frac{ d^{3} p }{ ( 2 \pi )^{3} } \frac{1}{2 E_{\mathbf{p}} } \left[ a_{\mathbf{p}} e^{-ip\cdot x} + a_{\mathbf{p}}^{\dagger} e^{ip\cdot x} \right]\tag{2}$$

Note the lack of square root here! In this case, I believe we'd have $$[a_{\mathbf{k}}, a_{\mathbf{p}}^{\dagger}]=(2\pi)^{3}2E_{\mathbf{k}}\delta^{(3)}(\mathbf{k} - \mathbf{p}).$$

$\ $

Both of these seem valid, and so to me it seems like the factor in front of the bracket is free for us to choose (as long as it makes $\phi$ Lorentz invariant). Is this correct? Or, what is going on?


1 Answer 1


Pretty much any QFT book you read has different prefactors that differ in factors of $2\pi$ or $E_{\mathbf{k}}$.

The point is that one can redefine the operator $a(\mathbf{p})$ by incorporating such factors to it. For example, if you want to consider the integral $(1)$ for $\phi(x)$ with the Lorentz-invariant measure


you can replace $a(\mathbf{p})$ with $\tilde{a}(\mathbf{p})=\sqrt{E_{\mathbf{p}}}a(\mathbf{p})$.

Or, if you want to get rid of the $2\pi$ factors in the commutation relations you can replace $a(\mathbf{p})$ with $a(\mathbf{p})/(2\pi)^{3/2}$. So its just redefining the operator $a(\mathbf{p})$ with a Lorentz-invariant measure.

  • $\begingroup$ The factors of $(2π)^n$ or $(\sqrt{2\pi})^n$ come from the convention used for the Fourier transform. But what about the $\sqrt{2E_\mathbf{p}}$? Since the measure is only Lorentz-invariant when inversely proportional to $E_\mathbf{p}$ (Srednicki p.39), are expressions for $\phi(x)$ that only integrate over $\frac{1}{\sqrt{2E_\mathbf{p}}}$ not Lorentz-invariant? Or has another $\frac{1}{\sqrt{2E_\mathbf{p}}}$ been integrated into $a_\mathbf{p}$? $\endgroup$ Jul 4, 2019 at 4:33

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.