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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 $$[a_{\mathbf{k}}, a_{\mathbf{p}}^{\dagger}]=(2\pi)^{3}\delta^{(3)}(\mathbf{k} - \mathbf{p}).$$

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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 $$[a_{\mathbf{k}}, a_{\mathbf{p}}^{\dagger}]=(2\pi)^{3}2E_{\mathbf{k}}\delta^{(3)}(\mathbf{k} - \mathbf{p}).$$

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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?

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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

$$\frac{d^3\mathbf{p}}{(2\pi)^3E_{\mathbf{p}}},$$

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.

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  • $\begingroup$ 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 $a_\mathbf{p}$? $\endgroup$ – alexchandel Jul 4 at 4:33

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