Wave Packets in QFT

In Peskin's book, it says that: $$|\phi\rangle =\int{\frac{\mathrm d^3k}{(2\pi)^3}\frac{1}{\sqrt{2E_{\vec{k}}}}}\phi\left(\vec{k}\right)\left|\vec{k}\right\rangle \tag{4.65}$$ Also, Eq(2.39) says that: $$(1)_\mathrm{1-particle}=\int{\frac{\mathrm d^3p}{(2\pi)^3}\left|\vec{p}\right\rangle\frac{1}{2E_{\vec{p}}}\left\langle\vec{p}\right|} \tag{2.39}$$

So, I tried:

$$|\phi\rangle=\int{\frac{\mathrm d^3p}{(2\pi)^3}\left|\vec{k}\right\rangle\frac{1}{2E_{\vec{k}}}\left\langle\vec{k}\bigg|\phi\right\rangle}=\int{\frac{\mathrm d^3p}{(2\pi)^3}\frac{1}{2E_{\vec{k}}}}\phi\left(\vec{k}\right)\left|\vec{k}\right\rangle,$$

which is different from Peskin's Eq(4.65) by a square root in the denominator.

Why is my equation wrong?

• It is probably because $\langle k|\phi\rangle = \sqrt{2E_k}\,\phi(k)$. I would however not worry too much about square roots and $2\pi$, generally, as different authors have different definitions. – gented Nov 5 '16 at 15:27
• Feynman: "If you do not get the factors of $2 \pi$ straight, you do not understand anything." – pfnuesel Nov 5 '16 at 15:48
• There is also some mismatch with powers of $2\pi$ – Valter Moretti Nov 5 '16 at 17:33
• Yes, there is a mismatch. Should be $(2\pi)^{3/2}$ in eq. 4.65 – Nogueira Nov 5 '16 at 17:50

There are different conventions according to the choice of the measure in the momentum space. (I disregard further conventions related to the powers of $2\pi$). Essentially everything depends on where you prefer to have a simple form for the action of Lorentz group (compare (1)-(1') and (3)-(3') below).

If one assumes that $$\left[a_{\vec{k}},a^*_{\vec{k}'}\right] = \delta^3\left({\vec{k}-\vec{k}'}\right)I\:,$$ namely $$\left\langle \vec{k}\bigg| \vec{k'} \right\rangle = \delta^3\left({\vec{k}-\vec{k}'}\right)\:,$$ then the unitary action of the Lorentz transformation is $$U_\Lambda \left|\vec{k}\right\rangle = \sqrt{\frac{E_{\vec{\Lambda k}}}{E_{\vec{k}}}}\left|\vec{\Lambda k}\right\rangle \:.\tag{1}$$ Using the fact that $$\frac{\mathrm d\vec{k}}{E_{\vec{k}}} = \frac{\mathrm d \vec{\Lambda k}}{E_{\vec{\Lambda k}}}\tag{2}$$ one easily sees that, assuming $$|\phi\rangle=\int{\frac{\mathrm d^3k}{(2\pi)^{3/2}}\frac{1}{\sqrt{2E_{\vec{k}}}}}\phi(\vec{k})\left|\vec{k}\right\rangle\:,$$ it arises $$\left(U_\Lambda \phi\right)\left(\vec {k}\right) = \phi\left(\vec{\Lambda^{-1}k}\right) \tag{3}$$ as expected. Moreover $$\phi\left(\vec{k}\right) = \sqrt{2E_{\vec{k}}} (2\pi)^{3/2} \langle \vec{k}| \phi \rangle\:.$$ Now the projector onto the one particle space is the standard $$P_\mathrm{1-particle}=\int \mathrm d\vec{p}\left|\vec{p}\right\rangle\left\langle\vec{p}\right|$$ In particular, using (1) and (2), one easily finds $$U_\Lambda P_\mathrm{1-particle} U^*_\Lambda = P_\mathrm{1-particle}$$ which must be evidently true since the one particle space is relativistically invariant and would be false in the presence of the factor $1/E_p$.

If one instead assumes that $$\left[a_{\vec{k}},a^*_{\vec{k}'}\right] = 2E_{\vec{k}}\delta^3\left({\vec{k}-\vec{k}'}\right)I\:,$$ namely $$\left\langle \vec{k}\bigg| \vec{k'} \right\rangle = 2E_{\vec{k}} \delta^3\left({\vec{k}-\vec{k}'}\right)\:,$$ then the unitary action of the Lorentz transformation is $$U_\Lambda \left|\vec{k}\right\rangle = \left|\vec{\Lambda k}\right\rangle \:.\tag{1'}$$ Using the fact that $$\frac{\mathrm d\vec{k}}{E_{\vec{k}}} = \frac{\mathrm d \vec{\Lambda k}}{E_{\vec{\Lambda k}}}\tag{2}$$ one easily sees that, assuming $$|\phi\rangle=\int{\frac{\mathrm d^3k}{(2\pi)^{3/2}}\frac{1}{\sqrt{2E_{\vec{k}}}}}\phi(\vec{k})\left|\vec{k}\right\rangle$$ it arises $$\left(U_\Lambda \phi\right)\left(\vec {k}\right) = \sqrt{\frac{E_{\vec{\Lambda^{1} k}}}{E_{\vec{k}}}}\phi\left(\vec{\Lambda^{-1}k}\right) \tag{3'}\:.$$ Moreover $$\phi\left(\vec{k}\right) = \left(\sqrt{2E_{\vec{k}}}\right)^{-1} (2\pi)^{3/2} \langle \vec{k}| \phi \rangle\:.$$ Now the projector onto the one particle space is $$P_\mathrm{1-particle}=\int \mathrm d\vec{p}\left|\vec{p}\right\rangle\frac{1}{2E_{\vec{k}}}\left\langle\vec{p}\right|$$ In particular, using (1') and (2), one easily finds $$U_\Lambda P_\mathrm{1-particle} U^*_\Lambda = P_\mathrm{1-particle}\:.$$

Let us assume again that $$\left[a_{\vec{k}},a^*_{\vec{k}'}\right] = 2E_{\vec{k}}\delta^3\left({\vec{k}-\vec{k}'}\right)I\:,$$ namely $$\left\langle \vec{k}\bigg| \vec{k'} \right\rangle = 2E_{\vec{k}} \delta^3\left({\vec{k}-\vec{k}'}\right)\:,$$ so that the unitary action of the Lorentz transformation is still $$U_\Lambda \left|\vec{k}\right\rangle = \left|\vec{\Lambda k}\right\rangle \:.\tag{1'}$$ But this time we assume $$|\phi\rangle=\int{\mathrm d^3k\frac{1}{2E_{\vec{k}}}}\phi(\vec{k})\left|\vec{k}\right\rangle\:.$$ With these choices it arises $$\left(U_\Lambda \phi\right)\left(\vec {k}\right) = \phi\left(\vec{\Lambda^{-1}k}\right) \tag{3}$$ and also $$\phi\left(\vec{k}\right) = \left\langle \vec{k}\bigg| \phi \right\rangle\:,$$ and eventually $$P_\mathrm{1-particle}=\int \mathrm d\vec{p}\left|\vec{p}\right\rangle\frac{1}{2E_{\vec{k}}}\left\langle\vec{p}\right|$$
Depend on the normalization of $|\vec{k}\rangle$. The identity operator revels that the normalization of a given state is acomplished or not. The $1/2E_p$ in the equation (2.39) tells you that the state are not normalized indeed and $$\sqrt{\langle\vec{k}|\vec{k}\rangle}=(2\pi)^{3/2}\sqrt{2E_k}.$$
Here the definition of $\phi (k)$ is in terms of the normalized $k$-state.