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I'm struggling to understand the following logic from Peskin and Schroeder, page 23:

The book defines $$ |\vec{p} \rangle = \sqrt{2E_p} a^\dagger_p |0 \rangle $$ so that the inner product $\langle \vec{p}| \vec{q}\rangle$ is a Lorentz invariant object. It then says that this above equation, which is a 'normalization condition', implies that $$ U(\Lambda) |\vec{p}\rangle = | \Lambda \vec{p} \rangle$$ or, if we prefer to think of this transformation as acting on the operator $a^\dagger_p$, we can write $$ U(\Lambda) a^\dagger_p U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda p}}{E_p}} a^\dagger_{\Lambda \vec{p}} $$ I don't understand how these two equations are arrived at. The book seems to suggest that it's obvious, but clearly I'm missing something. Thanks.

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  • $\begingroup$ I can't answer, but a slightly more intuitive (at least to me) derivation of the transformation law for the creation operator can be obtained from requiring that the scalar field $\phi(x)$ transform as $\phi(x)\rightarrow \phi'(x')=U^{-1}(\Lambda)\phi(x')U(\Lambda)=\phi(x)$. Then the transformation law for the creation operator you've given is the only transformation law compatible with Lorentz Invariance. From this transformation law we can then deduce that the normalization $|p\rangle=\sqrt{2E_p}a^\dagger |0\rangle$ gives a Lorentz invariant scalar product $\langle q|p\rangle$. $\endgroup$
    – Okazaki
    Jan 4, 2016 at 19:26

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The first statement

$| p \rangle$ is an eigenstate of $\hat{P}^{\mu}$ operators with eigenvalue $p^{\mu}$. It's natural then that $$ U(\Lambda )| p \rangle $$ is also eigenstate of $\hat{P}^{\mu}$: $$ \hat{P}^{\mu}U(\Lambda )| p \rangle = U(\Lambda )\left[ U(\Lambda^{-1})\hat{P}^{\mu}U(\Lambda )\right]| p \rangle = U(\Lambda )\Lambda^{\mu}_{\ \nu}\hat{P}^{\nu}| p \rangle = \Lambda^{\mu}_{\ \nu}p^{\nu}U(\Lambda )| p \rangle . $$ In general (for non-scalar case when $| p \rangle = | p ,\sigma \rangle$) from this follows that $$ U(\Lambda )| p, \sigma \rangle = \sum_{\sigma {'}}C_{\sigma {'} \sigma}| \Lambda p, \sigma {'}\rangle $$ and for all states $| p, \sigma \rangle $ states $U(\Lambda )| p, \sigma \rangle$ also belongs to the Hilbert space of eigenstates, so if in the beginning $| p, \sigma \rangle $ belongs to some definite orbit of the Lorentz group, then $U(\Lambda )| p, \sigma \rangle$ will also belong to this orbit: $$ | p, \sigma \rangle = |\mathbf p , \sigma \rangle_{p_{0} = f(\mathbf p , m)} \Rightarrow U(\Lambda )| p, \sigma \rangle = \sum_{\sigma {'}}C_{\sigma {'} \sigma}| \mathbf {\Lambda p}, \sigma {'}\rangle_{p_{0} = f(\mathbf p , m)}. $$ The invariance condition of $\langle p | q \rangle $ (since $\langle p| q\rangle = C(p) \delta (\mathbf p - \mathbf q) $ lorentz-invariance condition means that $C(p) \sim E_{p})$ means that $C_{\sigma \sigma {'}}$ satisfies $C_{\sigma \sigma{''}}C^{\dagger}_{\sigma {''}\sigma{'}} = \delta_{\sigma \sigma {'}}$. In simplest scalar case this directly leads to $$ \tag 1 U(\Lambda )| \mathbf p \rangle = e^{i\alpha}| \mathbf {U(\Lambda ) p}\rangle , $$ where $\alpha$ is some parameter of transformation given by $\Lambda$. It can be set to zero.

The second statement

It automatically follows from the first statement (given by eq. $(1)$) if we use definition $$ | \mathbf p \rangle = \sqrt{2E_{p}}a^{\dagger}(\mathbf p) | \rangle $$ and eq. $(1)$, we'll get $$ U(\Lambda )| \mathbf p \rangle = U(\Lambda )\left[\sqrt{2E_{p}} a^{\dagger}(\mathbf p)|\rangle\right] = \sqrt{2E_{p}}[U(\Lambda )\hat{a}^{\dagger}(\mathbf p)U^{-1}(\Lambda) ]U(\Lambda ) | \rangle = $$ $$ = \left| U(\Lambda ) | \rangle = | \rangle \right| = \sqrt{2E_{p}}[U(\Lambda )\hat{a}^{\dagger}(\mathbf p)U^{-1}(\Lambda) ] | \rangle = \sqrt{2E_{\Lambda p}}\hat{a}^{\dagger}(\mathbf {\Lambda p })| \rangle \Rightarrow $$ $$ U(\Lambda )\hat{a}^{\dagger}(\mathbf p)U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda p}}{E_{p}}}\hat{a}^{\dagger}(\mathbf {\Lambda p }). $$

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  • $\begingroup$ Clearly completely obvious. $\endgroup$
    – theage
    Dec 23, 2014 at 18:59
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    $\begingroup$ @Andrew McAddams : Is the invariance of the vacuum $U(\Lambda)|0\rangle=|0\rangle$ an assumption, or can it be proved? $\endgroup$
    – user7154
    Dec 23, 2014 at 20:27
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    $\begingroup$ @StephenBlake : formally it is postulate. $\endgroup$ Dec 23, 2014 at 21:02
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    $\begingroup$ @gj255 : it seems that Weinberg's QFT Vol. 1 contains derivarion you've asked. $\endgroup$ Dec 23, 2014 at 23:40
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    $\begingroup$ @StephenBlake: That's a physical input. If the state called "vacuum" wasn't invariant under Poincare transformations (among them Lorentz transformations), then we'd notice all kinds of wacky effects like non-conservation of momentum and/or angular momentum. To avoid such behaviour, the vacuum state better be homogeneous, isotropic and boost-invariant. $\endgroup$
    – Siva
    Dec 24, 2014 at 1:00

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