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.

  • $\begingroup$ 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

1 Answer 1


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 $$ U(\Lambda )| \mathbf p \rangle = U(\Lambda )\left[\sqrt{2E_{p}} a^{\dagger}(\mathbf p)|\rangle\right] = \sqrt{2E_{p}}[U(\Lambda )\hat{a}^{\dagger}(\mathbf p)U^{-1}(\Lambda) ]U(\Lambda ) | \rangle = $$ $$ = \left| U(\Lambda ) | \rangle = | \rangle \right| = \sqrt{2E_{p}}[U(\Lambda )\hat{a}^{\dagger}(\mathbf p)U^{-1}(\Lambda) ] | \rangle = \sqrt{2E_{\Lambda p}}\hat{a}^{\dagger}(\mathbf {\Lambda p })| \rangle \Rightarrow $$ $$ U(\Lambda )\hat{a}^{\dagger}(\mathbf p)U^{-1}(\Lambda) = \sqrt{\frac{E_{\Lambda p}}{E_{p}}}\hat{a}^{\dagger}(\mathbf {\Lambda p }). $$

  • $\begingroup$ Clearly completely obvious. $\endgroup$
    – theage
    Dec 23, 2014 at 18:59
  • 1
    $\begingroup$ @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
  • 1
    $\begingroup$ @StephenBlake : formally it is postulate. $\endgroup$ Dec 23, 2014 at 21:02
  • 1
    $\begingroup$ @gj255 : it seems that Weinberg's QFT Vol. 1 contains derivarion you've asked. $\endgroup$ Dec 23, 2014 at 23:40
  • 3
    $\begingroup$ @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

Your Answer

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

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