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Hello I am a beginner of QFT. I have some questions about the behavior of the momentum ladder operators under Lorentz Transformation.

For simplicity, let's consider the momentum ladder operators of a real scalar singlet $\phi$. We know that the ladder operators $a_{p}$ transform according to $$U(\Lambda)a^\dagger_{p}U^{-1}(\Lambda)=a^\dagger_{\Lambda p} , $$ where $U(\Lambda)$ is the unitary operator representing a given Lorentz Transformation $\Lambda$ (e.g Peskin p.59).

I tried to derive this identity in the following manner. Let's consider a typical situation in special relativity. Alice has a set of inertia coordinate $x$, Bob has another set of inertia coordinate $x^\prime = \Lambda x$. If a classical particle is observed to have momentum $p$ in the frame of Alice, it will be observed to have momentum $\Lambda p$ by Bob. In quantum mechanics, Alice observes the particle to be in a state $|p\rangle$ while Bob observes the particle to be in a state $|\Lambda p\rangle$, and the two single particle states are related by a unitary operator $U(\Lambda)$ representing a given Lorentz Transformation $\Lambda$. In summary, $$|\Lambda p\rangle=U(\Lambda)|p\rangle . $$ This then implies $$ a^\dagger_{\Lambda p}|0\rangle=U(\Lambda)a^\dagger _{p}|0\rangle , $$ where $|0\rangle$ on both sides is the ground state of the free field theory. We are then led to the conclusion that $$ a^\dagger_{\Lambda p}=U(\Lambda)a^\dagger_{p} , $$ in contrast to the standard result. Is there something wrong in my reasoning?

I observe that if we assume the vacuum ket $|0\rangle$ to be Lorentz invariant, i.e $$ U(\Lambda)|0\rangle=|0\rangle , $$ then we may write $$ a^\dagger_{\Lambda p}|0\rangle = U(\Lambda)a^\dagger_{p}|0\rangle = U(\Lambda)a^\dagger_{p}U^{-1}(\Lambda) U(\Lambda)|0\rangle = U(\Lambda)a^\dagger_{p}U^{-1}(\Lambda)|0\rangle . $$ Then the standard result $$ a^\dagger_{\Lambda p} = U(\Lambda)a_{p}U^{-1}(\Lambda) $$ is recovered. Is the assumption of the Lorentz invariance of the vacuum state of free field theory correct?

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  • $\begingroup$ I'm not totally sure, but I believe the answer is that your reasoning only proves that $a_{\Lambda p}^\dagger=U(\Lambda)a_p^\dagger$ when acting on the vacuum $\lvert0\rangle$. Generally speaking, $A\lvert0\rangle=B\lvert0\rangle$ does not imply $A=B$ (which is obvious if you think about it in terms of regular vectors). The reason the operators are equal when acting on the vacuum is precisely what you note: that the vacuum is Lorentz invariant. $\endgroup$ – glS Dec 24 '17 at 8:52
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You have more or less answered your own question. Yes indeed, the vacuum must be Lorentz invariant, otherwise the Lorentz invariance would be spontaneously broken.

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