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Assume a Lorentz transformation $\Lambda$ is to be implemented as the unitary operator $U(\Lambda)$ in the Hilbert space of quantum states of the Fock representation upon which the scalar Klein-Gordon field acts: $$ \varphi(x)=\int\frac{d^3k}{\sqrt{2}k_0}\left(a(\mathbf{k})e^{-ik\cdot x}+a^\dagger(\mathbf{k})e^{+ik\cdot x}\right)\equiv \int d\Omega_m\left(a(\mathbf{k})e^{-ik\cdot x}+a^\dagger(\mathbf{k})e^{+ik\cdot x}\right) $$ where $d\Omega_m$ is the Lorentz-invariant measure element. How do the annihilation and creation operator transform? How can I prove that $$ U(\Lambda)\varphi(x)U^\dagger(\Lambda) = \varphi(\Lambda x)? $$

Brightsun
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    A downvote seems harsh. I'm not sure why the question was posted in an "answer your own question" form, but a lot of effort has gone into it. Will downvoting and/or closing it really make the world a better place? – John Rennie Jan 27 '15 at 20:43
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    Given that this is a homework question and we have time & again said that homework-like questions & check my-work questions are off-topic, I see no reason why this should receive a special pass just because the OP answered it. – Kyle Kanos Jan 27 '15 at 21:04
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    @Kyle Kanos I was indeed in doubt whether to post it or not: I realized the mistake I was making while writing it down. So I checked the "Question and Answer style" linke below and I found "To be crystal clear, it is not merely OK to ask and answer your own question, it is explicitly encouraged"... – Brightsun Jan 27 '15 at 21:26
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    Yes, Asking & Answering is acceptable (I've done it myself). However, your question falls afoul of the "Do this calculation for me" type questions that we regard as off-topic. – Kyle Kanos Jan 27 '15 at 21:33
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    related: http://physics.stackexchange.com/q/154673/58382 – glS Jan 27 '15 at 21:40
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    @Kyle Kanos Are you suggesting that I should fake a doubt on a specific physical concept involved in the topic, in order to conform to the criteria listed in the link? Or that I should remove the post? I thought it might help me and other users to find this detailed calculation on the site, so I posted it in Q&A style. I mean no offense: my doubts are genuine. – Brightsun Oct 28 '15 at 18:57
  • Just for the record, I came upon this question because I wondered about it while studying the theory, and it was not a homework problem; rather a conceptual one about the creation/annihilation operators. This post and the solution were helpful to me. It might be beneficial to the site to reopen this question. @JohnRennie – doublefelix Nov 16 '18 at 16:01

1 Answers1

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Physically the creation of a particle with momentum $\mathbf{p}$ will be affected by the Lorentz group in the following manner: since $$ U(\Lambda)|\mathbf{p}\rangle = |\Lambda\mathbf{p}\rangle $$ $$ |\mathbf{p}\rangle=a^\dagger(\mathbf{p})|0\rangle $$ we get $$ U(\Lambda)a^\dagger(\mathbf{k})U^\dagger(\Lambda)=a^{\dagger}(\Lambda\mathbf{k}). $$ Indeed any transition amplitude gives: $$ \langle \Lambda\mathbf{p}| \Lambda \mathbf{q}\rangle=\langle0|a(\Lambda\mathbf{p})a^\dagger(\Lambda\mathbf{q})|0\rangle\\ \langle \Lambda\mathbf{p}| \Lambda \mathbf{q}\rangle= \langle \mathbf{p}|U^\dagger(\Lambda)U(\Lambda)|\mathbf{q}\rangle= \langle0|a(\mathbf{p})U^\dagger(\Lambda)U(\Lambda)a^\dagger(\mathbf{q})|0\rangle=\\ \langle0|U^\dagger(\Lambda) U(\Lambda) a(\mathbf{p})U^\dagger(\Lambda)U(\Lambda)a^\dagger(\mathbf{q})U^\dagger(\Lambda) U(\Lambda)|0\rangle=\langle0|U(\Lambda)a(\mathbf{p})U^\dagger(\Lambda)U(\Lambda)a^\dagger(\mathbf{q})U^\dagger(\Lambda)|0\rangle; $$ comparison yields the transformation formula for $a,a^\dagger$, where we have used the postulate: $U(\Lambda)|0\rangle = |0\rangle$. Then by taking the adjoint of the above:$$ U(\Lambda)a(\mathbf{k})U^\dagger(\Lambda)=a(\Lambda\mathbf{k}). $$ Now $$ U(\Lambda)\varphi(x)U^\dagger(\Lambda)= U(\Lambda)\int d\Omega_m\left(a(\mathbf{k})e^{-ik\cdot x}+a^\dagger(\mathbf{k})e^{+ik\cdot x}\right) U^\dagger(\Lambda)=\\ \int d\Omega_m\left(U(\Lambda)a(\mathbf{k})U^\dagger(\Lambda)e^{-ik\cdot x}+U(\Lambda)a^\dagger(\mathbf{k})U^\dagger(\Lambda)e^{+ik\cdot x}\right)=\\ \int d\Omega_m\left(a(\Lambda\mathbf{k})e^{-ik\cdot x}+a^\dagger(\Lambda\mathbf{k})e^{+ik\cdot x}\right) $$ changing variable and recalling $d\Omega_m$ is invariant under such change, which is in fact a boost, $\mathbf{k}=\Lambda^{-1}\mathbf{k}'$: $$ \int d\Omega'_m\left(a(\mathbf{k}')e^{-i(\Lambda^{-1}k')\cdot x}+a^\dagger(\mathbf{k}')e^{+i(\Lambda^{-1}k')\cdot x}\right)=\\ \int d\Omega'_m\left(a(\mathbf{k}')e^{-i(\Lambda^{-1}k')\cdot (\Lambda^{-1}x')}+a^\dagger(\mathbf{k}')e^{+i(\Lambda^{-1}k')\cdot (\Lambda^{-1}x')}\right) $$ where $x'=\Lambda x$. But the $\cdot$ product is invariant under $\Lambda$ so: $$ U(\Lambda)\varphi(x)U^\dagger(\Lambda) = \int d\Omega'_m\left(a(\mathbf{k}')e^{-ik\cdot x'}+a^\dagger(\mathbf{k}')e^{+ik'\cdot x'}\right)=\varphi(x'=\Lambda x). $$

rainman
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Brightsun
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    how are you obtaining your third equation? From the first two: $$ U(\Lambda) a^\dagger(p) |0\rangle = U(\Lambda) | p \rangle = | \Lambda p \rangle = a^\dagger(\Lambda p)|0\rangle,$$ from which one would conclude that $$ U(\Lambda)a^\dagger(p) = a^\dagger(\Lambda p).$$ – glS Jan 27 '15 at 21:46
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    added further explanation for $U(\Lambda)a^\dagger(p)U^\dagger(\Lambda)=a^\dagger(\Lambda p)$ – Brightsun Jan 27 '15 at 22:05
  • thanks. So you are saying my reasoning is wrong because $U(\Lambda)| 0 \rangle = | 0 \rangle$, right? Also, you could maybe specify that your $\textbf p$ are 4-momenta... the bold notation makes one think at 3-vectors – glS Jan 27 '15 at 22:11
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    Yes, I don't think there is a contradiction between my formula and yours, I just like mine better because usually operators transform by conjugacy... And yes, my $\mathbf{p}$ are 3-momenta since the physical state can be labeled by the three components of spacial momentum, and add $p_0$ by mass-shell when needed. – Brightsun Jan 27 '15 at 22:28
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    ok, but then how do you interpret $\Lambda \textbf p$ if $\textbf p$ is a 3-vector and $\Lambda$ could be (e.g.) a boost? – glS Jan 27 '15 at 22:56
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    $\Lambda\mathbf{p}=\mathbf{\Lambda\mathbf{p}}$ i.e. the spacial part of $\Lambda p$ where $p$ is a four vector; it's just a way to emphasize that only three of those parameters characterize the quantum state. – Brightsun Jan 27 '15 at 22:58
  • Let us continue this discussion in chat. – Brightsun Jan 27 '15 at 23:08
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    Hi there. I think there is more solid ground on which to justify your first assumptions. I'll write it here and if you like you can edit your answer.

    Let us investigate the operator $U(\Lambda)a^\dagger U^\dagger (\Lambda)$, note that $U^\dagger (\Lambda)=U(\Lambda^{-1})$. An operator's action is defined by its action on all basis vectors. In fock space, this means we need to evaluate

    $$U(\Lambda)a^\dagger (p) U^\dagger|p_1 ... p_N>$$

    The $U$ transform every momentum by $\Lambda$. Apply all 3 operators:

    $$=|p_1 ... \Lambda p ... p_N >$$ $$=a^\dagger (\Lambda p) |p_1 ... p_N>$$

     – doublefelix Nov 16 '18 at 16:44
  • So in the answer it has been proved that $U a_\textbf{p}^{\dagger} U^{\dagger} = a_{\Lambda \textbf{p}}^{\dagger}$ and in the first comment it is shown that $U a_\textbf{p}^{\dagger} = a_{\Lambda \textbf{ p}}^{\dagger}$. So both of them are correct? – rainman Nov 15 '21 at 00:05