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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 $$ U(\Lambda)\varphi(x)U^\dagger(\Lambda) = \varphi(\Lambda x)? $$

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closed as off-topic by ACuriousMind, Brandon Enright, Kyle Kanos, Martin, JamalS Jan 28 '15 at 6:58

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    $\begingroup$ 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? $\endgroup$ – John Rennie Jan 27 '15 at 20:43
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    $\begingroup$ 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. $\endgroup$ – Kyle Kanos Jan 27 '15 at 21:04
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    $\begingroup$ @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"... $\endgroup$ – Brightsun Jan 27 '15 at 21:26
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    $\begingroup$ 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. $\endgroup$ – Kyle Kanos Jan 27 '15 at 21:33
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    $\begingroup$ related: physics.stackexchange.com/q/154673/58382 $\endgroup$ – glS Jan 27 '15 at 21:40
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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 $a,a^\dagger$. 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: $$ U(\Lambda)\varphi(x)U^\dagger(\Lambda) = \int d\Omega'_m\left(a(\mathbf{k}')e^{-ik\cdot x'}+a^\dagger(\mathbf{k}')e^{+ik'\cdot x'}\right)=\varphi(x'=\Lambda x). $$

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  • $\begingroup$ 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 $$ U(\Lambda)a^\dagger(p) = a^\dagger(\Lambda p).$$ $\endgroup$ – glS Jan 27 '15 at 21:46
  • $\begingroup$ added further explanation for $U(\Lambda)a^\dagger(p)U^\dagger(\Lambda)=a^\dagger(\Lambda p)$ $\endgroup$ – Brightsun Jan 27 '15 at 22:05
  • $\begingroup$ 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 $\endgroup$ – glS Jan 27 '15 at 22:11
  • $\begingroup$ 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. $\endgroup$ – Brightsun Jan 27 '15 at 22:28
  • $\begingroup$ 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? $\endgroup$ – glS Jan 27 '15 at 22:56

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