0
$\begingroup$

I am reading Takagi's paper `Vacuum noise and stress induced by uniform accelerator'.

I will attach a screenshot of something I am confused about (from page 31). I have a simple question: what is meant by the state $|n\rangle^{(+)}$?

enter image description here

Here he is proving the Thermalization theorem (yielding the Unruh effect). In doing so, he is defining a state; $$ |n\rangle^{(+)} = \frac{(b_{ml \mathbf{k}}^{(+)\dagger})^{n}}{\sqrt{n!}} | 0_{\mathrm{R}} \rangle $$ where $| 0_{\mathrm{R}} \rangle$ is the Rindler vacuum. The operator $b_{\Omega \mathbf{k}}^{(+)\dagger}$ is the creation operator for particles in the Right Rindler wedge (with mode parameters $(\Omega,\mathbf{k}) = (\Omega, \mathbf{k}) \in \mathbb{R}_{>0}\times \mathbb{R}^2$ for the particle created). The related operator $b_{ml \mathbf{k}}^{(+)\dagger}$ is the `smeared' operator $$ b_{ml \mathbf{k}}^{(+)\dagger} = \int_{0}^{\infty} d\Omega \ f_{ml}(\Omega) b_{\Omega \mathbf{k}}^{(+)\dagger} $$ where $f_{ml}(\Omega)$ is a wave-packet, with $m,l\in\mathbb{Z}$

What exactly is this state? I am thinking it must be the following $$ |n\rangle^{(+)} = \frac{b_{m_1 l_1 \mathbf{k}_1}^{(+)\dagger}b_{m_2 l_2 \mathbf{k}_2}^{(+)\dagger} \cdots b_{m_n l_n \mathbf{k}_n}^{(+)\dagger} }{\sqrt{n!}} | 0_{\mathrm{R}} \rangle $$ since you can't have all the particles in the same state (so we've got distinct sets $\{m_j,n_j,\mathbf{k}_j\}_{j=1}^n$)

Am I right and he is just abusing notation? I can't think of any other option.

$\endgroup$
  • $\begingroup$ I imagine so, because he is suppressing all indices besides the particle number. He’s probably just saying “here’s what an $n$ particle state looks like”. $\endgroup$ – knzhou Aug 5 '18 at 9:15
  • $\begingroup$ I don't think that's what the author meant. For one thing, if the labels are different then you should not divide by $\sqrt{n!}$ (this factor being for overcounting the same label). I haven't read the article, but I believe the author is either considering a compact manifold (so that momentum is discrete) or just being sloppy with the notation, and aiming at a qualitative description rather than a precise/rigorous derivation. $\endgroup$ – AccidentalFourierTransform Aug 5 '18 at 13:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.