Is the momentum distribution of excitations in a BEC symmetric? Even if there is a step potential (which I think should not make a difference because this is in real space)? I think yes, because excitations are 'symmetric' (with a $+k$ you always have a $-k$ excitation, right?).
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$\begingroup$ That is also the reason that in many papers they only plot the k>0 side, right? $\endgroup$– MrQCommented Feb 18, 2022 at 14:43
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$\begingroup$ What sort of excitations? What do you mean a step potential - if you add a non-translationally invariant potential to the system then momentum will stop being a good quantum number. $\endgroup$– jacob1729Commented Feb 18, 2022 at 18:24
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$\begingroup$ Bogoliubov excitations or in general the momentum distribution of particles n_k $\endgroup$– MrQCommented Feb 18, 2022 at 18:25
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$\begingroup$ I mean, how are you creating the excitation? This may be true at finite temperature but it doesn't mean you can't create a state with an arbitrary momentum distribution, it will just be strongly non thermal. $\endgroup$– jacob1729Commented Feb 18, 2022 at 18:33
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$\begingroup$ @jacob1729; I assume zero temperature and see the Gross-pitaevskii equation $\endgroup$– MrQCommented Feb 18, 2022 at 18:37
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1 Answer
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Because the real space spectrum $n(x)$ is real, $n(-k)$ will be equal to $n(k)$ complex conjugated.
