In this paper the authors state that the inequality near the bottom of page 2 reduces to inequality (1) when $N=1$. However, I am struggling to get that result, as I have an extra minus sign in front of the integrals. Can anyone try this for themselves and see if they get the correct result?

$$\frac{n}{4}N(1+\ln{\pi})-\frac{1}{2}N^{-1} \int d^n \mathbf{r}|\Psi(\mathbf{r})|^2 \ln{|\Psi(\mathbf{r})}|-\frac{1}{2}N^{-1} \int d^n \mathbf{k}|\tilde{\Psi}(\mathbf{k})|^2 \ln{|\tilde{\Psi}(\mathbf{k})}|+N \ln{N \geq 0}$$

should reduce to

$$-\langle \ln{\rho}\rangle - \langle \ln{\tilde{\rho}}\rangle \geq n(1 + \ln{\pi}) $$


$\rho (\mathbf{r})= |\Psi(\mathbf{r})|^2$, $\tilde{\rho}(\mathbf{k})= |\tilde{\Psi}(\mathbf{k})|^2$ and $\langle \rangle$ denotes mean value, so that $\langle \ln{\rho} \rangle = \int d^n \mathbf{r} \,\rho (\mathbf{r})\ln{\rho(\mathbf{r})}$

  • $\begingroup$ Could you write out the relevent inequalities in your question. We generally encourage question on this site to be self contained if at all possible. People are also more likely to answer your question if they don't have to dig through a pdf to find the relevant informaiton $\endgroup$ Feb 8 at 18:36
  • $\begingroup$ Hello! Please read How do I ask homework questions on Physics Stack Exchange? and edit your question accordingly. Thanks! $\endgroup$
    – Jonas
    Feb 8 at 19:06

Due diligence first. The first term of the first equation you wrote is missing a sign, as you should have easily checked yourself, from the coefficient of $\ln \pi$.

That is, write down the parent $W(q)$ right above the equation you start with, around your minimum q =2, i.e. $q\equiv 2(1+\epsilon)$, so that $p= 2(1-\epsilon) +O(\epsilon^2)$, dismissing any higher orders of $\epsilon$.

The relevant term for the derivative w.r.t. $2\epsilon$ then is $\pi^{-n2\epsilon/4} N$, so you may catch that paper's typo.

  • $\begingroup$ Thank you, you are correct that I should have checked the derivative first. $\endgroup$
    – Bruno
    Feb 9 at 3:13

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.