I'm trying to follow this paper in Section 3.1, but I'm having trouble with a comment they make in the final paragraph of that section.

First, we start with a pure state:

$$|\psi\rangle = \sum_i^m \sum_j^n a_{ij} |i\rangle |j\rangle.$$

Then, they consider the complex numbers $a_{ij}$ being uniformly distributed on a hypersphere, so they have probability density:

$$P(a) \sim \delta \left( \sum_i^m \sum_j^n \lvert a_{ij} \rvert^2 - 1 \right),$$ where $\delta()$ is the Dirac distribution. Then, in the last paragraph before Section 3.2, they write:

Second, notice that the exact result of Lubkin can be estimated by relaxing the normalization constraint in the distribution, and replacing it with a product of independent Gaussian distributions, $P(a) = \prod_{i,j} (nm / \pi) \exp \left( -nm \lvert a_{ij} \vert^2 \right)$, with $\langle a_{ij} \rangle = 0$ and $\langle \lvert a_{ij} \rvert ^2 \rangle = 1/nm$.

They then go on to say that the central limit theorem says that this distribution tends to a Gaussian in $\sum_i^m \sum_j^n \lvert a_{ij}\rvert^2$ centered at $1$, with variance $1/\sqrt{nm}$.

Basically, I don't see why we get this distribution. From what I can tell, we need a sum of random variables (which we have when you take the product of those exponentials), but I'm not seeing the mean of $1$, nor the (few) steps being taken to get this end result. If someone could explain, that would be great. Also, I would like to understand what the assumption behind replacing our initial normalization with the Gaussians was.


I think that the gaussian distribution just emerges from the definition of the Dirac delta, but I'm only able to understand the approximation in a reversed way (starting from the delta and ending up with the product of gaussians). The Dirac function is defined as: $$\delta(x)=\lim_{\sigma\rightarrow 0}\frac{1}{\sqrt{2\pi}\sigma}e^{-x^2/2\sigma^2},$$ where $\sigma^2$ is the variance (or width) of the gaussian distribution, centered around zero in this case.

Using this definition, you can have: $$P(\alpha)\sim \delta (\sum^m_i\sum^n_j|\alpha_{ij}|-1)=\lim_{\sigma\rightarrow 0}\frac{1}{\sqrt{2\pi}\sigma}e^{-(\sum^m_i\sum^n_j|\alpha_{ij}|-1)^2/2\sigma^2},$$ where $\sigma$ must be $\sigma=1/mn$ since it is the parameter that controls the width of the distribution (in the limit of $mn\rightarrow \infty$, your system should have a perfect delta). Then they "relax" the normalization assuming a finite but very large dimension, so: $$\delta (\sum^m_i\sum^n_j|\alpha_{ij}|-1)\simeq\frac{nm}{\sqrt{2\pi}}e^{-nm(\sum^m_i\sum^n_j|\alpha_{ij}|-1)^2/2}.$$ Finally, they apply the limit central theorem to write the distribution as a product of distributions. For me, this derivation is a bit weird, but it is the only way I find to explain it. I hope it can help you!


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.