How many real numbers are required to describe density matrix for $n$ qubits?

Question

(All of these coming from the topic of simulation of quantum systems) A density matrix $\rho$ Which describe state of $n$ qubits will have $2^{n} \times 2^{n}$ size. We have couple of conditions like

1. $\mathrm{tr}(\rho) = 1$
2. $\rho$ is positive.

Then in this case we only need to specify $\dfrac{2^{n}(2^{n}-1)}{2} $ for off-diagonal elements and for the diagonal ones we need $2^{n}-1$ Terms. So total terms required $\dfrac{2^{n}(2^{n}-1)}{2} + 2^{n}-1 $. Now each of these terms can be complex numbers. And for defining a complex number we need one real number (assuming $e^{i\theta}$ form and defining theta to a good enough approximation).

Total number of independent real numbers - $\dfrac{2^{n}(2^{n}-1)}{2} + 2^{n}-1 = 4^{n}+\dfrac{2^{n}}{2}-1$

But in Nielsen Chuang it was asked to proof

Exercise 4.46: (Exponential complexity growth of quantum systems) Let $\rho$ be a density matrix describing the state of $n$ qubits. Show that describing $ρ$ requires $4^{n} − 1$ independent real numbers.

Who is correct here?

Answer 1

NC is correct. You are making two mistakes here,

1. The diagonal terms of a density matrix are always real (it is a hermitian operator).
2. You need two real numbers to specify a complex number. In polar form it is $r e^{i \theta}$, $r> 0$ and $\theta \in (-\pi,\pi]$. Or in Cartesian form $x + i y$ for $x,y\in \mathbb{R}$.

After taking into account these two points you should get $4^n - 1$.