What is the maximum number of linearly independent density matrices in a 2D space?

Question

I think for a generic 2 by 2 matrix we would require 4 linearly independent matrices to span the whole space, 1 per element. Then I would assume that the two constraints for a legitimate density matrix (trace equals 1, Hermitian) would each subtract one from this number, leaving me with 2 linearly independent matrices.

This logic definitely doesn't constitute as a proof however (or is even particularly convincing), so, firstly am I correct in saying that there are only 2 linearly independent density matrices in 2D? And secondly, if I am correct, does anybody have a more convincing proof for it?

Also if it is only 2 linearly independent density matrices in 2D, is it $(d^2-2)$ linearly independent ones in $d$ dimensions?

P.S. I'm not a quantum information guy or a mathematician so please speak slowly.

Answer 1

To parametrize a complex $2\times 2$ matrix you need four complex numbers: $$ M=\left(\begin{array}{cc}z& w \\u & v \end{array}\right) $$ Hermiticity imples that $u=w^*$ and $z,v\in\mathbb{R}$, whereas $\operatorname{tr}M=1$ gives the condition $v=1-z$. So a hermitian trace-one $2\times 2$ matrix is parametrized by three real numbers $x$, $y$ and $z$ in the following way $$ M=\frac{1}{2}\left(\begin{array}{cc}1+x& y-iz \\ y+iz & 1-x \end{array}\right) $$ In fact, such matrices are usually written in terms of the Pauli matrices $\vec{\sigma}=(\sigma^1,\sigma^2,\sigma^3)$ and the $2\times 2$ identity matrix $I_2$ as $\boxed{M=(I_2+\vec{x}\cdot\vec{\sigma})/2}$, where $\vec{x}=(x,y,z)\in\mathbb{R}^3$.

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For a $N\times N$ hermitian matrix the independent parameters are: two real numbers for each element of the upper triangle plus one real number for each element of the diagonal. Substracting one because of the trace-one condition we get $$ 2\frac{N(N-1)}{2}+N-1=N^2-1 $$ independent real parameters.