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

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.

• Why do you think that Hermiticity substracts one degree of freedom? If the matrix had to be equal to the identity matrix, would this also substract one degree of freedom? – Norbert Schuch Mar 17 '17 at 10:31
• Also: Linearly independent over $\mathbb R$ or over $\mathbb C$? – Norbert Schuch Mar 17 '17 at 10:32
• I guess I figured that the Hermiticity has to relate off diagonal elements. So whilst you could have $\begin{pmatrix} 1/2&1\\0&1/2\end{pmatrix}$ for one of your matrices if we are just talking about a general 2 by 2 matrix, this could not be the case if the matrix also has to be legit density matrix. – user148980 Mar 17 '17 at 10:45
• Both real and complex. Does it make a difference? Sorry I'm probably really showing my ignorance here. – user148980 Mar 17 '17 at 10:46
• Real and complex makes a difference. For instance, the numbers $1$ and $i$ are linearly dependent over $\mathbb C$ but linearly independent over $\mathbb R$. – Norbert Schuch Mar 17 '17 at 10:48

## 1 Answer

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 $\vec{x}=(x,y,z)\in\mathbb{R}^3$.

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.

• One observation: density matrices are only the positive subset of the unit trace hermitian matrices. For the 2 x2 case this means $|{\vec x}| \le 1$, a.k.a. the Bloch sphere. – udrv Mar 19 '17 at 5:45