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.

  • $\begingroup$ 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? $\endgroup$ – Norbert Schuch Mar 17 '17 at 10:31
  • $\begingroup$ Also: Linearly independent over $\mathbb R$ or over $\mathbb C$? $\endgroup$ – Norbert Schuch Mar 17 '17 at 10:32
  • $\begingroup$ 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. $\endgroup$ – user148980 Mar 17 '17 at 10:45
  • $\begingroup$ Both real and complex. Does it make a difference? Sorry I'm probably really showing my ignorance here. $\endgroup$ – user148980 Mar 17 '17 at 10:46
  • 1
    $\begingroup$ Real and complex makes a difference. For instance, the numbers $1$ and $i$ are linearly dependent over $\mathbb C$ but linearly independent over $\mathbb R$. $\endgroup$ – Norbert Schuch Mar 17 '17 at 10:48

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.

  • $\begingroup$ 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. $\endgroup$ – udrv Mar 19 '17 at 5:45

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.