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Let $A \cong \mathbb{C}^{n}$ be a Hilbert space $A,$ and let $\operatorname{Herm}(A)$ be the Hilbert space consisting of all Hermitian matrices on $A$. Give an example of a basis (not necessarily orthogonal) of Herm (A) consisting of pure density matrices in $\mathfrak{D}(A)$.

$\mathbf A\mathbf t\mathbf t\mathbf e\mathbf m\mathbf p\mathbf t$:

I Started with the case $n=2$ and considered the vectors $|0\rangle,|1\rangle,|+\rangle$ and $|+i\rangle$ , Because I think $\left|0\right\rangle\left\langle 0\right|$, $\left|1\right\rangle\left\langle 1\right|$, $\left|+\right\rangle\left\langle +\right|$, and $\left|+i\right\rangle\left\langle +i\right|$ are a non-orthogonal basis and all of them are pure density matrices. And as you know,

$|+\rangle=\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$

$|+i\rangle=\frac{1}{\sqrt{2}}(|0\rangle+i|1\rangle)$.

But my question is how can I prove or show that they are form a basis for $\operatorname{Herm}(A)$, in this case $A \cong \mathbb{C}^{2}$ and how can I expand it to $A \cong \mathbb{C}^{n}$? I mean how can I determine a basis for $A \cong \mathbb{C}^{n}$ which are pure density matrices and how can I show that it is a basis?

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For $ n = 2 $, it is known that the Pauli matrices together with the identity matrix $ I $ form a basis. Now observe that we can write:

  • $ I = |0 \rangle \langle 0| + |1 \rangle \langle 1| $
  • $ \sigma_z = 2 \cdot |0 \rangle \langle 0| - I $
  • $ \sigma_x = 2 \cdot |+ \rangle \langle +| - I $
  • $ \sigma_y = 2 \cdot |+i \rangle \langle +i| - I $

This means that also the pure density matrices $|0 \rangle \langle 0|, \hspace{0.3em} |1 \rangle \langle 1|, \hspace{0.3em} |+ \rangle \langle +|, \hspace{0.3em} |+i \rangle \langle +i| $ are a basis (not orthogonal).

For the general case, the matrices $ H_{a,b} $, with $ 1 \leq a,b \leq n $, form an orthogonal basis for Herm$(A)$ (see section 1.4.2) $$ \ H_{a,b} = \begin{cases} E_{a,a} & \text{if $a = b $} \\ E_{a,b} + E_{b,a} & \text{if $a < b$} \\ i (E_{a,b} - E_{b,a}) & \text{if $a > b$} \end{cases} \ $$ where $ E_{a,b} = |e_a \rangle \langle e_b| $ and $ |e_a \rangle $ a state with 1 in the $a$-th entry and all other entries zeros.

Now define the states: $$ \ |\psi_{a,b} \rangle = \begin{cases} |e_a \rangle & \text{if $a = b $} \\ \frac{1}{\sqrt{2}} (|e_a \rangle + |e_b \rangle) & \text{if $a < b$} \\ \frac{1}{\sqrt{2}} (i|e_a \rangle + |e_b \rangle) & \text{if $a > b$} \end{cases} \ $$ and the pure density matrices $ \rho_{a,b} = |\psi_{a,b} \rangle \langle \psi_{a,b}| $. After some calculations we get

  • $ H_{a, a} = \rho_{a,a} $
  • $ H_{a, b} = 2 \rho_{a,b} - \rho_{a,a} - \rho_{b,b} $

so $ \rho_{a,b} $ form a basis.

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  • $\begingroup$ Neat answer! Is there some intuition/proof for why $H_{a,b}$ form a basis for all Hermitian operators? The result is stated without proof in Watrous' notes $\endgroup$ Oct 11 '20 at 21:54
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    $\begingroup$ I think that you should be able to verify that $ M = \sum_{a=1}^{n} m_{aa} H_{a,a} + \sum_{a < b} Re\{ m_{ab} \} H_{a,b} + \sum_{a > b} Im\{ m_{ab} \} H_{a,b} $ for any Hermitian matrix $ M = [m_{ab}] $ $\endgroup$
    – tsgeorgios
    Oct 12 '20 at 7:30
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I will try to take a stab at it from my understanding of your question:

The basis for the space of $2 \times 2$ Hermitian matrices over $\mathbb{R}$ is:

\begin{equation} \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} \ \ \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix} \ \ \begin{pmatrix} 0 & 1\\ 1 & 0 \end{pmatrix} \ \ \begin{pmatrix} 0 & i\\ -i & 0 \end{pmatrix} \end{equation}

But from my understanding, you want to restrict the basis set to consist of only rank 1 matrices. Is that right? You are considering the basis set

\begin{equation} |0\rangle\langle 0| = \begin{pmatrix} 1 & 0\\ 0 & 0 \end{pmatrix} \ \ \ \ |1\rangle\langle 1| = \begin{pmatrix} 0 & 0\\ 0 & 1 \end{pmatrix} \ \ \ \ |+\rangle\langle +| =\dfrac{1}{2}\begin{pmatrix} 1 & 1\\ 1 & 1 \end{pmatrix} \ \ \ \ |i\rangle\langle i| =\dfrac{1}{2}\begin{pmatrix} 1 & i\\ i & 1 \end{pmatrix} \end{equation}


Well, if we take $H$ to be the Hermitian matrix

$$ H = \dfrac{1}{2}\begin{pmatrix} 1 & i\\ -i & 1 \end{pmatrix}$$

Can you form this Hermitian matrix $H$ from your supposedly basis set?


update: As commented, I made a wrong calculation, as $|i\rangle \langle i|$ should be

$$ |i\rangle\langle i| =\dfrac{1}{2}\begin{pmatrix} 1 & -i\\ i & 1 \end{pmatrix} $$ and therefore it can be written as $H = \dfrac{1}{2}|0\rangle\langle 0| + \dfrac{1}{2}|1\rangle\langle 1| - |i\rangle\langle i | $

And it turns out that the basis set in consideration is actually correct as now pointed out by the other answer! Thanks for bringing up this problem though.

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  • $\begingroup$ Yes, maybe you are right and my attempt is wrong, but the hint of the problem says that we can use the $|0\rangle,|1\rangle,|+\rangle$ and $|+i\rangle$ to form new basis which are pure density matrices(Positive semi-definite with rank 1). $\endgroup$
    – 299792458
    Oct 11 '20 at 7:41
  • $\begingroup$ The $H$ you give there is just $-|+\! i \rangle\langle +i|$, so not a good "counterexample". $\endgroup$
    – Kall
    Oct 11 '20 at 8:23

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