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What is a basis (not necessarily orthogonal) of Herm(A) consisting of pure density matrices in D(A)?($A cong mathbb{C}^{n}$)

Quantum Computing Asked on January 25, 2021

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 $mathfrak{D}(A)$.

$mathbf Amathbf tmathbf tmathbf emathbf mmathbf pmathbf t$:

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

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

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

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?

2 Answers

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.

Correct answer by tsgeorgios on January 25, 2021

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} |0ranglelangle 0| = begin{pmatrix} 1 & 0 0 & 0 end{pmatrix} |1ranglelangle 1| = begin{pmatrix} 0 & 0 0 & 1 end{pmatrix} |+ranglelangle +| =dfrac{1}{2}begin{pmatrix} 1 & 1 1 & 1 end{pmatrix} |iranglelangle 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 $|irangle langle i|$ should be

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

Answered by KAJ226 on January 25, 2021

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