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

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?

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.

• 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 Oct 11, 2020 at 21:54
• 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 $M = [m_{ab}]$ Oct 12, 2020 at 7:30

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.

• 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). Oct 11, 2020 at 7:41
• The $H$ you give there is just $-|+\! i \rangle\langle +i|$, so not a good "counterexample".
– Kall
Oct 11, 2020 at 8:23