# Why can any density operator be written this way? (quantum tomography)

From page 24 of the thesis "Random Quantum States and Operators", where $$(A,B)$$ is the Hilbert-Schmidt inner product:

\begin{aligned} \rho &=\left(\frac{1}{\sqrt{2}} I, \rho\right) \frac{1}{\sqrt{2}} I+\left(\frac{1}{\sqrt{2}} X, \rho\right) \frac{1}{\sqrt{2}} X+\left(\frac{1}{\sqrt{2}} Y, \rho\right) \frac{1}{\sqrt{2}} Y+\left(\frac{1}{\sqrt{2}} Z, \rho\right) \frac{1}{\sqrt{2}} Z \\ &=\frac{(I, \rho) I+(X, \rho) X+(Y, \rho) Y+(Z, \rho) Z}{2} \\ &=\frac{I+\operatorname{tr}(X \rho) X+\operatorname{tr}(Y \rho) Y+\operatorname{tr}(Z \rho) Z}{2} \end{aligned}

This is used for explaining quantum tomography. Can someone please explain each step clearly? I have pretty basic QC and linear algebra knowledge.

• The notation $(A,B)$ is referring to the Hilbert Schmidt inner product which will just be $(A,B) = \mathrm{tr}(A^{\dagger}B) = \mathrm{tr}(AB)$ as every operator here is Hermitian. After that, it's just basic simplifications. Mar 8 '21 at 21:13
• As to why a qubit can be written this way, the operators $I, X,Y,Z$ form a basis for the real vector space of $2\times 2$ Hermitian matrices. Mar 8 '21 at 21:17
• see quantumcomputing.stackexchange.com/q/5993/55 and links therein
– glS
Mar 8 '21 at 21:51

From linear algebra, if $$v_1, \dots, v_n$$ is a basis of the vector space $$V$$ then every vector $$v\in V$$ can be written as a linear combination

$$v = a_1 v_1 + \dots + a_n v_n\tag1$$

where the coefficients $$a_k$$ belong to the underlying scalar field. Moreover, if $$V$$ is an inner product space and $$v_1, \dots, v_n$$ is an orthonormal basis then the coefficients $$a_1, \dots, a_n$$ can be computed using the inner product as $$a_k = \langle v, v_k\rangle$$. This is easy to prove by acting with $$\langle\,.,v_k\rangle$$ on both sides of $$(1)$$.

Now, it turns out that the set of Hermitian matrices with complex entries is a real vector space with inner product defined as

$$\langle A, B \rangle = \mathrm{tr} (A^\dagger B).\tag2$$

Also, it is easy to check that $$I, X, Y, Z$$ are orthogonal with respect to the inner product $$(2)$$ and since the space of $$2\times 2$$ Hermitian matrices is four-dimensional the matrices form a basis. By normalizing we can turn it into an orthonormal basis $$I/\sqrt{2}, X/\sqrt{2}, Y/\sqrt{2}, Z/\sqrt{2}$$. Then, every $$2\times 2$$ Hermitian matrix $$\rho$$ can be written as

$$\rho = a_I \frac{I}{\sqrt{2}} + a_X \frac{X}{\sqrt{2}} + a_Y \frac{Y}{\sqrt{2}} + a_Z \frac{Z}{\sqrt{2}}\tag3$$

and

$$a_I = \frac{1}{\sqrt{2}}\mathrm{tr}(\rho) \\ a_X = \frac{1}{\sqrt{2}}\mathrm{tr}(\rho X) \\ a_Y = \frac{1}{\sqrt{2}}\mathrm{tr}(\rho Y) \\ a_Z = \frac{1}{\sqrt{2}}\mathrm{tr}(\rho Z).$$

Assuming $$\rho$$ is a state, then $$a_I = \frac{1}{\sqrt{2}}$$ because the trace of the density matrix representing the state $$\rho$$, denoted as $$\mathrm{tr}(\rho)$$, always equals 1. Finally, substituting into $$(3)$$, we get

$$\rho = \frac{I + \mathrm{tr}(\rho X)X + \mathrm{tr}(\rho Y)Y + \mathrm{tr}(\rho Z)Z}{2}.$$

• Thank you again. :) Mar 9 '21 at 16:43