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.

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    $\begingroup$ 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. $\endgroup$
    – Rammus
    Commented Mar 8, 2021 at 21:13
  • 1
    $\begingroup$ 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. $\endgroup$
    – Rammus
    Commented Mar 8, 2021 at 21:17
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    $\begingroup$ see quantumcomputing.stackexchange.com/q/5993/55 and links therein $\endgroup$
    – glS
    Commented Mar 8, 2021 at 21:51

1 Answer 1


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

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


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

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

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

  • 1
    $\begingroup$ Thank you again. :) $\endgroup$ Commented Mar 9, 2021 at 16:43

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