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Every density matrix of $n$ qubits can be written in the following way $$\hat{\rho}=\frac{1}{2^n}\sum_{i_1,i_2,\ldots,i_n=0}^3 t_{i_1i_2\ldots i_n} \hat{\sigma}_{i_1}\otimes\hat{\sigma}_{i_2}\otimes\ldots\otimes\hat{\sigma}_{i_n},$$ where $-1 \leq t_{i_1i_2\ldots i_n} \leq 1$ are real numbers and $\{\hat{\sigma}_0,\hat{\sigma}_1,\hat{\sigma}_2,\hat{\sigma}_3\}$ are the Pauli matrices. In particular for one particle ($n=1$) it is the Bloch representation.

Such representation is used e.g. in a work by Horodecki arXiv:quant-ph/9607007 (they apply $n=2$ to investigate the entanglement of two qubit systems). It is called decomposition in the Hilbert-Schmidt basis.

The question is if there is any good reference for such representation for qubits - either introducing it for quantum applications or a review paper? I am especially interested in the constrains on $t_{i_1i_2\ldots i_n}$.

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  • $\begingroup$ Just having read the question, and possibly not understanding correctly yet, What additional constraints should there be on $t_{i_1i_2\ldots i_n}$ other than leading to a normalised $\hat\rho$? $\endgroup$
    – qubyte
    Nov 22, 2011 at 15:25
  • $\begingroup$ @Mark: Normalization is equivalent to $t_{00\ldots0}=1$. All other constrains are related to the nonnegative definiteness (e.g. for $n=1$ there is $t_1+t_2+t_3 \leq 1$). $\endgroup$ Nov 22, 2011 at 15:35
  • $\begingroup$ With you now. Apparently my brain has already turned in for an early (earlier than me anyway) night. $\endgroup$
    – qubyte
    Nov 22, 2011 at 15:39
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    $\begingroup$ @AlexV I know that it has nothing to do with the Schmidt decomposition. $\endgroup$ Nov 22, 2011 at 17:35
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    $\begingroup$ This seems to be what you're looking for. $\endgroup$
    – Mateus Araújo
    Nov 22, 2011 at 23:53

3 Answers 3

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I use this decomposition all the time, but I have never read a paper solely devoted to the topic. From my experience a complete characterization of the constraints on $t_{i_{1}, t_{2},..t_{n}}$ is tricky, and so if you want to be sure $\rho$ is physical you should calculate the density matrix and its eigenvalues.

However, there are a lot of necessary conditions that have a useful form in this decomposition. For example, for a positive unit-trace Hermitian operator $\rho$ is follows that

$|t_{i_{1}, i_{2},.. i_{n}}| \leq 1$

$tr ( \rho^{2} ) =\frac{1}{2^{n}} \sum_{i_{1}, i_{2},.. i_{n}} t_{i_{1}, i_{2},.. i_{n}}^{2} \leq 1 $

The above condition tells us that if we think of $t$ as a vector in a real vector space, then the physical states live within the unit sphere. This is a bit like the Bloch sphere for 1 qubit but for many qubits we have some other constraints that take the form of hyperplanes. For every $\vert \psi \rangle$ expressed in the same form $\vert \psi \rangle \langle \psi \vert = \frac{1}{2^{n}} \sum_{i_{1},i_{2},... i_{n}} Q_{i_{1},i_{2},... i_{n}}\sigma_{i1} \otimes \sigma_{i2}... \sigma_{in}$ we require that

$\langle \psi \vert \rho \vert \psi \rangle \geq 0 $ and so $\sum Q_{i_{1},i_{2},... i_{n}}t_{i_{1},i_{2},... i_{n}}\geq 0$ which defines a hyperplane.

The problem is you have a hyperplane for every $\psi$ so that requiring $t$ to satisfy every inequality one of the infinite hyperplanes is impossible to check by brute force. If you want sufficient conditions for positivity of $\rho$ I suspect you have to calculate eigenvalues.

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  • $\begingroup$ +1 from me. I use it a lot too, but couldn't think of anything interesting to say! $\endgroup$ Nov 25, 2011 at 11:36
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Claudio Altafini studies precisely this subject, in Tensor of coherences parameterization of multiqubit density operators for entanglement characterization and some follow-ups.

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A good starting point, I have checked just chapter 4 but there is more, is

R. R. Puri, Mathematical Methods of Quantum Optics, Springer (2001) (see here).

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  • $\begingroup$ Thank you, Jon. In the meantime I found papers about discrete phase space, where $\hat{\sigma}_{\vec{i}}$ is a shift operator (Gibbons, Hoffman, Wootters quant-ph/0401155, Paz, Roncaglia, Saraceno quant-ph/0410117 and Sec 4.2 of Chris Ferrie's 1010.2701). $\endgroup$ Nov 24, 2011 at 11:57

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