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  Hilbert-Schmidt basis for many qubits - reference

+ 9 like - 0 dislike

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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asked Nov 22, 2011 in Theoretical Physics by Piotr Migdal (1,260 points) [ no revision ]
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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 $\hat\rho$?

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It is simply decomposition of $4^n$ dimensional “vector” with an orthogonal basis. Vector space is space of $2^n \times 2^n$ Hermitian matrices with respect to norm $(A,B) =Tr(AB) = Tr(AB^*)$. But I doubt, it could be called Hilbert-Schmidt decomposition because it is defined for any $n$ and for $n=2$ produces up to 16 terms instead of 4.

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Googling leads me into progressively more mathematical territory. There is a small chance that this is a result well known enough to mathematicians that it jumped into QI sideways as a standard tool used by a mathematical physicist. I've seen the Hilbert-Schmidt basis used without even labelling it. See for example the [answer I gave to my own question](http://theoreticalphysics.stackexchange.com/q/537/569) a couple of days ago. I used it without thinking of the name, as did the paper I was referring to in that post. This unfortunately only compounds the stupidity shown in my first comment...

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@AlexV I know that it has nothing to do with the Schmidt decomposition.

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So the term due to Hilbert-Schmidt inner product for matrices?

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@MateusAraújo: Don't forget to post your links as an answer! :)

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Maybe the discrete Wigner function is a bit other story, because they need to use trace on $GF(2^n)$ and exchange components in Hilbert-Schmidt scalar product.

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3 Answers

+ 3 like - 0 dislike

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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answered Nov 25, 2011 by Earl (405 points) [ no revision ]
+1 from me. I use it a lot too, but couldn't think of anything interesting to say!

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+ 1 like - 0 dislike

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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answered Nov 24, 2011 by JonLester (345 points) [ no revision ]
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](http://arxiv.org/abs/quant-ph/0401155), Paz, Roncaglia, Saraceno [quant-ph/0410117](http://arxiv.org/abs/quant-ph/0410117) and Sec 4.2 of Chris Ferrie's [1010.2701](http://arxiv.org/abs/1010.2701)).

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+ 1 like - 0 dislike

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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answered Nov 25, 2011 by Mateus Araújo (270 points) [ no revision ]

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