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In this PDF https://www-m5.ma.tum.de/foswiki/pub/M5/Allgemeines/MichaelWolf/QChannelLecture.pdf on page 14, the author says right after equation (1.11)

Note that by using a basis of Hilbert-Schmidt orthogonal unitaries $\{U_j\}_{j=1,...,d^2}$ we can construct an orthonormal basis of maximally entangled states

In this document, he considers the Hilbert space $\mathcal{M}_d(\mathbb{C})$ where each element is a $d \times d$ matrix. The scalar product between two elements of this Hilbert space is:

$$ \langle A | B \rangle = Tr(A^{\dagger} B)$$

My question is: How do we know we can have a basis of $\mathcal{M}_d(\mathbb{C})$ composed of unitary matrices ? For me asking the element of the basis are unitary matrices is a very strong condition. How is it possible ?

I would like answer not too technical, my main focus is quantum information, not general theory around Hilbert spaces. I have good basis in linear algebra but for example the Hilbert Schmidt inner product for matrices is a new concept for me.

StarBucK
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    See https://mathoverflow.net/questions/326767/is-there-always-a-complete-orthogonal-set-of-unitary-matrices – Eric Wofsey Dec 16 '19 at 16:21
  • Note that the Hilbert-Schmidt inner product is a really simple concept: it's just the usual inner product on $\mathbb{C}^{d^2}$ applied to the entries of the matrices. – Eric Wofsey Dec 16 '19 at 16:21
  • @EricWofsey thank you for your answer. In the related post I do not get why there are $n^2$ matrix that generates $e_i \rightarrow e_{i+1}$. For me there are only $n$ shift possible as there are $n$ $e_i$ vectors ? I looked at the wikipedia page but it didn't help me. – StarBucK Dec 16 '19 at 16:33
  • @EricWofsey also not sure to get your second comment ? Do you mean that if for all $|i \rangle \langle j |$ I associate a vector $|i,j\rangle$, then the Hilbert-Schmidt inner product is the usual inner product on this vector ? Maybe this remark is related to my confusion in my previous comment. – StarBucK Dec 16 '19 at 16:39
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    There are $n$ shifts, but there are also $n$ possible powers of the diagonal entries $(1,\omega,\omega^2,\cdots,\omega^{n-1})$ you can use (you can take $(1,\omega^j,\omega^{2j},\cdots,\omega^{(n-1)j})$ for any $j$ from $0$ to $n-1$). So you can take any of these $n$ sequences of diagonal entries, but with the diagonal shifted by any amount. – Eric Wofsey Dec 16 '19 at 16:58
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    As for your second question, yes, the matrices $|i \rangle \langle j |$ with a single nonzero entry that is $1$ are orthonormal with respect to the Hilbert-Schmidt inner product. – Eric Wofsey Dec 16 '19 at 16:59

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