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Is density matrix a tensor? Would it change if we represent it in another basis as a tensor would?

Is there any difference in this regard between pure and mixed quantum states?

(I definitely can say, e.g., that density matrix of a fully mixed state is a scalar - i.e. it doesn't change regardless of the basis).

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  • $\begingroup$ What, exactly, do you mean by a tensor? What if e.g. you had a tensor with nontrivial indices that didn't change regardless of the basis? $\endgroup$ Oct 9 '17 at 17:47
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    $\begingroup$ It's a linear automorphism of your Hilbert space, hence it's a multi-(read, 1)-linear operator on your Hilbert space, hence it's a tensor. In particular it lives in $H \otimes H^*$. $\endgroup$
    – zzz
    Oct 9 '17 at 18:13
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In general the density matrix is a reducible tensor in the sense that it can be expanded as a sum of irreducible tensors. In the case of a $(2S+1)\times (2S+1)$ density matrix describing a mixture of states of angular momentum $S$ $$ \rho = \sum_{L=0}^{2S} \sum_{M=-L}^L \rho_{LM} T^L_M \tag{1} $$ where $T^L_M$ is the component of an irreducible tensor of angular momentum L $$ T^L_M=\sqrt{\frac{2L+1}{2S+1}}\sum_{mm'} C_{Sm;LM}^{Sm'} \vert Sm'\rangle \langle Sm\vert\, , \tag{2} $$ such that $$ \rho_{LM}=\hbox{Tr}\left(\rho (T^L_M)^\dagger\right)\, . \tag{3} $$ Moreover $$ R(\Omega) T^L_M R^{-1}(\Omega)=\sum_{M'} T^L_{M'}D^L_{M'M}(\Omega)\, . $$ Note the irreducible tensors of (2) are orthonormal in the sense of $$ \hbox{Tr}\left((T^{L_1}_{M_1})^\dagger T^{L_2}_{M_2}\right)=\delta_{L_1L_2}\delta_{M_1M_2}\, . \tag{4} $$ Eq.(3) follows from combining (1) and (2) and taking the trace.

If you are thinking of a more general $N\times N$ density matrix, and ask about the tensorial properties under a general change of basis in $U(N)$, then the same general argument applies in the sense that you can expand your $\rho$ in terms of the unit matrix and the generalized Gell-Mann matrices (or any other set of generators of $u(N)$). The generators transform by the adjoint representation of $u(N)$ so that, in general, the density matrix would also be a $U(N)$-reducible tensor.

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