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).

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.