A unitary matrix U is a matrix such that the conjugate transpose of U, when multiplied on the right with U, yields identity. My question is, is it possible to obtain the transpose of any density matrix using some unitary operation? (I read somewhere that the transpose operation is an 'anti-unitary' operation, but I don't think that means that a unitary operation can NOT simulate a transpose operation).
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2$\begingroup$ I guess this question would fit better in math.SE. Anyway, fact is a anti-unitary operation is not a unitary operation so the conjugate-transpose is not unitary. To see that note that $(\alpha_1 U_1 + \alpha_2 U_2)^\dagger = \alpha_1^* U_1^\dagger + \alpha_2^* U_2^\dagger$ whereas for a unitary operation due to linearity you will never get the conjugate on $\alpha_1$ and $\alpha_2$. $\endgroup$– FabianDec 3, 2012 at 5:22
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2 Answers
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The transpose $\rho^T$ of a density matrix $\rho$ is again a density matrix, furthermore it has the same eigenvalues as $\rho$ (including multiplicities). Therefore $\rho^T$ is unitarily equivalent to $\rho$, i.e. there exists a unitary matrix $U_\rho$ such that $\rho^T= U^*_\rho \rho U_\rho$.
But this matrix depends on $\rho$. If there existed a unitary matrix $U$ such that $\rho^T=U^*\rho U$ for all density matrices, then this would imply that the map $\rho\mapsto\rho^T$ is completely positive, which it is not (as Juan already pointed out.
Note also that taking the transpose depends on the choice of a basis!
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$\begingroup$ Sorry, taking the transpose of a matrix does NOT depend on the choice of the basis. $\endgroup$– MartinoFeb 2, 2013 at 12:45
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4$\begingroup$ When you write down a density matrix, you have to choose a basis for your Hilbert space. If you choose a basis. where your density matrix is diagonal, then taking the transpose doesn't change it. If you represent your density matrix in another basis, where it is not diagonal, then of course taking the transpose changes it. Therefore clearly taking the transpose depends on the basis, it is not a physical operation that you can define without fixing a basis (writing a density operator as a matrix of course also requires fixing a basis). $\endgroup$– UwFFeb 2, 2013 at 13:27
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The transpose is not a unitary map. One simple way to understand this is by recalling that the transpose is not a completely positive map. More precisely, let $T$ be the transpose map. Then the state $\rho_1=(\hat{I}\otimes T)\cdot\rho$ in general will not be a valid density natrix. However, for maps of the form $\hat{I}\otimes U$ for some unitary $U$, the state $\rho_2=(\hat{I}\otimes U)\cdot\rho$ will always be a valid density matrix. Therefore, the transposition map is not unitary.
In some cases, the transpose map might be equivalent to some unitary map for a particular state, but this does not mean that the map itself can be represented as a unitary.
answered Dec 3, 2012 at 23:00