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I work on a Quantum Information Science II: Quantum states, noise and error correction MOOC by Prof. Aram Harrow, and I do not understand which property of tensor products is used in one of the transitions in the videos.

Let's consider an isometry $V: A \to B \otimes E$ ($E$ is a subspace to be thrown away at the end).

Let's fix and orthonormal basis $\{ |e\rangle \}$ in $E$ and partially expand the isometry $V$ as $V = \sum_e V_e \otimes |e\rangle$, where each $V_e$ is a linear operator from $A$ to $B$.

The Stinespring form of a quantum operation is a partial trace applied after an isometry: $N(\rho) = \mathrm{tr}_E [V \rho V^\dagger]$.

Now, if we expand that with our representation of $V$, we get $$ N(\rho) = \mathrm{tr}_E \left[ \sum_{e_1} \sum_{e_2} \left( V_{e_1} \otimes |e_1\rangle \right) \rho \left( V_{e_2}^\dagger \otimes \langle e_2| \right) \right]. $$

My question is how to get from here to the next step $$ N(\rho) = \mathrm{tr}_E \left[ \sum_{e_1} \sum_{e_2} (V_{e_1} \rho V_{e_2}^\dagger) \otimes |e_1 \rangle \langle e_2| \right]? $$

(BTW, eventually, we end up with the Kraus operator decomposition of a channel: $N(\rho) = \sum_e V_e \rho V_e^\dagger$.)

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    $\begingroup$ One way to see it is to think about $\rho$ as being the same as $\rho\otimes 1$ (i.e., $\rho$ tensored with the scalar 1). $\endgroup$ – John Watrous Jun 1 '18 at 21:15
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As pointed out in a comment, what you wrote as $\rho$ should more precisely be written as $\rho\otimes\mathbb 1$ (although the Kraus operator decomposition can be obtained similarly with any initial ancilla state, in which case you have $\rho\otimes|\phi\rangle\!\langle\phi|$).

The standard algebraic properties of tensor product spaces then tell you that $$(A\otimes B)(C\otimes D)=(AC)\otimes(BD),$$ from which you immediately get your result.

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  • $\begingroup$ Is that what you meant: (I) $\rho \left( V_{e_2}^\dagger \otimes \langle e_2| \right) = (\rho \otimes 1) \left( V_{e_2}^\dagger \otimes \langle e_2| \right) = \left( \rho V_{e_2}^\dagger \right) \otimes \langle e_2|$; (II) $(V_{e_1} \otimes |e_1 \rangle) \left(\left( \rho V_{e_2}^\dagger \right) \otimes \langle e_2|\right) = V_{e_1} \rho V_{e_2}^\dagger \otimes |e_1\rangle \langle e_2|$? $\endgroup$ – Alexander Pozdneev Jun 8 '18 at 19:29
  • $\begingroup$ @AlexanderPozdneev yes exactly $\endgroup$ – glS Jun 9 '18 at 12:21

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