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Suppose we have a partially separable 3-qubit state

$$φ = \left(a_0\left|0\right\rangle + a_1\left|1\right \rangle\right) \otimes \left(b_{00}\left|00\right \rangle + b_{01}\left|01\right \rangle + b_{10}\left|10\right \rangle + b_{11}\left|11\right \rangle\right)$$

i.e. the second and third qubits are entangled with each other, but the first qubit is separable.

The unseparated state is given by

$$φ = c_{000}\left|000\right\rangle + c_{001}\left|001\right\rangle + c_{010}\left|010\right\rangle + c_{100}\left|100\right\rangle+ c_{011}\left|011\right\rangle + c_{101}\left|101\right\rangle + c_{110}\left|110\right\rangle + c_{111}\left|111\right\rangle $$

with $c_{ijk} = a_i b_{jk}$.

Now suppose I want to apply a unitary transformation to the first two qubits, given by a 4x4 unitary matrix:

$U = (u_{nm})$

Q: What does the overall 8x8 matrix look like which will also calculate the effect on the third qubit (which won't leave the interaction unchanged)?

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  • $\begingroup$ It looks like you'll get the Kronecker product $U\otimes I_{2}$ $\endgroup$ – Omnomnomnom Jul 25 '16 at 11:52
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The entanglement of the qubits does not matter. All you need to do is tensor $U$ with $\begin{pmatrix}1&0 \\ 0 & 1\end{pmatrix}$, which is the effect of $U$ on the third qubit. You'll get $U' = \begin{pmatrix}U&0 \\ 0 & U\end{pmatrix}$ or a permutation thereof.

As a linear transformation, $U' = U \otimes \text{id}$, so you'd have $$U' \left|xyz\right\rangle = U'(\left|xy\right\rangle \otimes\left|z\right\rangle) = U\left|xy\right\rangle \otimes \left|z\right\rangle.$$

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  • $\begingroup$ Which permutation? What does it depend on? $\endgroup$ – Hans-Peter Stricker Jul 25 '16 at 12:31
  • $\begingroup$ Well, every index corresponds to one eigenstate, right? But there's no canonical ordering of the eigenstates. What matrix you get depends on what order you use for the eigenstates. This isn't an issue if you write it directly in terms of linear transformations instead. $\endgroup$ – Anon Jul 25 '16 at 12:40
  • $\begingroup$ How would it look like "in terms of linear transformations"? $\endgroup$ – Hans-Peter Stricker Jul 25 '16 at 12:59
  • $\begingroup$ Edited answer to explain. $\endgroup$ – Anon Jul 25 '16 at 13:06
  • $\begingroup$ Ah! It's so easy! Thanks a lot. $\endgroup$ – Hans-Peter Stricker Jul 25 '16 at 13:07

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