# Unitary transform of a partially entangled 3-qubit state

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)?

• It looks like you'll get the Kronecker product $U\otimes I_{2}$ – Omnomnomnom Jul 25 '16 at 11:52

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 $$U' \left|xyz\right\rangle = U'(\left|xy\right\rangle \otimes\left|z\right\rangle) = U\left|xy\right\rangle \otimes \left|z\right\rangle.$$