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What happens when an operator is applied only to some bits of a mixed state? For instance, assume $\vert x\rangle\vert f(x)\rangle$ is entangled. Then what is the result of $\vert Ux\rangle\vert f(x)\rangle$ (how to compute the amplitudes) ? What if U is Grover's diffusion? Will it still work (without uncomputing $f(x)$) ?

Update $y$ replaced with $f(x)$

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$|x\rangle|y\rangle = |x\rangle\otimes |y\rangle$ is the notation for disentangled state. Entangled state can't be written this way. In general, every pure state (entangled or disentangled) on a bipartite system is a linear combination of disentangled states $$ |\phi\rangle_{AB} = \sum_i \alpha_i |x_i\rangle_A\otimes|y_i\rangle_B $$

Application of $U$ on the first subsystem is equivalent to application of $U \otimes I$ on the whole system. The result will be $$ (U\otimes I) |\phi\rangle_{AB} = \sum_i \alpha_i U|x_i\rangle_A\otimes|y_i\rangle_B $$ Mixed state is a different thing (do not confuse it with entangled state). Mixed state can be seen as probability distribution over pure states: $\{\{p_i,|\phi_i\rangle\}\}, p_i>0, \sum_ip_i=1$. It has the corresponding density matrix $\rho=\sum_ip_i|\phi_i\rangle\langle\phi_i|$. Note that every $|\phi_i\rangle$ can be entangled.
The result of application of $U$ on the first subsystem of a mixed state is the probability distribution $\{\{p_i,(U\otimes I)|\phi_i\rangle\}\}$, or, in terms of density matrices, $(U\otimes I) \rho (U^\dagger\otimes I)$.

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  • $\begingroup$ Thank Danylo. Please take another look, I've updated the question. $\endgroup$ – mike_dole_z3 May 29 at 9:15
  • $\begingroup$ $|x\rangle \otimes |f(x)\rangle$ is also disentangled. So the result will be exactly $ (U|x\rangle) \otimes \vert f(x)\rangle$. You can compute the amplitudes $\alpha_i$ of $U|x\rangle = \sum_i \alpha_i |i\rangle$ and then take the product $\otimes$ with $|f(x)\rangle$. $\endgroup$ – Danylo Y May 29 at 9:44
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$\newcommand{\ket}[1]{|#1\rangle}\ket{x}\ket{y}$ is a pure state, not mixed, and is a product state, which is not entangled by definition, so your example is rather confusing.

To answer the question in your first sentence, applying a unitary operator $U_A$ to one subsystem of a bipartite system is equivalent to applying the operator $U_A\otimes\mathbb{1}_B$ to the whole system, where $\mathbb{1}_B$ is the identity for subsystem $B$, so you can treat $U_A\otimes\mathbb{1}_B$ as one big unitary and apply it how you normally would. If you have a pure state $\ket\psi = \sum_i \psi_i \ket{i}_A \otimes \ket{i}_B$, this means that applying such unitary gives $\sum_i \psi_i (U_A \ket{i}_A) \otimes \ket{i}_B$, and analogously for a mixed state.

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  • $\begingroup$ Thanks. Please take another look, I've updated the question. $\endgroup$ – mike_dole_z3 May 29 at 9:18

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