What happens when an operator is applied only to some bits of a mixed state?

Question

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 $f(x)$

Answer 1

$|x\rangle|y\rangle = |x\rangle\otimes |y\rangle$ 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, $(U\otimes I) \rho (U^\dagger\otimes I)$.

Answer 2

$\newcommand{\ket}[1]{|#1\rangle}\ket{x}\ket{y}$ 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.