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

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 $$f(x)$$

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

• Thank Danylo. Please take another look, I've updated the question. – mike_dole_z3 May 29 at 9:15
• $|x\rangle \otimes |f(x)\rangle$ 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 $|f(x)\rangle$. – Danylo Y May 29 at 9:44

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

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