1
$\begingroup$

Given two entangled qubits in the state ${|00⟩+|11⟩\over \root \of 2}$, I have trouble getting the way of applying a 1-qubit gate (let's say a Hadamard) to 1 of the entangled qubits (let's say the first one). Actually I have several questions:

  • Following this link, Ran G. says that "You can apply a quantum gate on any subset of the qubits."
    On the other, hand Agent Gotse says here that "You can not compute the Hadamard gate on only one qubit, as the state is entangled and cannot be factored into two independent qubits."
    Do the answers contradict themself?

  • I understand part of the principle of the entangled pair, it a superposition of bipartite system. But given for example an entangled pair $|00⟩+|11⟩ = |0⟩ \otimes |0⟩ + |1⟩ \otimes |1⟩$, and two unitary operators $A\ B$ to appy to the system.
    Why can't we juste do $A|0⟩ \otimes B|0⟩ + A|1⟩ \otimes B|1⟩$?
    I mean what is the mathematical meaning of that?

  • Finally, the superdense coding use an entangled pair, and in the procedure given in the link, Alice apply some operator to her qubit without changing Bob's one (because it's probably far far away...). Quantum Computing textbooks say that in the case of an entangled pair, Alice has to apply $U \otimes I$, with $U$ the chosing operator.
    But Alice has only 1 qubit, and clearly $U \otimes\ I$ acts on 2 qubits (it's a $4*4$ matrix), how it's works physically?

$\endgroup$
1
$\begingroup$

I think you're confusing people talking about what must be done theoretically (to compute operations on qubits on paper) with people talking about what must be done experimentally (to operate on qubits in the real world).

Do the answers contradict themself?

No. The person who appeared to be saying "you can't apply the operation" just meant that the algebra you do to apply the operation won't start by factoring the state. When doing algebra you can't apply the operation in the simple way that you'd use when the qubits factor.

Why can't we just do $A|0\rangle \otimes B |0\rangle + A|1\rangle \otimes B |1\rangle$?

You can! $A|0\rangle \otimes B |0\rangle + A|1\rangle \otimes B |1\rangle$ is equivalent to $(A \otimes B) (|00\rangle + |11\rangle)$. To apply a Hadamard operation to one qubit while leaving the other alone, set $A=H$ and $B=I$.

Alice has only 1 qubit, but U⊗I acts on 2 qubits. How does it work physically?

Alice just does whatever she would do to apply $U$ to her single qubit. That applies the operation $U \otimes I$ to the system of two qubits.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.