0
$\begingroup$

enter image description here

For question 7, Suppose that there exist such a reversible quantum process such that $$\lvert\varphi^{\perp}\rangle=U\lvert\varphi\rangle$$ where $U$ is a unitary matrix. Then,$$ 0 = \langle\varphi\lvert\varphi^{\perp}\rangle=\langle\varphi\lvert U \lvert\varphi^{\perp}\rangle$$ However, I dont know how to arrive a contradiction after that

For question 8, note that the the outcomes of the two measurements are different and all of the bases are ONB. For example in basis {$\lvert0\rangle$,$\lvert1\rangle$}, if alice get the state 0 in the first measurement, then the chance for her to get the different state(state 1) in the next measurement on the same qubit is 0(as the basis is ONB). Therefore, she definitely preformed the two measurements on two different qubits.

Would really appreaciate if anyone could tell me some hints or my attempts are correct or not.

$\endgroup$
2
  • $\begingroup$ What is the source of these exercises? $\endgroup$ – uniquesolution Mar 11 '19 at 20:34
  • $\begingroup$ What are you doing here: $\langle\varphi\lvert\varphi^{\perp}\rangle=\langle\varphi\lvert U \lvert\varphi^{\perp}\rangle$? I don't think you can simply pop in $U$. $\endgroup$ – Jens Wagemaker Mar 11 '19 at 21:00
0
$\begingroup$

For exc 7:

Since $U$ is a unitary operator it has an eigenvector $|\phi>$ with eigenvalue $\alpha$ such that $|\alpha| =1$.

Now: $ 0 = |<\psi|\psi^\perp>| = |<\psi|U\psi>| = |<\psi|\alpha\psi>| =|<\psi|\psi>|\;|\alpha| = 1$, contradiction.

For exc 8, suppose it where two qubits and we measure in the base $|0>, |1>$. We get different outcomes for each of the two measurements everytime. This means that the state will be of the form $\alpha |01> +\beta |10>$. I think that if you transfer this to one of the other bases (I don't know which, but you can consult the wiki on Bell states) you get a state where they are perfectly correlated. I.e. you get the same value for both measurements, which is a contradiction.

So if it are not two qubits, then what is happening? I think that after the first measurement. The qubit is put through a quantum NOT gate. Which then gives the opposite result in a second measurment.

$\endgroup$

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.