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3

Gates don't depend on states they act on. So it doesn't matter if the state is entangled or not. The gate you've described (let's call it $U$) acts on a standard basis by this rules $$ U |0x0\rangle = |0x0\rangle $$ $$ U |0x1\rangle = |1x0\rangle $$ $$ U |1x0\rangle = |0x1\rangle $$ $$ U |1x1\rangle = |1x1\rangle $$ for each $x=0$ and $x=1$. You can ...


2

So is it possible that a chip manufacturer (who wishes to spy) can pretty much entangle all the particles on the chip so that what ever processing goes on inside that chip is mirrored miles away by the spy manufacturer? No, not really. The big reason this can't be done is that quantum entanglement doesn't allow for communication. Suppose that there's a ...


3

Yes, and no. (Helpful answer, I know) If you rely on buying the entire package from a single supplier, then yes. This doesn’t need to have anything to do with quantum at all. A quantum computation starts and ends with normal classical bits on a normal classical computer. If that was supplied as part of the package, it can easily have some malware built in ...


1

It is not very easy to explain how a quantum computer works. Your question is very broad. It can take even whole book to explain everything you want to know clearly, so this forum is probably not the best place to do so. I would recommend reading this book: Hidden In Plain Sight 10: How To Program A Quantum Computer. I am self learner in quantum computing ...


5

Quantum states are highly unstable and subject to noise, so with current levels of technology it is unlikely that this entanglement would last for any useful amount of time. But let's assume that you have two entangled fault-tolerant qubits (one owned by a malicious manufacturer), and that they are in superposition (in the Z-basis): $$ \frac{1}{\sqrt{2}} (|...


2

Just some notes: 1) A symbol for tensor product is $\otimes$, so I edited your answer so. 2) After CNOT, you do not have to calculate firstly tensor product $H \otimes I$ and then $I \otimes H$ (by the way, you calculated $H \otimes I$ instead of $I \otimes H$ in step you denoted 2nd H) and simply use $H \otimes H$ 3) Regarding your comment "two Hadamards ...


0

Your end state is not entangled. As you have written, the state vector after the gate is the same as the one before the gate. And your starting state is obviously not entangled. As for the rest of your question, what would be the 'something' you are looking for in terms of 0's and 1's?


7

If the simulator is saying that state 00 occurs 75% of the time then the simulator has a bug. Reordering measurements can't make certain outcomes more likely in that way. It would violate the no communication theorem.


1

The fastest way I know to check if a given qubit is not entangled is to check if all its conditioned-on-others states are parallel. A conditioned-on-others state is a subset of your state vector where the other qubits are equal to some specific value. For example, the first two entries in your column vector are the state of one qubit conditioned on the ...


1

Correct, the final state is not an entangled state. 1/2(|00> + |01> + |10> + |11>) Therefore it represents a tensor product of two qubits' state. 1/sqrt(2)(|0> + |1>) <tensorproduct_symbol> 1/sqrt(2)(|0> + |1>) For both qubits, alpha = beta = 1/sqrt(2) This wouldn't have been possible if the qubits were entangled.


3

Let’s say that Alice, Bob, and Charlie hold random bits, which are either all or all (so, they’re classically correlated). If all three of them get together, they can see that their bits are correlated, and the same is true if only two of them are together. If Alice, Bob, and Charlie all have either the bits $\{1_A, 1_B, 1_C\}$ or the bits $\{0_A, 0_B, 0_C\}...


3

To measure, observe that you are simply projecting a quantum state onto some basis set of vectors. First, I will note that this state is not normalized. Let us first define the following quantum state. $$|\psi_i\rangle = \begin{pmatrix}1\\-1\\0\\0\end{pmatrix}.$$ Then, calculating the corresponding probability yields: $$|\langle \psi_i|\psi_i\rangle|^2 = (...


3

[Can't comment and so, writing an answer] As @Sanchayan aptly pointed the theoretical/mathematical arguments which could indicate that superposition is not just an assumption. A couple more thoughts: You're right in observing that the superposition collapses to 0 or 1 when observed and that's expected due to the collapse of the wave function. A quick ...


4

Whether or not quantum superposition is a "truth", is a philosophical question. Quantum theory is simply an axiomatic mathematical model of the universe that happens to give correct experimental predictions (at least, for a fairly broad range) for several physical phenomena. It certainly might be possible to come up with a different mathematical model of the ...


1

"Assume that there is a setup with Bob which creates some sort of interference of the qubits he receives." I think you need to be more specific on this point. From Bob's perspective his qubit is simply a mixed state in the computational basis $$\rho_B=\frac{1}{2}|0\rangle\langle 0|+\frac{1}{2}|1\rangle\langle1|$$ so he will not be able to perceive any ...


1

Bob does not "see" the interference in the sense you assume. To "see" the interference (actually the states of his qubits) Bob needs to measure the qubits, and the measurement gives Bob no information whether Alice measured her qubits or not.


4

Note that $H | 0 \rangle = | + \rangle = \frac{1}{\sqrt{2}}(| 0 \rangle + | 1 \rangle)$. So, after the two Hadamard gates the state will be $$ | 0 \rangle \otimes H| 0 \rangle \otimes H| 0 \rangle = \frac{1}{2} | 0 \rangle \otimes (| 0 \rangle + | 1 \rangle)\otimes (| 0 \rangle + | 1 \rangle) = $$ $$ = \frac{1}{2} (| 000 \rangle + | 001 \rangle + | 010 \...


3

Yes you are absolutely right with your understanding but here is the game: All the Qubits are starting with the |0> i.e. value 0 until the Hadamard gate is applied. The Hadamard gates just makes the probability of the Qubit being 0 or 1 to be 50-50 i.e $\frac{1}{\sqrt{2}}|\mathbf{0}> + \frac{1}{\sqrt{2}}|\mathbf{1}>$ Now in the second example you ...


3

As indicated by Danylo in his anwser, eq. (32) in arXiv: 1103.2030 presents the sixteen vectors ("ignoring overall phases and normalisation") \begin{equation} \left( \begin{array}{cccc} x & 1 & 1 & 1 \\ x & 1 & -1 & -1 \\ x & -1 & 1 & -1 \\ x & -1 & -1 & 1 \\ i & x & 1 & -i \\ i & x & -...


3

You can find it here Symmetric Informationally Complete Quantum Measurements or here SIC-POVMs: A new computer study, in the appendix B. Update Given a single fiducial vector $v = (a_1,a_2,a_3,a_4)^T \in \mathbb{C}^4$ it's pretty easy to write down all SIC-POVM vectors. They are just $C^kS^lv$ for $k,l \in \{0..3\}$, where $C$ and $S$ are clock and shift ...


2

Entanglement is a quantum physical phenomenon, demonstrated in practical experiments, mathematically modeled in quantum mechanics. We can come up with several creative speculations of what it is (philosophically), but at the end of the day we just have to accept it and trust the math. From a statistics point of view we can think of it as a complete ...


1

The way that I approach the calculation is quite different. I think of $$ e^{i\frac{\theta}{2}\sum_n X_i}=\prod_ne^{i\frac{\theta}{2}X_i}. $$ So, this means that you can consider the term $$ e^{i\frac{\theta}{2}\sum_n X_i}S_Me^{-i\frac{\theta}{2}\sum_n X_i} $$ on a qubit-by-qubit basis. You'll have terms like $$ e^{i\frac{\theta}{2}X}Ze^{-i\frac{\theta}{2}X},...


2

Here is a circuit for preparation of state $\frac{1}{\sqrt{2}}(|000\rangle + |111\rangle)$. And here one for state $\frac{1}{\sqrt{3}}(|001\rangle + |010\rangle+ |100\rangle)$


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