New answers tagged

2

No, there is no way to do this with probability better than 1/2. Basically, you are saying that, ideally, you'd like a protocol where $$ |0\rangle\mapsto|0\rangle,\quad |1\rangle\mapsto|0\rangle,\quad \frac{|0\rangle+e^{i\theta}|1\rangle}{\sqrt2}\mapsto|1\rangle\ \forall\theta. $$ We can get an upper bound on the achievable fidelity using the operator $$ R=\...


3

First, in order to express $|H'\rangle$, $|V'\rangle$, $|R\rangle$ and $|L\rangle$ in terms of $|H\rangle$ and $|V\rangle$, add and subtract the pair of equations $(2)$ and add and subtract the pair $(3)$, to get $$ |H'\rangle + |V'\rangle = \sqrt2|H\rangle \\ |H'\rangle - |V'\rangle = \sqrt2|V\rangle \\ |R\rangle + |L\rangle = \sqrt2|H\rangle \\ |R\rangle - ...


1

Given a quantum state $|\psi\rangle$, you can performa a basis change on this state using the $I$, the Identity operator, which of course can be expanded as $\sum_{i}|i\rangle\langle i|$, where $|i\rangle$ is simply a particular basis from a complete set of basis states. Applying $I$ to $|\psi\rangle$, we get $$I|\psi\rangle=\sum_{i}|i\rangle\langle i|\psi\...


2

Personally, I would do the calculation a little differently. Start by writing $$ \langle\psi|X^0_1Z_2X^0_1|\psi\rangle+ \langle\psi|X^1_1Z_2X^1_1|\psi\rangle= \langle\psi|X^0_1Z_2X^0_1+X^1_1Z_2X^1_1|\psi\rangle $$ Next, think about a term like $X^0_1Z_2X^0_1$, but expand out the tensor product. This is just $(X^0X^0)\otimes Z$. But since $X^0$ is a projector,...


1

From this paper, suppose you have three bases $S_x$, $S_y$ and $S_z$. And suppose you define the states of the $S_x$ and $S_y$ bases on the $S_z$ basis as follows: $$ |+x\rangle = \frac{1}{\sqrt{2}}(|+z\rangle + |-z\rangle) \\ |-x\rangle = \frac{1}{\sqrt{2}}(|+z\rangle - |-z\rangle) \\ |+y\rangle = \frac{1}{\sqrt{2}}(|+z\rangle + i|-z\rangle) \\ |-y\rangle = ...


4

Like you said, one reason why it's hard to do this in practice is simply due to the number of gates. Each gate comes with limited fidelity (never exactly 100%), so needing to do many of them will compound the total gate error. Moreover, for the GHZ state, notice that the whole state will dephase if even a single qubit dephases. Generally, the larger the ...


1

From quantiki we have $$\rho=p_{\mathrm{sym}}\frac{2\hat{P}_{\mathrm{sim}}}{d^2+d}+(1-p_{\mathrm{sym}})\frac{2\hat{P}_{\mathrm{anti-sym}}}{d^2-d}, $$ where $p_{\mathrm{sym}}$ is the probability of being in the symmetric subspace, $d$ is the dimension of the state vectors, the projectors onto the symmetric and anti-symmetric subspaces are given by $$\hat{P}_{\...


2

To understand what entanglement is, as a crude description it means if you measure one part of the system, then knowing what the measurement outcome is will change what you expect for measurement outcomes on the remainder of the system. This is true for both your examples, in both cases if you don't measure qubit 1, you expect to measure 0 or 1 with equal ...


5

All of your math is correct. Both of the states that you calculated are entangled states. An entangled state is one in which there exist measurable properties of subsystems that are correlated. In (1), measurements of the two qubits in the computational basis are correlated; it just happens to be a negative correlation. In (2), measurements in the ...


3

I'm going to label my 3 parties by A, B, C (just to be awkward). If we consider a bipartition of A|BC, we know that a general state can be written in the form $$ U_A\otimes I_{BC}(\alpha|0\rangle_A|\psi_0\rangle_{BC}+\beta|1\rangle_A|\psi_1\rangle_{BC}) $$ where $|\psi_i\rangle$ are mutually orthogonal and may be entangled. This is just the Schmidt ...


2

Another interesting example is a single-photon state in a superposition of two different spatial (or any other type of degree of freedom really) modes: $$\frac{1}{\sqrt2}(a_1^\dagger + a_2^\dagger)|0\rangle\equiv \frac{1}{\sqrt2}(|1\rangle+|2\rangle).$$ This type of states is sometimes not considered "entangled", as there is only a single particle ...


2

You may be interested in this paper Effcient quantum algorithms for GHZ and W states, and implementation on the IBM quantum computer. The paper provides general method how to prepare $n$-qubit W state. Concerning your question on how place the state's qubits to different location, you can prepare W state on qubits from $q_{1}$ to $q_n$ and then use SWAP ...


2

It absolutely depends on the subdivision of the spaces. Take the 3-qubit system (qubits A, B and C) in a state $$ |0\rangle_A(|00\rangle+|11\rangle)_{BC} $$ We can partition these qubits in various different ways. Clearly the $A|BC$ partition has no entanglement across the partition, while the $AB|C$ partitioning is maximally entangled. Note that if you go ...


1

I believe your problem is that you're splitting up your measurement observable too much. Yes, $M\otimes I$ has four eigenvalues, $-1,-1,1,1$. BUT, when you make the measurement, you do not get 4 difference answers. You only get 2. To see why what you've done cannot work, note that the choice of eigenvectors is not unique. So how can you sub-divide the ...


4

Recall that the W state may be defined as: $$\vert W\rangle=\frac{1}{\sqrt 3}(\vert 001\rangle+\vert 010\rangle+\vert 100\rangle.$$ Given three qubits, initially to prepare such a state local operations (wherein at least two of the three qubits are at the same location) will need to be performed. Depending on your background, see, for example, this question ...


2

If I may, I wanna share an algorithmic generalization for the inspired @Danylo Y answer. This considers also The Algorithmic Method in this answer here related to the CNOT gate. In order to algorithmically build a 2-qubit SWAP gate to operate swap within n-qubits, one could define an array containing n ID gates, i.e. ID=((1,0),(0,1)) matrices, named here as ...


1

For an arbitrary input product pure state $|\psi_1\rangle\otimes|\psi_2\rangle$, using the notation $|\psi_i\rangle\equiv\alpha_i|0\rangle+\beta_i|1\rangle$, we have \begin{align} |\psi_1\rangle\otimes|\psi_2\rangle &= \alpha_1\alpha_2 |00\rangle + \alpha_1\beta_2 |01\rangle + \beta_1\alpha_2 |10\rangle + \beta_1\beta_2 |11\rangle, \\ &= (\alpha_1|0\...


3

The CNOT matrix is the same regardless of the input (i.e. regardless of which vector is involved in the matrix-vector multiplication). Your question indicates some discomfort with how CNOT works when the control qubit is in the state: $H|0\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$, but it works just the same as CNOT does with any other control ...


3

Vidal proved that, pure-state quantum computation is efficiently simulable classically if the quantum computer’s state at every time step has amount of entanglement (measured by Schmidt rank) polynomially-bounded (theorem 1 in the linked paper) And if the amount of entanglement grows subexponentially, then it can be classically simulated with sub-exponential ...


4

The definition of separability does not require any restrictions on the purity of the state. However, if $\rho$ is pure then it is separable on some bipartition $AB$ iff it can be written as $$ \rho = \rho_A \otimes \rho_B $$ where both $\rho_A$ and $\rho_B$ are pure states. Here's a sketch of why that's true. Note that if $\rho$ is separable we have $$ \rho ...


2

No, we do not require $\rho_A^i$ and $\rho_B^i$ to be pure. The state is separable any time it can be written in this form with positive weights $p_i$. As a basic example, we can choose a single $p_i$ to be nonzero; this is equivalent to saying that $$\rho=\rho_A^i\otimes \rho_B^i$$ is separable. This is true regardless of the purities of $\rho_A^i$ and $\...


5

I might say: $$\vert\phi\rangle=\frac{1}{\sqrt 2}(\vert 0_A0_B0_C\rangle +\vert 1_A0_B1_C\rangle)= \frac{1}{\sqrt 2}(\vert 0_A0_C\rangle +\vert 1_A1_C\rangle)\otimes \vert 0_B\rangle,$$ because I think that might be intuitively understandable without much parsing. It also might help to label each qubit if I have to give the qubits to Alice, Bob, and Charlie ...


3

One idea is to do polarimetry. By using a polarizing beam splitter, the polarization qubit can have each of its polarization components directed to a different detector for photon counting (ideally a single-photon detector, here). A polarizing beam splitter might send horizontally polarized photons in one direction and vertically polarized photons in another....


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