# Tag Info

5

The quantum circuit model is one of the possible models for quantum computation. It is the most studied, because it is quite practical. Back to the origins, you can look at papers by David Deutsch, like: Quantum computational networks Proc. R. Soc. Lond. A 425, 73-90 (1989) For a slow start, I suggest to read Nielsen and Chuang's book and maybe Scott ...

5

"Why" is quite a nebulous concept that, ultimately, comes back to your understanding of quantum mechanics. For me, the way that I set up QM is with a set of postulates. The first postulate is that quantum states are represented by vectors in a Hilbert space of length 1. A second postulate is essentially that operations are linear. (This is not how ...

4

As Michele Amoretti correctly says, Quantikz is built on top of Tikz. It cannot do anything that Tikz cannot. It is simply intended to provide a wrapper that makes it more convenient for doing the specific tasks associated with drawing quantum circuits. More specifically, it's done in a way that I find most convenient (as the package author!). Mostly it came ...

4

Yes, this observation can be generalized. To start with, let's notice why is Hadamard the transformation required to measure a state $| \psi \rangle$ in the $\sigma_{x}$ basis. This is because it is the unitary intertwiner'' connecting the $\sigma_{x}$ basis to the $\sigma_{z}$ basis (a.k.a. computational basis). Recall that the $\sigma_{x}$ eigenvectors ...

3

We can implement $CT$ using the following circuit: This solutions uses an extra gate which isn't available above $R_I(\pi/8) = \sqrt{T}$ Explanation: We know that $T = \sqrt{S} = Z^{\frac{1}{4}}$. Since $S,T,Z$ are all diagonal matrices hence their controlled versions will also be diagonal matrices $CZ,CS,CT$. Thus $CT=CZ^{\frac{1}{4}}$. If we can implement ...

3

The black (white) dot means a condition that the corresponding qubit should be in $|1\rangle$ state ($|0\rangle$ state) in order to apply the gate. The first circuit implements the Hadamard gate only if the first qubit is in $|1\rangle$ state and the second qubit is in $|0 \rangle$ state (similar discussions can be found here). In other words, if the ...

3

I think your explanation based on the circuit is perfectly adequate. For a more rigorous "proof", why not simply take the output of the circuit? Substitute in $x_i=0$ for all $i$ and see that all the outputs are $(|0\rangle+|1\rangle)/\sqrt{2}$ for $i\neq 1$ and $(|0\rangle+(-1)^{x_1}|1\rangle)/\sqrt{2}$ for $i=1$, exactly as it would be for the ...

3

No, it's not possible to extract digits of the phase like that. It would violate the Holevo bound. In general there's no way to "amplify" single small phase differences into big phase differences, because of linearity.

3

Let's see what QFT does on two qubit (and then on three qubit) computational basis states and try to gain some insights. The QFT action on $|j\rangle$ basis state: $$QFT |j\rangle = \frac{1}{2^{\frac{n}{2}}} \sum_{k=0}^{2^n -1} e^{2 \pi i \frac{jk}{2^n}} |k\rangle$$ where $n$ is the qubit number. Now suppose $n=2$, then: \begin{align*} QFT |00\rangle &= ...

3

It is referring the the "function variable" register of figure 1. It consists of $\log_2N$ qubits, all prepared in the $|0\rangle$ state.

3

Noise effects introduce classical uncertainty in what the underlying state is. A mixed state is a statistical ensemble of several quantum states $|\psi_i\rangle$ (not necessarily orthogonal), with respective probabilities $p_i$. With the state vector you can represent pure states, not mixed ones. Instead, with the density operator you can represent both pure ...

3

As noted by Mateus in the comments, the transformation you are looking for is non-linear. This cannot be done with any matrix transformation. Thus, you will need more qubits, and your solution shows two (+1 scratch qubit) is sufficient. I guess you might wonder if a two-qubit unitary can do it, though? The problem is that the transformation you want to ...

3

The SWAP operator has been widely used to linearize'' polynomial functions of the density matrix. To understand this carefully, consider the following setup: Let $\mathcal{H} = \mathcal{H}_{A} \otimes \mathcal{H}_{A'}$, where $\mathcal{H}_{A} \cong \mathcal{H}_{A'}$, that is, we take two copies of the Hilbert space. The SWAP operator acting on this doubled ...

2

The XOR operation is not a well defined action in quantum computing since it is non-reversible. (For example: $|0 \oplus 0\rangle = |1 \oplus 1\rangle = |0\rangle$) However, XOR is implicit in the CNOT operation, as CNOT$(|a,b\rangle) = |a, a \oplus b\rangle$ Hence to answer my own question: CNOT$(|\psi \otimes \phi\rangle) = \begin{pmatrix} a_1b_1 \\ ... 2 The trick is that you don't need to calculate the inverse of$B$. What you really want to evaluate is $$(\langle 0|\otimes I)(B^\dagger \otimes I)(\text{select}(V))(B\otimes I).$$ So, the point is that you only need$\langle 0|B^\dagger$which is the Hermitian conjugate of$B|0\rangle$, which you know. 2 You do not need to use the density matrix approach. However, as the most general representation of a quantum state, doing so has several advantages. You can simulate noise using just statevectors using probabilistic approaches, eg wavefunction monte-carlo, that converge to the density matrix results in the limit of many repetitions. Along this same thread of ... 2 As I explained in my answer on a previous question of yours, the depolarizing channel is not really 'physical' - actual quantum systems don't really behave that way. So for simulations where you, for instance, investigate the performance of some code against the depolarizing channel, it doesn't really matter what the exact value of$p$is in your simulations.... 2 You can use Toffolis and an ancilla to remove the need for the control on the T, then decompose the Toffolis into T+H+CNOT. Since the second Toffoli is uncomputing the ancilla, it can be replaced by a measurement based uncomputation. 2 In general you need more than$U$and CNOT to implement a controlled-$U$. One approach to constructing a controlled-$U$gate, for arbitrary$U \in \mathbf{U}(2)$, from single qubit and CNOT gates begins by parameterizing$U$in terms of$U(\alpha,\beta,\gamma,\delta)$according to$$U = e^{i\alpha}\begin{bmatrix} \exp\left({-i\frac{\beta+\delta}{2}}\right)\... 2 If you don't have the ability to perform the controlled-not directly between a pair of qubits, then you simply need to swap the qubits to place them onto a pair of qubits which can have a controlled-not applied to them. 2 For interactions between non-nearest neighbour qubits, ancilla qubits are required, together with SWAP gates. The state of one of the (in this case) two qubits is swapped with the ancilla. This operation is repeated until the qubits are NN, and then the interaction can take place. After this is done, then the state of the ancilla is swapped back with the ... 2 This is a very interesting question. Indeed, CP maps - and this includes the operations used in the error correction (measurement and subsequent unitaries) - will always decrease the trace norm. The answer is that if you take a (strictly) contractive map on, say, a qubit, and consider how it acts if you apply it to many qubits, there will always be some ... 1 Very briefly, gate fidelity refers to a way to compare how "close" two gates, or more generally operations, are to each other. As discussed e.g. in (Magesan et al. 2012), if one wants to compare the action of an operation$\mathcal E$and a gate$\mathcal U$on a given state$\rho$, one can define their "gate fidelity" as the quantum ... 1 Fidelity is a distance measure between quantum states. The gate fidelity uses fidelity to decide how noisy a quantum gate is. Take two copies of a state, apply your implementation of a gate on one copy and apply the ideal gate on another copy (this can be done on paper, not in a lab) and compute the fidelity between the two outputs. This is the gate fidelity.... 1 This problem is caused by noise in quantum circuit. When you use a simulator, usually there is no noise (unless it is simulated), so you have ideal quantum computer. However, in reality, a quantum processor is influenced by thermal and electromagnetic noise. Current challenge for quantum computer designers it to shield the noise from qubits. Qubits also ... 1 My understanding of the OP's question is that there is some restriction imposed that a gates can only act on Adjacent Qubits. While this isn't necessary, we can still work with this restriction using SWAP gates to make non-adjacent qubit adjacent. If the Control qubits are$i$and$j$; and target qubit is$k$. Such that$i+1<j$and$j+1<k$. Then we can ... 1 The gate/operator in the brackets is the non-local CNOT gate, frequently used to create bipartite entanglement. Given it itself is a 2 qubit gate, then the tensor of this with the identity is simply a gate that acts on 3 qubits. This gate will take a 3 qubit state and flip the second qubit of this state when the first is$|1\rangle$It will take$|110\rangle ...

1

Some thoughts: A theoretical perspective From a theoretical perspective, the depolarizing channel is the 'standard' (if there is such a thing) or by some means the most applicable. Because the Paulis (together with the identity operator) form a basis for $SU(2)$, if a code can correct the $X, Y$ and $Z$ flips on a certain qubit (and it it is able to ...

1

A Toffoli gate is in fact NAND gate (when target qubit is in state $|1\rangle$, othwerwise it implements AND gate). Since NAND gate allows to implement any classical logical function and hence any classical algorithm, any gate based quantum computer is universal from classical point of view. Note that Toffoli gate can be realized with CNOT, H, S and T gates. ...

1

Not all quantum operations are reversible. Consider for example the quantum channel $\Phi$ defined as $\Phi(\rho)=\operatorname{tr}(\rho)\sigma$. This is a channel that gives as output the same state regardless of the input. This is not a reversible operation, as there is no information in the output that allows you to infer what the input was (you can ...

Only top voted, non community-wiki answers of a minimum length are eligible