17

Specific Circuit The first gate is a Hadamard gate which is normally represented by $$\frac{1}{\sqrt{2}}\begin{bmatrix}1&1\\1&-1\end{bmatrix}$$ Now, since we're only applying it to the first qubit, we use a kronecker product on it (this confused me so much when I was starting out - I had no idea how to scale gates; as you can imagine, it's rather ...


8

For any matrix $A$ we can write $$ A =\sum_{i,j,k,l}h_{ijkl}\cdot \frac{1}{4}\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l, $$ where $$ h_{ijkl} = \frac{1}{4}\text{Tr}\big((\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l)^\dagger \cdot A\big) = \frac{1}{4}\text{Tr}\big((\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l) \cdot A\big) $$ ...


7

From IBM Q Documentation (the link is hard to find) here is the definition of the generic gate: $$ U(\theta, \phi, \lambda) = \begin{pmatrix} \cos\left(\frac{\theta}{2}\right) & -e^{i\lambda} \sin\left(\frac{\theta}{2}\right) \\ e^{i\phi} \sin\left(\frac{\theta}{2}\right) & e^{i(\lambda + \phi)} \cos\left(\frac{\theta}{2}\right) \end{pmatrix} $$ ...


7

The Hilbert space dimension of $n$ qudits is $d^n$, where $d$ is the dimension of the qudit ($d=2$ for qubit, $d=3$ for qutrit, etc). So three qubits have an $8$ dimensional space, two qutrits have a $9$ dimensional space, and one $d=6$ qudit has a six dimensional space. As such, we cannot regard them as equivalent. I guess you meant to compare situations ...


6

The CNOT gate is a 2-qubit gate, and consequently, its operation cannot be expressed by the tensor product of two one-qubit gates as the example you gave with the Hadamard gates. An easy way to check that such matrix cannot be expressed as the tensor product of two other matrices is to take matrices $A =\begin{pmatrix}a & b \\ c & d\end{pmatrix}$ $...


6

This is actually a much easier problem. In the case of states, you're trying to use the PPT criterion, or others, to distinguish if $\rho$ can be written in the form $$ \rho=\sum_ip_i\sigma^A_i\otimes\sigma^B_i, $$ where $\sum_ip_i=1$ and the $\sigma^A_i$ and $\sigma^B_i$ are valid states on single sites. The difficulty actually comes from the freedom that ...


6

The Hadamard gate has close ties to the discrete Fourier transform. Consider the DFT for an $N$-level system: $$\vert j \rangle = \frac{1}{\sqrt{N}} \sum\limits_{k=0}^{N-1} e^{\frac{i2 \pi j k}{N}} \vert k \rangle.$$ For $N=2$ this is simply $$\vert j \rangle = \frac{1}{\sqrt{2}}\begin{bmatrix}1 & 1 \\ 1 & -1 \end{bmatrix} \, \vert k \rangle = H \...


6

$\newcommand{\bs}[1]{{\boldsymbol #1}} \newcommand{\tildebssigma}{\tilde{\bs\sigma}} \newcommand{\bssigma}{{\bs\sigma}}$Yes, products of Pauli matrices form a basis for the set of Hermitian matrices (of dimensions that are powers of $2$). More specifically, fix an integer $n$ and let $N\equiv 2^n$, define $\bssigma\equiv(\sigma_x,\sigma_y,\sigma_y)$, and $\...


5

Since your desired operation is a non-injective function, you need a third qubit and a unitary acting on all three qubits. Using an operator on your two input qubits and tensoring this with ${\rm I}_2$ on the third qubit is not going to work as you might as well forget about the third qubit completely if that were the case. By the way, the two matrices you ...


5

All quantum operators must be unitary. Unitary means the conjugate-transpose of the operator is its inverse. In your case: $UU^{\dagger} = \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0\\ 0 & ...


5

The insight that suggests that sparse matrices are useful goes along the lines of: for any $H$, we can decompose it in terms of a set of $H_i$ whose individual components all commute (making diagonalisation straightforward), $$ H=\sum_{i=1}^mH_i. $$ If the matrix is sparse, then you shouldn't need too many distinct $H_i$. Then you can simulate the ...


5

Immediately, we can see that $$ A = |1\rangle\langle0| + |0\rangle\langle1|. $$ If the input and out bases are $\{|0\rangle, |1\rangle\}$, then $$ |0\rangle = \begin{pmatrix} 1 \\ 0 \end{pmatrix}, \quad |1\rangle = \begin{pmatrix} 0 \\ 1 \end{pmatrix} \quad\textrm{and}\quad \langle0| = \begin{pmatrix} 1 & 0 \end{pmatrix}, \quad \langle1| = \begin{pmatrix}...


5

A matrix is positive if and only if it is Hermitian (and thus unitarily diagonalizable) and all its eigenvalues are positive (that they are real follows automatically from it being Hermitian). If this is not the way you define a positive operator, then you need to specify how you do so that we can prove the equivalence. In other words, $A$ is positive, $A\...


5

That's not the right way to look at it. In quantum mechanics, time evolutions are considered to be unitary and any unitary evolution can be written as a sequence of unitary operators $U_1, U_2, U_3,\ldots$ acting on a quantum state $|\Psi\rangle$. Any single-qubit unitary operation is a $2\times 2$ matrix of the form: $$U=\begin{pmatrix}a&b\\-e^{i\phi}b^...


5

Yes, in the circuit the qubit "enters" to the left, and exits to the right, but when applying the gates to a state you must apply the one on the far left first, then the next and so on, so concretely you do write them down right to left, but it's just a consequence of writing the operator that we want to apply on the left of the vector, while our natural way ...


5

You specifically ask about qubits, so I'll keep it to that. Imagine you have a state $$ |\psi\rangle=\sum_{x\in\{0,1\}^n}a_x|x\rangle. $$ You can choose to look at each qubit. I'll take the first qubit for the sake of simplicity. We have that $$ |\psi\rangle=|0\rangle\sum_{y\in\{0,1\}^{n-1}}a_{0y}|y\rangle+|1\rangle\sum_{y\in\{0,1\}^{n-1}}a_{1y}|y\rangle $$ ...


5

Here, what you need to do is to understand writing CNOT gate based on the control qubit. Your first CNOT gate has qubit 1 as control and qubit 2 as target. So, what this means is the second qubit will not be flipped until qubit 1 is set to zero. I am going to use computational basis for this $CNOT_1\left|00\right> = \left|00\right>$, $CNOT_1\left|01\...


5

All tensor products of $n$ Pauli operators $\{I,X,Y,Z\}$ (that is $4^n$ combinations) form an orthogonal basis for the vector space of $2^n \times 2^n$ complex matrices. Hence, for every matrix there is a unique decomposition as a linear combination of tensor products of Pauli unitaries. Same is true if we fix some other unitary basis. If we not fix the ...


4

Mostly I'm confused over whether the common convention is to use +i or -i along the anti-diagonal of the middle 2x2 block. The former. There are two $+i$'s along the anti-diagonal of the middle $2\times 2$ block of the iSWAP gate. See page 95 here[$\dagger$]. [$\dagger$]: Explorations in Computer Science (Quantum Gates) - Colin P. Williams


4

You can use Python with Qiskit. Say your string representation is written using OpenQASM syntax. qasm = """ OPENQASM 2.0; include "qelib1.inc"; qreg q[2]; h q[0]; t q[1]; cx q[0], q[1]; """ You can build a circuit out of this and simulate it on a unitary simulator: import qiskit as qk import numpy as np circuit = qk.load_qasm_string(qasm) result = qk....


4

Consider the linear maps $A: V\to W$ and $B: W\to X$. The composition $BA$ is a linear map from $V$ to $X$. Now, how can $\mathcal{M}(BA)$ be computed from $\mathcal{M}(B)$ and $\mathcal{M}(A)$? $\mathcal{M}(A)$ is the $n\times p$ matrix representation of the linear map $A$ w.r.t the basis $\{v_1,...,v_p\}$ and $\{w_1,...,w_n\}$. $\mathcal{M}(B)$ is the $m\...


4

$$ CNOT = \begin{bmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \\ \end{bmatrix} $$But what does this matrix mean? The above matrix means: on a two qubit system (such as $\left|00\right>$, $\left|10\right>$, $\left|11\right>$, etc.) if the first qubit is a one,...


4

This is called Sylvester's Criterion. There's plenty of information available once you have the name. The linked wikipedia article contains a proof. Strictly, Sylvester's Criterion requires that $W_2,W_3,W_4> 0$ for the state to be positive under the partial transpose. However, for a density matrix, $W_2$ is always positive semi-definite. This is because ...


4

Recall that the trace is both linear and invariant under cyclic permutation of the operators $$ \mathrm{Tr}(\Phi(\rho))=\mathrm{Tr}\left(\sum_j F_j^\dagger \rho F_j\right)=\sum_j\mathrm{Tr}\left( F_j^\dagger \rho F_j\right)=\sum_j \mathrm{Tr}\left(F_jF_j^\dagger \rho \right)= \mathrm{Tr}\left(\sum_jF_jF_j^\dagger \rho \right)$$ You can clearly see that if $...


4

Consider that a control qubit is $q_k$ and a target qubit is $q_{k+n}$ and you want to apply operator $U$ on the target qubit. Denote $N=2^{n+1}$. Then matrix representation of this controlled $U$ is \begin{equation} CU= \begin{pmatrix} I_{\frac{N}{2}} & O_{\frac{N}{2}} \\ O_{\frac{N}{2}} & I_{\frac{N}{4}} \otimes U \\ \end{pmatrix} \end{equation} ...


3

If we write $$ U=\sum_{i,j}U_{ij}|i\rangle\langle j|\quad V=\sum_{kl}V_{kl}|k\rangle\langle l|, $$ and $$ |x\rangle=\sum_jx_j|j\rangle\quad |y\rangle=\sum_ly_l|l\rangle, $$ then we can evaluate both sides of the equation $$ (U\otimes V)(|x\rangle\otimes|y\rangle)=(U|x\rangle)\otimes(V|y\rangle) $$ using the definition of the tensor product as $$ U\otimes V=\...


3

I will give you a few elements for the demonstration on real vectors which you can extend to complex. Let {$e_i$} be the standard basis for the space where $U (n*n)$ is defined . Let {$e_j$} be the standard basis for the space where $V (m*m)$ is defined. First, it is a property that the basis {$e_i \otimes e_j$} is a basis for the n*m-matrices space. $ U \...


3

$$ | \psi_3 \rangle = a | 0 1 1 \rangle + b | 1 1 1 \rangle\\ $$ Because the 1 on B and C criterion is met. $$ | \psi_4 \rangle = a | 0 0 1 \rangle + b | 1 1 1 \rangle\\ $$ Because only the first term meets the criterion for the controls so it is the only part affected to flip the B index. $$ | \psi_5 \rangle = a | 0 0 0 \rangle + b | 1 1 1 \rangle\\ $$ ...


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