9

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 $$ h_{ijkl} = \frac{1}{4}\text{Tr}\big((\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l)^\dagger \cdot A\big) = \frac{1}{4}\text{Tr}\big((\sigma_i\otimes\sigma_j\otimes\sigma_k\otimes\sigma_l) \cdot A\big) $$ ...


8

Marsl is correct, and his "hint" is really more a sketch of a solution than a hint. Rather than viewing the question or its solution as just formal algebra, you can also approach his same solution more conceptually. The conceptual reasoning is really identical to the algebra, just phrased differently. You can rely on the following two facts: 1) Trace ...


7

Imagine you have a vector that can be written in the form $$ |\psi\rangle=\sum_{i=0}^{d_A-1}\sum_{j=0}^{d_B-1}c_{ij}|i\rangle|j\rangle. $$ The coefficients can be arranged as a $d_A\times d_B$ matrix $C$, with the elements $c_{ij}$ (in your special case, you're talking about setting $d_A=d_B=\sqrt{m}$). Now, if you calculate $\rho_A=CC^\dagger$, this is ...


7

First of all, note that the statement, as written, is wrong (or rather, it is correct only as long as the "$\equiv$" symbol is taken to mean "equal up to a phase"). An easy way to see it is by computing the determinant of $H=e^{i\pi H/2}$, which gives $-1=1$ (using $\det[\exp(A)]=\exp[\operatorname{Tr}(A)] $ for all $A$ and $\operatorname{Tr}(H)=0$). Now, ...


7

Hint: To make your induction work, write $$\eqalign{p^{\otimes n} - q^{\otimes n} & = & \left(p^{\otimes(n-1)}\otimes p \right)-\left(q^{\otimes (n-1)} \otimes q\right)\\ & = & \left(p^{\otimes(n-1)}-q^{\otimes (n-1)} \right)\otimes p+\left(q^{\otimes (n-1)} \right) \otimes (p-q)}$$ Then, use triangle inequality and finally the fact that ...


7

$\newcommand{\bs}[1]{{\boldsymbol #1}} \newcommand{\tildebssigma}{\tilde{\bs\sigma}} \newcommand{\bssigma}{{\bs\sigma}}$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 $\...


7

A couple of points: The ground state is by definition the eigenvector associated with the minimum valued eigenvalue. Lets consider the Pauli Z matrix as you have. First, \begin{align*} Z = \begin{pmatrix}1 & 0\\ 0 & -1 \end{pmatrix}. \end{align*} As this matrix is diagonal, we can immediately see that the eigenvalues are the values on the main ...


6

In the paper I called it the Burnside decomposition, but it looks like the standard name is the Wedderburn decomposition. That might simply have been a mistake in terminology on my part. Anyway, there are two good ways to get the summands to be $V \otimes V^*$. (Of course they are closely related.) 1) You can interpret $\mathbb{C}[G]$ as an associative ...


5

For questions like this, the conventional physics notation is easier to work with than the QIT gate notation. Define $\vec \sigma = (\sigma_1,\sigma_2,\sigma_3)$ to represent the three Pauli matrices $$\sigma_1 = X = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}, \;\;\; \sigma_2 = Y = \begin{bmatrix} 0 & -i \\ i & 0 \end{bmatrix}, \;\;\; \...


5

It appears that you have some confusion regarding the basic notions of density operators and "dimension". Why $d^2$ dimensions "are required to describe" a density matrix isn't the right question to ask; density matrices are $d^2$ dimensional objects in the same sense that vectors in $\Bbb R^3$ are 3-dimensional (i.e., the cardinality of any basis set of $\...


5

A matrix function $f(A)$ for normal matrix $A$ is defined as follows \begin{equation} f(A)=\sum_{i=1}^{n}f(\lambda_i)v_iv_i^T \end{equation} where $\lambda_{i}$ is an eigenvalue and $v_{i}$ is coresponding eigenvector (note: transposed vector $v_{i}$ is a row vector). In your case: $f(A) = \mathrm{e}^A$ and $A = -i\frac{\phi}{2}\sigma_{2}$.


5

Quantum computing, in general, does not use Heisenberg's matrix mechanics or Dirac's interaction picture. I suggest that you carefully read the Wikipedia page on dynamical pictures. The differences between the Heisenberg picture, the Schrödinger picture and Dirac (interaction) picture are well summarized in the following chart. The development of matrix ...


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 ...


5

Inner products on Hilbert spaces are linear in their first argument and conjugate linear in their second argument. So Hilbert spaces also have the property $$\langle f, \, cg\rangle = c^\ast \langle f, \, g \rangle.$$ As you said, a bra represents the dual space to a ket. So to move to an inner product in Dirac notation simply note that for $c_1, \, c_2 \...


5

For any problem which is described by an infinite dimensional Hilbert space, you can regard any state as a superposition of an infinite number of states. The only real question is thus whether infinite-dimensional Hilbert space are "used in practice".


5

The shift operator takes his name from the fact that it shifts the position of its input, as in, it sends $1\to2$, $2\to3$ etc, with the last computational basis element being sent back to the first one: $d\to 1$ (or the same thing starting with $0$, depending on notation). As per the "boost" operator $Z$, I have usually seen those referred to as "clock ...


5

The orthonormal basis $|j\rangle$ of the $d$ dimensional finite Hilbert space corresponds to a configuration space of equally spaced clockwise ordered $d$ points on a circle $S^1$ or equivalently, the vertices of a $d$-dimensional regular polygon. One may think of a point as a discrete location of a particle, then the shift operator $X$ shifts the particle ...


4

A quantum bit can be represented by a two-level quantum mechanical system, and described by a state vector in two-dimensional Hilbert space. Traditionally, Dirac, or bra-ket notation has been used to represent them. The two computational basis states are therefore often written as |0〉 and |1〉 (pronounced: 'ket 0' and 'ket 1' respectively). A pure qubit state ...


4

You seem to be overcomplicating this somewhat! You are right to split it up into the two terms $H_1$ and $H_2$. So, we have $$ e^{-i(H_1+H_2)t}=e^{-iH_1t}e^{-iH_2t}. $$ Now, straightforwardly, $$ e^{-iH_1t}=I+(e^{-i\delta t}-1)|00\rangle\langle 00|. $$ Next, we need to think about the $e^{-iH_2t}$ term. Of course, it maps $|00\rangle$ to $|00\rangle$. So, ...


4

If you wish to distinguish two states $|\psi\rangle$ and $|\phi\rangle$, you can only guarantee to do this if $\langle\psi|\phi\rangle=0$. You do this by measuring in a basis defined by the two states (alternatively, you apply a unitary $U$ such that $$ U|\psi\rangle=|0\rangle,\qquad U|\phi\rangle=|1\rangle, $$ and then measure in the standard $Z$ basis. ...


4

The reason is relatively straightforward. Consider an $m$ dimensional vector space $V$ with basis $\lbrace \vert v_1 \rangle,...,\vert v_m \rangle \rbrace$, and an $n$ dimensional vector space $W$ with basis $\lbrace \vert w_1 \rangle,...,\vert w_n \rangle \rbrace$. As your intuition suggests, we can naturally express any element $A \in V \otimes W$ in the ...


4

Suppose $\lambda_0 = 1$ and the rest are $0$. $$ F_Q [\rho,A] = 2 \sum_{k,l} \frac{(\lambda_k-\lambda_l)^2}{\lambda_k + \lambda_l} | \langle k |A| l \rangle |^2\\ = 2 \sum_{k=0,l \neq 0} \frac{(1-0)^2}{1 + 0} | \langle 0 |A| l \rangle |^2 + 2 \sum_{k\neq 0,l = 0} \frac{(0-1)^2}{0 + 1} | \langle k |A| 0 \rangle |^2\\ = 4 \sum_{l \neq 0} | \langle l |A| 0 \...


4

If you write $Z^aX^b$, there's an implicit "add a phase $i$ to make it Hermitian if necessary", although I guess there are a couple of different conventions you might use the determine the sign used. So long as you're clear about the convention it doesn't really matter because you've got the extra $\pm 1$ freedom to add in to adjust for it. As for an ...


4

The metric you're looking for is known as entangling power. Here are some references: Entangling power of two-qubit gates on mixed states (Guan et al., 2014) Entangling power of two-qubit unitary operations (Yi Shen & Lin Chen, 2018) I will expand on this later.


4

The model's accuracy is purely empirical observation. The error trend (Fig 4, or 41:50 in the video) demonstrates that the error of the system (cross entropy fidelity with respect to simulated results) is tracked closely by the "high school probability" model he mentions. The way this basic model would work is to assume 1- and 2-qubit gate errors are ...


4

Depends on what you mean by SWAPN, that is what qubits are swapped. Your SWAP3 gate in Dirac notation is $$|000\rangle\langle 000|+|001\rangle\langle100|+|010\rangle\langle010|+|011\rangle\langle110|+|100\rangle\langle001|+|101\rangle\langle101|+|110\rangle\langle011|+|111\rangle\langle111|$$ that is the first and third qubits are swapped; assuming SWAPN ...


4

$$ \rho = \begin{pmatrix} a & 0 & 0\\ 0 & 1-a & 0\\ 0 & 0 & 0\\ \end{pmatrix}\\ \sigma = \begin{pmatrix} b & 0 & 0\\ 0 & c & 0\\ 0 & 0 & 1-b-c\\ \end{pmatrix}\\ Tr (\rho^2 ) = 2 a^2 - 2a + 1\\ Tr (\sigma^2 ) = b^2 + c^2 + (1-b-c)^2 $$ Let's pick $a=1/2$ so $2a^2-2a+1=\frac{1}{2}$. $$ b=\frac{2}{3}\\ c=\frac{1}...


4

The other answer already gave a counterexample. From a geometrical point of view, the question is about the intersection of hyperplanes with hyperspheres. Indeed, the purity of a state $\rho$ with eigenvalues $(p_i)_i$ is $\sum_i p_i^2=\|\boldsymbol p\|^2$, whereas these are probabilities and therefore also constrained by $\sum_i p_i=1$. The set of ...


4

As you say, start by expanding $e^{-iS\Delta t}=\cos(\Delta t)I-i\sin(\Delta t)S$, so you'd be calculating $$ (\cos(\Delta t)I-i\sin(\Delta t)S)\rho\otimes\sigma(\cos(\Delta t)I+i\sin(\Delta t)S). $$ If you multiply out all the terms, then the $\cos^2(\Delta t)$ comes from the two $I$ terms, leaving you with $\rho\otimes\sigma$. If you trace out the first ...


4

Your statements about bases are correct. For a qubit, any two distinct states can be used as a basis such that linear combinations of them can describe any state that you want. (If you want to describe something larger, you need more states.) First, a comment about the term "distinct". Are states $$ \frac{|0\rangle+|1\rangle}{\sqrt{2}},\qquad{\text{and}}\...


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