10
I suggest two different ways of trying to solve this, which will give you experience of different bits of the formulation of Quantum Information Theory. I'll give examples that are closely related to the question you asked, but are not what you asked so that you still get the value of answering the question yourself.
Long-hand Method
Represent the kets as ...
9
One way order to perform Z rotations by arbitrary angles is to approximate them with a sequence of Hadamard and T gates. If you need the approximation to have maximum error $\epsilon$, there are known constructions that do this using roughly $3 \lg \frac{1}{\epsilon}$ T gates. See "Optimal ancilla-free Clifford+T approximation of z-rotations" by Ross et al.
...
9
$$ CNOT =
\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 1 & 0 \\
\end{bmatrix}
$$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,...
8
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
$A =\begin{pmatrix}a & b \\ c & d\end{pmatrix}$
$...
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
The quantum states that differ by a global phase (i.e., by a complex number multiple which has absolute value of 1) are considered the same quantum state, since they can not be distinguished using any operations or measurements.
Thus, the eigenstates for hω are
$|+\rangle = \frac{1}{\sqrt2}(|0\rangle + |1\rangle)$,
$-|+\rangle = \frac{1}{\sqrt2}(-|0\...
7
Yes, the set of tensor products of all possible $n$ Pauli operators (including $I$) forms an orthogonal basis for the vector space of $2^n \times 2^n$ complex matrices. So see this first we notice that the space has a dimension of $4^n$ and we also have $4^n$ vectors ( the vectors are operators in this case). So we only need to show that they are linearly ...
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
A unitary $U$ and $e^{i\phi}U$, which differs from it by a phase, act exact identically on any quantum state. Thus, they should really be considered the "same" unitary in terms of their action.
You can therefore use $X$ instead of your unitary, which is $-X$. This will have exactly the identical action in any circuit.
(Why is this? There are different ...
6
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, ...
6
What $\sigma^z_i$ means is that you've got a Pauli-$Z$ applied to qubit $i$, and nothing else on the other qubits (i.e. the identity). So, you could expand it as
$$
I^{\otimes(i-1)}\otimes\sigma^z\otimes I^{n-i}
$$
if your system has $n$ qubits. A term such as $\sigma^z_i\sigma^z_j$ is then a product of two of these, which is equivalent to the tensor product ...
5
Expanding on Jalex
Look at what happens on the possible terms.
\begin{eqnarray*}
\mid 0 \rangle \otimes \mid + \rangle &\to& \mid 0 \rangle \otimes \mid + \rangle\\
\mid 0 \rangle \otimes \mid - \rangle &\to& \mid 0 \rangle \otimes \mid - \rangle\\
\mid 1 \rangle \otimes \mid + \rangle &\to& \mid 1 \rangle \otimes \mid + \rangle\\
\...
5
The difference in definitions is from either taking the unitary group or the projective unitary group. That accounts for the constant prefactors of $\pm i$ that are missing.
In lieu of a tikz commutative diagram
\begin{align}
<X,Y,Z> & \hookrightarrow & U(2)\\
\downarrow & & \downarrow\\
<[X],[Z]> & \hookrightarrow & PU(...
5
You get two decompositions for your matrix (let's call it $A$) because you are using two different operatorial bases.
In the first case you are considering the matrix as acting in a space of dimension $3\times 2$, that is, using the operatorial basis $\{\lambda_i\sigma_j\}_{ij}\equiv\{\lambda_i\otimes\sigma_j\}_{ij}$.
In other words, you are computing the ...
5
I like to use the projectors
$$
P_0=|0\rangle\langle 0|\equiv\left(\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right)\qquad P_1=|1\rangle\langle 1|\equiv\left(\begin{array}{cc} 0 & 0 \\ 0 & 1 \end{array}\right)
$$
to express the intuition of the controlled-not when constructing it:
$$
CNOT=P_0\otimes\mathbb{I}+P_1\otimes X.
$$
Here you can ...
5
This does indeed work because this construction works for any matrix $H$ where $H^2=\mathbb{I}$.
$$
R_\theta(H)=e^{-i\theta H/2}=\cos\frac{\theta}{2}\mathbb{I}-i\sin\frac{\theta}{2}H
$$
There are several ways that you could prove this. I think the easiest way is to realise that because $H^2=\mathbb{I}$, then the eigenvalues of $H^2$ are 1, and hence the ...
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 $A = -i\frac{\phi}{2}\sigma_{2}$.
4
Let's say you have a Hamiltonian of the form
$$
H=\sigma_1\otimes\sigma_2\otimes\sigma_2\otimes\ldots\otimes\sigma_n
$$
There's a straightforward circuit construction that lets you implement its time evolution $e^{-iHt}$. The trick is basically to decompose the state that you're evolving into the components that are in the $\pm 1$ eigenspaces of $H$. Then, ...
4
This looks essentially similar to the property of non-commutativity of the Kronecker product: $X\otimes \lambda_6\neq \lambda_6\otimes X$:
$$X\otimes\lambda_6 = \begin{pmatrix}0&1 \\1&0\end{pmatrix}\otimes \begin{pmatrix}0&0&0 \\0&0&1 \\0&1&0\end{pmatrix} =
\begin{pmatrix}0&0&0&0&0&0 \\
0&0&0&...
4
There is a very closely related representation of the tableau representation of Aaronson (and Gottesman), which works not only for qubits but for qudits of arbitrary finite dimension, which works particularly well for purely Clifford circuits (i.e. at most one terminal measurement).
In this alternative representation, one has tableaus describing how ...
4
Note that the second definition actually doesn't make more sense in the context of the stabiliser formalism, as neither of $\pm i Y$ have a +1 eigenspace. That means that you can only describe states which are stabilised by operators with two or more factors of $\pm i Y$. This is only enough to simulate real stabiliser circuits, which is what you're noticing....
4
This phenomenon is sometimes known as a discretization of errors. It is a property of certain error correcting codes that allows it to work. It is described (somewhat briefly) in Section 10.2 of Nielsen and Chuang.
Suppose that we have an arbitrary error that affects just one qubit, and suppose that we represent this error by a channel $\Phi$ mapping one ...
4
The state $\psi$ (this is denoting the density matrix, even though it's a pure state) can be described as a sum of all the products of the stabilizers. We are promised that $X_i$ is not in the stabilizer, so every term in the sum of $\psi$, when multiplied by $X_i$, returns a tensor product of terms that is not just identity. Hence, it has zero trace. Thus, $...
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
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}, \;\;\;
\...
4
The phase of -1 generated on the ancilla is just a global phase. You can shift it to any qubit without affecting the statistics of the system. Nothing to do with entanglement.
Quantum mechanics is a mathematical framework that describes reality insofar that it predicts the correct observable statistics of a system. Shifting around a global phase factor from ...
4
Generally, quantum states are determined up to a phase - i.e. up to multiplication by scalar. So $\frac{\left|0\right\rangle -\left|1\right\rangle }{\sqrt{2}},\frac{-\left|0\right\rangle +\left|1\right\rangle }{\sqrt{2}}$ are essentially the same state.
It is a good question how the statevector simulator chooses to normalize its statevector representation.
3
The Pauli matrices form an orthogonal basis of $\mathcal{M}_2$, this vector space can be endowed with a scalar product called the Hilbert-Schmidt inner product
$$ \langle A,B\rangle=\mathrm{Tr}(A^\dagger B)$$
since the Pauli matrices anticommute, their product is traceless, and since they are Hermitian this implies that they are orthogonal with respect to ...
3
No, the matrix, its inverse or its conjugate transpose don't have to equal the identity matrix to be unitary (that would make the class of unitary matrices rather restricted).
The definition of unitary matrix is that its conjugate transpose should equal its inverse, i.e., $U U^\dagger = U^\dagger U = I$. You can check that this is indeed the case for the ...
Only top voted, non community-wiki answers of a minimum length are eligible
