7
Yes. Intuitively, the set of pure product states has lower dimension than the set of all pure states. Therefore, almost all pure two-qubit states are entangled.
Let $\mathcal{F}$ denote the set of all pure states of two qubits and $\mathcal{S}$ denote the set of all pure product states of two qubits. Note that $\mathcal{S}$ can be thought of as the Cartesian ...
5
I've seen claims that a single ancillary bit suffices to make the Toffoli gate universal; is there a good reference for that?
See this blog post:
With a single ancilla, you can do an operation that swaps two states in the phase space. This lets you build any permutation, i.e. you get universality. There is the relevant caveat that I'm assuming that if you ...
3
I don't think the last step holds. For example, choose $A$ to be the permutation matrix
$$A=\begin{pmatrix}0&0&1\\1&0&0\\0&1&0\end{pmatrix},$$ which indeed satisfies $A^3=\mathbb{I}$. Mathematica can perform both of the following:
$$A^{1/n}=\left(
\begin{array}{ccc}
\frac{1}{3} \left(2 \cos \left(\frac{2 \pi }{3 n}\right)+1\right) &...
3
Consider the state $|\Psi\rangle$. This has a Schmidt decomposition
$$
|\Psi\rangle=U_A\otimes U_B\sum_i\lambda_i|ii\rangle.
$$
Its reduced density matrix is
$$
\rho_A=U_A\left(\sum_i\lambda_i|i\rangle\langle i|\right)U_A^\dagger.
$$
It must be that if $|\Phi\rangle$ has the same reduced density matrix, the density matrices have the same spectrum and hence $|...
3
It turns out Toffoli + NOT is universal for alternating permutations of bit-strings for $n \ge 4$. The construction of a $\mathrm{C}^n\mathrm{NOT}$ gate with one borrowed bit (starting in unknown state, and returning there) generates a permutation that is a pair of flips on any 2-d face of the hypercube $\{0,1\}^n$. This strictly contains the group $IGL(\...
3
Once you apply the rotations to change the basis back to the computational basis ($Z$ basis) then it's just the parity check, odd parity gives $-1$ and even parity gives $1$. So in this case, where you are measuring $\langle ZZ \rangle$ (after applying the $H$ to the first qubit and $S^\dagger H$ to the second qubit), you have
$$ |00\rangle, |11\rangle \...
3
I believe Neilsen and Chuang were the first to use this particular notation. Previous work had referred to $S$ and $T$ as $\sigma_z^{1/2}$ and $\sigma_z^{1/4}$, respectively (Boykin et al. 1999). The use of $S$ may have been inspired by Deutsch's "S-matrix" (Deutsch 1989), though this was really a root-of-NOT gate. The use of $T$ may have been ...
3
Showing that $e^{i \sigma_z \otimes \sigma_z t} = \text{CNOT}(I \otimes e^{i \sigma_zt})\text{CNOT}$
If your question is only regarding why $| 0 \rangle \langle0 | \otimes \sigma_z - | 1 \rangle \langle 1 | \otimes \sigma_z$ ; you can simply factor it given that trivially: $\sigma_z = | 0 \rangle \langle0 | - | 1 \rangle \langle 1 | $
$$| 0 \rangle \langle0 | \otimes \sigma_z - | 1 \rangle \langle 1 | \otimes \sigma_z$$
$$=\big( | 0 \rangle \langle0 | - | ...
1
Both limits are dealt with in a fair amount of detail in the work that originally defined the sandwiched entropies: On quantum Renyi entropies: a new generalization and some properties. In particular, you'll find the relevant results in section IV.C.
1
As hinted in the highlighted text, $E_T$ arises from the normalization factor $A(\tau)$. Specifically, differentiating $(1)$, we get two terms
$$
\begin{align}
\frac{\partial|\psi(\tau)\rangle}{\partial\tau} &= \frac{\partial}{\partial\tau}\left(A(\tau)e^{-H\tau}|\psi(0)\rangle\right) \\
&= \frac{\partial A(\tau)}{\partial\tau}e^{-H\tau}|\psi(0)\...
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