# Tag Info

10

Correct, unitarity is a sufficient and necessary condition. From Nielson and Chuang page 18: Amazingly, this unitary constraint is the only constraint on quantum gates. Any unitary matrix specifies a valid quantum gate! The interesting implication is that in contrast to the classical case, where only one non-trivial single bit gate exists - the NOT gate - ...

8

Transposing a matrix is trace preserving since for $\rho = \sum_{a,b} \rho_{a,b} | a \rangle \langle b |$: $$\text{Tr}(\rho)= \sum_c \langle c| \big( \sum_{a,b} \rho_{a,b} | a \rangle \langle b | \big) | c \rangle = \sum_{a,b,c} \rho_{a,b} \delta_{a,c} \delta_{b,c} = \sum_c \rho_{c,c}$$ $$\text{Tr}(\rho^T)= \sum_c \langle c| \big( \sum_{a,b} \rho_{a,b} | b \... 7 As is the case with ordinary multiplication, tensor product distributes over addition, so we can pull |0\rangle on the first qubit out in front$$ \begin{align} |\Psi⟩ &= \frac{1}{\sqrt{2}}|\color{red}{0}0\rangle+\frac{i}{\sqrt{2}}|\color{red}{0}1\rangle \\ &= \frac{1}{\sqrt{2}}\color{red}{|0\rangle}\otimes|0\rangle+\frac{i}{\sqrt{2}}\color{red}{|0\...

7

Right. But when you build a quantum computer, you want to have a certain set of gates that you want to implement, and all other gates (unitary matrices) can be built from that set of gates. This is known as a universal set. Surprisingly (or not), it is quite small. One example of a universal set is: $G = \{H, S, T, CNOT \}$ where $$H = \dfrac{1}{\sqrt{2}}\... 7 Trace preservation The trace must be preserved in the transpose of a matrix, because the trace is the sum of the diagonal elements. When transposing a matrix, you do not change the diagonals at all! Only the off-diagonals change when you transpose a matrix. "Positive" preservation "A positive matrix is a matrix in which all the elements are ... 7 Yes. Intuitively, the set of pure product states has lower dimension than the set of all pure states. Therefore, almost all pure two-qubit states are entangled. Let \mathcal{F} denote the set of all pure states of two qubits and \mathcal{S} denote the set of all pure product states of two qubits. Note that \mathcal{S} can be thought of as the Cartesian ... 6 The 1-norm decreases under partial trace and so we have an upper bound of 1 when the states are normalized,$$ \|\mathrm{Tr}_B[|\psi_1\rangle \langle \psi_2|]\|_1 \leq \||\psi_1\rangle \langle \psi_2|\|_1 = 1. $$This bound cannot be improved upon without extra information about the states. Here is a counterexample. Take |\psi_1 \rangle = |00\rangle ... 6 This is a very particular application of Adiabatic Quantum Computing so I think it's worth briefly mentioning some context. Roughly speaking, one wants to show that given a quantum circuit defined as a sequence of unitary gates U_1,U_2,\ldots,U_L it is possible to define a version of the quantum adiabatic algorithm that reproduces (a state with a large ... 6 TL;DR: Yes, ignoring the unobservable global phase, every single-qubit unitary corresponds to a unique rotation of \mathbb{R}^3 and vice versa. Single-qubit unitaries and rotations Let us first pin down the two objects in question. The first one - the set of single-qubit unitaries - is sometimes imprecisely described as the group U(2) of 2 \times 2 ... 5 There are different ways to prove what you want to prove, including the solution tsgeorgios has suggested, but for the sake of gaining greater intuition I would suggest starting with the recognition that the trace norm of any matrix is equal to the sum of its singular values. Once you have this, the inequality you are trying to prove follows pretty easily. ... 5 Yes, such a measurement is possible and the outcomes can be mapped to the outcomes of a computational basis measurement. This is accomplished by the use a unitary to transform from the basis we wish to measure in to the basis we know how to measure in. In the example from the question, we wish to measure in the X basis$$ |+\rangle = \frac{|0\rangle + |1\...

5

In general, the knowledge of the marginals $\rho_A$ and $\rho_B$ and the operators $A$ and $B$ is insufficient to compute $\mathrm{tr}_A((A\otimes B)\rho_{AB})$. Indeed, we can find two different density matrices $\rho_{AB}$ and $\sigma_{AB}$ with the same marginals for which $$\mathrm{tr}_A((A\otimes B)\rho_{AB}) \ne \mathrm{tr}_A((A\otimes B)\sigma_{AB}).$$...

5

The terms $II$ and $ZZ$ do not uniquely specify the state $|11\rangle$ because you could equally have the state $|00\rangle$. Indeed, you should not include the identity term in your stabilizer. Thus, you need to add a second term, which could be either $-ZI$ or $-IZ$. Either way, you can easily see how to make a product $-ZI$ out of your stabilizers.

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 ...

5

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: $\sigma_z = | 0 \rangle \langle0 | - | 1 \rangle \langle 1 |$ $$| 0 \rangle \langle0 | \otimes \sigma_z - | 1 \rangle \langle 1 | \otimes \sigma_z$$ $$=\big( | 0 \rangle \langle0 | - | ... 4 The \dfrac{1}{\sqrt{2}} is the normalization constant to make sure the state/eigenvector is a unit vector. Note that: if |\psi \rangle = \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ 1 \end{pmatrix}  then \bigg| \bigg| |\psi \rangle \bigg| \bigg| = |1/\sqrt{2}|^2 + |1/\sqrt{2}|^2 = 1 . The reason for this is because in quantum mechanics, states are always ... 4 It is just the convention that people use the notation |1 \rangle  to represent the vector \begin{pmatrix} 0 \\ 1 \end{pmatrix} and |0 \rangle to represents the matrix \begin{pmatrix} 1 \\ 0 \end{pmatrix}. Similarly, people use the notation |i\rangle  to represent the vector \dfrac{1}{\sqrt{2}}\begin{pmatrix} 1 \\ i \end{pmatrix} . I could have ... 4 First, note that by definition:$$Tr(X) = \sum_i X_{ii} = \sum_i \langle i| X | i\rangle $$then now let Y = |\psi\rangle \langle\psi|  then this implies that$$Tr(YX) = Tr(XY) = \sum_i \langle i|XY|i\rangle = \sum_i \langle i|X |\psi\rangle \langle\psi |i\rangle = \sum_i \langle \psi| i \rangle \langle i|X |\psi\rangle = \langle\psi|X|\psi\rangle = \...

4

Note that usually, the state $|0+\rangle$ can denote apply the operator $(I \otimes H)$ to the state $|00\rangle$. Hence, it would be the circuit: $\hspace{8cm}$ Figure 1: $\hspace{6.5 cm}$ and also in a standard textbook, this can be written as $I \otimes H$. But because qiskit uses little-endian convention, the state $|0+\rangle$ actually actually means ...

4

ANSWER MADE CW Arguably the one thing that a quantum computer can do quickly, as long as the matrix is defined in the appropriate manner, is the quantum Fourier transform (QFT). Much as the classical fast Fourier transform (FFT) is the workhouse of many classical matrix algorithms, the QFT is the basis for many such quantum matrix algorithms. For example, ...

4

Firstly, how rude of them to use the word 'obviously' in their solution! If it was obvious then we wouldn't be trying their exercises in the first place! But anyway, Adam Zalcman described the solution very elegantly and succinctly! But for those wanting a more in-depth explanation of his answer, here it is: To understand the textbook solution a bit more, ...

4

Generalization of Schmidt decomposition The Schmidt decomposition $|\psi_{AB}\rangle = \sum_i\lambda_i|i_A\rangle|i_B\rangle$, with $\lambda_i$ positive real numbers and $|i_A\rangle$ and $|i_B\rangle$ (possibly incomplete) orthonormal bases, of a bipartite state $|\psi_{AB}\rangle$ can be thought of as a type of singular-value decomposition. Specifically, ...

4

When you have $\langle \varphi | I \otimes Z | \varphi \rangle$ It means you are calculating the expectation of the operator $I \otimes Z$ with respect to some state $|\varphi \rangle$. Since $I$ is on the first qubit, we would not need to do anything there, no need to do any rotation or even measurement. The eigenspace is decomposed into two halves ...

4

Think about that projector $$\rho=|\psi\rangle\langle\psi|.$$ Note that this is Hermitian, $\rho^\dagger=\rho$. Take the transpose, $${\rho^\dagger}^T=\rho^T$$ but since the hermitian conjugate is the complex conjugate transpose, ${\rho^\dagger}^T=\rho^\star$. If you want to see what pure state $\rho^\star=|\phi\rangle\langle\phi|$ corresponds to, think ...

4

You are correct: the units must indeed match. If we take a standard evolution with unitary $U=\exp(-i H \theta)$, then the units of $H$ and $\theta$ must match such that $H\theta$ is unitless. For a pure state $\rho_\theta=U|\psi\rangle\langle\psi|U^{\dagger}$ with unitary evolution, the quantum Fisher information takes the form F_Q[\psi,H]=\langle\psi|H^2|...

4

This is because we are using a spherical coordinate system. You can think of this coordinate system as specifying a radius, then two angular coordinates $\theta$ and $\phi$. Pure single-qubit states have a Bloch vector with radius equal to unity, so they only need two more coordinates to fully specify the direction in which the vector points. When you look ...

4

No this is not possible. When people talk about the distribution of eigenvalues they mean the expected eigenvalues when the unitary matrices are sampled with respect to some measure. If you have a fixed unitary matrix $U$ then you can have absolutely any distribution of eigenvalues that lie on the unit circle in $\mathbb{C}$. For instance $\alpha I$ for any ...

4

(General framework) The general form of this problem is the following. Take a generic Hermitian, unit-trace operator, $X\in\operatorname{Herm}(\mathcal X), \operatorname{Tr}(X)=1$, acting on some $N$-dimensional complex vector space $\mathcal X$. The set of such operators is an affine space of dimension $N^2-1$, embedded in the $N^2$-dimensional real vector ...

4

Your reasoning is correct, this is indeed the resulting state they would get. More generally, you can think about it as applying a $X\otimes I$ gate on the whole system, where $I$ is the identity gate. However, do note that if Alice and Bob do not communicate, there is nothing Alice can do to influence the measurement she will get, since its qubit is in the ...

4

Let me try to reformulate your question: Given a Universal Set of Quantum Gates $\mathcal{G}$; and some $n$-bit Unitary $U$. Can we find some $q$ such that $q$ is the minimum number of gates selected from $\mathcal{G}$ to have the effect of $U$ on $n$ qubits? (Note: I changed some variables since typically $n$ is the number of bits and $N=2^n$ is the ...

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