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7

Not every tensor is a simple tensor. The simple tensors of the form $a\otimes b\in A\otimes B$ span the tensor product, which means that a general element $t$ of the tensor product is a linear combination of these simple tensors, i.e. $$ t = \sum c_{ij} a_i\otimes b_j$$ for some basis $a_i,b_j$ of $A$ and $B$ respectively. If $A$ itself is a space of ...


6

No. Counterexample: choose a basis in 2D quantum space, so the basis states are written $\left(\begin{array}{c}1\\0\end{array}\right)$ and $\left(\begin{array}{c}0\\1\end{array}\right)$. Now think of a mixture, with a state's having probability $p$ to be in state 1. The density matrix is then $\mathrm{diag}(p,\,1-p)$, which is not singular for ...


5

The answer to your question is NO. The simplest counter-example is the identity matrix. There is no solution to $|a\rangle\langle b| = I$. For example, try to solve $\begin{bmatrix} x \\ y \end{bmatrix} \begin{bmatrix} z & t \end{bmatrix} = \begin{bmatrix} 1&0 \\0&1 \end{bmatrix}$. Since $xt=0$ we know that $x=0$ or $t=0$. But clearly $x$ ...


4

The operations aren't faster, they're more flexible. From that flexibility comes power. For example: in a classical computer there is no single-bit operation that, when applied twice, flips a bit. There's no boolean function $f$ such that $f(f(x)) = \overline{x}$. But quantum computers do have such an operation, represented by the matrix $M = \frac{1}{2} ...


4

No, $|\phi\rangle$ doesn't need to be orthogonal to $|0\rangle$. Neither does $|\psi\rangle$. They only need to be orthogonal to each other. Example Suppose you know that you will be given a state $|\omega\rangle$ that will either equal $|\phi\rangle = \frac{1}{\sqrt 2}|0\rangle + \frac{1}{\sqrt 2}|1\rangle$ or equal $|\psi\rangle = \frac{1}{\sqrt ...


2

The title of the paper is "Optimum Unambiguous Discrimination Between Linearly Independent Symmetric States". Random guessing is an ambiguous discrimination procedure. It can return the wrong answer without telling you that it failed. More generally, unambiguous discrimination procedures are less likely to succeed because removing any chance of accidental ...


2

A crucial hypothesis is missed in your construction. Each $\phi_i$ must also satisfy $\phi_i \not \perp \psi_i$, otherwise $\langle \psi_i |E_i \psi_i\rangle >0$ is false. This point provides an answer to your last question as well. If $\psi$ is an added further vector, linearly dependent on the vectors $\psi_i$, the construction you made cannot be ...


2

The answer is superposition. Let's take one classical bit. It can be in two states: $|0\rangle$ or $|1\rangle$. Now let's consider a qubit: its general state will be $$a |0\rangle + b |1\rangle$$ where $a,b$ are complex numbers with the constraint $$|a|^2+|b|^2=1$$ Now you should be able to see the difference. A state like $$\frac{|0\rangle + ...


2

It seems I have figured out an answer for 2 terms in the original state. Suppose that the state is $$\rho = a |\alpha \rangle \langle \alpha | + (1-a) |\beta\rangle \langle \beta|$$ We need to write it in a basis, which is $$|+\rangle = \frac{|\alpha\rangle + |\beta\rangle}{\sqrt{2}}; \quad |-\rangle = \frac{|\alpha\rangle - |\beta\rangle}{\sqrt{2}}. $$ ...


2

This hand waving assumption you are making is the crux: At the mirrors A and B (and also at the half-transparent ones not considered) the photon interacts with one or more electrons of the mirror, transferring momentum The photon is not interacting with one or more electron on its way, it is interacting with the lattice of atoms. This means that the ...


1

You are misunderstanding (and thus misquoting) the book of Nielsen and Chuang. What they do is to first define the reduced density matrix $\rho_A$ of a bipartite state $\rho_{AB}$ as $$\rho^{A} =\mathrm{tr}_B(\rho^{AB})$$ Then, they explain what a partial trace is: It is the linear map defined by its action $$ \mathrm{tr}_B \left( |a_1 \rangle \langle a_2| ...


1

The basis states are represented as four-dimensional vectors, like so: \begin{align} |00\rangle&=[1, 0, 0, 0]^T \\ |01\rangle&=[0, 1, 0, 0]^T \\ |10\rangle&=[0, 0, 1, 0]^T \\ |11\rangle&=[0, 0, 0, 1]^T \end{align} You then simply apply the matrix transformation to these vectors. For example, a quantum state may be expanded as a linear ...


1

You can use circuit simulators to help with understanding small circuits like this, and to check your work. For example, here's your circuit in Quirk: The key things you need to understand for this circuit are: How to apply an operation on paper to an entangled state (for the H after the CNOT). Group by the uninvolved bits, and apply the operation ...


1

A full quantum mechanical description of the interferometer is complicated because it's not an isolated system. But we can do a thought experiment where we imagine it to be made out of mirrors that are floating in free space. Then as Anna has explained, the photon will interact with the entire system, it will not excite vibrational modes of the lattice. This ...


1

Particles are described by quantum fields, and the quantum field determines the mass, spin and charge. So for example all electrons (and positrons) have the same mass, spin and (magnitude of) charge because they are all excitations of the electron quantum field. Individual electrons can have different energy and momenta, but I'm guessing you wouldn't ...



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