15

This answer is the opinion of someone who is essentially an outsider to "CQM" (= Categorical Quantum Mechanics), but a broadly sympathetic outsider. It should be interpreted as such. The motivations of CQM The motivations of Categorical quantum mechanics are not computation as such, but logic; and not quantum dynamics as such, but foundations of physics. ...


10

Quantum xor games are a method of greatly simplifying the ideas behind Bell's theorem, which states that no physical theory of local hidden variables can ever reproduce all of the predictions of quantum mechanics. Basically, when two qubits are entangled, measurements on them appear correlated even if they are vastly far apart. The question then is whether ...


8

Every quantum channel has many Kraus representations that may differ in the number of Kraus operators. For example, for any positive integer $n$ and numbers $p_i$ with $i=1,\dots,n$ and $\sum_{i=1}^np_i=1$ the matrices $E_i=\sqrt{p_i}I$ form a valid, if impractical, Kraus representation of the identity channel with $n$ Kraus operators. This example also ...


7

Yes, it is possible to conceive theories with "stronger correlations" than those given by quantum mechanics. One way to make this statement precise is to consider some kind of "measurement apparatus" (you can think of it as simply a black a box with some buttons that you can push and different LEDs that correspond to different possible outputs), and analyse ...


6

Semantics aside, I’m assuming that your question is essentially “why do we use a matrix formulation of quantum mechanics rather than a continuous variable/differential equation/integral formulation” (I may be wrong and would welcome clarification) and nothing to do with the interaction picture and the like, which some other answers seem to be touching upon. ...


5

Choi operator of a linear map $\mathcal{E}$ is defined as $$ J(\mathcal{E}) = \sum_{ij} \mathcal{E}(|i\rangle\langle j|)\otimes |i\rangle\langle j|.\tag1 $$ Substituting $\mathcal{E}(\rho)=\sum_k E_k\rho E_k^\dagger$ into $(1)$, we have $$ \begin{align} J(\mathcal{E}) &= \sum_{ijk} \left(E_k|i\rangle\langle j| E_k^\dagger\right)\otimes |i\rangle\langle j|...


5

The need for quantum fields There are two seemingly unrelated conceptual steps between quantum mechanics (QM) and quantum field theory (QFT). One step reconciles QM with special relativity (SR) and the other replaces a fixed finite number of particles with fields that have infinite degrees of freedom. These two steps appear unrelated, but in order to arrive ...


5

The difference is as follows: the original Bell inequality requires that outcomes from the same setting are always perfectly anti-correlated. It says nothing about the case where they are even marginally different. by contrast, in CHSH, the ideal (giving maximum violation) is that outcomes from the same setting would be anti-correlated, but it is not ...


4

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 ...


4

There is no "standard" method to implement XNOR, but it can be logically obtained by attaching a NOT gate (often called an X gate in quantum computing) to a logical XOR (which you know is implemented using CNOT). The X gate is applied to the target qubit of the CNOT. To answer your question more directly, there is no standard "quantum gate" that is ...


4

I think the best way to understand this is by showing that one can violate (ontological) locality while respecting (operational) no-signalling. Take the case of Bohmian mechanics. In it the result of Alice's measurement is deterministic, and it will depend Bob's choice of measurement, so it is clearly nonlocal. Nevertheless, the "quantum equilibrium&...


4

No signaling means can't be used for communication. Can't be used to move messages from a sender to a receiver. Local means doesn't require communication to implement. Some processes, such as phenomena violating bell inequalities, don't enable communication but also can't be emulated classically without communication. They take without giving back. They are ...


4

It depends on what exactly you mean by "affected". First of all, I'd say from your description that your "a" does not actually enter the discussion. The question is then essentially: suppose Alice (A) and Bob (B) have entangled states, and A performs a measurement; does this affect B's state? The knowledge acquired by A with her ...


4

Here is an approach that requires no specific knowledge about $|\psi\rangle$ whatsoever. In your description you implied that each $H_i$ has the same maximum and minimum eigenvalues $\lambda_m$ and $\lambda_M$ respectively so I will assume this in the derivation. The process of measuring $\langle h \rangle$ empirically can be thought of running a series of ...


3

Let $M\in\mathrm{Lin}(\mathcal Y\otimes\mathcal X)$ be some linear operator whose input and output spaces are both $\mathcal Y\otimes\mathcal X$, for some pair of finite-dimensional Hilbert spaces $\mathcal X,\mathcal Y$. Moreover, suppose $M$ is positive semidefinite: $M\ge0$. It being positive semidefinite implies it admits a decomposition of the form $M=\...


3

Let's start with the Bell state $\frac1{\sqrt2}(|01\rangle+|10\rangle)$. Let's further assume that the entire universe (or at least the parts of relevance) are pervaded by a static magnetic field along the $z$-axis, which I relate to the unitary evolution $\exp(-itH_z)$. Now let's place one part of the entangled state next to a massive object, such that ...


3

Gate superposition Superposition is a complex linear combination of pure states with a physical interpretation of coefficients. One aspect of this interpretation concerns the probability of measurement outcomes on a superposition state. Gates are not pure states and generally cannot be measured. Therefore, strictly speaking, there is no such thing as a ...


2

By a quantum gyroscope, it is usually meant a device or a sensor capable of measuring the same quantities as a classical gyroscope, namely angular velocities or orientations but with extremely higher precision, limited only by the Heisenberg uncertainty due to the exploitation of the quantum nature of the sensor. The principle of measurement accuracy ...


2

The state of Bob's pair is not $1/2 (|0\rangle\langle0| + |1\rangle\langle1|)$. This is only his reduced density matrix. By definition, it is a representation of the locally accessible information Bob has. It's perfectly natural that it will change when Bob gets some nonlocal information from Alice. The question remains, however, what is Bob's state. The ...


2

Just to sort of "talk out" the conceptual difficulty of a "superposition of gates" would be (which was discusssed in a more abstract way by @Adam Zalcman), let's try to conceptually think of an experiment where it would be reasonable to say there's a "superposition of two gate operations." Consider the following experiment: Our ...


2

This is nothing but a controlled-unitary gate. The only difference to an "ordinary" controlled-unitary is that now, the control qubit is encoded in the overall state of the device that implements the unitary.


1

If I were you, I'd ignore the matrix $R$ and instead work with the matrix $Q$. They give you a conversion between vectors in the two different representations. First, I'm going to simplify things a bit by working with $$ \tilde Q=\left(\begin{array}{cc} e^{i\phi/2} & 0 \\ 0 & e^{-i\phi/2} \end{array}\right)Q. $$ You'll have to compensate for this in ...


1

Modes are governed by eigenfunctions, I agree. In quantum optics, we need more than just eigenfunctions to describe a state of light: we need to know how many photons have properties corresponding to each eigenfunction. This is somewhat beyond what an eigenfunction describes, so we need a new term. For example, we can have a two-photon state of light where ...


1

Putting something as massive as even the simplest gate-device into a superposition goes beyond currently understood physics. At a "device-scale" general relativistic considerations may become significant because the two device-states will have different mass-distributions and distinct gravitational self-energies. Roger Penrose has worked through ...


1

Remember that mixed states can be a subjective description of a quantum state. In a teleportation operation, where Alice has made the measurement, but Bob has not yet received the measurement result, then Alice and Bob have different information, and therefore they have different descriptions. Alice knows exactly the state that Bob holds. Bob has no idea, ...


1

The quantum circuit model does not use the Heisenberg picture of quantum mechanics. Also, what is "Dirac's Matrix Mechanics"? Quantum computing widely uses Dirac's bra-ket notation. Quantum circuit model considers how state vectors of qubits are transformed by quantum gates which follows the Schrödinger picture of quantum mechanics. PS: Maybe I need to say ...


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