Tag Info

Hot answers tagged

8

I think it is a mistake, in this case, to think of entropy as "a description of our ignorance." Rather, I would suggest that you think of entropy as a well-defined, objective property provided that you specify which degrees of freedom in the universe are inside and outside of your system. The content of this statement isn't really different, but it ...


6

There is a difference between the 'fine grained' and the 'coarse grained' entropy. If we start with a pure state (zero entropy), and we time-evolve it, the entropy indeed stays zero by unitary of time evolution. The fine grained entropy did not change. The coarse grained entropy is what we usually call the thermal entropy, and is the thing that always ...


5

The "Kronecker product", better known as the tensor product, is the natural notion of a product for spaces of states, when these are considered properly: A space of states is not a Hilbert space $\mathcal{H}$, but the projective Hilbert space $\mathbb{P}\mathcal{H}$ associated to it. This is the statement that quantum states are rays in a Hilbert space. ...


2

ACuriousMind's Answer pretty much summed up the reasons, which are essentially mathematical. If you want to grasp the "physical significance", then I suggest you should work through an example: think of two quantum systems, each with three base states: $\left.\left|1\right.\right>$, $\left.\left|2\right.\right>$ and $\left.\left|3\right.\right>$. ...


2

What makes you think that information can be transferred in this way in the first place? I cannot tell you why your method won't work if I don't know what method you have in mind. In any case, I can take a guess as to what you're referring to. Suppose we have a pair of electrons in the state (up to normalization) $\left| +\right> \left| -\right> ...


2

The density matrix formalism follows naturally from a state vector formalism when we average over a part of variables. It is not a "more general", but a more common approach in practical applications. Sometimes it is the only possible formulation due to experimental restrictions.


2

Density matrix is NECESSARY when the quantum system is not in a PURE state, i.e. it cannot be described by a wave-function. Such a case is when a beam of freely moving electrons is not spin-polarized, i.e. there is no direction in space along which all the electrons have the same spin-projection. (For finding whether the electrons are polarized, and on ...


2

State vectors define pure quantum states of a system, and, for an isolated system, evolve with time following the Schrödinger equation (in the Schrödinger picture; in the Heisenberg picture they are constant. Density matrices define classical statistical mixtures of pure quantum states. These can arise in two ways: When we have incomplete information ...


2

Yes, as it turns out there are distinct types of multipartite entanglement not witnessed by the spectra of reduced states. The simplest example is probably the one from Bennett et al., "Exact and asymptotic measures of multipartite pure-state entanglement", Phys. Rev. A 63, 012307 (2000). For the case of three 4-dimensional systems, consider the states ...


2

I'll answer my own question and hope this information is useful to someone. I'll take $\hbar = 1$ and will deal with systems of one degree of freedom (the generalisation should be obvious). The normalisation factors are very confusing, so I'll omit most of the reasons for them to be what they are. First, let's take a look at the good old continuous Wigner ...


2

The short answer to the first question is yes. You need enough power above the noise level of your systems (and space) to even receive any signal from somewhere else. You can look up articles on the Shannon-Hartley theorem, which discusses the relationship between information transfer and power, bandwidth, etc. This is one of the most limiting factors ...


2

The issue is that when a wavefunction collapses it has an inherent randomness to it. Since entropy is fundamentally related to information, I'll start with information to explain why this is significant. Information, Randomness, and Entropy If you think of the information of a state as what is needed to completely define a system something interesting ...


1

I join the comments of Julian Fernandez. Just some hints for some of your ideas: Spacetime is relative i.e. observer-depending. It seems that behind relative spacetime there is a system of absolute time (= proper time) and absolute space (=space without considering time dilation and length contraction which are relative) where spacetime is "mixing up" ...


1

Let me try to (partially) answer the following question then: Can I buy or build a quantum system, where I can perform certain quantum computations (maybe not universal) at home? Tl;dr: The biggest caveat is the amount of control necessary to perform any controlled computation at all. You simply don't have it, except by studying a system with knew ...


1

The conjecture is false. A counterexample for a single-qubit state and a single-qubit ancilla is: $$P_1:=\left(\begin{array}{rrrr} \frac{1}{2} & 0 & 0 & \frac{1}{2} \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \frac{1}{2} & 0 & 0 & \frac{1}{2} \end{array}\right), \qquad P_2:=\left(\begin{array}{rrrr} \frac{1}{2} ...


1

Can this scenario be considered as erasure of information as referred to in Landauer's principle? No. The reason is that the whole process is reversible and Landauer's principle specifically needs an irreversible step - otherwise no "erasure" can have taken place. Where's the problem? It's here: The state $$\frac{1}{\sqrt{2}}(|0\rangle+|1\rangle)$$ ...



Only top voted, non community-wiki answers of a minimum length are eligible