Hi guys. I'm studying an article on the measurements of entanglement in a pure bipartite state.
I don't understand the definition of the Schmidt rank. It is equal to the rank of the reduced density matrix, isn't it?
Is the Schmidt rank continuous and/or additive? I have no found on the net any articles or information about this.
Could you help me?
Suppose you have a quantum system involving two parties, Alice and Bob. Alice has an ##n##-level quantum system, and Bob has an ##m##-level quantum system. The combined system can be described by ##n m## amplitudes. Arrange those amplitudes into a grid with ##n## rows and ##m## columns, where each row corresponds to one of Alice's system's levels and each column corresponds to one of Bob's system's levels. For example, if they both have a qubit then the grid would be laid out like:
Code:
| Alice qubit OFF | Alice qubit ON
--------------+-----------------+----------------------
Bob qubit OFF | amplitude_00 | amplitude_01
Bob qubit ON | amplitude_10 | amplitude_11
and if you treat it like a matrix, and perform a singular value decomposition to turn it into ##U \cdot S \cdot V##, then you'll find that Alice's operations affect ##U## and Bob's operations affect ##V## but neither of them can affect ##S##. Because ##S## is a pretty simple matrix, with nothing but real non-negative entries in descending order along its diagonal, it makes sense to think of the entries on that diagonal as a measure of the entanglement between Alice and Bob.
The terms on the diagonal of ##S## might be what you mean by "schmidt rank"?
Fully entangled systems have an equal value along the whole diagonal, making the system act a whole lot like a unitary matrix. Non-entangled systems have a single non-zero entry in the top left cell, and act like the tensor product of two vectors.
Thank you Strilanc for your explanation. I don't know if the terms on the diagonal are what I mean "Schmidt rank". I think so. I have just read that for all quantum system, the sum of all eigenvalues is essentially finite and the whole entangled system becomes finite dimensional. I don't know if it is useful to consider "not continuos" the Schmidt rank ( however it is possibile use continuous variables and define an infinite Schmidt rank, cit. Lewenstein). I should demonstrate that the rank is neither continuos nor additive.
Hi atyy. I read an article ( maybe lecture notes) where the Schmidt rank is defined as the number of terms in the Schmidt decomposition, and it is also equal to the dimension of the support of the reduced density matrix. That's all.
Could the Schmidt rank be a measurement of entanglement ( being neither continuos nor additive)? For systems with local dimensional 2, I think that it is not possible. I don't know how to explain it...
Insights auto threads is broken atm, so I'm manually creating these for new Insight articles.
Towards the end of the first lecture for the Qiskit Global Summer School 2025, Foundations of Quantum Mechanics, Olivia Lanes (Global Lead, Content and Education IBM) stated...
Source: https://www.physicsforums.com/insights/quantum-entanglement-is-a-kinematic-fact-not-a-dynamical-effect/
by @RUTA
If we release an electron around a positively charged sphere, the initial state of electron is a linear combination of Hydrogen-like states. According to quantum mechanics, evolution of time would not change this initial state because the potential is time independent. However, classically we expect the electron to collide with the sphere. So, it seems that the quantum and classics predict different behaviours!
The wavefunction of an atomic orbital like ##p_x##-orbital is generally in the form ##f(\theta)e^{i\phi}## so the probability of the presence of particle is identical at all the directional angles ##\phi##. However, it is dumbbell-shape along the x direction which shows ##\phi##-dependence!