Representation of linear operator using series ?

V0ODO0CH1LD
Messages
278
Reaction score
0
representation of linear operator using "series"?

I was looking into the progression of quantum states with respect to time. From what I understood the progression of a state ## \left|\psi(t)\right> ## is given by:
$$ \left|\psi(t)\right> = U(t)\left|\psi(0)\right> $$
I'm not sure if that's right. But it's okay if I haven't got that yet.. That was just to give some context to my actual question.

What is this representation of the linear operator ## U(t) ## at ## t = \epsilon ##, where ## \epsilon ## is an infinitesimal?
$$ U(\epsilon) = I - i\epsilon H $$
Where ## i ## is the imaginary unit, ## I ## is the identity matrix and I think ## H ## is the hamiltonian.
It also apparently has more terms of order ## \epsilon^2 ## and so on. What "series" is this? Is it some first order approximation of ## U(t) ##? What should I look into to understand where those terms are coming from?
 
Physics news on Phys.org
V0ODO0CH1LD said:
I was looking into the progression of quantum states with respect to time. From what I understood the progression of a state ## \left|\psi(t)\right> ## is given by:
$$ \left|\psi(t)\right> = U(t)\left|\psi(0)\right> $$
I'm not sure if that's right.
It's right.

V0ODO0CH1LD said:
What is this representation of the linear operator ## U(t) ## at ## t = \epsilon ##, where ## \epsilon ## is an infinitesimal?
$$ U(\epsilon) = I - i\epsilon H $$
Where ## i ## is the imaginary unit, ## I ## is the identity matrix and I think ## H ## is the hamiltonian.
It also apparently has more terms of order ## \epsilon^2 ## and so on. What "series" is this? Is it some first order approximation of ## U(t) ##? What should I look into to understand where those terms are coming from?

Those are the first two terms of the Taylor series expansion for e^{-i\mathbf{H}\epsilon}. It just so happens that U(t) = e^{-i\mathbf{H}t/\hbar}. This can be related to the idea that E = \hbar \omega \rightarrow \omega = E/\hbar. A state \mid n\rangle with definite energy E_n has the following time evolution.
\mid n(t)\rangle = e^{-i\omega t}\mid n(0)\rangle = e^{-iE_n t/\hbar}\mid n(0)\rangle

A function of an operator like e^{-i\mathbf{H}t/\hbar} has the same eigenvectors as the operator, but the eigenvalues are replaced with the corresponding function of the eigenvalues.

Here are some relevant video lectures:
https://www.youtube.com/watch?v=cVbB6wFNqYc&list=PL9LRV0x7N1NCpd-ZerPxiTzm97y9VGPLp
https://www.youtube.com/watch?v=31XrxGMRwtw&list=PL9LRV0x7N1NCpd-ZerPxiTzm97y9VGPLp
 
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!
Back
Top