Measured result is equal to expectation value

BUI TUAN KHAI
Messages
12
Reaction score
0
Can I ask a basic question. This was a question in a test, I could not solve this.
When is it true that the result of a single measurement for a dynamical variable is equal to the expectation value of the operator corresponding to that dynamical variable?
Thank you for your help.
Sincerely yours.
Khai.
 
Physics news on Phys.org
In general the result of a single measurement will not be equal to the expectation value, but if you make a large number of measurements the average is likely to be close the expectation value; the more measurements you make, the closer the average will be.

That's "in general". In the specific case of a system prepared in an eigenstate of the observable that you are measuring, the result will in principle be the corresponding eigenvalue every time and the expectation value is exactly equal to that eigenvalue.
 
  • Like
Likes bhobba and BUI TUAN KHAI
As I think,
What we observe from the measurement is the eigenvalue.
According to the question, it means that the eigenvalue = expectation value ?
If we have an observable A, and a physical state |a>, we have A|a> = a|a>.
<A> = <a|A|a> = a<a|a> = a
Thus, this case only happens if:
  • |a> is the eigenket of A (we measure A in physical state, which is also the eigenstate of A)
  • A has real eigenvalue
 
Nugatory said:
In general the result of a single measurement will not be equal to the expectation value, but if you make a large number of measurements the average is likely to be close the expectation value; the more measurements you make, the closer the average will be.

That's "in general". In the specific case of a system prepared in an eigenstate of the observable that you are measuring, the result will in principle be the corresponding eigenvalue every time and the expectation value is exactly equal to that eigenvalue.
Thank you for you explanation.
 
BUI TUAN KHAI said:
  • A has real eigenvalue
If A is an observable, it can only have real eigenvalues.
 
  • Like
Likes bhobba and BUI TUAN KHAI
Not an expert in QM. AFAIK, Schrödinger's equation is quite different from the classical wave equation. The former is an equation for the dynamics of the state of a (quantum?) system, the latter is an equation for the dynamics of a (classical) degree of freedom. As a matter of fact, Schrödinger's equation is first order in time derivatives, while the classical wave equation is second order. But, AFAIK, Schrödinger's equation is a wave equation; only its interpretation makes it non-classical...
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
Is it possible, and fruitful, to use certain conceptual and technical tools from effective field theory (coarse-graining/integrating-out, power-counting, matching, RG) to think about the relationship between the fundamental (quantum) and the emergent (classical), both to account for the quasi-autonomy of the classical level and to quantify residual quantum corrections? By “emergent,” I mean the following: after integrating out fast/irrelevant quantum degrees of freedom (high-energy modes...
Back
Top