What is the Meaning of Expectation and Deviation of an Operator?

esornep
Messages
5
Reaction score
0
What does the expectation and deviation of an operator mean??

The way I understood it was every observable has a operator to it and the expectation of the observable uses the operator to calculate the deviation ...

for ex :: <p>=integral( (si)* momentum operator (si) ) dx ... so what does the standard deviation and expectation of an operator mean and is my understanding right??
 
Physics news on Phys.org
The expectation value of an operator is the "expected value" for an experiment determining the observable which corresponds to that operator's value.

In other words. Given a large number of identically prepared systems, if I make a measurement of the observable corresponding to the specific operator, and I average the results I get, I should get the expectation value of the operator.

Same thing for the deviation.
 
Matterwave said:
The expectation value of an operator is the "expected value" for an experiment determining the observable which corresponds to that operator's value.

...although it's probably better to call it the "average expected value", since it's fairly trivial to find combinations of operators and wavefunctions where the expectation value falls right where there's 0% probability density of actually getting a result.

For instance, an electron with nonzero angular momentum will never be found where it's "expected" at the center of an atom.
 
I put quotes around "expected value" because it has a specific meaning which I described later in my post.
 
"Average expected value" is redundant. That's exactly what expectation value means. Don't confuse it with something like the most frequent value, which would be like the mode.

The standard deviation is a measure of the spread of a distribution.
 
The expectation is the sum of (possible outcome * probability of that outcome). The integral is essentially this sum, because the eigenfunctions of the operator are the possibilities, and the squares of the coefficients of the wavefunction, when expressed as a linear combination of the eigenfunctions of the operator, are the respective probabilities.

So when you work out <psi|X|psi>, if psi = c1x1 + c2x2 + ... (x1,x2 are eigenfunctions of X), you end up with |c1|^2.x1 + |c2|^2.x2 + ... because all the cross terms (ie. <c1x1|c2x2>) equal zero. It's a mathematical trick.

The standard deviation is the square root of the expectation of the squared deviation from the expected value, ie.
sqrt(<psi|(X - <psi|X|psi>)^2|psi>).
 
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