I Why Can We Use This Mathematical Formula for Mean Value of Measurement?

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The discussion centers on the mathematical justification for the mean value of a measurement in quantum mechanics, expressed as <T>=<phi|T|phi>. This formulation is closely tied to the Born rule, which connects measurement probabilities to the corresponding operator and the wave function in the eigenstate basis. The conversation highlights two key postulates of quantum theory: observables are represented by self-adjoint operators and pure states are described by vectors in a Hilbert space. The expectation value of a measurement is derived from the completeness of the eigenstates of the operator. Overall, the discussion emphasizes the foundational principles of quantum mechanics that underpin this mathematical expression.
ofirg55
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mean value of measurement
Hi,
I'm new to the quantum world, and would like to know why mathematically can we say that for mean value of measurment:
<T>=<phi|T|phi>
?
 
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ofirg55 said:
Summary:: mean value of measurment

Hi,
I'm new to the quantum world, and would like to know why mathematically can we say that for mean value of measurment:
<T>=<phi|T|phi>
?
That's quite close to an axiom of QM. It's related to the Born rule that identifies the probability of measurement outcomes with the operator representing the measureable and the wave-function expressed in the basis of eigenstates of that operator.

Does what I've written make sense to you?
 
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It rests on two fundamental postulates of quantum theory.

To describe a physical system within quantum theory you have a Hilbert space.

(1) An observable ##T## of the system are described by self-adjoint operators ##\hat{T}##. The possible results of measuring accurately the observables are the eigenvalues of that operator. The eigenvectors span a complete set of orthornormal vectors.

(2) A pure state of a system is described by a vector ##|\psi \rangle## with ##\langle \psi|\psi \rangle## (modulo a phase factor). If ##t## is an eigenvalue of ##T## and ##|t,\lambda \rangle## an orthonormal set of eigenvectors for this eigenvector, then
$$P(t) =\sum_{\lambda} |\langle t,\lambda|\psi \rangle|^2$$
is the probability to obtain ##t## when measuring ##T##.

Form this it follows that the expectation value for the outcome of measurements of ##T## given that the system is prepared in the pure state described by ##|\psi \rangle## must be
$$\langle T \rangle = \sum_t t P(t)=\sum_{t,\lambda} t \langle \psi | t,\lambda \rangle \langle t,\lambda|\psi \rangle=\sum_{t,\lambda} \langle \psi|\hat{T}|t,\lambda \rangle \langle t,\lambda \psi = \langle \psi |\hat{T}|\psi \rangle,$$
where in the last step I used the completeness of the eigenstates of ##\hat{T}##,
$$\sum_{\lambda,t} |\lambda, t \rangle |\langle \lambda,t |=\hat{1}.$$
 
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PeroK said:
That's quite close to an axiom of QM. It's related to the Born rule that identifies the probability of measurement outcomes with the operator representing the measureable and the wave-function expressed in the basis of eigenstates of that operator.

Does what I've written make sense to you?
@ofirg55:
Not sure what you mean by "T". I assume "phi" is the more usual "psi".
So let me paraphrase what I think you're saying: Given a (1-dimensional) wave function ## \psi (x) ## we state that the probability of a measured particle's position is ## \psi (x) \cdot \psi^* (x) ## normalized to unity, and yes that is practically a QM postulate. I say "practically" because it can really be derived from a different postulate, but that is a fine point for introductory QM. In other words, for introductory QM you can consider it a postulate but for advanced QM it's derivable from a more general postulate involving any combination of position, momentum and/or energy measurements.
 
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ofirg55 said:
Summary:: mean value of measurement

Hi,
I'm new to the quantum world, and would like to know why mathematically can we say that for mean value of measurment:
<T>=<phi|T|phi>
?
Are you also new to the classical statistical physics world? Do you know why in the classical world the mean value is ##\langle T(x,p)\rangle=\int dxdp\, T(x,p)f(x,p)##? If not, then probably neither of the answers above will make sense to you.
 
PeroK said:
That's quite close to an axiom of QM. It's related to the Born rule that identifies the probability of measurement outcomes with the operator representing the measureable and the wave-function expressed in the basis of eigenstates of that operator.

Does what I've written make sense to you?
yes, thank you!
 
thank you all!
 

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