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

[ofirg55]

TL;DR

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>
?

 

[PeroK]

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?

 

[vanhees71]

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 $\hat{T}$,
$$\sum_{\lambda,t} |\lambda, t \rangle |\langle \lambda,t |=\hat{1}.$$

 

[rude man]

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.

 

[Demystifier]

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.

 

[ofirg55]

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!

 

[ofirg55]

thank you all!