Will a uniform magnetic field be considered as measurment?

[Adel Makram]

How did you come to this conclusion? I don't know, what this might represent. The notation doesn't make any sense.

Look up here under "Discrete, nondegenerate spectrum" https://en.wikipedia.org/wiki/Measurement_in_quantum_mechanics

 

[vanhees71]

So what's the problem then? That's standard QT, as I've written above (I considered the general case that usually applies, i.e., degenerate eigenstates of a single observable operator).

 

[Adel Makram]

So what's the problem then? That's standard QT, as I've written above (I considered the general case that usually applies, i.e., degenerate eigenstates of a single observable operator).

No great problem, just I am not familiar with this expression of probability. The probability is, in general, written in term of a value over the sum of all values representing the system. So the sum should be in the denominator. And the value in our case is the square of the projection of the general state on the eigen state (the square of the dot product).

 

[vanhees71]

I still don't understand your problem.

Let's take the simplest case of a nondegenerate discrete spectrum. Then the operator $\hat{A}$ representing the observable $A$ has a complete set of orthonormal vectors $|a \rangle$, i.e., each eigenspace is onedimensional. Now let the system be prepared in a pure state, represented by a unit ray in Hilbert space and let $|\psi \rangle$ be a representant of this ray. Then the probability to find the value $a$ when measuring the observable $A$ is [corrected in view of #35]
$$P(a|\psi)=|\langle a|\psi \rangle|^2.$$
These are non-negative real numbers. Since the vector $|\psi \rangle$ by definition is normalized and since the orthonormal set $|a \rangle$ is complete, you have
$$1=\langle \psi |\psi \rangle=\sum_a \langle \psi|a \rangle \langle a|\psi \rangle=\sum_a |\langle a|\psi \rangle|^2=\sum_a P(a|\psi),$$
i.e., $P(a|\psi)$ are indeed a correct set of probabilities for the outcome of measurements of observable $A$ provided the system is prepared in the said pure state.

 

[Adel Makram]

I still don't understand your problem.

Let's take the simplest case of a nondegenerate discrete spectrum. Then the operator $\hat{A}$ representing the observable $A$ has a complete set of orthonormal vectors $|a \rangle$, i.e., each eigenspace is onedimensional. Now let the system be prepared in a pure state, represented by a unit ray in Hilbert space and let $|\psi \rangle$ be a representant of this ray. Then the probability to find the value $a$ when measuring the observable $A$ is
$$P(a|\psi)=|\langle a|\psi \rangle|.$$
These are non-negative real numbers. Since the vector $|\psi \rangle$ by definition is normalized and since the orthonormal set $|a \rangle$ is complete, you have
$$1=\langle \psi |\psi \rangle=\sum_a \langle \psi|a \rangle \langle a|\psi \rangle=\sum_a |\langle a|\psi \rangle|^2=\sum_a P(a|\psi),$$
i.e., $P(a|\psi)$ are indeed a correct set of probabilities for the outcome of measurements of observable $A$ provided the system is prepared in the said pure state.

Although this is excursion from the main point of the thread, I still think the first probability must be $P(a|\psi)=|\langle a|\psi \rangle|^2$ not $|\langle a|\psi \rangle|$ . Example, spin-1/2 has the state $|\psi \rangle=\frac{1}{\sqrt2} |up \rangle+\frac{1}{\sqrt2} |down\rangle$ So the probability that the system in anyone of those base states is 1/2

 

[vanhees71]

Sure, that was a typo. Sorry! I've corrected it.