Will a uniform magnetic field be considered as measurment?

  • #26
vanhees71
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This is the probability to find the value ##a## when measuring the observable ##A## (as explained in #24) provided the measured system is prepared in the state ##\hat{\rho}##.
 
  • #27
PeterDonis
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the point is that unitary operators have complete sets of (generalized) orthonormal eigenvectors

But with eigenvalues that might not be real, which is why operators representing observables are normally required to be Hermitian.

The paper you linked to does not appear to actually claim that the operator ##\exp(i \hat{\phi})## represents the phase angle observable. Rather, it appears to be saying that there is no "phase angle" observable conjugate to the number operator, but there are "cosine" and "sine" observables (i.e., Hermitian operators with real eigenvalues) that can be used instead.
 
  • #28
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This is the probability to find the value ##a## when measuring the observable ##A## (as explained in #24) provided the measured system is prepared in the state ##\hat{\rho}##.
I thought it is in the form ##\frac{\mid\langle a,\beta|\rho\rangle\mid^2}{\sum \mid\langle a_k,\beta|\rho\rangle\mid^2}##
 
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  • #29
vanhees71
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But with eigenvalues that might not be real, which is why operators representing observables are normally required to be Hermitian.

The paper you linked to does not appear to actually claim that the operator ##\exp(i \hat{\phi})## represents the phase angle observable. Rather, it appears to be saying that there is no "phase angle" observable conjugate to the number operator, but there are "cosine" and "sine" observables (i.e., Hermitian operators with real eigenvalues) that can be used instead.
Yes, sure, and instead of using ##\cos## and ##\sin## you can use ##\exp## to define an angle or phase operator. The eigenvalues of unitary operators are of course all of the form ##\exp(\mathrm{i} \varphi)## with ##\varphi \in \mathbb{R} \; \text{mod} \; 2 \pi##.
 
  • #30
vanhees71
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I thought it is in the form ##\frac{\mid\langle a,\beta|\rho\rangle\mid^2}{\sum \mid\langle a_k,\beta|\rho\rangle\mid^2}##
How did you come to this conclusion? I don't know, what this might represent. The notation doesn't make any sense.

##\hat{\rho}## is the Statistical Operator uniquely representing the quantum state the system is prepared in. For a pure state, there is a normalized vector ##|\psi \rangle## such that ##\hat{\rho}=|\psi \rangle##. Then, of course, the probability discussed is
$$P(a|\psi)=\sum_{\beta} |\langle a,\beta|\psi \rangle|^2.$$
Here I assumed, you just measure the observable ##A## and not additionally any other compatible observables. Then you have to sum (integrate) over all degeneracies.
 
  • #32
vanhees71
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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).
 
  • #33
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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).
 
  • #34
vanhees71
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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.
 
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  • #35
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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 any one of those base states is 1/2
 
  • #36
vanhees71
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Sure, that was a typo. Sorry! I've corrected it.
 

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