Unobserved measurement, eigenvalues, and entanglement.

[AbstractDreamer]

So I was out running again.  Sometime during the run, I discovered my heartrate monitor had fallen off my chest strap.  Now I knew when I had last checked it, and this meant it must have fallen off on one of three paths I had run up.  So I went back and searched two of paths for it, and didn't find anything.  At this point I knew it was on the third path.  At this point it also occurred to me that I had measured the position of my heartrate monitor without actually making a measurement or an observation of it.

Now here's the giant leap of faith to some relevance.  Had I known its momentum before hand, I would have information on both its momentum and position at the same time.  Had it been a quantum particle, I would have the avoided the problem of the Observer Effect - by deducing the value of the observable, by measuring all the states that it is not in except one and finding nothing, instead of making a direct measurement of the state it is in.

Clearly the Uncertainty Principle remains inviolable.  But if quantum mechanics assigns observables as operators and values as eigenvalues of the operator, how does it model unobserved values or rather values for a operator that remains unobserved?

Now leaping over the fence, how might this unobserved measurement be involved when considering entanglement, and "action at a distance".  Does the unobserved measurement of a quantum state of an entangled particle still collapse the waveform simply by deducing information without measurement?

PS.  I did find my heartrate monitor, and yes it was on the third path.

 

[swansont]

45 minutes ago, AbstractDreamer said:

Clearly the Uncertainty Principle remains inviolable.  But if quantum mechanics assigns observables as operators and values as eigenvalues of the operator, how does it model unobserved values or rather values for a operator that remains unobserved?

Amplitudes and superposition 

45 minutes ago, AbstractDreamer said:

Now leaping over the fence, how might this unobserved measurement be involved when considering entanglement, and "action at a distance".  Does the unobserved measurement of a quantum state of an entangled particle still collapse the waveform simply by deducing information without measurement?

There has to be a measurement.

In your example, as an analogy, the paths are entangled. Even though you didn’t find the monitor after your first two measurements, as you note, it gave you information. You made a measurement. It just returned zero as an answer.

Not getting a photon can be a measurement, as in some quantum zeno effect experiments. If you’ve prepared an atom to be in one of two states, knowing it’s not in one state tells you it’s in the other.

 

[MigL]

As Swansont explains, your HR monitor had to be in one of three possible places, eliminating the two possibilities, leaves the third option as the only possibility. Entanglement has the same 'correlation'; Of the two possible spin directions, measuring the state of one, tells you the state of the other must be in.
Some people make the argument that, since neither state is known before the first measurement, some 'information has to be 'sent' to fix the second measurement, bringing all sorts of complications. But I'm a simple guy who doesn't like complications.
( only complicated women )

As for circumventing the HUP, if you had known the momentum of your HR monitor before you lost it, you would have needed to know its position on that third running path such that the product of its momentum standard deviation, and position along the path standard deviation is greater/equal to hbar/2. IOW for large macroscopic objects, almost exact.

[AbstractDreamer]

The confusion for me is that I can't see the similarity of position state of a photon with say electron spin.

The spin of an electron is a property of the electron, I can't see a how you can measure this property without probing the electron.

The position of photon is a little different.  I can measure an area to the left and an area to the right, and if its not there, its position must be in the middle.  I can deduce and obtain the value of this property by NOT probing the photon.

What am I not understanding?

[joigus]

1 hour ago, swansont said:

There has to be a measurement.

In your example, as an analogy, the paths are entangled. Even though you didn’t find the monitor after your first two measurements, as you note, it gave you information. You made a measurement. It just returned zero as an answer.

Not getting a photon can be a measurement, as in some quantum zeno effect experiments. If you’ve prepared an atom to be in one of two states, knowing it’s not in one state tells you it’s in the other.

Exactly. This is the basis for the Elitzur-Vaidmann bomb tester. If you have a detector, even if the detector is unaffected, there is a measurement. It's also called "interaction-free measurement."

P. G. Kwiat; H. Weinfurter; T. Herzog; A. Zeilinger; M. A. Kasevich (1995). "Interaction-free Measurement". Phys. Rev. Lett. 74 (24): 4763–4766. Bibcode:1995PhRvL..74.4763K

ABSTRACT:

Quote

We show that one can ascertain the presence of an object in some sense without interacting with it. One repeatedly, but weakly, tests for the presence of the object, which would inhibit an otherwise coherent evolution of the interrogating photon. The fraction of ``interaction-free'' measurements can be arbitrarily close to 1. Using single photons in a Michelson interferometer, we have performed a preliminary demonstration of some of these ideas.

 

[swansont]

45 minutes ago, AbstractDreamer said:

The spin of an electron is a property of the electron, I can't see a how you can measure this property without probing the electron.

In a magnetic field, the two spin states have different energy (this is the case in most atoms). A null result from one state means it must be in the other.

[AbstractDreamer]

6 minutes ago, swansont said:

In a magnetic field, the two spin states have different energy (this is the case in most atoms). A null result from one state means it must be in the other.

I'm still none the clearer.  The two energy states is like a coin its either heads or tails for the electron.  I get that.  If you have something that checks for heads, you don't have to check for a tail. But you still need to check the coin.

The position of a photon is not like a two sided coin.  Its like a millions coins, and only one of them is heads.  You can check 999,999 of them and find tails.  You don't have to check the last one.

[swansont]

10 hours ago, AbstractDreamer said:

I'm still none the clearer.  The two energy states is like a coin its either heads or tails for the electron.  I get that.  If you have something that checks for heads, you don't have to check for a tail. But you still need to check the coin.

The position of a photon is not like a two sided coin.  Its like a millions coins, and only one of them is heads.  You can check 999,999 of them and find tails.  You don't have to check the last one.

If the electron is in an atom, then there are transitions that can be interrogated, and owing to this energy difference, might be quite distinct from each other. In Hydrogen, for example, the ground states are 1.4 GHz apart. If you know the atom is in one of the ground states, and interrogate one transition (representing one spin state) and it's not there, you know it's in the other state.