Can Dirac Delta Functional Observables Be Regarded as True Quantum Observables?

In summary, the conversation discusses an issue with expressing certain types of operators, which is solved by decomposing them with the help of a specific class of functionals. These functionals, defined by a Dirac delta acting on the Hilbert space, are linear and self-adjoint, making them suitable as operators. They are also compatible with each other, allowing for measurements of all of them at once. The conversation also mentions the use of indicator functions to fix issues with ensembles and the potential for deterministic measurements in quantum mechanics. Finally, there is a brief discussion about the linearity of operators and the importance of adhering to the rules of PF discussion."
  • #1
Killtech
344
35
I ran into an issue of expressing certain types of operators and after a little thinking managed to do it by decomposing them with the help of a weird class of functionals:
$$|Φ⟩\rightarrowδ_{\{|Ψ⟩\}}(|Φ⟩)$$
for any state ##|\Psi\rangle##. And yes, it's a Dirac delta acting on the the Hilbert space itself, so the entire functional is defined there too. One can immediately see that this definition is linear and therefore it represents an operator. It is also self-adjoint since ##\langle \delta_z x | y \rangle = \langle x | \delta_z y \rangle##. Albeit very untypical it renders it to formally satisfy all requirements of being called an "observable". I doubt this corresponds to anything that can be experimentally measured, but within the framework of the QT axioms, it is allowed to treat it as such.

Now these operators have a very peculiar property: they do commute with every other operator, in particular they do commute amongst each other. So if we were to go purely by the axioms of QT we would have to conclude they can all be measured at once. Furthermore the von Neuman measurement scheme makes it so that the measuring of any of these observables leaves the state exactly unchanged and the outcome is always deterministic, never producing a true ensemble of different states.

Notice that:
$$δ_{\{|i⟩\}}(a|i⟩+b|j⟩) =
\begin{cases}
1 & \text{if } b = 0 \\
0 & \text{if } b \neq 0
\end{cases}
$$
So these operators would have to be interpreted as measuring a system to be exactly in the state ##|\Phi\rangle## and any ever so minor superposition of that state with something else will render the measurement yield a null result. Since all of them are compatible with each other they formally allows us to measure them all and thus determine the exact state the system is in, perfectly within the limits of Heisenbergs uncertainty principle for general operators, since if we calculate their uncertainty we simply get
$$\sqrt {\langle\delta_x^2\rangle -\langle\delta_x\rangle^2} = \sqrt {\langle\delta_x\rangle -\langle\delta_x\rangle} = 0$$
as it must be for something that always has a deterministic result.

I stumbled upon those when i had some trouble dealing with ensembles which could be any composition of states like ##|i\rangle##, ##|j\rangle##, ##\frac 1 {\sqrt 2} (|i\rangle + |j\rangle)## and ##\frac 1 {\sqrt 2} (|i\rangle - |j\rangle)##. Density operators do not seem like dealing with ensembles for which the composite sates don't map onto a basis of pure states and where the rate coefficients ##p_{i}##, ##p_{j}##, ##p_{i\oplus j}## and ##p_{i\oplus -j}## have to be freely set and treated as independent constituents of the ensemble. My issue is that an uniform ensemble made out of all 4 states yields the identical density operator as one made of the first two 2 pure states. However these do not act the same in all cases, needing me to distinguish between them. Making use of indicator functions (which are also operators) fixes the issue. The indicator functions onto a single state are the weird operators mentioned here.
 
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  • #2
Just to clarify the Dirac symbol in the above definition is meant to be the Dirac measure, not the Dirac delta function, which is the density function of that measure. A measure always induces an operator and in this case it works like a Kronecker symbol though written as a function. Wasn't sure how well everyone is still familiar with analysis and measure theory from the early semesters, so i tried to refrained from using indicator functions which i don't remember seeing in physics.

But this got me thinking. QT does not actually disallow deterministic measurements at all. Sure, this one may be exotics, but you can also point out that measuring a particles spin can always be made deterministic if one can set the measured axis such that the particle is in a pure state for that basis. Such measurement still extracts information if the state was up or down, so it's actually a measurement. This works in general, we just have to pick our observables such that they make sense for the quantum systems state, i.e. it must become pure in that basis.

All deterministic measurements have the neat property that they never change the sate thus we can do as many of them one after the other on the very same system as we want because we don't disturb it at all (theoretical). So we don't need many identical copies of the system for measurement, since the theory allows us to get all we want with a single instance. It's all just a question to build the right observables.

So on it's own grounds it would seem that QT is a fully deterministic theory, with an extension to measurements that intend to extract information that a state doesn't have forcing it to come up with a random answer. Come to think of it, this sounds like measuring by torture... even if someone doesn't know something, one can still make him answer - just don't expect information obtained that way to be accurate, and don't expect the one being questioned to survive let alone be in the same state as before measurement. Okay, the analogy doesn't work entirely since there is no guarantee torture will give you the truth, in the case the interogee knows it. The method is unreliable in general as there is too little benefit in not lying (to someone that cannot verify it).
 
  • #3
Your operators are not linear, hence do not qualify.
 
  • #4
Killtech said:
I ran into an issue of expressing certain types of operators
In the course of doing what? It looks like you are doing personal research. That is off limits for PF discussion.
 
  • #5
A. Neumaier said:
Your operators are not linear, hence do not qualify.
Oh, right. The integral ##\Psi \rightarrow \int \Psi d\delta## defined by a measure is a linear operator, but the measure itself is just sigma additive in the set. Doing that i end up with the projection operator onto ##|\Psi\rangle##. I just mistaken the definition and ended up with the wrong thing.

PeterDonis said:
In the course of doing what? It looks like you are doing personal research. That is off limits for PF discussion.
Just wanted to know how any possible ensemble distribution can be properly depicted in QT and i am just struggling to understand how density operators can do the job. Okay, i figured there is a purification method... but in order be able to depict some simple distributions (e.g. uniform distribution over all superpositions spanned by a finite state basis) i would have to take a limit over a sequence of purifications and even then i don't know if the limit even covers everything (or exists).

Usually in the most general case of an ensemble can be simply represented via a probability measure over the state space. In classical mechanics the state space is small enough so that you have a Lebesgue-measure on it, therefore you can work with its Lebesgue density function to represent the ensemble. It's measure is then given by the Lebesgue integral over the density.

So i wondered how the classical case compared to this situation and if its technical definition wasn't the more general one.
 
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Related to Can Dirac Delta Functional Observables Be Regarded as True Quantum Observables?

1. What is a weird type of observable?

A weird type of observable is a phenomenon or event that is unusual, unexpected, or difficult to explain based on current scientific understanding. It may challenge existing theories and require further investigation to fully understand.

2. How do scientists study weird types of observables?

Scientists use various methods such as observation, experimentation, and mathematical modeling to study weird types of observables. They may also collaborate with other scientists and use advanced technology to gather data and analyze the phenomenon.

3. Can weird types of observables be explained by known scientific laws?

In some cases, weird types of observables can be explained by known scientific laws, but in other cases, they may require new theories or modifications to existing ones. This is why scientific research and exploration are important to expand our understanding of the world around us.

4. Are weird types of observables only found in outer space?

No, weird types of observables can be found in various areas of study, including biology, chemistry, and physics. They can occur in different environments, such as outer space, deep sea, or even in a laboratory setting.

5. Why are weird types of observables important in science?

Weird types of observables are important in science because they challenge our current understanding and push the boundaries of knowledge. By studying them, scientists can gain new insights and potentially discover groundbreaking discoveries that can advance our understanding of the world.

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