- #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.
$$|Φ⟩\rightarrowδ_{\{|Ψ⟩\}}(|Φ⟩)$$
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.
