As it is claimed in this question, the identity operator is an hermitian operator, but not an observable.

However, if I were to build a device, which only measures the existence of an electron in a given region - say in Hydrogen atom. Wouldn't this correspond to the observable identity ?

Because, what I'm observing is the existence of particle, and if it exists, I get 1, if not, I get 0.

This question is, actually, a way of asking the following question; if I have an hermitian operator, what are the criteria for it to be an observable ? For example, as in this case, to say that "Yes, indeed, the identity operator correspond to the observable that I constructed above", what should one check?

  • $\begingroup$ Your operator cannot be the identity operator because it has a zero eigenvalue. $\endgroup$ – Ryan Thorngren Jan 21 at 6:38
  • $\begingroup$ @RyanThorngren What is the eigenfunction then ? $\endgroup$ – onurcanbektas Jan 21 at 6:39
  • $\begingroup$ You just said you can measure 0. The identity operator always measures 1. $\endgroup$ – Ryan Thorngren Jan 21 at 6:39
  • $\begingroup$ @RyanThorngren I can measure zero if there is no particle, in which case, there is no state to talk about, which corresponds to the zero state (which cannot be an eigenfunction) $\endgroup$ – onurcanbektas Jan 21 at 6:41
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    $\begingroup$ This is not the proper way to discuss quantum systems with variable particle number. Look up "second quantization". $\endgroup$ – Ryan Thorngren Jan 21 at 10:10

By definition of the identity, "measuring the identity operator" always yields 1. Always. A sheet of paper with a big '1' written on it is a perfect measurement apparatus for the identity operator, and it doesn't even need to be connected to your system! It's the best and the worst observable.

If you have an operator that measures 1 in some states and 0 in others, it's not the identity. If the state "the system does not exist" is a state you are modelling, then your actual system is a meta-system whose state of space decomposes into a part "no system exists" and a part "the system exists".

However you are not wholly wrong in that one naturally can use the identity operator on a "subsystem" to get the "existence" operator in a larger system:

Suppose we have a quantum system described by a Hilbert space $H_1$ that embeds into some larger system $H = H_0 \oplus H_1 \oplus H_2$. Then the identity operator on $H_1$ naturally extends to the projection operator onto $H_1$ as an operator on $H$. Measuring this operator now tells you "how much" of your state is in $H_1$.

However, the direct sum (instead of the tensor product) there means that we are not using the standard notion of a quantum subsystem, but rather the notion of a "sector" (as it occurs most prominently in "superselection sector") - a subspace of the space of states that is interesting for some reason but is not a true subsystem.

A common example might be a system with a variable number of particles - its Hilbert space is a direct sum of spaces with fixed particle number. If you measured the projector onto one of these spaces, a result of 1 would mean that your resultant state is now a state with definite particle number corresponding to the subspace, a result of 0 means the resultant state does not include a state of this particle number. That is, the projectors onto the spaces with definite particle number approximate the number operator.

(If $H_n$ is the space with $n$ particles and $P_n$ its projector, then the true number operator is $N = \sum_n n P_n$.)


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