3
$\begingroup$

I'm a little confused about a recent discussion, where a one dimensional wave-function $\Psi(x)$ was considered, associated with a state

$$|\Psi\rangle = \int dx \, \Psi(x) |x\rangle .$$

Now consider a projection operator, somehow narrowing x to be between a and b:

$$P = \int_a^b dx \, |x\rangle \langle x | .$$

The resulting state would be

$$P |\Psi\rangle = \int dx \int_a^b \, dx' |x'\rangle \langle x'|x\rangle \Psi(x)= \int_a^b d x' \Psi(x') |x'\rangle . $$

I would say the result of applying this operator is to project onto a reduced subspace of $|x\rangle$, defined by the interval $[a,b]$.

Now the confusion: is that operator at the same time a measurement? If yes, on what? More precise: what is the observable? I think it cannot define a "measurement" because in that case a reduction of the wave function must occur. Here there is just a reduction of the original Hilbert space to a space containing only vectors $|x\rangle$ with $a \le x \le b$, but not a collapse onto a particular wave function. So in my opinion the projection operator P itself can not be regarded as a measurement.

However, from textbook knowledge, after a measurement the state of the system must be an Eigenstate of a Hermitian operator.

Is it correct to say that P is Hermitean and has Eigenvalues 0 and 1, for vectors $\int dx \Psi(x) |x\rangle $ with $\Psi(x)$ vanishing for any x outside or inside $[a,b]$ respectively? Then the measurement we talk about would be the binary decision whether we find $x$ inside or outside $[a,b]$ and the measured value is the binary information "inside" or "outside" ? In my opinion such measurement does not need to end up in a single $|x\rangle$ as the reduced wave function as it would be in the case of position measurement; I can only say that after measurement the resulting state is with a $\Psi(x)$ either completely vanishing inside or outside $[a,b]$.

Can I say that such process of "incomplete" reduction is a measurement at all? I have some doubts, because this kind of measurement would end up in sub state, that is still a superposition of base vectors and does, in general, not reduce to a particular one.

Please correct me where I'm not correct.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

What you wrote here is exactly right:

Is it correct to say that $P$ is Hermitean and has Eigenvalues $0$ and $1$, for vectors $\int dx\ \psi(x)|x\rangle$ with $\psi(x)$ vanishing for any $x$ outside or inside $[a,b]$ respectively? Then the measurement we talk about would be the binary decision... "inside" or "outside"? ...does not need to end up in a single $|x\rangle$ ... I can only say that after measurement the resulting state is with a $\psi(x)$ either completely vanishing inside or outside $[a,b]$.

Real position-measurements don't have infinite precision, so this is actually a better model of a real position measurement than the typical textbook collapses-to-$|x\rangle$ model.

The projection operator $P$, together with its complement $1-P$, constitute a dichotomic observable — an observable that, when measured, has only two possible outcomes. For convenience, we can use a single-operator representation of this observable, like $A=\alpha P+\beta(1-P)$ where $\alpha$ and $\beta$ are distinct real numbers used to label the different outcomes, but that's just for convenience. Nature doesn't care how we label things.

In general, the effect of an idealized measurement is to project onto a subspace that is not one-dimensional. For example if you perfectly measure the polarization of an electron in the up-down basis, you only have two possible outcomes: up and down. But to describe the world, you need a Hilbert space with many more than two dimensions (nominally infinitely many). So the effect of the electron-polarization measurement is to project the original state-vector into one of the two complementary infinite-dimensional subspaces: one in which the electron has spin up (and the rest of the world is whatever), and one in which the electron has spin down (and the rest of the world is still whatever).

Applying a projection operator, whatever the dimensionality of the corresponding subspace may be, is just a mathematical operation that may or may not be associated with any measurement. The question of how to recognize whether or not a given observable has actually been measured is one that is usually treated only empirically. In principle, it can be treated mathematically in a model that includes a microscopic description of the measurement equipment and its surroundings, modulo the same ambiguities in the definition of "measurement" that are present in the purely-empirical case, but the math is difficult. It is rarely attempted, except in relatively simple cases.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.