Meaning of eigenvalue of projection operator/projector

[LightPhoton]

TL;DR

Measurement of eigenvalues of projector

Consider an energy eigenstate. We know that measuring the energy of this state we will get the eigenvalue of it as our measurement value.
But what is the meaning of the eigenvalues of the projector? Since the projector is Hermitian, its eigenvalue is observable, but how do we measure it?

 

[DrClaude]

Not all observables are easy to observe in reality. But a projector is simply a detector that clicks when it is hit by a particle in the corresponding state, and doesn't click otherwise.

 

[PeterDonis]

the projector

What projector? There is more than one.

 

[LightPhoton]

What projector? There is more than one.

$P = \sum_j \ket{j} \bra{j}$, having an eigenvalue of 1

 

[PeterDonis]

$P = \sum_j \ket{j} \bra{j}$, having an eigenvalue of 1

$1$ is not the only eigenvalue of a projector; it also has an eigenvalue of $0$.

Also, that's all abstract, so asking what the meaning of the eigenvalues is doesn't connect to any actual experiment. If you pick a specific projector and ask your question about it, I think it will be easier to see what the meaning of the eigenvalues is.

 

[LightPhoton]

$1$ is not the only eigenvalue of a projector; it also has an eigenvalue of $0$.

Also, that's all abstract, so asking what the meaning of the eigenvalues is doesn't connect to any actual experiment. If you pick a specific projector and ask your question about it, I think it will be easier to see what the meaning of the eigenvalues is.

Lets say i have a generic state, $\ket{\psi}$ and i apply $P$ to it, getting $P\ket\psi=(1)\ket\psi$. How would i measure this "1"?

 

[PeterDonis]

Lets say i have a generic state, $\ket{\psi}$ and i apply $P$ to it, getting $P\ket\psi=(1)\psi$. How would i measure this "1"?

There is no way to answer this question as you ask it because you are not telling us anything about what the state describes. What actual, real, physical system are you measuring? Without that information your question is pointless.

 

[LightPhoton]

Without that information your question is pointless.

How exactly? If I can say

Consider an energy eigenstate. We know that measuring the energy of this state we will get the eigenvalue of it as our measurement value.

without ever mentioning anything about the state except the fact that it is an energy eigenstate, and it makes perfect sense, then why doesn't saying it's a generic eigenstate of projector valid?

 

[renormalize]

If I can say...
without ever mentioning anything about the state except the fact that it is an energy eigenstate, and it makes perfect sense, then why doesn't saying it's a generic eigenstate of projector valid?

Because in quantum mechanics a generic energy eigenstate satisfies $\hat{H}\psi_{E}=E\psi_{E}$ where the Hamiltonian operator is the specific operator $\hat{H}=-\frac{\hslash}{2m}\nabla^{2}+V\left(x\right)$ that represents the physical quantity "energy". But for $P$ an unspecified projector, the eigenvalue equation $P\psi=\lambda\psi$ has no physical content unless and until you explicitly write the specific form of the projection operator. Can you do that for some physical projection of interest to you?

 

[PeterDonis]

why doesn't saying it's a generic eigenstate of projector valid?

Because "energy" tells us what physical observable is involved, and that tells us what we're measuring and how to measure it. We know how to measure energy and what measuring energy means.

A "generic" state of a generic projector doesn't give any of that information. It could be anything. So how can you possibly expect us to tell you anything about how to measure it, or what measuring it means, if we don't even know what it is?

 

[Nugatory]

without ever mentioning anything about the state except the fact that it is an energy eigenstate, and it makes perfect sense, then why doesn't saying it's a generic eigenstate of projector valid?

An “energy eigenstate” isn’t a generic eigenstate, it’s an eigenstate of a specific operator, the Hamiltonian.

 

[LightPhoton]

Because "energy" tells us what physical observable is involved, and that tells us what we're measuring and how to measure it. We know how to measure energy and what measuring energy means.

A "generic" state of a generic projector doesn't give any of that information. It could be anything. So how can you possibly expect us to tell you anything about how to measure it, or what measuring it means, if we don't even know what it is?

Consider a photon having the following state:
$$\ket\psi=\frac{1}{\sqrt 2}(\ket v+\ket h)$$
where $\ket v$ is the state of the photon in vertical polarisation and $\ket h$ in horizontal.
Let the projection operator be:
$$\hat P=\ket v\bra v+\ket h\bra h$$
We want to measure "1" in
$$\hat P\ket\psi=(1)\ket\psi$$

 

[martinbn]

Consider a photon having the following state:
$$\ket\psi=\frac{1}{\sqrt 2}(\ket v+\ket h)$$
where $\ket v$ is the state of the photon in vertical polarisation and $\ket h$ in horizontal.
Let the projection operator be:
$$\hat P=\ket v\bra v+\ket h\bra h$$
We want to measure "1" in
$$\hat P\ket\psi=(1)\ket\psi$$

In this case you can measure polarisation at a 45degree angle. But in general why do you think that an eigenstate to a projection will relate to something that can be measured?

 

[LightPhoton]

Not the eigenstate, but the eigenvalue. As I said in the original post " Since the projector is Hermitian, its eigenvalue is observable, but how do we measure it?"

 

[martinbn]

Not the eigenstate, but the eigenvalue. As I said in the original post " Since the projector is Hermitian, its eigenvalue is observable, but how do we measure it?"

You are confused by the word observable. It doesn't mean that you can measure it in the laboratory. So, why do you think that a general eigenvalue of a projector will relate to something you can measure?

 

[PeroK]

TL;DR Summary: Since the projector is Hermitian, its eigenvalue is observable, but how do we measure it?

An observable is represented by a Hermitian operator, but it doesn't logically follow that every Hermitian operator corresponds to a physically realistic observable.

 

[PeterDonis]

We want to measure "1" in
$$\hat P\ket\psi=(1)\ket\psi$$

Is $\ket{\psi}$ an eigenstate of $P$ with eigenvalue $1$?

 

[Demystifier]

You never measure the observable directly. Instead, you have a complicated apparatus with a macroscopic pointer, which is designed so that the position of the pointer is correlated with the value of the observable. For a simple model of the pointer for an arbitrary observable see the von Neumann theory of quantum measurement. A nice simple explanation can be found in Sec. 6.1 of https://arxiv.org/abs/1406.5535 . Since it works for arbitrary observable, it can also be applied to a projector as a special case.