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Assume that initial wave function had the form of $\psi(x)= u_1(x) + u_2(x)$ where $u_1$ and $u_2$ are eigenfunctions of $\psi(x)$ to an observable operator $S$. The eigenvalues of $u_1$ and $u_2$ are 1 and 0 respectively.

Then let the observable $S$ act upon the initial wave function. Then, the final wave function has the form of either $u_1$ or $u_2$, but $u_2$'s eigenvalue is zero!

If the resultant wave function was indeed $u_2$, then because its eigenvalue is zero, isn't this equivalent to a zero state? In other words, what does a zero eigenvalue mean to its corresponding eigenstate?

(I'm confused because in simple systems I've learned so far, the most prominent observable, which is Hamiltonian, never yielded zero eigenvalue, and eigenvalues corresponds to the energy level.)

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    $\begingroup$ You're mixing up the concepts of "measuring an observable $S$" and "acting with the operator $S$ on the state of the system $\psi$ to make $S\psi$". These two things have nothing to do with each other! We almost never care about 'acting with observables'. $\endgroup$
    – knzhou
    Apr 21, 2016 at 6:13
  • $\begingroup$ @knzhou if we were to measure an observable $S$, then shouldn't we act the operator $S$ onto whatever system we want to (a wavefunction) and read its eigenvalues and eigenfunctions? $\endgroup$
    – VladeKR
    Apr 21, 2016 at 6:21
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    $\begingroup$ We do care about the eigenvalues and eigenfunctions of $S$. But the final state of the system after measurement is not $S\psi$. $\endgroup$
    – knzhou
    Apr 21, 2016 at 6:23
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    $\begingroup$ No, the final state is $u_n(x)$ with probability $|a_n|^2$, and your measuring device reads $\lambda_n$ in that case. $\endgroup$
    – knzhou
    Apr 21, 2016 at 6:29
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    $\begingroup$ @VladeKR There's nothing special about a zero eigenvalue! For example, you can always replace the Hamiltonian $H$ with $H - E_k$ where $E_k$ is the energy of state $k$. Then state $k$ has zero eigenvalue. $\endgroup$
    – knzhou
    Apr 21, 2016 at 6:51

3 Answers 3

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Then let the observable S act upon the initial wave function. Then, the final wave function has the form of either u1 or u2, but u2's eigenvalue is zero!

But this simply isn't true; acting on a general state with the observable $S$ does not collapse the state to an eigenstate of $S$.

To be sure, if the system is in a general state $|\psi\rangle$, the state $S|\psi\rangle$ is not an eigenstate of $S$. For example:

$$S(c_1 |s_1\rangle + c_2 |s_2\rangle) = c_1 s_1 |s_1\rangle + c_2 s_2 |s_2\rangle)$$

But the measurement postulate states that, immediately after a measurement of $S$, the system is in an eigenstate of $S$. Thus, measurement of $S$ is not operating with $S$.

What does a zero eigenvalue mean to its eigenstate?

If $S|s_0\rangle = 0$ then, if a measurement of $S$ yields $0$, the state of the system immediately after the measurement is $|s_0\rangle$.

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A zero Eigenvalue for the system means that the "physical quantity" observed yielded zero. This is easiest thought as of spin or charge or even quantum numbers such as charm and strange.

S would measure and give either 0 or 1 for charm in this chase. This means that we can measure between the particle having a charm and having no charms. The analogy holds for spins of charges too.

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    $\begingroup$ I understand the meaning of eigenvalues, but what I'm not really getting is the corresponding eigenstate. $\endgroup$
    – VladeKR
    Apr 21, 2016 at 5:58
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Reading the comments it looks like your confusion is in the measurement process.

If you start with an initial state $\psi(x) = \sum_n a_n u_n(x)$ where $u_n(x)$ are the normalized Eigenfunctions of the measurement operator with Eigenvalues $\lambda_n$, then measurement returns the value $\lambda_n$ with probability $|a_n|^2$, after which the system is in state $\psi'(x)=u_n(x)$ (up to an unimportant phase). The wavefunction is not multiplied by the Eigenvalue after measurement, so whether the Eigenvalue is zero or not makes no difference to the measurement process.

It is clear that this should be true because (for example), shifting the Hamiltonian by a constant cannot affect the observable behaviour of the system.

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