1
$\begingroup$

Suppose that the state of a system can be represented as a superposition of finite position eigenstates $|\Psi \rangle = \sum_{i}c_{i}|x_{i}\rangle$, where each $|x_{i}\rangle$ corresponds to a unique subregion $X_{i}$ of a larger region $X$ so that $\cup X_{i} = X$.

Let A be the observable corresponding to the question "Is the system in the region $X$?"

  1. How do you construct A in terms of $|x_{i}\rangle$?
  2. Are there any conditions for $|\Psi\rangle$ to be an eigenstate of A?
  3. Is A commutable with the position operator?
$\endgroup$

1 Answer 1

2
$\begingroup$

Let's ignore the complications that arise with a continuous basis, since this is really a general question about more general quantum projective measurements. So, let's pretend that the basis $|x\rangle$ is actually discrete.

Then, since the measurement doesn't distinguish between different values of $x$ that lie in $X$ and doesn't distinguish between different values of $x$ that lie outside of $X$, we can say that there are only two outcomes of the measurement, 1 corresponding to $x\in X$ and 0 corresponding to $x \notin X$. By doing that, we can write down the operator $$ \hat{A} = \sum_{i\in X} (1)|x_i\rangle\langle x_i| + \sum_{i\notin X} (0)|x_i\rangle\langle x_i| =\sum_{i\in X} |x_i\rangle\langle x_i|\,. $$ This operator represents the "physical observable" that we are measuring here. Notice that:

  1. It's explicitly written in the $|x\rangle$ basis.
  2. The conditions for $|\Psi\rangle$ to be an eigenstate of $\hat{A}$ is that it is either a linear combination of states $|x\rangle$ where $x\in X$ or it's a linear combination of states $|x\rangle$ where $x\notin X$.
  3. It clearly commutes with $\hat{x}$ because it is diagonal in the eigenbasis of $\hat{x}$.

For continuous variables, you really can't talk about the probability of getting a finite (or even countable) subset of the values of $x$. To get a feeling for what you have to do in this case, we usually think about intervals on the real line. That is, let's suppose we want an operator that corresponds to the question, "Is the particle in $[a,b]$ or not"? Then, the corresponding operator is $$ \hat{A} = \int_a^b |x\rangle\langle x|\,dx\,. $$ This also commutes with $\hat{x}$, and as long as a wave function $\Psi(x)$ has support completely contained in $[a,b]$ or completely disjoint from $[a,b]$, then the corresponding state $|\Psi\rangle$ will be an eigenstate of $\hat{A}$ (in the loosey-goosey physics sense of eigenvectors of operators like $\hat{x}$, because, strictly speaking, they don't actually have eigenvectors in the strict mathematical sense, but that's another story).

$\endgroup$
15
  • $\begingroup$ Thanks! Does the 'strict mathematical sense' in which $|\Psi\rangle$ is not an eigenstate of $A$ have to do with the continuous nature of the position states? $\endgroup$
    – Lory
    Apr 14, 2023 at 0:52
  • 2
    $\begingroup$ @Torpido Yes. My understanding is that for eigenvalues in the continuous part of an operator's spectrum, there is no eigenvector, but there are approximate eigenvectors. The issue, for instance, is that the "eigenvector" of the position operator is $\delta(x-x')$ in position space, which is not part of the Hilbert space. $\endgroup$
    – march
    Apr 14, 2023 at 1:11
  • $\begingroup$ The solution is kinda easier than that. Quantum states live in Hilbert space. The eigenvectors corresponding to the continuous eigenvalues of any observable (Hermitian) operator lives in Rigged Hilbert space. We are allowed to do mathematical manipulations using these eigenvectors living in Rigged Hilbert space, as long as we remember that, when we need to deal with quantum states, they cannot be one of these eigenvectors, only wavepackets of them. $\endgroup$ Apr 14, 2023 at 7:43
  • $\begingroup$ Now, position eigenstates is actually even worse than that. We love to work with position eigenstates in NR QM, but actually SR QFT disallows that. The concept of position eigenstates turn out to violate Lorentz covariance, and this is supremely annoying for everybody. $\endgroup$ Apr 14, 2023 at 7:44
  • $\begingroup$ Thanks for the addition! Now, I understand that the operator A you defined commutes with the position operator, but is it quite often the case that a system is in an eigenstate of A but not in an eigenstate of B, where [A, B] = 0? $\endgroup$
    – Lory
    Apr 17, 2023 at 6:22

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.