I am reading chapter 3 of Quantum Mechanics - A Modern Development by Leslie E Ballentine, where he derives the operators for the common dynamical variables from space-time symmetry considerations.

At the start, he states that for each space-time transformation there must be a transformation of observables, $A \to A'$, and of states, $|\Psi\rangle \to |\Psi'\rangle$, following certain relations:

  1. If $A|\phi_n\rangle = a_n|\phi_n\rangle$, then $A'|\phi'_n\rangle = a_n|\phi'_n\rangle$.

  2. $|\psi\rangle = \sum_n c_n|\phi_n\rangle \to |\psi'\rangle = \sum_n c'_n|\phi'_n\rangle$, where $\left\{|\phi_n\rangle\right\}$ and $\left\{|\phi'_n\rangle\right\}$ are the eigenvectors of $A$ and $A'$ respectively. The two state vectors must obey $|c_n|^2 = |c_n'|^2$; that is, $|\langle\phi_n|\psi\rangle|^2 = |\langle\phi'_n|\psi'\rangle|^2$.

He then continues with Wigner's theorem, and so on. My issues begin with point 1. For some operators and transformations this makes intuitive sense to me, but not for others. Take for example the position operator $Q$ and a space translation $\mathbf x \to \mathbf x' = \mathbf x + \mathbf a$. If a particle was localized about $\mathbf x$ before the translation, it would be localized about $\mathbf x' = \mathbf x + \mathbf a$ after it. How does that correspond to

$$Q'|\mathbf x'\rangle = \mathbf x |\mathbf x'\rangle,$$

as implied by point 1 above? (Now, I know $|\mathbf x\rangle$ does not represent a particle at $\mathbf x$, but still.) My intuition would instead tell me that $Q'|\mathbf x'\rangle = \mathbf x' |\mathbf x'\rangle$, so apparently I am missing something.


1 Answer 1

  • Maybe the wikipedia article could help you : https://en.wikipedia.org/wiki/Symmetry_in_quantum_mechanics

  • You could think of it as $|x'\rangle=T|x\rangle$ with some translation operator $T$ that maps $|x\rangle$ onto $|x'\rangle$ and $T^{-1}$ the mapping back $T^{-1}|x'\rangle=|x\rangle$. We could then take $Q'=TQT^{-1}$ and evaluate the action of $Q'$ on a state $|x'\rangle$ as $$Q'|x'\rangle=TQT^{-1}T|x\rangle=TQ|x\rangle=xT|x\rangle=x|x'\rangle$$

  • So by the symmetry transformation you change you states $|x'\rangle\rightarrow|x\rangle$ but you also change your operators (this is the important point).

  • This does not mean, that $Q$ is in our case invariant under the transformation as it is modified to $Q'$.

  • An operator $A$ would be invariant under a symmetry transformation ($\Omega$-Operators) if $A\psi=A'\psi$ or in other words $A\Omega\psi=\Omega A\psi$

  • As you've correctly states the position operator is not invariant under translations.

  • We could show that for example the momentum operator with the plane wave basis of momentum states $e^{-ikx}$ is invariant under translations $x'=x+a$. $$pTe^{-ikx}=-i\hbar \nabla T e^{-ikx}=-i\hbar \nabla e^{-ik(x+a)}=\hbar k e ^{-ik(x+a)}=T\hbar k e^{-ikx}=Tpe^{-ikx}$$.

  • $\begingroup$ Sorry for responding to your answer so late. I pretty much realized what my main misunderstandings were and then forgot that I had posted this question. Anyhow, I will keep the question up despite its flaws, so that you get credit for your answer. Thank you for responding! $\endgroup$
    – ummg
    Sep 23, 2020 at 1:11
  • $\begingroup$ @ummg I see that you say here that you understand what Ballentine meant. I know it's been a long time, but would you consider letting me know at this question (physics.stackexchange.com/questions/745785/…) if you understood what Ballentine meant? I am not sure I agree that the answer here suffices. In particular, Ballentine seems to say that the properties (1) and (2) you give are what we must demand, and then goes on to show in (3.2) that this implies the transformation rule for operators... $\endgroup$
    – EE18
    Jan 24, 2023 at 16:46
  • $\begingroup$ ...This is different than what is done here, wherein the operator transformation rule (3.2) is shown to be consistent with requirement (1). It doesn't answer why requirement (1) is fundamental. Also, if you're OK with it, I would love to have the chance to ask you some questions about some points in Ballentine -- I've not found anyone available to study it with. I can be reached at [email protected], though I understand if you prefer to remain anonymous and not reach out. $\endgroup$
    – EE18
    Jan 24, 2023 at 16:48

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