How can the position operator be displacement invariant?

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.

• 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 $$x'=x+a$$. $$pTe^{-ikx}=-i\hbar \nabla T e^{-ikx}=-i\hbar \nabla e^{-ik(x+a)}=\hbar k e ^{-ik(x+a)}=T\hbar k e^{-ikx}=Tpe^{-ikx}$$.

• 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!
– ummg
Sep 23 '20 at 1:11