[Physics] Completeness relations of eigenstates in the Heisenberg picture

Question

I've been reading Srednicki's introduction to path integrals and I'm slightly unsure of the notation that he uses for the completeness relation of position eigenstates in the Heisenberg picture. In particular, when he writes $$\mathbf{1}=\int_{-\infty}^{\infty}dq_{j}\lvert q_{j}\rangle\langle q_{j}\rvert$$ is this the instantaneous completeness relation for the position eigenstates in the Heisenberg picture? i.e. Given the instantaneous position eigenstate $\lvert q_{j},t_{j}\rangle =\hat{U}^{\dagger}\lvert q_{j}\rangle$ at time $t_{j}$, then we have $$\int_{-\infty}^{\infty}dq_{j}\lvert q_{j},t_{j}\rangle\langle q_{j},t_{j}\rvert =\int_{-\infty}^{\infty}dq_{j}\hat{U}^{\dagger}\lvert q_{j}\rangle\langle q_{j}\rvert\hat{U}= \hat{U}^{\dagger}\left(\int_{-\infty}^{\infty}dq_{j} \lvert q_{j}\rangle\langle q_{j}\rvert\right)\hat{U} =\hat{U}^{\dagger}\hat{U} =\mathbf{1}$$ where $\lvert q_{j}\rangle$ are the constant position eigenstates corresponding to those in the Schrödinger picture at time $t_{j}$, and thus satisfy the completeness relation $\int_{-\infty}^{\infty}dq_{j}\lvert q_{j}\rangle\langle q_{j}\rvert =\mathbf{1}$. (We've also used that $\hat{U}^{\dagger} \hat{U}=\mathbf{1}$.)

Answer 1

I) Let us for clarity use a subscript "$S$" (and "$H$") to denote the Schrödinger (Heisenberg) picture, where bras and kets evolve (are unchanged) and operators are unchanged (evolve), respectively.

Moreover, let us assume that the two pictures coincide at the instant $t_0$ (which Ref. 1 assumes is $t_0=0$).

II) Recall first of all the possibly confusing fact that the Heisenberg instantaneous position eigenstate $|q,t\rangle_H $ does not evolve in time but does depend on the time parameter $t$, such that

$$\hat{Q}(t)_H ~|q,t\rangle_H~=~q~|q,t\rangle_H , \qquad \hat{Q}(t)_H~\equiv~e^{i\hat{H}(t-t_0)/\hbar} \hat{Q}_S ~e^{-i\hat{H}(t-t_0)/\hbar}. $$

The Heisenberg instantaneous position eigenstates satisfy

$$\int_{\mathbb{R}}\! \mathrm{d}q~| q,t\rangle_H ~{}_H\langle q,t|~=~{\bf 1}$$

for all $t$. Moreover

$$|q,t_2\rangle_H~=~e^{i\hat{H}(t_2-t_1)/\hbar} |q,t_1\rangle_H.$$

III) Let us now return to OP's question about whether the ket $|q\rangle $ in Ref. 1 is in the Schrödinger or in the Heisenberg picture? The ket $|q\rangle $ satisfies

$$ \hat{Q}_S~|q\rangle~=~q~|q\rangle. $$

The ket $|q\rangle $ does not evolve in time. Hence it is by definition in the Heisenberg picture. In fact, it is the Heisenberg instantaneous position eigenstate $|q,t_0\rangle_H $. Be aware that $|q\rangle $ often is called a Schrödinger picture position eigenket, cf. Ref. 2, because ... that's what it also is!

References:

1. M. Srednicki, QFT, 2007; Chapter 6. A prepublication draft PDF file is available here.

2. R. Dick, Advanced Quantum Mechanics: Materials and Photons, 2012; p.243.

3. J.J. Sakurai, Modern Quantum Mechanics, 1994; Chapter 2.