Completeness relations of eigenstates in the Heisenberg picture

Question

I've been reading Srednicki's introduction to path integrals (chapter 6 in "QFT") 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

1. 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$^1$.

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

2. Recall first of all the possibly confusing fact that the Heisenberg instantaneous position eigenstate $|q,t\rangle_H $ does not evolve in time but it does depend on the time parameter $t$, such that $$\hat{Q}_H(t) ~|q,t\rangle_H~=~q~|q,t\rangle_H , \qquad \hat{Q}_H(t)~\equiv~e^{i(t-t_0)\hat{H}_S/\hbar} \hat{Q}_S ~e^{-i(t-t_0)\hat{H}_S/\hbar}. $$ 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 $$|q,t_2\rangle_H~=~e^{i(t-t_0)\hat{H}_S/\hbar} |q,t_1\rangle_H.$$

3. 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.

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$^1$ In this answer we assume for simplicity that the Hamiltonian $\hat{H}_S$ in the Schrödinger picture has no explicit time dependence. Then we avoid the issue of time-ordering, cf. e.g. this Phys.SE post.