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As far as I understood, In Heisenberg's picture, eigen vectors are time independent and Schrodinger's, eigen vectors are time dependent, also they are equivalent. Let $${\lvert q(t) \rangle}_S=e^{-i\hat Ht/\hbar }{\lvert q \rangle}_H$$ be the relation between Heisenberg's and Schrodinger's eigen state basis vectors. Also the completeness property is defined as $$\int_{\mathbf -\infty}^\infty {\lvert q(t) \rangle}_S {\langle q(t)\lvert}_S dq=1 .$$ My question is

  1. Wheather the time-independent eigen basis vector (Heisenberg's view) follow this completeness property?$$\int_{\mathbf -\infty}^\infty {\lvert q(t) \rangle}_H {\langle q(t)\lvert}_H dq=1 .$$ If yes

  2. How can we prove that completeness of Schrodinger's eigen state basis vectors equivalent to completeness Heisenberg's eigen state basis vectors. That is, How can we prove, $$\int_{\mathbf -\infty}^\infty {\lvert q(t) \rangle}_S {\langle q(t)\lvert}_S dq=1 \implies \int_{\mathbf -\infty}^\infty {\lvert q(t) \rangle}_H {\langle q(t)\lvert}_H dq=1$$

I attempted it by writing $$\int_{\mathbf -\infty}^\infty {\lvert q(t) \rangle}_S {\langle q(t)\lvert}_S dq=1 \implies \int_{\mathbf -\infty}^\infty e^{-i\hat Ht/\hbar }{\lvert q(t) \rangle}_H {\langle q(t)\lvert}_H e^{i\hat Ht/\hbar } dq=1 $$ and I was not able to eliminate the unitary operator. Can someone help me to proceed.

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    $\begingroup$ Multiply by $e^{iHt/\hbar}$ on the left and $e^{-iHt/\hbar}$ on the right and you will get what you want. $\endgroup$ – StarBucK Jun 1 at 7:59
  • $\begingroup$ How can we include a position dependent quantity in the integral? Hamiltonian operator depends on the the position operator too. Thus it may affect the integral. $\endgroup$ – walber97 Jun 1 at 9:15
  • $\begingroup$ @walber97 no, it doesn't. $q$ here is simply a label for the basis states. $H$ is a basis independent object (even if it is often convenient to represent it in the position basis) $\endgroup$ – By Symmetry Jun 1 at 9:28
  • $\begingroup$ I am sorry that I still have confusion with this. $\hat H $ is depends on position operator. Thus action of position operator on position eigen vector will change the eigen value and thus the integrand. Is it like that? $\endgroup$ – walber97 Jun 2 at 4:39

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