Amplitude's time dependence in Heisenberg representation

[LayMuon]

<br /> <br /> A = \langle q_f(t) \mid q_i(t) \rangle = \langle q_{f,H} \mid e^{iH(t_0-t)} e^{-iH(t-t_0)} \mid q_{i,H} \rangle = \langle q_{f,H} \mid q_{i,H} \rangle<br /> <br />

This means that A is time-independent, and depends only on the reference point $t_0$. How is it possibly? From Schoedinger picture it does depend on time! Thanks.

 

[vanhees71]

You have to be careful about the meaning of your vectors. From what you write, I conclude that |q(t) \rangle are the (generalized) eigenvectors of some operator representing an observable in the Heisenberg picture. Then, from the Heisenberg equation of motion
\mathrm{d}_t \hat{O}=\frac{1}{\mathrm{i}}[\hat{O},\hat{H}].
Assuming we have time independent Hamiltonian and that implies
\hat{O}(t)=\exp(\mathrm{i} t \hat{H}) \hat{O}(0) \exp(-\mathrm{i} t \hat{H}). For sake of simplicity, I've set t_0=0.

Now by definition |q(t) \rangle is the eigenvector of \hat{O}(t) with the fixed eigenvalue q:
\forall t: \quad \hat{O}(t) |q(t) \rangle=q |q(t) \rangle.
This implies that
|q(t) \rangle=\exp(\mathrm{i} t \hat{H}) |q(0) \rangle.
So up to a sign this implies your formula.

Of course, because the time evolution is unitary, the scalar product is time independent
\langle q(t)|q&#039;(t) \rangle=\langle q(0) | q&#039;(0) \rangle=\delta(q,q&#039;),
where the \delta stands for a Kronecker \delta (discrete eigenvalues) or a Dirac \delta distribution (continuous eigenvalues).