# Amplitude's time dependence in Heisenberg representation

1. Apr 6, 2013

### LayMuon

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

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.

2. Apr 6, 2013

### 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
$$\langle q(t)|q'(t) \rangle=\langle q(0) | q'(0) \rangle=\delta(q,q'),$$
where the $\delta$ stands for a Kronecker $\delta$ (discrete eigenvalues) or a Dirac $\delta$ distribution (continuous eigenvalues).