# A Time correlation of an observable in the Schrödinger picture

1. Mar 7, 2016

### kith

I think that an interpretation of this in the Schrödinger picture should be possible at least in the style of Feynman. This would go something like this.

$\begin{eqnarray*} \langle A(s)A(t)\rangle_\psi &=& \langle \psi| U^\dagger (s) A U(s) U^\dagger(t) A U(t)|\psi\rangle\\ &=& \langle \psi(s)| A U(s-t) A |\psi(t)\rangle\\ &=& \langle \psi(s)| \left( \sum_i |a_i\rangle \langle a_i| \right) A U(s-t) A \left( \sum_j |a_j\rangle \langle a_j| \right) \psi(t)\rangle\\ &=& \sum_{i,j} a_i a_j \langle \psi(s)| a_i\rangle \langle a_i| U(s-t) |a_j\rangle \langle a_j| \psi(t)\rangle\\ \end{eqnarray*}$

For each term in the sum, we have an initial probability amplitude $\langle a_j| \psi(t)\rangle$, a propagator-like quantity $\langle a_i| U(s-t) |a_j\rangle$ and a final probability amplitude $\langle \psi(s)| a_i\rangle$, and we sum over all possibilities weighted with the product of the corresponding eigenvalues. This doesn't look completely meaningless to me, although I'm scratching my head a bit about what it's actual meaning could be. ;-)

Any thoughts?

2. Mar 7, 2016

### A. Neumaier

Maybe one can relate it to consistent histories?

3. Mar 7, 2016

### kith

I am not really familiar with this approach but it seems to be a second step to me. I don't yet understand what one such a history or path is, i.e. how the elements in the mathematical expression above could be interpreted. And maybe how this relates to the path integral about which I also have only superficial knowledge.

4. Mar 7, 2016

### Jilang

I think that the next step would be to break down the propagator into many terms and inserting many complete terms of energy eigenstates.

5. Mar 8, 2016

### vanhees71

Very nice, but you should really indicate the quantities in different pictures with different symbols. Obviously you start with the Heisenberg picture with a system prepared in the pure state represented by the state vector
$$|\psi \rangle_H$$
which is constant in time. The observable $A$ is represented by the operator $\hat{A}_H(t)$ obeying the equation of motion
$$\dot{\hat{A}}_H(t)=\frac{1}{\mathrm{i}} [\hat{A}_H(t),\hat{H}_H],$$
where I assume that $\hat{A}$ is not explicitly time dependent and that $\hat{H}_H$ is time independent (closed system, energy conserved). The solution reads
$$\hat{A}_H(t)=\hat{U}(t) \hat{A}_H(0) \hat{U}^{\dagger}(t)=\hat{U}(t) \hat{A}_S \hat{U}^{\dagger}(t)$$
with
$$\hat{U}(t)=\exp(\mathrm{i} \hat{H}_H t).$$
I assumed that the Schrödinger picture coincides with the Heisenberg picture at $t=0$. We obviously have $\hat{H}_H=\hat{H}_S$ in this case. The unitary transformation from one to the other picture is given by $\hat{U}$:
$$\hat{A}_H=\hat{U} \hat{A}_S \hat{U}^{\dagger}, \quad |\psi,t \rangle_H= |\psi,0 \rangle_H=|\psi,0 \rangle_{S}=\hat{U}(t) |\psi,t \rangle_S.$$
Now indeed you have
$$C_{AA}(s,t)=_H\langle \psi,0|\hat{A}_H(s) \hat{A}_H(t)|\psi,0 \rangle_H=_S \langle \psi,t|\hat{U}^{\dagger}(t) \hat{U}(s) \hat{A}_S \hat{U}^{\dagger}(s) |\hat{U}(t) \psi,t \rangle= _S \langle \psi,t| \hat{U} (s-t) \hat{A}_S \hat{U}^{\dagger}(s-t)|\psi,t \rangle_S.$$
This is the picture independent expression. You have to read $\hat{U}$ as the unitary matrix transforming from the arbitrary picture you like to use (here the Schrödinger picture) to the Heisenberg picture. The direct derivation of this is a bit lengthy but as simple as this one for the special case of looking at the Schrödinger and Heisenberg pictures.

I've given the formulation of the transformations of states and observable-operators as well as transformations between arbitay pictures in

https://www.physicsforums.com/threa...n-the-heisenberg-picture.859478/#post-5393704

Last edited: Mar 10, 2016
6. Mar 9, 2016

### kith

Please excuse my sloppiness. ;-) I think omitting the indices is ok as long as we don't deal with operators which are time-dependent even in the Schrödinger picture.

Hmm, how did you arrive at the second identity?

7. Mar 9, 2016

### kith

My next step would be to consider a simple spin-1/2 example but unfortunately, I'm really busy right now.