1. Sep 13, 2015

### Sir Beaver

Hi all,

I have a severe confusion about the time-ordering operator. It is the best thing ever, I think, since it simplifies many proofs, due to the fact that operators commute (or anti-commute, but let's take bosonic operators for simplicity) under the time-ordering.

However, sometimes I feel uneasy using it, and, well, I think I have narrowed down why. I think the relevant question is: Do I have to work in a specific picture in order to time-order my objects? I seem to run into inconsistencies if I switch pictures, for example by moving from a Schrödinger picture to a Heisenberg one. Below is the simplest example I have found, which illustrate the issue.

In the usual way, the Heisenberg operator is defined as (here I assume a time-independent Hamiltonian)
$$\hat{A}_H (t) = e^{i\hat{H} t} \hat{A}_S (t) e^{-i\hat{H} t}.$$
By taking the time ordering on both sides, I obtain
$$\hat{A}_H (t) = T [ \hat{A}_H (t) ] = T [ e^{i\hat{H} t} \hat{A}_S (t) e^{-i\hat{H} t}]$$
And since on the right-hand side, everything commutes under the time-ordering, the exponentials cancel, and we end up with the (upsetting) equality
$$\hat{A}_H (t) = \hat{A}_S (t)$$
My question is: where did I go wrong? I ran into the same type of inconsistencies when trying to prove some things on the Keldysh contour, but there it was way less obvious. It seems to be that time is kind of different in different pictures. Could someone confirm this, or point out some mistake I made along the path?

Cheers!

2. Sep 13, 2015

### Heinera

What exactly do you mean by "everything commutes under the time-ordering"?

3. Sep 13, 2015

### Sir Beaver

I mean, that if we have bosonic operators, $A$ and $B$, the equality $$T[ \hat{A} (t) \hat{B} (t') ] = T [ \hat{B} (t') \hat{A} (t) ]$$ holds, even if $A$ and $B$ themselves do not commute. Thus, if the operators are under the time-ordering operator, they can be treated (formally) as if they commute. I should probably have been a bit more precise.

4. Sep 13, 2015

### Heinera

Yes, but from $$T[ \hat{A} (t) \hat{B} (t') ] = T [ \hat{B} (t') \hat{A} (t) ]$$ you cannot conclude that $\hat{A} (t) \hat{B} (t') = \hat{B} (t') \hat{A} (t)$ since the time-ordering operator is not one-to-one. This is what you erroneously have used in your argument above.

5. Sep 13, 2015

### Sir Beaver

I agree with the point that it is not one-to-one in general, and this solves the issue with a string of operators of two or more.
However, is it really what I use here? Considering the equality above (which I think we agree on then)
$$T [ \hat{A}_H (t) ] = T [ \hat{A}_S (t)],$$
is it not weird that I cannot say such a simple statement such as $T [ \hat{A}_H (t) ] = \hat{A}_H (t)$ ? Somewhat more like an identity operator.

Thanks a lot for your time!

6. Sep 13, 2015

### Heinera

But the problem starts way before that. Implicitly, you have $$T [ e^{i\hat{H} t} \hat{A}_S (t) e^{-i\hat{H} t}] = T [ e^{i\hat{H} t}e^{-i\hat{H} t} \hat{A}_S (t) ],$$ but to get the exponentials to vanish you must now assume $$e^{i\hat{H} t} \hat{A}_S (t) e^{-i\hat{H} t} = e^{i\hat{H} t}e^{-i\hat{H} t} \hat{A}_S (t) = \hat{A}_S (t).$$And it is this step from the first to the second equation that is incorrect. Just because the time-ordered results are equal, you can't assume the original arguments are equal.