# Time evolution with time dependent Hamiltonians

1. Dec 1, 2013

### wotanub

I understand that in general, it's not true that in the case of a time dependent hamiltonian, the exponential map of the Hamiltonian is not a unitary transformation/the time evolution operator?

$U(t) \ne e^{-i \frac{H(t)}{\hbar} t}$

Is this thing allegedly not unitary or is it just not time evolution?

Why exactly is this the case and how, in general, is the time evolution operator constructed then? Is there a "recipe" in the Schrodinger or Heisenberg picture or is this where the Dirac picture has to come into play?

2. Dec 1, 2013

### Bill_K

Just integrate the time-dependent Schrodinger equation. Where you have H(t) t in the exponent, you'll get instead ∫H(t) dt.

3. Dec 1, 2013

### Chopin

In the case of a time-dependent Hamiltonian, the time evolution operator becomes $U(t) = T\:e^{-i/\hbar \int dt\(t)}$, where $T$ is the time-ordering operator. You need this because the Hamiltonian operator at different times do not commute with each other, so a simple integration of the Schrodinger equation is no longer sufficient.

4. Dec 1, 2013

### rubi

In the case of a time-dependent Hamiltonian, it's not good to use the symbol $U(t)$ anymore, since the time-evolution depends not only on the difference of the initial and final times, but also on their absolute values:
$$U(t_2,t_1) = T\exp\left(-\frac{\mathrm i}{\hbar}\int_{t_1}^{t_2} H(\tau)\mathrm d\tau\right)$$
Strictly speaking, $T$ isn't an operator, but rather just a symbol that indicates how this expression is to be understood. Instead of being a symmetry ($U(t_1)U(t_2)=U(t_1+t_2)$), the time-evolution satisfies only a less general rule ($U(t_3,t_2)U(t_2,t_1)=U(t_3,t_1)$).

Last edited: Dec 1, 2013
5. Dec 5, 2013

### Sugdub

Is it so obvious that the time symbol t points to the same physical quantity in Schrödinger's evolution equation and in the time-dependent Hamiltonian?
In many basic illustrations of Schrödinger's equation the evolution of the state vector actually results from a change in the distance between a “source” and a “detector”. From an operational perspective, the Schrödinger's evolution equation could as well be dependent on a space variable x, the time variable t being due to a metaphor (I don't challenge it but still it is an interpretation and not an experimental fact) whereby a “particle” is assumed to travel at constant speed between the source and the detector.
So in such cases, can one assume that the time symbol in the Schrödinger's equation and the time symbol in the temporal evolution of the Hamiltonian represent different physical quantities and therefore might be assigned different mathematical symbols?

6. Dec 5, 2013

### Chopin

I'm not sure I see how they could. The Schrodinger equation involves the time in both senses that you have used it in this description:$$i\hbar\frac{d}{dt}\psi(t) = H(t)\psi(t)$$

The $t$ in $\psi(t)$ is how the particle evolves in time, and the $t$ in $H(t)$ is how the Hamiltonian evolves in time. The Schrodinger equation tells you that the particle's evolution at any point in time is determined by the value of the Hamiltonian at that same point in time. So the two have to be equivalent. Was that your question, or did I misinterpret it?

7. Dec 6, 2013

### Sugdub

My approach was more directed towards a “phenomenology”, irrespective of any hypotheses or postulates regarding what happens there inside the experimental device.
Assuming the change in experimental conditions consists in modifying the distance between two devices in the experimental setup, I expect that a space variable will drive the unitary evolution of the state vector (the statistical property of the data flow produced by the experiment). And if on top of this the measured evolution of the statistical distribution appears to be only dependent on the distance between successive locations of the detector (this position being fixed during the iterative measurement process), then I expect the evolution described by the Schrödinger's equation to be linear in respect to the space variable, not time.

This was the trigger for my question: it seemed possible to preserve linearity vs space in the Schrödinger's equation in spite of a time-dependent Hamiltonian. But on second thoughts there is no escape: the issue is now whether the Hamiltonian is dependent on the space variable. Sorry for that.