Explicit form of time evolution operator


by bluesunday
Tags: evolution, explicit, form, operator, time
bluesunday
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#1
Dec16-10, 08:30 AM
P: 8
The time evolution operaton may be written formally as:



This is an actual solution to:



only in the case that [H(t1),H(t2)]=0 (that is: the hamiltonian commutes in different instants of time) Of course, this includes the case of a time independent hamiltonian.

If this is not the case, the actual U(t) may be obtained with the Dyson series.

This is what is stated in Sakurai, for example. But I don't really understand WHY you cannot use the exponential form of U(t) in case the Hamiltonian doesn't commute at different times. I guess it may have something to do with its integrability, but I don't know.

Any hints?
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DrDu
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#2
Dec16-10, 08:55 AM
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Well, you may try it out. Assume that you have H(t)=H_1 for 0<t<t_1 and H(t)=H_2 for t_1< t<t_2. For t>t_2, the time evoltion operator will be a product of two exponentials of the form you gave. Can you write it as a single exponential of an integral if H_1 and H_2 don't commute?
DrDu
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#3
Dec16-10, 08:58 AM
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Btw. there are other representations than the Dyson series, e.g. the Magnus expansion.

bluesunday
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#4
Dec16-10, 09:17 AM
P: 8

Explicit form of time evolution operator


No, of course you can't because the product of exponentials of two operators can't be written as the exponential of the sum unless the operators commute!

Thank you!
A. Neumaier
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#5
Dec16-10, 10:24 AM
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Quote Quote by DrDu View Post
Well, you may try it out. Assume that you have H(t)=H_1 for 0<t<t_1 and H(t)=H_2 for t_1< t<t_2. For t>t_2, the time evolution operator will be a product of two exponentials of the form you gave. Can you write it as a single exponential of an integral if H_1 and H_2 don't commute?
Yes, since it is a unitary operator, and any unitary operator can be written as the exponential of i times a self-adjoint operator. There are even explicit formulas for
the rsult, but they are in terms of infinite series, and not very useful.
DrDu
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#6
Dec16-10, 10:28 AM
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Of course you can write it as a single exponential, but certainly not as the exponential of an integral over H(t).
xepma
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#7
Dec16-10, 10:40 AM
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The time evolution operator for Hamiltonians which are not constant in time is

[tex] U(t,t_0) = \mathcal{T}\exp\left(-\frac{i}{\hbar}\int_{t_0}^t dt' H(t')\right)[/tex]

where [tex]\mathcal{T}[/tex] is the time ordering symbol. This exponential is just a short-way of writing the following power series:

[tex]U(t,t_0) = 1 + \sum_{n=1}^{\infty} \frac{1}{n!}\left(-\frac{i}{\hbar}\right)^n} \int dt_1\cdtos dt_n \mathcal{T}\left[H(t_1)\cdots H(t_n)\right][/tex]

This series is what's actually the time evolution operator. You can look it up in e.g. the books by Rammer, Mahan or Fetter and Walecka.

Again, the proof relies on the fact that this power series is a unitary solution to the Schrodinger equation.
bluesunday
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#8
Dec16-10, 10:43 AM
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I think DrDu is right, when you integrate you are sort of adding hamiltonians at different t's, and then the exponential form doesn't hold anymore unless they all commute.
xepma
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#9
Dec16-10, 10:57 AM
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Quote Quote by bluesunday View Post
I think DrDu is right, when you integrate you are sort of adding hamiltonians at different t's, and then the exponential form doesn't hold anymore unless they all commute.
...which is why the time-ordering operator is introduced. It takes care of that problem.
bluesunday
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#10
Dec16-10, 11:01 AM
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Right xepma :) Thanks to all of you!
DrDu
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#11
Dec16-10, 01:47 PM
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xempa, yes, you can write it with time ordering, but this is nothing else than the Dyson series bluesunday mentioned in the first post.


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