What if Hamiltonian is not constant in time?

[jostpuur]

If the Hamilton's operator H(t) depends on the time parameter, what is the definition for the time evolution of the wave function \Psi(t)? Is the equation

<br /> i\hbar\partial_t\Psi(t) = H(t)\Psi(t)\quad\quad\quad (1)<br />

or the equation

<br /> \Psi(t) = \exp\Big(-\frac{it}{\hbar}H(t)}\Big)\Psi(0)\quad\quad\quad (2)<br />

These are not equivalent, because if the wave function satisfies the equation (2), then it also satisfies

<br /> i\hbar\partial_t\Psi(t) = H(t)\Psi(t) + t\big(\partial_tH(t)\big)\Psi(t)<br />

Because these alternatives are not equivalent now, I don't which one to believe in.

 

[Maxim Zh]

The equation (2) seems to be wrong. The right one is

<br /> \Psi(t) = \exp\Big(-\frac{i}{\hbar}\int_0^t \hat{H}(\tau)d\tau}\Big)\Psi(0)\quad\quad\quad (2&#039;)<br />

The more detailed explanation of quantum dynamics is in "Modern Quantum Mechanics" of J.J.Sakurai

 

[jostpuur]

The equation (2) seems to be wrong. The right one is

<br /> \Psi(t) = \exp\Big(-\frac{i}{\hbar}\int_0^t \hat{H}(\tau)d\tau}\Big)\Psi(0)\quad\quad\quad (2&#039;)<br />

The more detailed explanation of quantum dynamics is in "Modern Quantum Mechanics" of J.J.Sakurai

I see, and this is equivalent with the equation (1) of my first post.

But the next question is that "why"? If we know that equations (1) and (2) are equivalent when H is constant, and if we are then given a task of generalising the time evolution to a case where H is not a constant, how do we know that it is the equation (1) which can be generalised directly with a substitution H\mapsto H(t), and not the equation (2)?

 

[Maxim Zh]

...if we are then given a task of generalising the time evolution to a case where H is not a constant, how do we know that it is the equation (1) which can be generalised directly with a substitution H\mapsto H(t), and not the equation (2)?

What do you mean?

 

[SpectraCat]

I see, and this is equivalent with the equation (1) of my first post.

But the next question is that "why"? If we know that equations (1) and (2') are equivalent when H is constant, and if we are then given a task of generalising the time evolution to a case where H is not a constant, how do we know that it is the equation (1) which can be generalised directly with a substitution H\mapsto H(t), and not the equation (2)?

Equations 1 and 2' are always equivalent, whether H is constant in time or not ... equation 2' is just the propagator form to find the time evolution of the wavefunction from some initial state.

 

[jostpuur]

You misunderstood my post, because you accidentally replaced (2) with (2') when quoting me.

 

[Demystifier]

This is analogous to the following question in high-school physics.
When the velocity v is constant, then the path traveled during the time t is
s=vt
But what if v(t) depends on t?
Is it
s=v(t)t
or is it
s=Integral v(t) dt ?
The correct equation, of course, is the last one, but why?

I hope this analogy helps.

If not, then let me be slightly more explicit. Because the Hamiltonian, by definition, is the generator of INFINITESIMAL time translations. Not only in quantum physics, but in classical physics as well. See e.g. classical mechanics formulated in terms of Poisson brackets.

 

[jambaugh]

The equation (2) seems to be wrong. The right one is

<br /> \Psi(t) = \exp\Big(-\frac{i}{\hbar}\int_0^t \hat{H}(\tau)d\tau}\Big)\Psi(0)\quad\quad\quad (2&#039;)<br />

The more detailed explanation of quantum dynamics is in "Modern Quantum Mechanics" of J.J.Sakurai

This form is only valid if the Hamiltonians at different times commute:
[H(\tau),H(\tau&#039;)]=0

Otherwise you can't simply integrate (add up). Addition is commutative the operator products are not. You can't simply exponentiate the sum of the logs to get the product.
This is why we must use techniques such as Feynmann's path integrals.

Rather than a continuous sum (integral) we really need the notation of a continuous (ordered) product:

\Psi(t) =\prod_{\tau=t}^{0}\exp\left(-\frac{i}{\hbar}\hat{H}(\tau)d\tau}\right)\Psi(0)

with the product order understood to be written left to right. (or the reverse and reverse also the limits of multiplicative integration.)

Of course notation to write the product adds nothing to the problem of solving it.

 

[jostpuur]

This is analogous to the following question in high-school physics.
When the velocity v is constant, then the path traveled during the time t is
s=vt
But what if v(t) depends on t?
Is it
s=v(t)t
or is it
s=Integral v(t) dt ?
The correct equation, of course, is the last one, but why?

I hope this analogy helps.

If not, then let me be slightly more explicit. Because the Hamiltonian, by definition, is the generator of INFINITESIMAL time translations. Not only in quantum physics, but in classical physics as well. See e.g. classical mechanics formulated in terms of Poisson brackets.

Oh.. well this helped me noticing that my time evolution operator in the equation (2) did not depend on H(t&#039;) for 0\leq t&#039;\leq t at all I'm not sure what I was thinking...

 

[jostpuur]

continuous products from 2007-04-06

Makes you wonder what's the point in studying, when you can't remember what you have already learned...

 

[Ben Niehoff]

The usual physics notation for the time-ordered continuous product is

{\mathcal T} \exp \int_0^t H(t&#039;) \; dt&#039;

where \mathcal T is the so-called "time-ordering operation". However, I think the continuous product notation Jambaugh writes is more logical. Here is a paper I found that discusses product integrals, and uses a "curly Pi" notation for them:

http://www.math.leidenuniv.nl/~gill/prod_int_1.pdf

The product integral was originally conceived by Volterra to help in solving linear systems of differential equations. For ordinary functions, the product integral reduces to the ordinary integral, and so it has not garnered much mathematical interest.

For Lie algrebra valued functions, the product integral is useful, typically showing up in Wilson loops (or holonomy loops), where a function is integrated around a closed curve. In this case, it is possible to define a notion of surface-ordered product integration which respects a non-Abelian Stokes theorem.