- #1

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If the evolution operator is given as follows

[tex]U(t) = \exp[-i (f(p, t) + g(x))/\hbar][/tex]

where p is momentum, t is time. Can I conclude that the Hamiltonian is

[tex]H(t) = f(p, t) + g(x)[/tex]

if no, why?

- Thread starter KFC
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- #1

- 488

- 4

If the evolution operator is given as follows

[tex]U(t) = \exp[-i (f(p, t) + g(x))/\hbar][/tex]

where p is momentum, t is time. Can I conclude that the Hamiltonian is

[tex]H(t) = f(p, t) + g(x)[/tex]

if no, why?

- #2

- 649

- 3

[tex]

U=\exp (-i (f(p,t)+g(x))/\hbar )\approx \exp (-i(f(p,0)+\dot{f}(p,0)t+g(x))/\hbar )

[/tex]

This kind of looks like the normal form

[tex]

U=e^{-iHt/\hbar}

[/tex]

Now you have something next to [itex]t[/itex] and divided by [itex]\hbar[/itex], which at least has to have units of energy. What do you think?

- #3

- 488

- 4

[tex]

U=e^{-i\int Hdt/\hbar}

[/tex]

If that's true, how do we find the Hamiltonian if there should be integral in the exponential ?

[tex]

U=\exp (-i (f(p,t)+g(x))/\hbar )\approx \exp (-i(f(p,0)+\dot{f}(p,0)t+g(x))/\hbar )

[/tex]

This kind of looks like the normal form

[tex]

U=e^{-iHt/\hbar}

[/tex]

Now you have something next to [itex]t[/itex] and divided by [itex]\hbar[/itex], which at least has to have units of energy. What do you think?

- #4

- 649

- 3

[tex]

H=\frac{\partial f(p,t)}{\partial t}\bigg|_{t=0}

[/tex]

thus no integral expression is required.

- #5

Jano L.

Gold Member

- 1,333

- 74

I'll try to explain the relation between U and H as I see it.

The first notion is the Hamiltonian operator [itex]\hat H[/itex], which plays role in the equation

[tex]

\partial_t \psi = -\frac{i}{\hbar} {\hat H}(t) \psi.

[/tex]

There is an alternative description in terms of an evolution operator. The operator [itex]\hat U(t,t_0)[/itex] is called an evolution operator, if it changes the function at time [itex]t_0[/itex] to a function at later time t:

[tex]

\psi(t) = \hat U(t,t_0) \psi(t_0).

[/tex]

The evolution operator obeys the equation

[tex]

\frac{\partial \hat U}{\partial t}(t) = \hat H(t) \hat U(t) ~~~(*)

[/tex]

In case the Hamiltonian is time-independent, Schroedinger's equation gives [itex]\hat U(t,t_0) = e^{-i\hat H (t-t_0)/\hbar}[/itex].

In case the Hamiltonian is time-dependent, there is no simple formula but there is perturbative series called Dyson series: http://en.wikipedia.org/wiki/Dyson_series

In your case, you seek Hamiltonian from known U. There is, as far as I know, no simple way to proceed. You can try to guess correct form of H that will recover the equation (*); if you succeed, the expression in front of U is your Hamiltonian.

Warning: the derivative

[tex]

\frac{\partial U}{\partial t} \neq -i/\hbar (\dot f(p,t)) \hat U(t),

[/tex]

except for case when operators f(p,t) + g(x) at different times commute.

- #6

- 488

- 4

I'll try to explain the relation between U and H as I see it.

The first notion is the Hamiltonian operator [itex]\hat H[/itex], which plays role in the equation

[tex]

\partial_t \psi = -\frac{i}{\hbar} {\hat H}(t) \psi.

[/tex]

There is an alternative description in terms of an evolution operator. The operator [itex]\hat U(t,t_0)[/itex] is called an evolution operator, if it changes the function at time [itex]t_0[/itex] to a function at later time t:

[tex]

\psi(t) = \hat U(t,t_0) \psi(t_0).

[/tex]

The evolution operator obeys the equation

[tex]

\frac{\partial \hat U}{\partial t}(t) = \hat H(t) \hat U(t) ~~~(*)

[/tex]

In case the Hamiltonian is time-independent, Schroedinger's equation gives [itex]\hat U(t,t_0) = e^{-i\hat H (t-t_0)/\hbar}[/itex].

In case the Hamiltonian is time-dependent, there is no simple formula but there is perturbative series called Dyson series: http://en.wikipedia.org/wiki/Dyson_series

In your case, you seek Hamiltonian from known U. There is, as far as I know, no simple way to proceed. You can try to guess correct form of H that will recover the equation (*); if you succeed, the expression in front of U is your Hamiltonian.

Warning: the derivative

[tex]

\frac{\partial U}{\partial t} \neq -i/\hbar (\dot f(p,t)) \hat U(t),

[/tex]

except for case when operators f(p,t) + g(x) at different times commute.

- #7

Jano L.

Gold Member

- 1,333

- 74

No problem. Can you post a link? I am curious what result they found for H.

- #8

- 488

- 4

Yes, I send you the link via message. Check it out :)No problem. Can you post a link? I am curious what result they found for H.

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