Understanding Unitary Operator Evolution in Quantum Mechanics

[KFC]

Hi there,
I am reading a book in which the unitary evolution operator is
[tex]U = \exp(-i H/\hbar)[/tex]

where H is the given Hamiltonian. But in another book, I found that the evolution operator is general given as
[tex]U = \exp(-i \int H(t) dt / \hbar)[/tex]

which one is correct and why there are two expression? Thanks

 

[vanhees71]

The first expression has a typo, but it applies to time-independent Hamilton operators. It solves the operator equation for the time-evolution operator of states in the Schrödinger picture,
[tex]\frac{\mathrm{d} \hat{U}}{\mathrm{d} t}=-\frac{\mathrm{i}}{\hbar} \hat{H} \hat{U}.[/tex]
Since [itex]\hat{H}[/itex] is time-independent it trivially always commutes with itself at any instant of time, and you can formally integrate this equation as if you had a usual differential equation. Together with the initial condition [itex]\hat{U}(0)=1[/itex], you get
[tex]\hat{U}(t)=\exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H} t \right).[/tex]
The second equation is usually not generally correct. It only holds, if [itex]\hat{H}(t)[/itex] commutes with [itex]\hat{H}(t')[/itex] for any two times [itex]t,t'[/itex]. The correct solution for the equation in this case is
[tex]\hat{U}(t)=\mathcal{T}_c \exp \left (-\frac{\mathrm{i}}{\hbar} \int_0^t \mathrm{d} t' \hat{H}(t') \right ),[/tex]
where [itex]\mathcal{T}_c[/itex] is the time-ordering operator. For a detailed explanation and derivation of the formula, see

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

page 16. Don't worry that this is a script on quantum-field theory. The first chapter is a summary of usual non-realtivistic quantum mechanics.

 

[KFC]

Thanks vanhees. Suppose the Hamiltonian is somethings like [tex]H(t) = p^2/(2m) + \sin(x)[\delta(t-3) + \delta(t+3)][/tex]
where p is the momentum, x is the spatial variable, t is the time, so Hamiltonian is time-dependent so I should use the intergral form of evolution operator, right?

But in this case, how do I check H(t) and H(t') commutes or not? The delta function looks complicated to me.

The first expression has a typo, but it applies to time-independent Hamilton operators. It solves the operator equation for the time-evolution operator of states in the Schrödinger picture,
[tex]\frac{\mathrm{d} \hat{U}}{\mathrm{d} t}=-\frac{\mathrm{i}}{\hbar} \hat{H} \hat{U}.[/tex]
Since [itex]\hat{H}[/itex] is time-independent it trivially always commutes with itself at any instant of time, and you can formally integrate this equation as if you had a usual differential equation. Together with the initial condition [itex]\hat{U}(0)=1[/itex], you get
[tex]\hat{U}(t)=\exp \left (-\frac{\mathrm{i}}{\hbar} \hat{H} t \right).[/tex]
The second equation is usually not generally correct. It only holds, if [itex]\hat{H}(t)[/itex] commutes with [itex]\hat{H}(t')[/itex] for any two times [itex]t,t'[/itex]. The correct solution for the equation in this case is
[tex]\hat{U}(t)=\mathcal{T}_c \exp \left (-\frac{\mathrm{i}}{\hbar} \int_0^t \mathrm{d} t' \hat{H}(t') \right ),[/tex]
where [itex]\mathcal{T}_c[/itex] is the time-ordering operator. For a detailed explanation and derivation of the formula, see

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

page 16. Don't worry that this is a script on quantum-field theory. The first chapter is a summary of usual non-realtivistic quantum mechanics.

 

[vanhees71]

Where does such a strange Hamiltonian come from? What should it describe? It's pretty strange to say the least!

 

[KFC]

Where does such a strange Hamiltonian come from? What should it describe? It's pretty strange to say the least!

That's for kicking system. It is used for a kicking at time is t=3 and t=-3. For each kicking time, a potential sin(x) is applied