Is U(t)=exp(-iH/th) a Lie Group in Quantum Mechanics?

[Ratzinger]

Is U(t)=exp(-iH/th) a Lie group?

Is it an infinite dimensional Lie group?

To what 'family' of Lie groups does it belong?

thank you

 

[meopemuk]

Is U(t)=exp(-iH/th) a Lie group?

Is it an infinite dimensional Lie group?

To what 'family' of Lie groups does it belong?

thank you

U(t)=exp(-iHt/h) is a 1-dimensional (the parameter is t) Lie group of unitary operators in the Hilbert space.

In quantum physics it is more common to call U(t)=exp(-iHt/h) a "unitary representation" of the group of time translations. The latter group is usually considered as a 1-dimensional subgroup of the 10-dimensional Poincare group, which also includes space translations, rotations, and boosts.

Eugene.

 

[Ratzinger]

thanks meopemuk!

But the generator of time translation (the Hamilton) consists of differential operators, so do the generators of space translation and angular momentum. Does not that mean an infinite dimenensional Lie algebra?

 

[meopemuk]

thanks meopemuk!

But the generator of time translation (the Hamilton) consists of differential operators, so do the generators of space translation and angular momentum. Does not that mean an infinite dimenensional Lie algebra?

The dimension of a Lie group is the number of independent parameters of transformations. For time translations there is only one parameter - time, so this group is 1-dimensional. For general Poincare transformation there are 10 independent parameters, so the Poincare Lie group is 10-dimensional.

When you are talking about the Hamiltonian, angular momentum, etc., you are talking not about the Poincare group itself, but about its representation in a Hilbert space. Group elements are represented by unitary operators in an infinite-dimensional Hilbert space. So, the dimension of the corresponding matrices is, indeed, infinite, but this has nothing to do with the dimension of the Lie group or the Lie algebra.

Eugene.

 

[Ratzinger]

many thanks meopemuk!