Help understanding this Lie Algebra/Group stuff

1. Oct 1, 2012

genericusrnme

If someone could walk me through this example I'd be extremely grateful.

A Hamiltonian acts in a 2 j + 1 dimensional space through a set of three operators Jz , J± that obey angular momentum commutation relations. We wish to determine the evolution of some particular state $|j,m_j \rangle$ . The Hamiltonian is;

$H = \epsilon(t)J_z + \alpha (t) J_+ + \alpha * (t) J_- \overset{j \rightarrow \frac{1}{2}}{\rightarrow} \left( \begin{array}{ccc} \frac{1}{2} \epsilon(t) & \alpha (t) \\ \alpha * (t) & -\frac{1}{2}\epsilon (t) \end{array} \right)$

I'm having trouble on this step for a few reasons
1. why are we letting j go to $\frac{1}{2}$
2. That isn't necessarily unitary

The unitary operator acting in the 2 j + 1 dimensional space is a unitary representation of some operation in the group SU (2). It is simpler to determine how g(t) ∈ SU (2) evolves, and then construct its unitary representation, than it is to determine the time evolution of the (2 j + 1) × (2 j + 1) unitary matrix. Specifically, the equation of motion in the group is;

$\frac{d}{dt} \left( \begin{array}{ccc} a(t) & b (t) \\ -b * (t) & a* (t) \end{array} \right) = -\frac{i}{\hbar} \left( \begin{array}{ccc} \frac{1}{2} \epsilon(t) & \alpha (t) \\ \alpha * (t) & -\frac{1}{2}\epsilon (t) \end{array} \right) \left( \begin{array}{ccc} a(t) & b (t) \\ -b * (t) & a* (t) \end{array} \right)$

I understand this part, it's just Schrodinger but in group form and the g(t) ∈ SU(2) is just $\psi$ (I think). I still have the problem with
1. No explicit Det=1 requirment

After some algebraic manipulation this matrix equation reduces to two equations for the complex coefficients a(t) and b(t) or three equations for the real coefficients of the Pauli spin matrices σ1 , σ2 , σ3 . These are first order equations and can be solved by standard integration methods (e.g., RK4). The initial conditions are $a(t_i ) = 1$, $b(t_i ) = 0$. The final 2 × 2 unitary matrix is determined by $a(t_f )$, $b(t_f )$. This is a group operation in SU (2) that can subsequently be mapped into the (2 j + 1) × (2 j + 1) unitary irreducible representation of this group. At this point the problem is solved, independent of the initial state $|\psi(t_i )\rangle$

And here I'm not sure how exactly you map back into the original space

(Any book recommendations that would help with this are also appreciated!)

Last edited: Oct 1, 2012
2. Oct 2, 2012

vanhees71

Are you sure you quoted the problem correctly?

I guess what you have there is a spin operator, denoted with $\hat{\vec{J}}$ and a Hamiltonian built by the interaction of the spin with a homogeneous but time-dependent magnetic field, and you assume that the spin is 1/2. The Hamiltonian is already rewritten in terms of the spin-component in quantization direction (as usual chosen as the $z$ direction) and in terms of the "ladder operators" $\hat{J}_{\pm} = \hat{J}_x \pm \mathrm{i} \hat{J}_y$.

The spin-Hilbert space in this case is two-dimensional with a basis given by the eigenvectors of $\hat{J}_z$ with eigenvalues $m_j \in \{-1/2,1/2 \}$, which I'll denote as $|m_j \rangle$ for brevity.

The ladder operators act on these states as
$$\hat{J}_+ |1/2 \rangle=0, \quad \hat{J}_+ |-1/2 \rangle =|1/2 \rangle, \quad \hat{J}_- |1/2 \rangle =|-1/2 \rangle, \quad \hat{J}_- |-1/2 \rangle =0.$$
Thus the hamiltonian reads in the matrix representation wrt. this basis as given in your posting
$$\hat{H}=\begin{pmatrix} \epsilon(t)/2 & \alpha(t) \\ \alpha^*(t) & -\epsilon(t)/2 \end{pmatrix}.$$
Of course, $\epsilon(t) \in \mathbb{R}$, so that the Hamiltonian is self-adjoint as it must be. Why should it be unitary?

Then I'm not sure what you are asked to solve. Is it the time evolution for a (pure) state in the Schrödinger picture, i.e.,
$$\mathrm{i} \frac{\mathrm{d}}{\mathrm{d} t} |\psi(t) \rangle = \hat{H}(t) |\psi(t) \rangle$$
or is it the evolution of some observable in the Heisenberg picture? I guess it's the former task.

In that case, in matrix-vector notation you have to solve for
$$\mathrm{i} \frac{\mathrm{d}}{\mathrm{d} t} \begin{pmatrix} a(t) \\ b(t) \end{pmatrix}=\hat{H}(t) \begin{pmatrix} a(t) \\ b(t) \end{pmatrix}.$$
The formal solution is
$$\begin{pmatrix} a(t) \\ b(t) \end{pmatrix}=\hat{U}(t,t_i) \begin{pmatrix} a(t_i) \\ b(t_i) \end{pmatrix},$$
where the unitary time-evolution operator is formally given by
$$\hat{U}(t,t_i)=T_c \exp \left [-\mathrm{i} \int_{t_i}^t \mathrm{d} t' \hat{H}(t') \right].$$
Here $T_c$ denotes the time-ordering operator. In simple cases it is even possible to solve this in closed form, e.g., for a time-independent Hamiltonian.