Understanding how to determine the time evolved state vector for a unitary operator constructed from non-commutating operators

Question

Suppose we have a time independent hamiltonian $$H = \hbar g (\sigma_x + \sigma_y + \sigma_z)$$

I know that the unitary operator is as follows:

$$U(t) = exp(-iHt/{\hbar})$$ Sinnce the pauli spin operators in H pairwise anticommute we know that

$$exp(-itg(\sigma_x + \sigma_y + \sigma_z)) \neq exp(-itg\sigma_x) exp(-itg\sigma_y) exp(-itg\sigma_z)$$

My question is how do we use this information to deduce the time-evolved state vector for the particle $|\psi(t)\rangle = U(t)|\psi(0)\rangle$.

Suppose we know that $|\psi(0)\rangle = |\uparrow\rangle$

We can state that $$|\psi(t)\rangle = exp(-itg(\sigma_x + \sigma_y + \sigma_z))|\uparrow\rangle$$

I also know that for an operator A and scalar $\alpha$ we have the identity $$exp(iA\alpha) = cos(\alpha ) + isin(\alpha)A$$

Generally for these sort of problems the operators commute so i would try to solve something along the lines of $$ |\psi(t)\rangle = exp(-itg\sigma_x) exp(-itg\sigma_y) exp(-itg\sigma_z) |\uparrow \rangle$$ and then invoke the operator A relation. However since the pauli operators pairwise anti commute I cannot do this and am unsure how to solve simplify such as, $$|\psi(t)\rangle = exp(-itg(\sigma_x + \sigma_y + \sigma_z))|\uparrow\rangle$$

and so any helo would be appreciated:

Answer 1

There is a very useful identity for exponentials of Pauli matrices (see https://math.stackexchange.com/questions/3236998/exponential-of-pauli-matrices/3237834 for a proof):

$\begin{eqnarray} e^{i\theta \hat{\bf n}\cdot \sigma} = \cos \theta I + i(\hat{\bf n}\cdot \sigma) \sin \theta \end{eqnarray}$

For your Hamiltonian $\theta = -tg$ and $\hat{\bf n}\cdot \sigma = \sigma_x + \sigma_y + \sigma_z$, which gives:

$U(t) = \cos(tg)I - i\sin(tg)(\sigma_x + \sigma_y + \sigma_z)$

Note that $U(t)$ can also be conveniently represented as a $2x2$ matrix that acts on vectors representing the quantum state:

$U(t) = \begin{bmatrix}\cos(tg) - i\sin(tg)& (-1-i)\sin(tg) \\ (1-i)\sin(tg) & \cos(tg) + i\sin(tg)\end{bmatrix}$

So, with your example state,

$|\psi(t)\rangle = U(t)|\uparrow\rangle = (\cos(tg) - i\sin(tg))|\uparrow\rangle + (1-i)\sin(tg) |\downarrow\rangle$

You can alternatively follow the method from the above comments, where you take the matrix $H$, diagonalize this matrix to find the eigenvectors and eigenvalues, represent your initial state in this eigenbasis, and then time evolve the state using the eigenvalues. Both methods are equivalent, and the best method often depends on the application.