How do we find a time-independent Hamiltonian that generates a given unitary transformation?

Question

I know that for time independent Hamiltonians we can make the statement

$$U = e^{-iHt}\tag{1}$$

where $H$ is a time-independent Hamiltonian (divided by $\hbar$) and $U$ the unitary, also known as time evolution operator.

Now when studying I've met a unitary which corresponds to a two-level rotation defined as

$$U = \lvert i \rangle \langle j \rvert +\lvert j \rangle \langle i \rvert -\lvert i \rangle \langle i \rvert - \lvert j \rangle \langle j \rvert + \mathbb{I}\tag{2}$$

For clarity for a 4 state system it would look like this

$$ \begin{bmatrix} 0& 1 & 0 & 0 \\ 1& 0 & 0 & 0 \\ 0& 0 & 1 & 0 \\ 0& 0 & 0 & 1 \\ \end{bmatrix} $$

I'd like to understand how I can ever hope to use eq. (1) to obtain such a unitary eq. (2).

Answer 1

In general Unitary matrices are generated by Hermitian matrices, so you need a basis of Hermitian matrices $\left\{ \mathbb{I},T^a \right\} _{a=1,\dots ,N^2-1}$ for an $N$ dimensional Hilbert space (edit: see note at bottom).

For $N=2$ one usually chooses the Pauli matrices. For $N=3$ the Gell-Mann matrices.

In this case one need not be so methodical. First note that for a Pauli matrix:

$$ e^{-i \theta \sigma} = \mathbb{I}\cos{\theta}-i\sigma \sin{\theta} $$ $$ \Rightarrow e^{\frac{i \pi}{2}\sigma} = i\sigma$$ $$ \Rightarrow e^{\frac{i \pi}{2}(\sigma-\mathbb{I})} = \sigma$$ (prove this using a power series)

Now notice that your unitary operator can be thought of as "$\sigma_{x}^{i,j}$" + the identity for all other indices. Hence a good guess would be:

$$ e^{\frac{i \pi}{2}(\sigma_{x}^{i,j}-\mathbb{I}^{i,j})} \tag{1} $$

where here $\sigma_{x}^{i,j}$ is understood as having 0-s in all other indices, e.g.:

$$\sigma_{x}^{i,j}-\mathbb{I}^{i,j}= \begin{bmatrix} -1& 1 & 0 & 0 \\ 1& -1 & 0 & 0 \\ 0& 0 & 0 & 0 \\ 0& 0 & 0 & 0 \\ \end{bmatrix} \tag{2} $$

(Check that it works)

Edit: Actually there is a simpler way of finding such a Hamiltonian in general, without finding a basis!

Diagonalize the unitary - it will then have the form:

$$ \begin{bmatrix}e^{iE_{1}}\\ & e^{iE_{2}}\\ & & \ddots\\ & & & e^{iE_{N}} \end{bmatrix} $$

Now in this new basis the Hamiltonian is obvious.