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).

  • $\begingroup$ Are you familiar with exponentials of Pauli vectors such as $\exp(i\pi \sigma_1 /2 )=i \sigma_1$ ? $\endgroup$ – Cosmas Zachos Jan 22 '19 at 16:34

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.

  • $\begingroup$ Made some corrections. If you include $\mathbb{I}$ in your basis you can produce all elements of $U(N)$ and not just the special unitaries. $\endgroup$ – Tal Sheaffer Jan 22 '19 at 16:55
  • $\begingroup$ 1. Hey so first of all thank you for your time (: But to be honest I am not quite sure where are you getting at. Let me recap what I know: We have a particle in a some potential, which gives us the Hamiltonian. Now I can use that Hamiltonian to determine the energy-eigenstates and eigenvalues of said particle in said potential. I am thinking that the Hamiltonian is in the energy eigenbasis such that we have the eigenvalues on the diagonal $\mathrm{diag}=\{E_1,E_2,...\}$ And obviously this Hamiltonian can be used to derive the time evolution of a any state. $\endgroup$ – CatoMaths Jan 22 '19 at 17:05
  • $\begingroup$ 2. But suddenly you are trying to convince me that in order to generate such unitary I am asking for I need to consider that the hamiltonian can be described via the pauli matrix? Sorry I haven't seen that before. I am used to think of the Hamiltonian as Energy operator, so it makes no sense for me why I'd re-write him as pauli matrix. Can you explain a bit? $\endgroup$ – CatoMaths Jan 22 '19 at 17:07
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    $\begingroup$ @CatoMaths This is just a trick based on the properties of the Pauli matrices; given the final form of your unitary (it's diagonal in a $2\times 2$ subspace), it's likely the trick will work and indeed it does. Your Hamiltonian here would be (2) and you can verify that (1) actually produces the unitary you want. $\endgroup$ – ZeroTheHero Jan 22 '19 at 17:09
  • $\begingroup$ @CatoMaths any Hamiltonian is Hermitian and so can be written using a basis of Hermitian operators (e.g. Pauli matrices). One simply needs to pick the right Hamiltonian and evolve for the right 'time' (in this case angle) to get the desired unitary. $\endgroup$ – Tal Sheaffer Jan 22 '19 at 17:21

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