Hermitian matrices and great circles I am considering parameterised curves in an $n$-dimensional complex vector space, given by the solution to the discrete Schrödinger equation:
$$
|\psi\rangle(t) = e^{-iHt}|\psi_0\rangle,
$$
Where $H$ is a Hermitian matrix and the initial point $|\psi_0\rangle$ has magnitude 1, i.e $\langle \psi_0|\psi_0\rangle=1$.
Geometrically, these curves are confined to the surface of the unit $n$-sphere. A subset of these curves are geodesics, i.e. they trace the great circles of the unit $n$-sphere. I would like to know the answers to the following two questions:


*

*Given a matrix $H$, how could one tell whether it represents a great circle when plugged in to the above equation? I.e. do the matrices corresponding to geodesics have some special properties in terms of their elements, eigenvalues, etc.? Is there an accepted term for such matrices?

*More importantly: given the coordinates of two points on the unit $n$-circle, how can I find a Hermitian matrix $H$ that corresponds the great circle that passes through both points? I want to do this numerically, but of course if there's an analytical solution that would be great.
Note: my actual problem is regarding curves in a real vector space of the form $|\psi\rangle(t) = e^{At}|\psi_0\rangle$, where $A$ is antisymmetric. The complex version described above is a generalisation of this, and I used that in the question because it seemed more likely that the answer would be expressed in those terms if it's a known result. If a solution only exists for the real case I would be happy with that.
 A: In Question 1, whether or not $|\psi\rangle(t)$ traces a great circle depends not just on the matrix $A$ (or $H$), but on the initial condition $|\psi_0\rangle$.
If $A$ is an $n \times n$ skew-symmetric matrix, its eigenvalues come in pairs $\pm i\lambda_k$ with $\lambda_k$ real (and a single additional $0$ eigenvalue if $n$ is odd). If $|\psi_0\rangle$ lies in a generalized eigenspace of $A$ (i.e., in a real $2$-plane preserved by $A$), the curve $|\psi\rangle(t) = e^{At} |\psi_0\rangle$ traces a great circle in the sphere $S^{n-1} \subset \mathbf{R}^n$. Conversely, if $|\psi\rangle(t)$ traces a great circle $C$, the real $2$-plane containing $C$ is invariant by $A$.
(IIRC, if $H$ is an $n \times n$ Hermitian matrix, its eigenvalues $\lambda_k$ are real and $H$ is unitarily diagonalizable. If $|\psi_0\rangle$ lies in some eigenspace of $H$, the curve $|\psi\rangle(t) = e^{-iHt} |\psi_0\rangle$ traces a great circle in the sphere $S^{2n-1} \subset \mathbf{C}^n$.)
As for Question 2, if you have two non-antipodal points $p_1$ and $p_2$ on the unit sphere, it suffices (as Tom notes) to find an orthonormal basis $\{\mathbf{u}_{j}\}$ of $\mathbf{R}^n$ such that $\mathbf{u}_1 = p_1$ and $\text{span}\{\mathbf{u}_1, \mathbf{u}_2\} = \text{span}\{p_1, p_2\}$; this is a standard algorithmic computation (essentially Gram-Schmidt on the set $\{p_1, p_2\}$ union the standard basis of $\mathbf{R}^n$). 
If $U$ is the orthogonal matrix with your $\mathbf{u}_j$ as columns, you can take $A = UDU^t$, with $D$ a block-diagonal matrix, and $|\psi_0\rangle = p_1$.
A: I will not give you full answer, I will give you full answer only in 3d.
Ok Let's have a points $\mathbf{p_1,p_2}$ and you want rotation in $\mathbf{p_1,p_2}$ plane. Let $\mathbf{e_1,e_2}$ be orthonormal basis of $\mathbf{p_1,p_2}$ plane and $\mathbf{e_3} = \mathbf{e_1} \times \mathbf{e_2}$. Now if we have point $\mathbf{x}(t)$ rotating around axis $\mathbf{e_3}$ with $\omega$ angular velocity. Then $\mathbf{x}(t)$ satisfy equation:
$$
\dot{\mathbf{x}}(t) = \omega \mathbf{e_3} \times \mathbf{x}(t)
$$
Operation "$\mathbf{e_3} \times $" can be writen as matrix. So(using notation from wiki)
$$
\dot{\mathbf{x}}(t) = \omega [\mathbf{e_3}]_\times \mathbf{x}(t)
$$
Thus solution is
$$
\mathbf{x}(t) = e^{\omega [\mathbf{e_3}]_\times t} \mathbf{x}(0)
$$
And the matrix you are looking for is $\omega [\mathbf{e_3}]_\times$

Note that $\mathbf{e_3}$ can be calculated as $\mathbf{e_3} = \frac{ \mathbf{p_1} \times \mathbf{p_2}}{ \| \mathbf{p_1} \times \mathbf{p_2} \| }$. This will be useful in proceeding.

Generalization of preceding is not straightforward because it is not clear what it is cross product in more dimensions. 
But we have exterior product! So the next will be wild guessing but I think that antisymnetric matrix you are looking for is matrix of this 2-form $\omega \frac{\mathbf{p_1}\wedge \mathbf{p_2}}{\|\mathbf{p_1}\wedge \mathbf{p_2} \|}$. I'm not sure about this but I will think about it.

Edit:
Well that was easy :D
For start have a look at rotation in 2d. If we have point $x(t)$ rotating at angular speed $\omega$ than it satisfy this equation:
$$
\dot x(t) = \omega 
\begin{pmatrix}
  0 & -1 \\
  1 &  0 
 \end{pmatrix} x(t)
$$
This will be usefull later as we will se that in n dimensions the equation is the same.
Now in n-dimensiolan space. Let $e_1,\dots,e_n$ be orthonormal and we want rotation in $e_1,e_2$ plane. So if $x(t)$ is point rotating in $e_1,e_2$ plane it satisfy this equation
$$
\dot x(t) = \omega (e_2e^T_1 - e_1e^T_2) x(t)
$$
And I hope you can see the similarity of this equation and the equation in 2d.
So how to compute that? Calculate orthogonal basis $e_1,e_2$ of $p_1,p_2$ plane and the matrix you are looking for is $\omega(e_2e_1^T - e_1e_2^T)$.

Edit2: Well I really like this question. So I will add how to numericaly compute it. The are 3 options:
$$
A = \frac{p_2p_1^T-p_1p_2^T}{\sqrt{1-(p_1\cdot p_2)^2}}
$$
as Nathaniel proposed in comment,
or 
$$
A = e_2e_1^T-e_1e_2^T
$$
Where $e_1,e_2$ is obtained from $p_1,p_2$ via Gram-Schmidt. Or there is third possibility
$$
A =  \frac{p_2p_1^T-p_1p_2^T}{ \sqrt{\frac{\|p_2p_1^T-p_1p_2^T\|_F^2}{2}} }
$$
Where $\|\cdot\|_F$ is Frobenius norm of matrix. This idea comes from 3d example. There you calculate cross product and then normalize that vector. But vector entries corresponds to above diagonal entries in corresponding anti-symmetric matrix. 
So I did numerical test how they preform, I took two vectors $u,v$ and I brought them closer and closer(but keeping the plane, they span, the same) and observed how each algorithm preforms. Denote $A_i(u,v)$ the matrix obtained from vectors $u,v$ via $i$-th algorithm. And I ploted these values(as functions of k) $$\|A_3(u,v) - A_1(u,u + 10^{-k}(v-u))\|$$$$\|A_3(u,v) - A_2(u,u + 10^{-k}(v-u))\|$$$$\|A_3(u,v) - A_3(u,u + 10^{-k}(v-u))\|$$
and I got this picture

So Gram-Schmidt and normalizing with Frobenius norm beats the first option. And I think that Gram-Schidt is more effective.
Here is matlab code I used for generating it
clear
% construct points p1=u p2=v
n = 100
u0 = rand(n,1);
v0 = rand(n,1);
d = v0-u0;

% calculate desired antisymetric matrix
u = u0;
v = v0;
AREF = v*u' - u*v';
AREF = AREF/sqrt(sumsqr(AREF)/2);

% now calculate that matrix again but bring v gradualy closer to the u, but
% keep u,v plane the same
% and observe how will the matrix change
k = 15;
error = zeros(3,k);
for i=1:k
    u = u0;
    v = u0 + 10^(-i)*d; 

    % frst method, 
    A = v*u' - u*v';
    len = norm(u)*norm(v)* sqrt(1- (u'*v)^2/(sumsqr(u)*sumsqr(v)))
    A = A/len;

    %second method
    e1 = u/norm(u);
    e2 = (v-e1*(e1'*v))/norm(v-e1*(e1'*v));
    B = e2*e1' - e1*e2';

    %third method
    C = v*u' - u*v';
    C = C/sqrt(sumsqr(C)/2);

    error(1,i) = norm(AREF-A);
    error(2,i) = norm(AREF-B);
    error(3,i) = norm(AREF-C);
end

semilogy(error')
legend('normalizing via dot product','gram-schmidt','normalizing via Frobenius norm');

