Let $A(t)$ be a smooth function from $[-1,1]$ to the $n \times n$ complex matrices. Define the time ordered exponential $$\prod_{-1}^1 \exp(A(t) dt)$$ as in this question, as the limit of Riemann products $\prod_{i=1}^n \exp(f(t^{\ast}_i) \ \delta t_i)$.

The actual quantity I am interested in is $$B(r) = \prod_{-1}^1 \exp(r A(t) dt)$$ as $r \to \infty$.

As $r \to 0$, there is a known power series expansion for $B(r)$ called the Magnus series. As $r \to \infty$, I would expect there to be something like the stationary phase approximation, but I haven't been able to find it or figure it out.

I should mention that in my situation, $A(t)$ obeys $$A(-t) = A(t)^{\ast} \quad (\dagger)$$ where $\ast$ is conjugate transpose. Condition $(\dagger)$ implies that $B(r)$ is Hermitian. I don't know whether this is helpful in any way.

  • $\begingroup$ It's maybe too late to comment on this but your question is related (except for one important point) to the adiabatic theorem of quantum mechanics. You are looking for an expansion of $B$ for large $r$ where $B$ is solution of $(1/r)\dot{B} = A(t) B(t)$ (dot indicates differentiation with respect to time). This is also called singular perturbation theory because when $r=\infty$ formally the character of the ODE changes. In quantum mechanics $A(t)$ is antihermitian for all $t$ meaning its eigenvalues are purely imaginary. This in turn implies that $B$ is unitary. $\endgroup$ – lcv Dec 9 '16 at 6:45
  • $\begingroup$ Provided some smootheness of $A(t)$ and a no level crossing $B$ can be expanded into a series in $1/r$. One of the best reference on this is still Kato's classic journals.jps.jp/doi/abs/10.1143/JPSJ.5.435. The adiabatic theorem has been generalized to the case where $A(t)$ is a generator of a contraction semigroup for all $t$. This still implies nicely bounded solution and an adiabatic theorem in the same vein. You seem to be interested to a case where $A(t)$ can have eigenvalues with positive real part. $\endgroup$ – lcv Dec 9 '16 at 6:52
  • $\begingroup$ I guess that case can be somehow reduced to the previous one factoring out the growth bound of $A(t)$. $\endgroup$ – lcv Dec 9 '16 at 6:53

This question has a solution presented in this paper even if with the jargon and notation of theoretical physics. So, I will use a somewhat different notation and I will change

$${\bf A}(t)\rightarrow -i{\bf A}(t).$$

Then, I will compute the eigenvalues and eigenvectors of ${\bf A}(t)$ through


Now, you get a series with a leading order term

$${\bf B}(r)=\sum_n e^{i\gamma_n}e^{-ir\int_{-1}^1 dt\lambda_n(t)}|n;1\rangle\langle n;-1| \qquad r\rightarrow\infty$$

being $\gamma_n=\int_{-1}^1dt\langle n;t|i\partial_t|n;t\rangle$ known as geometric phase. Then, an expansion in the inverse of $r$ can be obtained with the matrix

$$\tilde {\bf A}(t)=-\sum_{n,m,n\ne m}e^{i(\gamma_n(t)-\gamma_m(t))}e^{-ir\int_{t_0}^tdt[\lambda_m(t)-\lambda_n(t)]}\langle m;t|i\partial_t|n;t\rangle|m;t_0\rangle\langle m;t_0|$$

being in this case

$$\tilde {\bf B}(r)=\prod_{-1}^1e^{-i\tilde {\bf A}(t)dt}$$

so that

$$B(r)=\sum_n e^{i\gamma_n}e^{-ir\int_{-1}^1 dt\lambda_n(t)}|n;1\rangle\langle n;-1|\tilde {\bf B}(r).$$

This represents a solution of the Schroedinger equation

$$-ir{\bf A}(t)B(r;t,t_0)=\partial_tB(r;t,t_0)$$

in the interval $t\in [-1,1]$ and $r\rightarrow\infty$.

An example:

$$ A(t) = \frac{1}{1+t^2} \begin{pmatrix} 2 & t\\ -t & -2 \end{pmatrix} $$

and one has to solve the problem $$ \dot U(t)=rA(t)U(t) $$ with $r\gg 1$. We want to apply the technique outlined above. We note that $A(t)$ is not Hermitian and so, solving the eigenvalue problem, we get $\lambda_{\pm}=\pm r\frac{\sqrt{4-t^2}}{1+t^2}$ and $$ v_+=\frac{1}{2}\begin{pmatrix} \sqrt{2+\sqrt{4-t^2}}\\ -\frac{t}{\sqrt{2+\sqrt{4-t^2}}}\end{pmatrix} \qquad v_-=\frac{1}{2}\begin{pmatrix}-\frac{t}{\sqrt{2+\sqrt{4-t^2}}} \\ \sqrt{2+\sqrt{4-t^2}}\end{pmatrix}. $$ But $v_+^Tv_-\ne 0$ and so these vectors are not orthogonal. We need to solve also the eigenvalue problem $u^T(A-\lambda I)=0$ producing the following eigenvectors $$ u_+=\frac{1}{2}\begin{pmatrix} \sqrt{2+\sqrt{4-t^2}}\\ \frac{t}{\sqrt{2+\sqrt{4-t^2}}}\end{pmatrix} \qquad u_-=\frac{1}{2}\begin{pmatrix} \frac{t}{\sqrt{2+\sqrt{4-t^2}}} \\ \sqrt{2+\sqrt{4-t^2}}\end{pmatrix}. $$ It is easy to see that $u_+^Tv_-=u_-^Tv_+=0$. It is important to note that $\lambda(t)=\lambda(-t)$ and $u_+(-t)=v_-(t)$ and $u_-(-t)=v_+(t)$ and so, these eigenvectors are just representing a backward evolution in time. Now, we want to study the time evolution of a generic eigenvector $$ \phi(t)=\begin{pmatrix}\phi_+(t) \\ \phi_-(t)\end{pmatrix} $$ and this can be done by putting $$ \phi(t)=c_+(t)e^{r\int_0^tdt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}}v_+(t)+ c_-(t)e^{-r\int_0^tdt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}}v_-(t) $$ that will produce the set of equations $$ \dot c_+=\gamma_+c_++e^{-2r\int_0^tdt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}}\frac{u_+^T\frac{dv_-}{dt}}{u_+^Tv_+}c_- $$

$$ \dot c_-=\gamma_-c_-+e^{2r\int_0^tdt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}}\frac{u_-^T\frac{dv_+}{dt}}{u_-^Tv_-}c_+ $$ having set $\gamma_+=\frac{u_+^T\frac{dv_+}{dt}}{u_+^Tv_+}$ and $\gamma_-=\frac{u_-^T\frac{dv_-}{dt}}{u_-^Tv_-}$. These equations are interesting because they provide the way time evolution is formed in a non-hermitian case. But this is also saying to us that each component may evolve in time differently: One can be really smaller than the other for $r\gg 1$. But we can also understand the form of the higher order corrections:

$$ c_+(t)=c_+(0)+\int_0^tdt'e^{\int_0^{t'}dt''(\gamma_+(t'')-\gamma_-(t''))}e^{-2r\int_0^{t'}dt''\frac{\sqrt{4-t^{''2}}}{1+t^{''2}}}\frac{u_+^T\frac{dv_-}{dt''}}{u_+^Tv_+}c_-(0)+\ldots. $$

Using a saddle point technique, we can uncover here that the correction is exponentially small and cannot be stated that is something like $e^{r}/r^k$ in the general case.

Now, we consider the simple case $c_+(0)=1$ and $c_-(0)=0$. The approximate solution will be

$$ \phi_+(t)=\frac{1}{2}\sqrt{2+\sqrt{4-t^2}}e^{r\int_0^{t}dt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}} \qquad \phi_-(t)=-\frac{1}{2}\frac{t}{\sqrt{2+\sqrt{4-t^2}}}e^{r\int_0^{t}dt'\frac{\sqrt{4-t^{'2}}}{1+t^{'2}}} $$

and solving numerically the set of differential equations for $r=50$ we get the following

     (source: Wayback Machine)

The agreement is strikingly good.

  • $\begingroup$ Thanks! This will take some time to digest, but I have the following immediate questions: In your equation for $\tilde{B}(r)$, there is no $r$ on the right hand side. Is that right? Also, you describe your answer as a series in inverse powers of $r$, but I don't see such a series appearing anywhere in your answer. Thanks again! $\endgroup$ – David E Speyer Dec 1 '11 at 13:22
  • $\begingroup$ David, you are welcome. I have fixed the equation for $\tilde {\bf A}(t)$ as $r$ appears to multiply the eigenvalues of course. Then $\tilde {\bf B}(r)$ is a series, analogous to the Magnus one (but we physicists prefer to call it Dyson series), that is meaningful in the limit $r\rightarrow\infty$. $\endgroup$ – Jon Dec 1 '11 at 13:38
  • $\begingroup$ I just started thinking about the problem that motivated this again. I am suspicious of this answer. The leading term is a linear combination of the terms $| n;1 \rangle \langle n; -1|$, meaning that the leading term takes the eigenvectors of $A(-1)$ to the eigenvectors of $A(1)$, right? Independent of whatever happens in the middle of the "integral"? I am pretty sure this isn't true, and that my problem suggests a counterexample. I'll try to compute one in a few days. $\endgroup$ – David E Speyer Apr 26 '12 at 22:09
  • $\begingroup$ I take it back. Your answer is right. It's just that I am going to need to get way inside those asymptotics to see the terms I care about. Thanks! $\endgroup$ – David E Speyer Apr 27 '12 at 18:59
  • $\begingroup$ Actually, I am still confused. But I am seeing similar formulas in other sources, so presumably this is right. Hope I can figure out how! $\endgroup$ – David E Speyer Apr 27 '12 at 20:01

I think I might see what was confusing me. This is really a comment, but it's too long for the comment thread. As my example, let's take $$A(t) = \frac{1}{1+t^2} \begin{pmatrix} 2 & t \\ -t & -2 \end{pmatrix}$$ So we want to solve the differential equation $U'(t) = r A(t) U(t)$, where $U$ is a $2 \times 2$ matrix with initial condition $U(-1) = \mathrm{Id}$.

We can actually compute the eigenvalues of $A(t)$ explicitly: They are $\sqrt{4-t^2}/(1+t^2)$. We compute $\int_{-1}^1 \pm \sqrt{4-t^2}/(1+t^2) dt \approx \pm 3.03022$. So your formula, as I understand it, is $$U(1) = e^{3.03022 r} u_1 v_1^T + e^{-3.03022 r} u_2 v_2^T + \cdots$$ where $u_i$ and $v_i$ are the eigenvectors of $A(-1)$ and $A(1)$.

What I think was confusing me is that it is somewhat misleading to call this the leading terms. The later terms in the series look like $e^{3.03022 r} r^{-k} (\mbox{stuff})$, right? So they actually dominate the $e^{-3.03022 r}$ term.

I wish I weren't having so much trouble getting good numerical data, it would probably clear up my confusion a lot. In the meantime, here is why I am worried.

Let $A(t)$, $B(t)$ and $C(t)$ be three $2 \times 2$ matrix-valued functions as above, with $A(1)=B(1)=C(1)$ (and hence the same at $-1$.) Let $X(r)$, $Y(r)$ and $Z(r)$ be the parallel transport from $-1$ to $1$ be the differential equations $\phi'(t) = r A(t) \phi(t)$, $\phi'(t) = r B(t) \phi(t)$ and $\phi'(t) = r C(t) \phi(t)$. As I understand it, your method gives asymptotic expansions $$X(r) \approx U \begin{pmatrix} e^{x_1 r} & 0 \\ 0 & e^{x_2 r} \end{pmatrix} V \quad Y(r) \approx U \begin{pmatrix} e^{y_1 r} & 0 \\ 0 & e^{y_2 r} \end{pmatrix} V \quad Z(r) \approx U \begin{pmatrix} e^{z_1 r} & 0 \\ 0 & e^{z_2 r} \end{pmatrix} V \quad (1)$$ where I have the SAME matrices $U$ and $V$ in each cases, because they depend only on the eigenvectors of $A(1)=B(1)=C(1)$ and of $A(-1)=B(-1)=C(-1)$.

Am I right about $(1)$?

If so, here is the issue. Look at the quadratic form $$\det(x X(r) + y Y(r) + z Z(r)) \approx \det(U) \left( e^{r x_1} x + e^{r y_1} y + e^{r z_1} z \right) \left( e^{r x_2} x + e^{r y_2} y + e^{r z_2} z \right) \det(V).$$

The matrix of this form has leading terms $$\begin{pmatrix} \exp(r(x_1+x_2)) & & \\ \exp(r\max(x_1+y_2, x_2+y_1)) & \exp(r(y_1+y_2)) & \\ \exp(r\max(x_1+z_2, x_2+z_1)) & \exp(r\max(y_1+z_2, y_2+z_1)) & \exp(r(z_1+z_2)) \\ \end{pmatrix}$$ as long as the approximations in $(1)$ are good enough that we don't get extra cross terms.

Unless I am very confused, I can construct $A(t)$, $B(t)$, $C(t)$ such that this quadratic form looks like $x^2+y^2+z^2 + (e^r+e^{-r}) (xy+xz+yz)$. And there are no real numbers $(x_1, x_2, y_1, y_2, z_1, z_2)$ with $x_1+x_2=y_1+y_2=z_1+z_2=0$ and $\max(x_1+y_2, x_2+y_1)=\max(x_1+z_2, x_2+z_1)=\max(y_1+z_2, y_2+z_1)=1$. So something is wrong...

  • $\begingroup$ David, you missed the geometric phases here. These terms go like $e^{\int_0^t dt'u_1\dot u_1^T}\, e^{\int_0^t dt'v_1\dot v_1^T}$ and should be included. I will take some time to work out this example. $\endgroup$ – Jon May 1 '12 at 8:54
  • $\begingroup$ Ok, I have found at least a couple of problems with your example. I think that at the foundation of your confusion lies the fact that you are not working with Hermitian self-adjoint matrices. This has the important implication that you must have left and right eigenvector, let us say $v_{\pm}$ and $u_{\pm}$ and so, the series takes eventually the form $$e^{kr}v_+^Tu_++e^{-kr}v_-^Tu_-+\ldots$$. Finally, you are systematically omitting the geometric contribution going like $\exp{\pm\int_0^tdt'v_\pm^T\frac{d}{dt'}u_\pm}$ and this cannot be done here. Do you need an explicit example? $\endgroup$ – Jon May 1 '12 at 10:28
  • $\begingroup$ Right, they are not Hermitian. As stated in the original question, they obey $A(-t) = A(t)^{\ast}$. This has the effect that the total transport along the curve is Hermitian, but it is made up out of a lot of non Hermitian things. $\endgroup$ – David E Speyer May 1 '12 at 12:02
  • $\begingroup$ Fine. Give me a few time to work out completely this example and expand my answer. For this I have completely evaluated eigenvectors and eigenvalues at leading order. It is new also for me as I have always applied this to quantum mechanics. $\endgroup$ – Jon May 1 '12 at 13:23
  • $\begingroup$ David, I cannot do statements about combined solutions of more differential equations but I can show you how precise is my approximation for the first example you gave. I solved numerically the equation amd compared with the approximate solution. The agreement is strikingly good. $\endgroup$ – Jon May 2 '12 at 9:35

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