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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.

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up vote 6 down vote accepted

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

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The agreement is strikingly good.

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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! –  David Speyer Dec 1 '11 at 13:22
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 $r\rightarrow\infty$. –  Jon Dec 1 '11 at 13:38
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. –  David Speyer Apr 26 '12 at 22:09
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! –  David Speyer Apr 27 '12 at 18:59
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! –  David Speyer Apr 27 '12 at 20:01
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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...

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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. –  Jon May 1 '12 at 8:54
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? –  Jon May 1 '12 at 10:28
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. –  David Speyer May 1 '12 at 12:02
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. –  Jon May 1 '12 at 13:23
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. –  Jon May 2 '12 at 9:35
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