# Integrating derivatives in matrix elements

#### jostpuur

First a little warm up problem. Suppose $g:\mathbb{R}^N\to\mathbb{C}$ is some fixed function, and we want to find $f:\mathbb{R}^N\to\mathbb{C}$ such that

$$g(x) = u\cdot\nabla_x f(x)$$

holds, where $u\in\mathbb{R}^N$ is some constant. The problem is not extremely difficult, and after some work the following formula can be found:

$$f(x) = f\Big(x - \frac{u\cdot x}{\|u\|^2}u\Big) + \int\limits_0^{(x\cdot u)/\|u\|^2} g\Big(x - \frac{u\cdot x}{\|u\|^2}u+ \nu u\Big) d\nu$$

If it is assumed that $f,g$ are differentiable, the formula can be checked to work. The idea is that $f$ can be chosen arbitrarily as initial condition in the orthogonal complement of $\langle u\rangle$.

Now, my real problem. Assume that $u$ is $NM^2$ component object, whose elements have been arranged so that first $u=(u^1,\ldots, u^N)$, and then each $u^n$ is a $M\times M$ matrix. Assume that $g:\mathbb{R}^N\to\mathbb{C}^M$ is some fixed function. Then we want to find $f:\mathbb{R}^N\to\mathbb{C}^M$ such that symbolically the same formula

$$g(x) = u\cdot\nabla_x f(x)$$

holds. Now the previous solution formula doesn't work anymore, and I don't know how to fix the situation.

I am in particular interested in the case $N=3,M=2$ and

$$u=\Big( \left(\begin{array}{cc} \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \end{array}\right),\; \left(\begin{array}{cc} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \end{array}\right),\; \left(\begin{array}{cc} 0 & i \\ -i & 0 \\ \end{array}\right)\Big)$$

So this case is the same as

$$g_1(x) = \frac{1}{\sqrt{2}}\big(\partial_1 f_1(x) + \partial_2 f_1(x) + \partial_1 f_2(x) - \partial_2 f_2(x) \big) + i \partial_3 f_2(x)$$
$$g_2(x) = \frac{1}{\sqrt{2}}\big(\partial_1 f_1(x) - \partial_2 f_1(x) - \partial_1 f_2(x) - \partial_2 f_2(x)\big) - i \partial_3 f_1(x)$$

I don't know if this special case is any easier than the general case, though.

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#### jostpuur

One possible way to start is to first consider the special case $g=0$. The equation $0=u\cdot\nabla_x f(x)$ with my $u$ is the same as

$$\left(\begin{array}{c} \partial_3 f_1 \\ \partial_3 f_2 \\ \end{array}\right) = -\frac{i}{\sqrt{2}} \left(\begin{array}{cc} \partial_1 - \partial_2 & -\partial_1 - \partial_2 \\ -\partial_1 - \partial_2 & -\partial_1 + \partial_2 \\ \end{array}\right) \left(\begin{array}{c} f_1 \\ f_2 \\ \end{array}\right)$$

Unfortunately, I was unable to learn anything about this equation either. A formal solution can be written as

$$\left(\begin{array}{c} f_1(x_1,x_2,x_3) \\ f_2(x_1,x_2,x_3) \\ \end{array}\right) = \exp\Big(-\frac{ix_3}{\sqrt{2}} \left(\begin{array}{cc} \partial_1 - \partial_2 & -\partial_1 - \partial_2 \\ -\partial_1 - \partial_2 & -\partial_1 + \partial_2 \\ \end{array}\right)\Big) \left(\begin{array}{c} f_1(x_1,x_2,0) \\ f_2(x_1,x_2,0) \\ \end{array}\right)$$

Unfortunately, the matrices

$$\left(\begin{array}{cc} 1 & -1 \\ -1 & -1 \\ \end{array}\right),\quad\quad \left(\begin{array}{cc} -1 & -1 \\ -1 & 1 \\ \end{array}\right)$$

cannot be diagonalized simultaneously, and therefore the exponential series cannot be written in terms of Taylor series after a linear transform. I cannot see what kind of transformation the series represents.

#### jostpuur

If it is assumed that $f,g$ are differentiable, the formula can be checked to work.
I think we'll have to assume that the partial derivatives of $g$ are bounded too. This is probably not relevant for the discussion, since I'm interested in finding formulas mostly, but anyway, I wanted to fix the original claim...