Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Integrating derivatives in matrix elements

  1. Nov 19, 2013 #1
    First a little warm up problem. Suppose [itex]g:\mathbb{R}^N\to\mathbb{C}[/itex] is some fixed function, and we want to find [itex]f:\mathbb{R}^N\to\mathbb{C}[/itex] such that

    [tex]
    g(x) = u\cdot\nabla_x f(x)
    [/tex]

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

    [tex]
    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
    [/tex]

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

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

    [tex]
    g(x) = u\cdot\nabla_x f(x)
    [/tex]

    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 [itex]N=3,M=2[/itex] and

    [tex]
    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)
    [/tex]

    So this case is the same as

    [tex]
    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)
    [/tex]
    [tex]
    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)
    [/tex]

    I don't know if this special case is any easier than the general case, though.
     
  2. jcsd
  3. Nov 21, 2013 #2
    One possible way to start is to first consider the special case [itex]g=0[/itex]. The equation [itex]0=u\cdot\nabla_x f(x)[/itex] with my [itex]u[/itex] is the same as

    [tex]
    \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)
    [/tex]

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

    [tex]
    \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)
    [/tex]

    Unfortunately, the matrices

    [tex]
    \left(\begin{array}{cc}
    1 & -1 \\ -1 & -1 \\
    \end{array}\right),\quad\quad
    \left(\begin{array}{cc}
    -1 & -1 \\ -1 & 1 \\
    \end{array}\right)
    [/tex]

    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.
     
  4. Nov 21, 2013 #3
    I think we'll have to assume that the partial derivatives of [itex]g[/itex] 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...
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Integrating derivatives in matrix elements
Loading...