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I'm wondering if there is a smart way of solving the system of equations

$$\frac{d\vec{n}}{dt} = \gamma (\vec{n} \times \vec{B}(t)),$$

where $\vec{n}(t) = \big(x(t),y(t),z(t) \big)$ is the Bloch vector and $\vec{B}(t) = (B_1\cos\omega t, B_1 \sin \omega t, B_2)$ is the external magnetic field.

By smart I mean faster than just writing it down in matrix form and using methods for solving a system of homogeneous linear equations with variable coefficients. I'm looking for some sort of reference where such a solution is shown.

Spine Feast
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  • possible duplicate of Deriving the Bloch vector time evolution equation – Phoenix87 Jan 22 '15 at 14:29
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    $\mathbf n\times\mathbf B$ is linear in $\mathbf n$, so there exists a matrix $\star\mathbf B$ that acts on $\mathbf n$. Hence your system is of the form $\dot{\mathbf n} = \star\mathbf B\mathbf n$. The solution has the form $\mathbf n = e^{\star\mathbf Bt}\mathbf n_0$ – Phoenix87 Jan 22 '15 at 14:31
  • Yes, I figured out the matrix form (btw. is there a reason You denote it with a star?), but the matrix is itself dependent on time, so I don't see how I can just exponentiate? – Spine Feast Jan 22 '15 at 14:38
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    oh right didn't notice time dependence, an integral should be taken then. The $\star$ is the Hodge dual. – Phoenix87 Jan 22 '15 at 14:39
  • What do you mean an integral should be taken? So the solution is $n = e^{\int_0^t dtB(t)} n_0$ ? B(t) is the matrix of course. – Spine Feast Jan 22 '15 at 14:42
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    yes but that is more an evocative symbol than a legitimate operation – Phoenix87 Jan 22 '15 at 14:52
  • How should I understand the matrix integration then? Elementwise? Why is it not a legitimate operation? – Spine Feast Jan 22 '15 at 14:56
  • @DepeHb You need to use a time-ordered exponential since the matrix may not commute with itself at different times, making the normal exponential ill-defined. Note that solving the general Bloch equations even with constant coefficients involves a fairly tedious calculation. You might want to look into trying to simplify matters by employing the rotating-wave approximation, if applicable. – Mark Mitchison Jan 22 '15 at 15:20
  • Hmm... Why is nothing about an ordered exponential mentioned here: http://en.wikipedia.org/wiki/Linear_differential_equation#Systems_of_Linear_Differential_Equations ? I don't know how to proceed if this turns out to be incorrect. I think exponentiating this (calculating eigenvectors) is murder anyway... that's why I was curious if there's some clever way to this. – Spine Feast Jan 22 '15 at 15:42
  • In the Wiki article it states that the normal exponential is applicable if $[A(x_1),A(x_2)] = 0$. If this is not true then one must use a path-ordered exponential. – Mark Mitchison Jan 22 '15 at 19:40
  • You're right, I didn't notice that. But, this won't be true in this case? So my question is, is this even doable analytically (within reason?) with undergraduate tools? – Spine Feast Jan 22 '15 at 20:10

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I can suggest a simple way which is not without equations, but it is with very simple equations. It goes as follows:

You can rotate your system of axes by $\omega t$ around the axis $z$, then by an angle $\theta = arctg(B_1/B_2)$ so as to lay your vector $\vec {B(t)}$ over a new axis $z'$ ? You will get a vector constant in time, of length $B = \sqrt {B_1^2 + B_2^2}$.

In the new axes $x', y', z'$ the system of equations will become

$\frac {d\ n_{x'}}{dt} = \gamma B \ n_{y'}$

$\frac {d\ n_{y'}}{dt} = -\gamma B \ n_{x'}$

and the solution is immediate,

(1) $n_{x'} = cos(\gamma B t)$,$ \ \ \ n_{y'} = -sin(\gamma B t)$.

Whatever is left after this is to reverse the two rotations, i.e. first rotate back around the axis $y'$ by $-\theta$, then, as the axis $z'$ is laid back on $z$, rotate by $-\omega t$ around the axis $z$ for which you have the rotation matrices.