ST-FMR

# ST-FMR

I’ve noticed that the useful calculations for spin torque ferromagnetic resonance (ST-FMR) are normally only glossed over in papers, although they can often be found buried in dissertations. Here’s a quick introduction to anyone who wants to do their own further calculations or analysis.  This covers the simple case of a rectangular geometry with an in-plane magnetic material, with the applied field also in-plane at some angle to the direction of the current, and includes the possibility of DC torques.

# Coordinate system

The unprimed coordinates are based on the orientation of the DC component of the moment, so that the time-averaged moment is parallel to the magnetic field in the y direction.  The primed coordinates match the geometry of the device, and the current flows in the y’ direction.  To save space the first part of the calculations are written in terms of $$H$$ and $$M$$, rather than $$B_{ext}$$ and $$\mu_0 M_{eff}$$.

# Landau–Lifshitz–Gilbert–Slonczewski Equation

In our geometry the unit moment is $$\hat m(t)=\vec M(t)/M_s=m_x(t)\hat x+\hat y+m_z(t)\hat z$$. The LLGS equation is1

$$\begin{equation*} \begin{gathered}\dot{\hat m}=-\gamma \hat m\times \vec H_{\mathit{eff}}+\alpha \hat m\times \dot{\hat m}+\tau _{\parallel}\frac{\hat m\times (\hat x’\times \hat m)}{\left|\hat x’\times \hat m\right|}+\tau _{\perp}\frac{\hat x’\times \hat m}{\left|\hat x’\times \hat m\right|}\\=-\gamma \hat m\times \left(\vec H_{\mathit{ext}}-\hat m\cdot \hat zM_{\mathit{eff}}\hat z\right)+\alpha \hat m\times \dot{\hat m}+(\tau _{\parallel,\mathit{RF}}+\tau _{\mathit{\parallel},\mathit{DC}})\frac{\hat m\times (\hat x’\times \hat m)}{\left|\hat x’\times \hat m\right|}+(\tau _{\perp,\mathit{RF}}+\tau _{\mathit{\perp},\mathit{DC}})\frac{\hat x’\times \hat m}{\left|\hat x’\times \hat m\right|}\end{gathered}\end{equation*}$$

where in the second line we have explicitly written out the contribution from the anisotropy and the possibility of both RF and DC torques.  Note that we have also included the normalizing terms which account for the change in the angle between the orientation of the spins and the moment, which is $$\begin{equation*} \left|\hat x’\times \hat m\right|=\sqrt{m_{\mathit{z0}}^2\sin ^2(\phi )+m_{\mathit{z0}}^2\cos ^2(\phi )+\cos ^2(\phi )+m_{\mathit{x0}}\sin (\phi )\cos (\phi )+m_{\mathit{x0}}^2\sin ^2(\phi )}\approx \left|\cos (\phi )\right|\end{equation*}$$.

To solve for the RF response we assume moments and RF torques of the form $$m_x=m_{\mathit{x0}}e^{i\omega t}$$ and ignore higher-frequency components to get

$$\begin{equation*}\begin{gathered}0=(H+M)m_{\mathit{z0}}\gamma -\tau _{\mathit{\parallel},\mathit{RF}}-\tau _{\mathit{\parallel},\mathit{DC}}-im_{\mathit{x0}}\omega +im_{\mathit{z0}}\alpha \omega +(m_{\mathit{x0}}\tau _{\mathit{\parallel},\mathit{DC}}-m_{\mathit{z0}}\tau _{\mathit{\perp},\mathit{DC}})\tan (\phi )\\{}\\0=-Hm_{\mathit{x0}}\gamma -\tau _{\mathit{\perp},\mathit{RF}}-\tau _{\mathit{\perp},\mathit{DC}}-im_{\mathit{z0}}\omega -im_{\mathit{x0}}\alpha \omega +(m_{\mathit{x0}}\tau _{\mathit{\perp},\mathit{DC}}+m_{\mathit{z0}}\tau _{\mathit{\parallel},\mathit{DC}})\tan (\phi )\end{gathered}\end{equation*}$$

The first thing to notice is that the effect of the DC out-of-plane torque is the same as the applied field, so we can simplify things by writing $$H’=H-\tau _{\perp,\mathit{DC}}\tan (\phi )/\gamma$$.  Solving for the moments we get

$$\begin{equation*}\begin{gathered}m_{\mathit{x0}}=\frac{-i\alpha \tau _{\perp,\mathit{RF}}\omega -\gamma (H’+M)\tau _{\perp,\mathit{RF}}+\tau _{\mathit{\parallel},\mathit{RF}}\tau _{\mathit{\parallel},\mathit{DC}}\tan (\phi )-i\tau _{\parallel,\mathit{RF}}\omega }{\gamma ^2H'(H’+M)+2i\alpha \gamma \omega H’-\omega ^2(1+\alpha ^2)+i\alpha \gamma \omega M+\tau _{\parallel,\mathit{DC}}^2\tan ^2(\phi )-2i\omega \tau _{\parallel,\mathit{DC}}\tan (\phi )}\\{}\\m_{\mathit{z0}}=\frac{i\omega (\alpha \tau _{\parallel,\mathit{RF}}-\tau _{\mathit{\perp},\mathit{RF}})+\gamma H’\tau _{\parallel,\mathit{RF}}+\tau _{\mathit{\parallel},\mathit{DC}}\tau _{\mathit{\perp},\mathit{RF}}\tan (\phi )}{\gamma ^2H'(H’+M)+2i\alpha \gamma \omega H’-\omega ^2(1+\alpha ^2)+i\alpha \gamma \omega M+\tau _{\parallel,\mathit{DC}}^2\tan ^2(\phi )-2i\omega \tau _{\parallel,\mathit{DC}}\tan (\phi )}\end{gathered}\end{equation*}$$

Only the real component of this has any physical meaning, and for certain measurements like ST-FMR we only care about the real component that is in-phase with the driving current.  The exact expression for the real component is very long but we can make a few generally-safe assumptions like that $$\alpha ^2\ll 1$$ and also that anything higher-order in the torques can be neglected (we’re in the linear response regime).  The simplified complex expression is then

$$\begin{equation*}\begin{gathered}m_{\mathit{x0}}=\frac{-i\alpha \tau _{\perp,\mathit{RF}}\omega -\gamma (H’+M)\tau _{\perp,\mathit{RF}}-i\omega \tau _{\parallel,\mathit{RF}}}{\gamma ^2H'(H’+M)+2i\alpha \gamma \omega H’-\omega ^2+i\alpha \gamma \omega M+2i\omega \tau _{\parallel,\mathit{DC}}\tan (\phi )}\\{}\\m_{\mathit{z0}}=\frac{\gamma H’\tau _{\parallel,\mathit{RF}}+i\omega (\alpha \tau _{\parallel,\mathit{RF}}-\tau _{\mathit{\perp},\mathit{RF}})}{\gamma ^2H'(H’+M)+2i\alpha \gamma \omega H’-\omega ^2+i\alpha \gamma \omega M+2i\omega \tau _{\parallel,\mathit{DC}}\tan (\phi )}\end{gathered}\end{equation*}$$

and making similar assumptions the real component is

$$\begin{gathered}\Re (m_{\mathit{x0}}e^{i\omega t})=\frac{\cos (\omega t)(-\gamma \omega ^2\tau _{\perp,\mathit{RF}}(H’+M)+\gamma ^3\tau _{\perp,\mathit{RF}}H'(H’+M)^2-\alpha \gamma \omega ^2(2H’+M)\tau _{\parallel,\mathit{RF}})}{(\omega ^2-\gamma ^2H'(H’+M))^2+\omega ^2(\gamma ^2\alpha ^2(2H’+M)^2-4\alpha \gamma \tau _{\parallel,\mathit{DC}}(2H’+M)\tan (\phi ))}\\{}\\+\frac{\omega \sin (\omega t)(-\omega ^2(\alpha \tau _{\perp,\mathit{RF}}+\tau _{\mathit{\parallel},\mathit{RF}})+\gamma ^2(H’+M)(\gamma H’\tau _{\parallel,\mathit{RF}}-\alpha \gamma \tau _{\mathit{\perp},\mathit{RF}}(H’+M)))}{(\omega ^2-\gamma ^2H'(H’+M))^2+\omega ^2(\gamma ^2\alpha ^2(2H’+M)^2-4\alpha \gamma \tau _{\parallel,\mathit{DC}}(2H’+M)\tan (\phi ))}\\{}\\\Re (m_{\mathit{z0}}e^{i\omega t})=\frac{\cos (\omega t)(-\gamma \omega ^2H’\tau _{\parallel,\mathit{RF}}+\gamma ^3H’^2\tau _{\parallel,\mathit{RF}}(H’+M)-\alpha \gamma \omega ^2\tau _{\perp,\mathit{RF}}(2H’+M))}{(\omega ^2-\gamma ^2H'(H’+M))^2+\omega ^2(\gamma ^2\alpha ^2(2H’+M)^2-4\alpha \gamma \tau _{\parallel,\mathit{DC}}(2H’+M)\tan (\phi ))}\\{}\\+\frac{\omega \sin (\omega t)(\omega ^2(\alpha \tau _{\parallel,\mathit{RF}}-\tau _{\mathit{\perp},\mathit{RF}})+\gamma H'(\alpha H’\tau _{\parallel,\mathit{RF}}+\gamma \tau _{\mathit{\perp},\mathit{RF}}(H’+M)))}{(\omega ^2-\gamma ^2H'(H’+M))^2+\omega ^2(\gamma ^2\alpha ^2(2H’+M)^2-4\alpha \gamma \tau _{\parallel,\mathit{DC}}(2H’+M)\tan (\phi ))}\end{gathered}$$

It’s now clear that the resonant field is given by $$\omega _0^2=\gamma ^2H'(H’+M)$$ and that the damping term is $$\omega ^2(\gamma ^2\alpha ^2(2H’+M)^2-4\alpha \gamma \tau _{\parallel,\mathit{DC}}(2H’+M)\tan (\phi ))$$.  Note that the contribution from the DC in-plane torque is not the same as a modification of $$\alpha$$!

# Mixing voltages

For any effect that causes the resistance to vary with the angle between the moment and the current flow (like the anisotropic magnetoresistance) the mixing voltage is

$$\begin{equation*}V_{\mathit{mix}}=\langle \delta I\cdot \delta R\rangle =\langle \delta I\cdot \left(\frac{dR}{d\phi }\right)\cdot \delta \phi \rangle =\langle I_{\mathit{RF}}\cos (\omega t)\cdot \left(\frac{dR}{d\phi }\right)\cdot \Re (m_x(t))\rangle\end{equation*}$$

In general for small excitations $$\Re (m_x(t))$$ oscillates sinusoidally with a phase and magnitude that varies across the resonance. Only the component proportional to $$\cos (\omega t)$$ contributes to the mixing voltage, and using the LLGS calculation above we see that

$$\begin{gathered}V_{\mathit{mix}}=I_{\mathit{RF}}\frac{\omega }{2\pi }\left(\frac{\mathit{dR}}{d\phi }\right)\int _0^{2\pi /\omega }\cos (\omega t)\Re (m_x(t))\mathit{dt}\\{}\\\mathit{}=\frac{I_{\mathit{RF}}} 2\left(\frac{\mathit{dR}}{d\phi }\right)\frac{\gamma \omega ^2\tau _{\perp,\mathit{RF}}(H’+M)-\gamma ^3\tau _{\perp,\mathit{RF}}H'(H’+M)^2-\alpha \gamma \omega ^2(2H’+M)\tau _{\parallel,\mathit{RF}}}{(\omega ^2-\gamma ^2H'(H’+M))^2+\omega ^2(\gamma ^2\alpha ^2(2H’+M)^2-4\alpha \gamma \tau _{\parallel,\mathit{DC}}(2H’+M)\tan (\phi ))}\end{gathered}$$

# Spin pumping

In spin pumping the important quantity is the pumped spin which is given by

$$\begin{equation*}j_s\hat s=\frac{\hbar }{4\pi }\Re (g_{\uparrow \downarrow}^{\mathit{eff}})\left(\vec m(t)\times \frac{d\vec m(t)}{dt}\right)\end{equation*}$$

where $$\Re (g_{\uparrow \downarrow}^{\mathit{eff}})$$ is the effective real component of the spin mixing conductance for the interface. Note that this does not include the contribution from the imaginary component of the spin mixing conductance (but it is easy to add in for the cases when it is not negligible).  The moment crossed with its time derivative has several components.  Crossing the y component of the unit moment ( $$\hat y$$) with $$dm_x(t)/\mathit{dt}$$ or $$dm_z(t)/\mathit{dt}$$ gives an AC pumped spin; this is the basis of the AC spin pumping effect. Crossing one of the time-dependent components with the time-derivative of the other gives a DC spin current in addition to higher-order effects; this is the basis of the DC spin pumping effect.

For DC spin pumping we have

$$\begin{equation*}\begin{gathered}\int _0^{2\pi /\omega }\left(\vec m(t)\times \frac{d\vec m(t)}{dt}\right)\mathit{dt}=\int _0^{2\pi /\omega }\left(m_z(t)\frac{dm_x(t)}{\mathit{dt}}-m_x(t)\frac{dm_z(t)}{\mathit{dt}}\right)\mathit{dt}\hat y\\{}\\=\frac{\gamma \omega ^2(H'(\tau _{\parallel,\mathit{RF}}^2+\tau _{\mathit{\perp},\mathit{RF}}^2)+M\tau _{\perp,\mathit{RF}}(\tau _{\mathit{\perp},\mathit{RF}}-\tau _{\mathit{\parallel},\mathit{RF}}))}{(\omega ^2-\gamma ^2H'(H’+M))^2+\omega ^2(\gamma ^2\alpha ^2(2H’+M)^2-4\alpha \gamma \tau _{\parallel,\mathit{DC}}(2H’+M)\tan (\phi ))}\hat y\end{gathered}\end{equation*}$$

The vector quantity represents the orientation of the spins being pumped, not their direction of travel.  DC spin pumping always produces spins that are aligned with the main component of the moment.

For AC spin pumping we have, making the same approximations as before,

$$\begin{equation*}\left(\vec m(t)\times \frac{d\vec m(t)}{dt}\right)_{\mathit{AC}}=\frac{dm_z(t)}{\mathit{dt}}\hat x-\frac{dm_x(t)}{\mathit{dt}}\hat z\end{equation*}$$

We don’t care about the phase of the resulting AC signal, the only thing that matters for measurements is the magnitude of the sinusoidal pumped spin in each direction which is

$$\begin{equation*}\begin{gathered}\left|\frac{dm_z(t)}{\mathit{dt}}\hat x-\frac{dm_x(t)}{\mathit{dt}}\right|\hat z=\\{}\\\omega \sqrt{\frac{\omega ^2(\alpha \tau _{\parallel,\mathit{RF}}-\tau _{\mathit{\perp},\mathit{RF}})^2+\gamma ^2H’^2\tau _{\parallel,\mathit{RF}}^2}{(\omega ^2-\gamma ^2H'(H’+M))^2+\omega ^2(\gamma ^2\alpha ^2(2H’+M)^2+4\alpha \gamma \tau _{\parallel,\mathit{DC}}(2H’+M)\tan (\phi ))}}\hat x\\{}\\+\omega \sqrt{\frac{\omega ^2(\tau _{\parallel,\mathit{RF}}+\alpha \tau _{\mathit{\perp},\mathit{RF}})^2+\gamma ^2(H’+M)^2\tau _{\perp,\mathit{RF}}^2}{(\omega ^2-\gamma ^2H'(H’+M))^2+\omega ^2(\gamma ^2\alpha ^2(2H’+M)^2+4\alpha \gamma \tau _{\parallel,\mathit{DC}}(2H’+M)\tan (\phi ))}}\hat z\end{gathered}\end{equation*}$$

Note that while the DC pumped spin is proportional to the torques squared, the AC pumped spin is linear in the torques (although it is proportional to $$m_x$$ which is large compared to the DC torques which scale as $$m_x^2$$ ).

### Disclaimer

If possible, use the exact forms for fitting rather than the approximate ones shown above.  If you see any typos or errors please let me know.

### References

1. Slonczewski, J. C. Current-driven excitation of magnetic multilayers. J. Magn. Magn. Mater. 159, L1–L7 (1996).