How a small tilt travels through a magnet as a spin wave

Small tilts around the up domain

Take the up domain, $\mathbf m\approx(m_x,m_y,1)$ with small $m_x$ and $m_y$ . Keep each field in the effective field (Abert 2019, Sec. 2) to first order in the tilt. The applied field and the effective anisotropy point along $z$ : $H+H_{K0}$ , with $\mu_0H_{K0}=2K_0/M_s=927\,\mathrm{mT}$ . The chain of the Lab also keeps the small hard-axis term of Chapter 4 in the domains, so $m_x$ feels an extra $-H_Km_x$ . Exchange adds $(2A/\mu_0M_s)\nabla^2\mathbf m$ .

For a plane wave $\mathbf m\propto e^{i(kx-\omega t)}$ the Laplacian becomes $-k^2$ , so exchange acts like an extra field $H_{\mathrm{ex}}=2Ak^2/(\mu_0M_s)$ pulling the tilt back. A shorter wave bends the magnetization more sharply and feels a stronger restoring field.

Interfacial DMI drops out at this order. Its energy density is $D(m_x\partial_xm_z-m_z\partial_xm_x)$ . The first term is second order in the tilt, because $m_z\approx1-(m_x^2+m_y^2)/2$ . The second is $-D\,\partial_xm_x$ to first order, a total derivative that integrates to zero. We derived this; it holds for waves along $x$ in a perpendicular domain and says nothing about other geometries.

Worked example

The dispersion relation and its gap

Without damping the LLG equation reduces to $\dot m_x=-\gamma_0H_2m_y$ and $\dot m_y=\gamma_0H_1m_x$ , with the two stiffness fields below. Differentiate the first once more and substitute the second: $\ddot m_x=-\gamma_0^2H_1H_2m_x$ . The tilt precesses at $\omega=\gamma_0\sqrt{H_1H_2}$ .

At $k=0$ and zero field, $\mu_0H_1=958\,\mathrm{mT}$ and $\mu_0H_2=927\,\mathrm{mT}$ , so the lowest frequency is $f_0=\omega/2\pi=26.4\,\mathrm{GHz}$ . A 40 mT field along the domain raises it to 27.5 GHz. Below the gap an antenna can only make a tilt that dies away next to it.

Now go the other way, from frequency to wavelength. At 60 GHz and zero field, solve $(\gamma_0H_1)(\gamma_0H_2)=\omega^2$ for $H_{\mathrm{ex}}$ : it needs $\mu_0H_{\mathrm{ex}}=1.20\,\mathrm{T}$ , so $k=0.203\,\mathrm{nm^{-1}}$ and the wavelength is 30.9 nm. A 10 nm wave needs an exchange field of 11.5 T and runs at 348 GHz.

\omega^2=\gamma_0^2H_1H_2,\qquad H_1=H+H_{K0}+H_K+H_{\mathrm{ex}},\quad H_2=H+H_{K0}+H_{\mathrm{ex}},\quad H_{\mathrm{ex}}=\frac{2Ak^2}{\mu_0M_s}

What the grid does to the wave

On a chain of cells of size $a$ the second difference replaces $k^2$ by $(2-2\cos ka)/a^2$ , which is smaller for short waves. With 1 nm cells the 60 GHz wave comes out at 30.86 nm instead of 30.92 nm, a 0.2 per cent change. At 150 GHz the gap between chain and continuum grows to 0.6 per cent. This is the same second-order mesh error as in Chapter 3, now seen as a shift of the wavelength.

The Lab checks the chain against its own discrete dispersion. A periodic chain of 512 cells, seeded with a small disturbance and run without damping, is Fourier transformed in space and time. Every peak lands within half a 1.95 GHz frequency bin of the formula.

k^2\;\longrightarrow\;\frac{2-2\cos ka}{a^2}=k^2\left(1-\frac{k^2a^2}{12}+\dots\right)

Worked example

Damping turns the wavenumber complex

An antenna drives the film at one real frequency. With Gilbert damping each $\gamma_0H$ in the precession equations becomes $\gamma_0H-i\alpha\omega$ , so the condition reads $(\gamma_0H_1-i\alpha\omega)(\gamma_0H_2-i\alpha\omega)=\omega^2$ . Write $X=\gamma_0H_{\mathrm{ex}}-i\alpha\omega$ . The condition is a real quadratic in $X$ , so $X$ is real, and $k^2$ picks up an imaginary part $\propto\alpha\omega$ .

The real part of $k$ sets the wavelength, $\lambda=2\pi/\mathrm{Re}\,k$ . The imaginary part sets how far the amplitude travels before falling by $e$ : $\ell=1/\mathrm{Im}\,k$ .

On the 1 nm chain at zero field and 60 GHz: with $\alpha=0.1$ the wave has $\lambda=30.7\,\mathrm{nm}$ and $\ell=55.1\,\mathrm{nm}$ ; with the film’s $\alpha=0.5$ , $\lambda=28.6\,\mathrm{nm}$ and $\ell=11.9\,\mathrm{nm}$ , less than half a wavelength. That is why the spin-wave preset lowers the damping to 0.1, an assumed value chosen so the wave stays visible.

In the Lab the antenna at 60 GHz launches a wave the chain measures at 30.75 nm, against 30.74 nm from the damped formula. The benchmark solves the same condition by Newton iteration instead of the closed form and matches the chain to 0.07 per cent in wavelength and 0.6 per cent in decay length at 40, 60, 100 and 150 GHz.

X=\frac{-(p+q)+\sqrt{(p-q)^2+4\omega^2}}{2},\qquad \frac{2A\gamma_0}{\mu_0M_s}\,k^2=X+i\alpha\omega

What this picture leaves out

The chain keeps the stray field only as local terms, the same approximation Thiaville et al. use for ultrathin films. The long-range part of a wave’s stray field is left out, and this course does not estimate how much it would change the dispersion. The antenna is a uniform field over a 5 nm strip, and the temperature is zero.

How a small tilt travels through a magnet as a spin wave

Small tilts around the up domain

The dispersion relation and its gap

What the grid does to the wave

Damping turns the wavenumber complex

What this picture leaves out

Predict, change, explain

Try explaining it yourself

1. What is the lowest frequency a spin wave can have in the up domain at 40 mT?

2. Without damping, what is the continuum wavelength at 100 GHz and zero field?

3. With damping 0.1 the 100 GHz wave decays over 48.5 nm. Why does a higher frequency decay over a shorter distance even though the imaginary part of $k^2$ grows only in proportion to $\omega$ ?

Sources to read alongside

Small tilts around the up domain

The dispersion relation and its gap

What the grid does to the wave

Damping turns the wavenumber complex

What this picture leaves out

Predict, change, explain

Try explaining it yourself

1. What is the lowest frequency a spin wave can have in the up domain at 40 mT?

2. Without damping, what is the continuum wavelength at 100 GHz and zero field?

3. With damping 0.1 the 100 GHz wave decays over 48.5 nm. Why does a higher frequency decay over a shorter distance even though the imaginary part of k2k^2k2 grows only in proportion to ω\omegaω?

Sources to read alongside

3. With damping 0.1 the 100 GHz wave decays over 48.5 nm. Why does a higher frequency decay over a shorter distance even though the imaginary part of $k^2$ grows only in proportion to $\omega$ ?