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The Raman Spectrum of Graphene

Whether graphene is one layer or two, whether the crystal has defects, and whether strain or doping has occurred. To investigate such information in detail, advanced measurements suited to the purpose are needed, such as observing atomic-scale structure or measuring electronic states and electrical properties.

Yet in graphene, simply shining a laser and measuring the scattered light makes it possible to read much of this information. The clue is the Raman spectrum.

Why can a single Raman spectrum provide so much different information?

01

The horizontal axis represents the energy lost by light

In Raman scattering, a small frequency difference arises between the incident and scattered light. If the angular frequencies of the incident and scattered light are $\omega_i$ and $\omega_s$, respectively, then in Stokes scattering $\omega_s<\omega_i$, and the light loses the energy corresponding to that difference.

In spectroscopy, the frequency difference is converted to the unit $\mathrm{cm}^{-1}$ and taken as the horizontal axis of the spectrum.

$$\tilde{\nu}\,[\mathrm{cm}^{-1}]=\frac{\omega_i-\omega_s}{2\pi c}\tag{1}$$

The same difference expressed as energy is

$$\Delta E=\hbar(\omega_i-\omega_s)=hc\tilde{\nu}\tag{2}$$

For example,

$$1580\ \mathrm{cm}^{-1}\longleftrightarrow196\ \mathrm{meV}$$

The Raman shift and the energy difference correspond one-to-one and can be converted into each other using Eq. (2). A peak farther to the right in the spectrum means that the energy transferred to graphene by the light is larger.

In the spectrum, large peaks appear near 1580 $\mathrm{cm}^{-1}$ and 2700 $\mathrm{cm}^{-1}$, while the peak near 1350 $\mathrm{cm}^{-1}$ is weak. The peak near 1350 $\mathrm{cm}^{-1}$ is called the D band, the peak near 1580 $\mathrm{cm}^{-1}$ the G band, and the peak near 2700 $\mathrm{cm}^{-1}$ the 2D band.

In the G, D and 2D bands, graphene's lattice vibrations receive the energy lost by light.

02

Connecting Raman peaks to phonon dispersion

Because graphene's carbon atoms are bonded to one another, they do not move independently one by one; many atoms vibrate together. The collective vibration in a crystal treated as a quantum is a phonon.

A single phonon with angular frequency $\omega$ has the energy

$$E_{\mathrm{ph}}=\hbar\omega$$

When a scattering process creates one phonon, the light's energy decreases by the amount that phonon receives. This is why the Raman shift is connected to the energy of a lattice vibration.

Vibrations in a crystal differ not only in energy but also in their spatial patterns. The wavevector $Q$ of a phonon represents how much the phase changes when moving from one unit cell to the next.

A plot of the relation between wavevector $Q$ and the frequency of a phonon is called the phonon dispersion. A graphene unit cell contains two carbon atoms, A and B, and each can move in three directions, so the phonon dispersion has six branches.

Γ

Here $\Gamma$ is the point where the wavevector is zero ($Q=0$), and the same vibration pattern repeats with the same phase in each unit cell. $\Gamma$, K and M are points in reciprocal space that also appear in the electronic band structure.

Let us correspond each Raman peak to the phonon dispersion. The G band involves an in-plane optical phonon near $\Gamma$, while the D band and 2D band involve in-plane optical phonons near K.

The wavevector of the phonon related to G is $Q\simeq0$, whereas the phonons related to D and 2D have large wavevectors. This difference matters because scattering must balance wavevector as well as energy.

03

G, D and 2D balance wavevector differently

In light scattering, not only energy but also wavevector must balance. The wavevector that light can transfer to graphene is determined by the difference between the incident and scattered light wavevectors, and when light with an excitation wavelength of 514 nm is used, even at its maximum estimate,

$$|\Delta k_{\mathrm{light}}|\lesssim\frac{4\pi}{514\ \mathrm{nm}}\simeq0.024\ \mathrm{nm}^{-1}\tag{3}$$

This remains only

On the other hand, the distance from $\Gamma$ to K in graphene is about

$$|\Gamma K|\simeq17\ \mathrm{nm}^{-1}$$

This is the distance.

The wavevector light can transfer is only about one seven-hundredth of the distance from $\Gamma$ to K. On the scale of reciprocal space, the wavevector that light can transfer is regarded as nearly zero.

Therefore, when one phonon is created without wavevector compensation by a defect or the like, the phonon's wavevector must also be nearly zero. The G band, which uses a phonon near $\Gamma$, meets this condition.

In contrast, the phonon related to the D band is near K, and creating only one phonon does not balance the wavevector. In a periodic crystal, the periodicity of the atomic arrangement strongly constrains wavevector conservation, but at defects and edges that periodicity is interrupted, so the constraint is relaxed and the large wavevector of phonons near K can be taken up by the scattering process.

If the wavevector of a phonon is $Q$ and the wavevector change brought to the scattering process by a defect or edge is $Q_d$, the D band can balance the wavevector for the whole scattering process as

$$Q+Q_d\simeq0$$

This is why the D band appears readily where defects or edges are present and becomes weak inside defect-poor graphene.

For the 2D band, two phonons take the place of the defect. If two phonons have nearly opposite wavevectors,

$$Q_1+Q_2\simeq0$$

the wavevectors can balance even without a defect. The two phonon wavevectors cancel, but the energies they receive are added. Therefore, the 2D band can appear even without a defect and has a peak in a Raman-shift region approximately twice that of the D band.

04

Wavevector conservation alone does not determine the spectrum

However, wavevector conservation tells us only whether a scattering process satisfies the wavevector condition. Even when multiple processes satisfy the same condition, they do not appear in the Raman spectrum with the same strength.

In graphene, electronic states also affect how readily Raman scattering occurs, so even when the same wavevector condition is satisfied, the scattering strength changes greatly depending on which electronic states are involved.

Here, the Dirac cones near K and K′ become important.

The phonon dispersion shows what frequency phonons exist at each wavevector, wavevector conservation narrows down scattering processes from the perspective of wavevector, and electronic states influence which scattering processes are likely to occur.

If the lattice or electronic states change, their effects appear in the position, intensity, width and shape of the peaks.

That is why one Raman spectrum can provide different kinds of information, such as layer number, defects, strain and doping.

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