Chapter 2 Two-dimensional electron system
2.2 Bound States- Electrons on liquid Helium
In order to study the escape of electrons from the helium surface it is very necessary to understand electrostatic barrier which is responsible for binding electrons to the surface. Here, we will discuss key mechanism i.e., binding potential, electronic states etc.
There are combination of three forces that confines electrons in the potential well. These are:
[I] An attractive force due to induced image charges in liquid helium: In principle, as soon as a charge q approaches close to the dielectric substrate like helium it polarizes the material i.e., helium atoms in the liquid, which results an attractive polarization force towards helium surface. This polarization can be explained as an image charge at a distance z under the surface as shown in Figure3.
Mathematically, the image potential (Coulomb potential) is as follows,
2
( ) 0, 0 V z =V z≤
where z is the distance normal to the surface. Q represents the strength of image charge, which is equal to 0.0069 (e). The size of the image charge depends on the material used through its dielectric constant [19], for 4He=1.057 (at 4.2K) and dielectric constant of vapour, εv=1. Where as ε0 represents dielectric constant of vacuum. Moreover, the inter-electron spacing is approximately two order of magnitude higher than the distance between electron and image charge. So that each electron finds itself in an externally created potential well whose arising from its own image charge [20].
+ eQ - e Z Vapor
z> 0
Liquid 4Helium z< 0
Figure3. A charge approaching dielectric material at z distance above surface polarizes the atoms of liquid 4He.
[II] A short range repulsive force, the potential barrier, which arises from the Pauli’s Exclusion principle. The image charge is positive and opposite in sign to the electron so that the result is an attractive to the helium surface, and it can be imagined that such an attractive force could pull the electron down to the liquid, but electron experiences a large potential barrier close to the surface. Due to the fact that the s-shell of the helium atoms of the liquid is completely filled, and the wave function of any excess electron must oscillate in the vicinity of each atom which requires minimum energy of the order of 1eV [11].That is the energy needed by an electron to enter into the substrate(liquid 4helium), and binding energy of the liquid is very small in comparison to the potential barrier of 1eV. Hence the size of V0 is much
more larger than any other energy in the system, such as potential energy due to image charge and the applied field. That’s why to a first approximation it is taken to be ∞, and the penetration of the electron wave function into the helium is zero. The image potential for motion in z-direction (first two states are shown) and potential barrier, V0 =1 eV are shown in Figure 4.
Figure 4. The image potential for motion in z-direction(first two states are shown), and potential barrier of 1 eV together with the ground state and first excited state of the electron on liquid 4He are shown.
[III] External field: An applied vertical electric field E⊥ gives a linear potential variation eE⊥z.
Bound states of the system is expected to be localized outside, very close to the surface, due to the fact that an electron is repelled when inside the dielectric surface and attracted when outside the dielectric surface.
Furthermore, we can explore the change, considering that the assumption is made (caused by V0) that the wave function vanishes at the surface of liquid 4He, an electron then behaves like a one-dimensional hydrogen atom with a reduced nuclear charge Q=0.007e, in the direction. The energy for the motion in z-direction will be quantized in the Rydberg series,
n 2
E R
= −n (2.2) Where, n=1,2, and R is the Rydberg constant given as,
2
2 7.6
R 2 K
= mbh = (2.3)
For F=0, the ground state and first excited state wave function for the potential well were given
ψ1(z) 3 / 22 zexp z b/ b
= − (2.4)
ψ2(z) 13/ 2 2 exp / 2 2
z b
z z
b b
⎛ ⎞ −
= ⎜⎝ − ⎟⎠ (2.5)
Here, b=7.64 nm for 4He and 9.9 nm for 3He, and denotes the effective Bohr radius of the problem, and explains the length scale of the wave functions. Wave functions for ground and first excited state are shown in Figure 5. The wave functions are without electric field. However, by applying electric fieldE⊥, the average distance to the liquid helium will be shifted closer to the surface.
Figure 5. Ground and first excited state wave function. Distance from the surface ( in units of beff=7.6 nm).
The form of the energy of the ground and first excited state for the vertical motion are given as,
E1=-0.66 meV = -7.6 K (2.6) E2=-0.17 meV =-1.9 K
⇒ ΔE = E2 - E1 = 5.7 K (2.7)
Due to the fact that the smallness of ground state energy justifies the assumption of infinite potential barrier (V0 =1 eV) at the surface.
The average distance of the electrons from the liquid helium surface is, ψ1(z)= 11.4 nm (2.8) ψ2(z)= 45.6 nm (2.9) The fermi temperature is given by,
f f ,
The maximum electron density on bulk liquid helium, n<2*109 cm-2; Tf is less than 50 mK. At the experimental temperature T >100mK, the Fermi temperature is always much smaller than the average thermal energy, k TB of the electrons.
In general, the electrons on liquid 4He form a two-dimensional, classical electron gas at a distance about 11.4 nm from the helium surface obeying Boltzmann statistics in energy distribution. Since all measurements performed here are in temperature range between 1.59 K and 1.95 K, therefore almost all electrons are in ground state. Ground state of the system is important for most of the experimental
and drags attention, like as direct spectroscopic observation of bound states by Grimes et al [21], where they observed transition between the ground state, n=1 and excited states up to n=7 at a driving frequency of 220 GHz (see article 2.4).