Amplitude Reflection Coefficient for an Electron Pure State, at a Single Step in Electric Potential and Magnetic Flux Density

The incident and transmitted electron waves are modelled (figure 2.3) as plane waves, allowing the well-defined wave-vector components $p$ in the plane of the interface, and $q_i$ perpendicular to the interface. $p$ must be the same for all the waves, in order to satisfy the boundary condition of continuity of the wave-function at the interface. The amplitude reflection coefficient is [30]

$$r_{01} = \frac{q_0-q_1}{q_0+q_1}\textrm{,}$$ (2.18)

or, for a general interface,

$$r_{ij} = \frac{q_i-q_j}{q_i+q_j}\textrm{.}$$ (2.19)

Next, I need to build an expression for the energy of the electrons. There will be kinetic energy terms, which, in the non-relativistic limit, are

$$\frac{\hbar^2p^2}{2m_e}\textrm{,}$$

and

$$\frac{\hbar^2q_i^2}{2m_e}\textrm{,}$$

along with an electrostatic potential energy

$$-eV_i\textrm{,}$$

and a term due to the torque, on the electron magnetic moment, in a magnetic field [30]

$$\frac{e\hbar{}B_i\cos{}S_i}{2m_e}\textrm{,}$$

where $S_i$ is the angle between the electron spin and the magnetic flux density. The form of this last term assumes a well-defined energy for all values of $S_i$. Strictly, only certain $S_i$ values are eigen-states of a Hamiltonian which includes a magnetic field; more about this later (section 2.5.) The total energy is

$$E = \frac{\hbar^2p^2}{2m_e}+\frac{\hbar^2q_i^2}{2m_e}-eV_i+\frac{e\hbar{}B_i\cos{}S_i}{2m_e}\textrm{,}$$ (2.20)

or, where $p$ is expressed as a fraction $\sin{}I$ of the total wave-number in the absence of any potential, $I$ being an angle of incidence like that in figure 2.2,

$$E = E\sin^2I+\frac{\hbar^2q_i^2}{2m_e}-eV_i+\frac{e\hbar{}B_i\cos{}S_i}{2m_e}$$ (2.21)

$$\Rightarrow{}q_i = \left(\frac{2m_eE\cos^2I}{\hbar^2}\right)^{1/2}\left(1+\frac{eV_i}{E\cos^2I}-\frac{e\hbar{}B_i\cos{}S_i}{2m_eE\cos^2I}\right)^{1/2}\protect$$

$$= \left(\frac{2m_eE\cos^2I}{\hbar^2}\right)^{1/2}(1+x_i)^{1/2}\textrm{,}$$ (2.22)

where $x_i = y_i+z_i\cos{}S_i$, $y_i = \frac{eV_i}{E\cos^2I}$, and $z_i = -\frac{e\hbar{}B_i}{2m_eE\cos^2I}$.

I now use a binomial expansion [48] for the case where the potential energy terms are much smaller than the total electron energy, where the dimensionless numbers I've just devised are small. The magnetic term associated with the Weiss field in a ferromagnet is [43] a few tenths of an electron-volt, and the electrostatic contact potentials in the metals which I study will not be more than a few volts, whereas, in my experimental set-up, the incident electron energies range from a few hundred to a few thousand electron volts, so this approximation seems reasonable.

$$q_i = \left(\frac{2m_eE\cos^2I}{\hbar^2}\right)^{1/2}\left(1+\frac{1}{2}x_i-\frac{1}{8}x_i^2+O(x_i^3)\right)\textrm{.}$$ (2.23)

The amplitude reflection coefficient is, therefore,

$$r_{ij} = \frac{1}{2}\left(\frac{1}{2}x_i-\frac{1}{2}x_j+\frac{1}{8}x_j^2-\frac{1}{8}x_i^2+O(\{x_i,x_j\}^3)\right)\protect$$

$$\times\left(1+\frac{1}{4}x_i+\frac{1}{4}x_j-\frac{1}{16}x_i^2-\frac{1}{16}x_j^2+O(\{x_i,x_j\}^3)\right)^{-1}\protect$$

$$= \frac{1}{2}\left(\frac{1}{2}x_i-\frac{1}{2}x_j+\frac{1}{8}x_j^2-\frac{1}{8}x_i^2+O(\{x_i,x_j\}^3)\right)\protect$$

$$\times\left(1-\frac{1}{4}x_i-\frac{1}{4}x_j+\frac{1}{8}x_i^2+\frac{1}{8}x_ix_j+\frac{1}{8}x_j^2+O(\{x_i,x_j\}^3)\right)\protect$$

$$= \frac{1}{2}\left(\frac{1}{2}x_i-\frac{1}{2}x_j-\frac{1}{4}x_i^2-\frac{1}{4}x_ix_j+\frac{1}{4}x_j^2+O(\{x_i,x_j\}^3)\right)\protect$$

$$= \frac{1}{4}x_i-\frac{1}{4}x_j-\frac{1}{8}x_i^2-\frac{1}{8}x_ix_j+\frac{1}{8}x_j^2+O(\{x_i,x_j\}^3)\textrm{.}$$ (2.24)