4. Universal conductance reduction

Another important case, which is going to be relevant to the transistor operation, occurs below the next energy transition where there is a dip in the transmission <sup>2</sup> coefficient, and the conductance decreases by exactly 2e =h.

Let the incoming particle energy be within the energy range EQp<sup>1</sup> < E < EQ ; for which case it is convenient to split the denominator of the coefficient, i.e., to rewrite Eq. (19) as

$$A\_{n,m} \equiv \lim\_{\rho \to 0} \frac{\frac{\chi\_{\mathfrak{a}}(\epsilon)\chi\_{m}^{\*}(\epsilon)}{ik\_{m}}}{\frac{\ln\left(\rho\_{0}/\rho\right)}{\pi} + \sum\_{q=1}^{Q-1} \frac{\chi\_{q}(\epsilon)\chi\_{q}^{\*}(\epsilon)}{i\sqrt{E-E\_{q}}} - \frac{\chi\_{\mathbb{Q}}(\epsilon)\chi\_{\mathbb{Q}}^{\*}(\epsilon)}{\sqrt{E\_{Q}-E}} - \sum\_{q=Q+1}^{\infty} \frac{\chi\_{q}(\epsilon)\chi\_{q}^{\*}(\epsilon)}{\sqrt{E\_{q}-E}} \exp\left(-\left|E - E\_{q}/\rho^{2}/4\right|\right) \tag{24}$$

or

$$A\_{n,m} \equiv \frac{\frac{\chi\_m^\*(\epsilon)\chi\_n(\epsilon)}{i\sqrt{E - E\_m}}}{\frac{\ln\left(\rho\_0/\mathbb{R}(\epsilon)\right)}{\pi} + \sum\_{q=1}^{Q-1} \frac{\chi\_q(\epsilon)\chi\_q^\*(\epsilon)}{i\sqrt{E - E\_q}} - \frac{\chi\_Q(\epsilon)\chi\_Q^\*(\epsilon)}{\sqrt{E\_Q - E}}} \tag{25}$$

where

$$\ln \mathcal{R}(\varepsilon) \equiv \lim\_{\rho \to 0} \left[ \ln \rho + \pi \sum\_{q=Q+1}^{\infty} \frac{\chi\_q(\varepsilon) \chi\_q^\*(\varepsilon)}{\sqrt{E\_q - E}} \exp \left( - \left| E - E\_q \right| \rho^2 / 4 \right) \right]. \tag{26}$$

The device's conductance can be evaluated as ([25, 26])

$$G = \frac{1}{\pi} \sum\_{m\_1 l\_{$$

where

$$T\_{n,m} = \left| \delta\_{n,m} - A\_{n,m} \right|^2 \frac{k\_m}{k\_n} = \begin{cases} \left| 1 - A\_{n,n} \right|^2 & n = m \\ \left| A\_{n,m} \right|^2 \frac{k\_m}{k\_n} & n \neq m \end{cases} \tag{28}$$

are the transmission coefficients (from the nth to the mth modes) of the wire with the impurity.

At the transition points, i.e., when E ¼ EQ (kQ ¼ 0), the conductance is exactly

$$G = \frac{1}{\pi}(Q - 1),\tag{29}$$

which is <sup>G</sup> <sup>¼</sup> <sup>2</sup> <sup>e</sup><sup>2</sup> ðQ � 1Þ in ordinary physical units. <sup>h</sup>

min On the other hand, at the minimum transmission point (<sup>E</sup> <sup>¼</sup> EQ <sup>&</sup>lt; EQ ), when the real part of the denominator of Eq. (25) vanishes, i.e., when

$$\frac{\ln\left(\rho\_0/\mathcal{R}(\varepsilon)\right)}{\pi} = \frac{\chi\_Q(\varepsilon)\chi\_Q^\*(\varepsilon)}{\sqrt{E\_Q - E\_Q^{\min}}},\tag{30}$$

then

$$A\_{n,m} \equiv \frac{\frac{\chi\_m^\*(\epsilon)\chi\_n(\epsilon)}{i\sqrt{E-E\_m}}}{\sum\_{q=1}^{Q-1} \frac{\chi\_q(\epsilon)\chi\_q^\*(\epsilon)}{i\sqrt{E-E\_q}}} = \frac{1}{\frac{\sqrt{E-E\_m}}{\chi\_m^\*(\epsilon)\chi\_n(\epsilon)}\sum\_{q=1}^{Q-1} \frac{\chi\_q(\epsilon)\chi\_q^\*(\epsilon)}{\sqrt{E-E\_q}}} = \frac{1}{\frac{k\_m}{\chi\_m^\*(\epsilon)\chi\_n(\epsilon)}\sum\_{q=1}^{Q-1} \frac{\chi\_q(\epsilon)\chi\_q^\*(\epsilon)}{k\_q}}.\tag{31}$$

Using the definition

$$\sigma \equiv \sum\_{q=1}^{Q-1} \frac{\chi\_q(\varepsilon)\chi\_q^\*(\varepsilon)}{\sqrt{E - E\_q}} = \sum\_{q=1}^{Q-1} \frac{\chi\_q(\varepsilon)\chi\_q^\*(\varepsilon)}{k\_q} \tag{32}$$

<sup>2</sup> km the conductance <sup>G</sup> <sup>¼</sup> <sup>π</sup> <sup>1</sup> <sup>∑</sup>n, <sup>m</sup> <sup>&</sup>lt; <sup>Q</sup> <sup>j</sup>δn,m � An,m<sup>j</sup> <sup>k</sup> is equal exactly to <sup>n</sup>

$$G = \frac{1}{\pi} \sum\_{n\_2, m < Q} \left| \delta\_{n, m} - \frac{\chi\_m^\*(\varepsilon) \chi\_n(\varepsilon)}{k\_m \sigma} \right|^2 \frac{k\_m}{k\_n} = \frac{1}{\pi} \sum\_{n\_2, m < Q} \left[ \delta\_{n, m} - 2 \frac{\chi\_m^\*(\varepsilon) \chi\_n(\varepsilon)}{k\_m \sigma} + \left| \frac{\chi\_m^\*(\varepsilon) \chi\_n(\varepsilon)}{k\_m \sigma} \right|^2 \right] \frac{k\_m}{k\_n} = \frac{Q - 2\varepsilon}{\pi} \tag{33}$$

which is <sup>G</sup> <sup>¼</sup> <sup>2</sup> <sup>e</sup><sup>2</sup> ðQ � 2Þ in ordinary physical units. <sup>h</sup>

Therefore, there is exactly a one unit of conductance reduction between the min transition energy EQ and the minimum point just below it EQ

$$
\Delta G = G\left(E\_Q\right) - G\left(E\_Q^{\min}\right) = \pi^{-1},\tag{34}
$$

which <sup>2</sup> is 2e =h in ordinary physical units.

### Figure 3.

The nanowire's conductance as a function of the particles' Fermi energy. The lower panel is a magnification of the transition zone. The dashed perpendicular line stands for the transition energy formula (Eq. (12)) E ¼ E2, and the dotted line corresponds to the zero-current energy formula (Eq. (30)) <sup>E</sup> <sup>¼</sup> <sup>E</sup>min. <sup>2</sup>

This result is a generalization of [17]. The probability density at the point of minimum conductance is presented in the lower panel of Figure 2, and the dependence of the conductance on the particles' Fermi energy is presented in Figure 3. The minima are clearly seen.

Moreover, the approximate analytical expressions of the transition energy Eq. (12) and the minimum energy Eq.(30) are presented by horizontal lines.
