Clarification of Landauer approach I am trying to understand the Landauer approach. Consider the setup: (left contact)-(conductor)-(right contact). For simplicity, the conductor is a 1d wire (the transverse part is not relevant for this question). The eigenstates of the conductor are 
$$\psi_n(x)=\sin(k_nx), \qquad k_n=\frac{n\pi}{L}$$ 
($L$ length of conductor). I also ignore disorder. In the Landauer approach you assume the left (right) contact populates the right(left)-going states $e^{ik_n x}$ where $k_n=\frac{2\pi}{L}$ with $n>0$ ($n<0$). Then you assume these "scattering states" are populated with a Fermi distribution and you compute the conductance.
I would like to understand why you should think in terms of these scattering states because they do not satisfy the boundary condition $\psi(0)=\psi(L)=0$.
 A: Your approach appears to be incompatible with your assumptions; namely, are not using the correct energy spectrum. Judging by the expression for the wave function that you have given ($\psi_{n}(x)=\sin(k_{n}x)$), it appears as if you are solving an infinite square well problem in 1-D. The different $n$’s label the discrete energy eigenstates in your square well like a hydrogen atom. This does not represent a crystalline material (i.e. electron hopping on a periodic lattice of ions); the 1-D wire, for which you are trying to solve the Landauer formula, falls into this category, at least in the non-interacting limit when you have Bloch bands. The spectrum that you are trying to calculate for could be viewed as that of a quantum dot; it is a 0-D system, except your confinement would be in all three dimensions. It is also worth mentioning that when one typically uses the Landauer formula one implicit assumption is that you are dealing with Fermi liquids. For your specific requirements, i.e. a 1-D wire, we will not be considering Luttinger liquids. To my knowledge I am not aware of any generalizations of the Landauer formula to Luttinger liquids.
Nevertheless, despite your attempts to compute the conductance of a 1-D wire, there is no reason why the Landauer approach should not work for the system you provided, i.e. a quantum dot. Besides it is pedagogical to start with a simple system (say) with just one discrete level. Say your conductor is a quantum dot with only one discrete energy level coupled to two metallic contacts (source and drain) which have an abundance of electronic states. As shown in the figure below, just like the flow of water, electrons from the source contact (with energy $>\varepsilon$) will jump into the discrete energy level. Then the electron which arrives in this discrete energy level then jumps into the drain contact. I'm sure it must be obvious that you must have $\mu_{1}>\varepsilon>\mu_{2}$ for this to work.

The quantities $\gamma_{1}$ and $\gamma_{2}$ are some interface parameters (with units of energy) which determine the coupling strength of this discrete level to the source and drain contacts respectively. Now, $\gamma_{1/2}/\hbar$ represent the rate at which an electron can hop into/out of the discrete energy level. By dimensional analysis you can convince yourself that $\gamma_{1/2}/\hbar$ is indeed a rate (i.e. units $s^{-1}$). I know this is a bit hand wavy, but $\gamma_{1/2}$ are phenomenological parameters, and can be determined such that they are consistent with our formalism. The quantities $f_{1/2}(\varepsilon)$ are $$f_{1/2}(\varepsilon)=\frac{1}{1+\exp\left({\displaystyle \frac{\varepsilon-\mu_{1/2}}{k_{B}T}}\right)}$$ are the Fermi-Dirac distributions in the source/drain contacts. Now, for $T=0$, we have $f_{1}(\varepsilon) = 1$ and $f_{2}(\varepsilon) = 0$. In other words, since a state in the source contact is always occupied at energy $\varepsilon$, it will jump into the discrete energy level. At the same time, since a state in the drain contact is always empty at energy $\varepsilon$, the electron, which jumped from the source to the discrete energy level, will now jump into the drain. The expression for the current, due to a single discrete energy level, can be written as
\begin{align}
I(\varepsilon)&=\frac{2e}{\hbar}\left(\gamma_{1}f_{1}(\varepsilon)-\gamma_{2}f_{2}(\varepsilon)\right)\\
&=\frac{2e}{\hbar}\gamma\left(f_{1}(\varepsilon)-f_{2}(\varepsilon)\right)
\end{align}
where in the second line we have assumed the simple case $\gamma_{1}=\gamma_{2}$ (the factor of 2 comes from spin degeneracy). However, if you now had more than one states, like (say) in the 1-D quantum wire, you can find the total current by
\begin{align}
I&=\int_{-\infty}^{\infty}d\varepsilon\; I(\varepsilon)D(\varepsilon)\\
&=\frac{2e}{\hbar}\gamma\int_{-\infty}^{\infty}d\varepsilon\;\left(f_{1}(\varepsilon)-f_{2}(\varepsilon)\right)D(\varepsilon)
\end{align}
where $D(\varepsilon)$ is the density of states of the 1-D wire. The plot of the density of states has been shown below

It can be noted that only the electronic states in the red box will conduct any current since they satisfy $\mu_{1}>E>\mu_{2}$ (please excuse the fact that I am using $E$ and $\varepsilon$ interchangeably). For a single sub-band the density of states in a 1-D nanowire goes as $D(E)\propto E^{-1/2}$. The above picture shows the density of states for a multi-subband nanowire.The derivations for these can be found in many excellent sources. For example, check section 9.5.3. of this reference.
Now, by noting that $\mu_{1}-\mu_{2}=eV$, where $V$ is the applied voltage between source and drain, can make the following approximation
\begin{align}
f_{1}(\varepsilon)-f_{2}(\varepsilon) &= \frac{1}{1+\exp\left({\displaystyle \frac{\varepsilon-\mu_{1}}{k_{B}T}}\right)}-\frac{1}{1+\exp\left({\displaystyle \frac{\varepsilon-\mu_{2}}{k_{B}T}}\right)} \\
&\approx \frac{\partial}{\partial\mu}\left\{ \frac{1}{1+\exp\left({\displaystyle \frac{\varepsilon-\mu}{k_{B}T}}\right)}\right\} \left(\mu_{1}-\mu_{2}\right) \\
\end{align}
However, by a simple change of variables, it is easy to show that $$\frac{\partial}{\partial\mu}\left\{ \frac{1}{1+\exp\left({\displaystyle \frac{\varepsilon-\mu}{k_{B}T}}\right)}\right\} =-\frac{\partial}{\partial\varepsilon}\left\{ \frac{1}{1+\exp\left({\displaystyle \frac{\varepsilon-\mu}{k_{B}T}}\right)}\right\} $$ Therefore, defining $$f(\varepsilon)=\frac{1}{1+\exp\left({\displaystyle \frac{\varepsilon-\mu}{k_{B}T}}\right)}$$ and $\mu_{1}\approx\mu_{2}\equiv\mu$, we can rewrite the expression for current as
\begin{align}
I &= \frac{2e}{\hbar}\gamma\int_{-\infty}^{\infty}d\varepsilon\;\left(-\frac{\partial f(\varepsilon)}{\partial\varepsilon}\right)\left(\mu_{1}-\mu_{2}\right)D(\varepsilon) \\
&= \frac{2e}{\hbar}\gamma\int_{-\infty}^{\infty}d\varepsilon\;\left(-\frac{\partial f(\varepsilon)}{\partial\varepsilon}\right)\left(eV\right)D(\varepsilon) \\
\frac{I}{V} &= \frac{2e^{2}}{h}\left(\gamma\int_{-\infty}^{\infty}d\varepsilon\;\left(-\frac{\partial f(\varepsilon)}{\partial\varepsilon}\right)D(\varepsilon)\right)
\end{align}
As $T \rightarrow 0$ it is easy to see that $\partial f(\varepsilon)/\partial \varepsilon$ acts as a delta function at $\varepsilon = \mu$ and the quantity in the round parenthesis is simply a number proportional to the density of states at $\varepsilon = \mu$ i.e. $D(\mu)$. We can then write down the conductance as $$\sigma = \frac{2e^{2}}{h}M(\mu)$$ where $M(\varepsilon)$ is the "density of modes" at energy $\varepsilon$.
Now, if you want to incorporate elastic scattering into this formalism you simply need to stick in a factor of $$P(\varepsilon)=\frac{\Lambda(\varepsilon)}{\Lambda(\varepsilon)+L}$$ into the integral over $\varepsilon$ above. $P(\varepsilon)$ represents the probability that an electron with energy $\varepsilon$ will get back scattered. $\Lambda(\varepsilon)$ is obviously the average distance the electron (with energy $\varepsilon$) travels between successive collisions (i.e. mean free path); note that since these collisions are assumed to be elastic the same value of $\Lambda(\varepsilon)$ can be used for this particular electron throughout the transport process, from source to drain. Now, when an electron undergoes a collision, it can either scatter forward or backward. By doing a statistical average you can show that you scatter forward and backward with half probability each. Now, if my device length $L$ was equal to the mean free path $\Lambda(\varepsilon)$, or in other words, if there was only one scatterer in my device, then I would get $$P(\varepsilon)=\frac{\Lambda(\varepsilon)}{\Lambda(\varepsilon)+L} = \frac{1}{2}$$ which is exactly what we expect.
Also, to answer you last question, we normally used the Born–von Karman boundary conditions in crystalline solids and not Dirichlet.
