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If we solve the time-independent Schrödinger equation in 1D

$$\frac{d^2\psi(x)}{dx^2} + \frac{2m}{\hbar^2} \left[E - V(x)\right] \psi(x) = 0$$

for a periodic potential $V(x)$ with wave function $\psi(x)$ subject to a periodic boundary condition: $\psi(x) = \psi(x + Ga)$, where $a$ is the period of $V(x)$ and $G$ is a positive integer so that the lattice lenght $L = G a$, then the general form of $\psi(x)$ is given by $$ \psi_k(x) = e^{ikx} u_k(x), $$ with $u_k(x) = u_k(x+a)$ and $k = \frac{2\pi g}{Ga}, g \in \mathbb{Z}$. This is the famous Bloch theorem.

I am reading a textbook The Wave Mechanics of Electrons in Metals by Stanley Raimes where the author then introduces a new quantum number $l$, relabel $\psi$ and $u$ as $\psi_{kl}(x)$ and $u_{kl}(x)$ and plug $\psi_{kl}(x) = e^{ikx} u_{kl}(x)$ in the Schrödinger equation. Then he takes complex comnugates of the Schrödinger equaiton and compare it with the same equation but $k$ replaed by $-k$. These operations yield the following familiar equations:

$$u_{-kl}(x) = u_{kl}^*(x),$$

and

$$E_l(-k) = E_l(k).$$

Question

Why do we need to introduce a new quantum number $l$ in the first place? If we wanted to solve the Schrödinger equation for a specified potential $V(x)$, then we would have seen how a new parameter $l$ makes its way into the wave function and energy expression, if needed. But here we are not doing that, so what is the reason?

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  • $\begingroup$ I do not have access to the book, but could not $l$ be the label of the individual energy bands? The Bloch wavenumber $k$ is not enough to label all the states. You need another discrete index to label the bands. $\endgroup$
    – mike stone
    Commented Dec 26, 2019 at 16:20
  • $\begingroup$ You are right, but how do we know this information in the first place? Is that an assumption? $\endgroup$
    – rainman
    Commented Dec 26, 2019 at 16:24
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    $\begingroup$ It's not an assumption, but something one can deduce from looking at how, for any energy $E$, the linearly independent pair of solutions $\psi_1(x)$ with boundary conditions $\psi_1(0)=1$, $\psi'(0)=0$ and $\psi_2(x)$ with $\psi_2(0)=0$, $\psi'_2(0)=1$ map under the shift $x\to x+a$. If we set $\Psi=(\psi_1,\psi_2)^{\rm transpose}$ we must have $\Psi(x+a) = T \Psi(x)$ for some transfer matrix $T$. Consideration of ${\rm det}(T)$ gives you the bands. This is described in many books $\endgroup$
    – mike stone
    Commented Dec 26, 2019 at 17:09
  • $\begingroup$ Sorry! I meant ${\rm tr}(T)$ not ${\rm det}(T)$. The latter is alwyas unity. $T$ depends on $E$ and ${\rm tr}(T(E))= 2\cos k$ gives the relation between $E$ and $k$. So the energy range where ${\rm tr}(T(E))\le 2$ are the energy bands and the range where ${\rm tr}(T(E))\ge 2$ are the band gaps. $\endgroup$
    – mike stone
    Commented Dec 27, 2019 at 15:11

1 Answer 1

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We have the Bloch's theorem to get the ansatz for the wavefunction, $ψ_{k}(x)=e^{ikx}u_{k}(x)$ to solve the Schrodinger's equation. We know that $u_{k}(x)$ is a periodic function in x with a period equal to the lattice constant, but we are still left with the task of finding what exactly $u_{k}(x)$ is. Substituting our ansatz into the Schrodinger equation now gives us a differential equation in $u_{k}(x)$ : $$ \left[ \frac{1}{2m}\left(\frac{ℏ}{i} \nabla + ℏk \right)^{2} + V(x) \right] u_{k}(x) = E u_{k}(x)$$

Now for this equation, for the given potential $V(x)$ and the given boundary condition, we can have different functions $u_{k}(x)$ that satisfy it, giving a spectrum of energies. Hence, we need a new quantum number to label these different states, which correspond to eigenstates of different energies. In a solid crystal, these correspond to the different energy bands.

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