91影视

• A differential equation that describes matter in quantum mechanics in terms of the wave-like properties of particles in a field. Its answer is related to a particle's probability density in space and time.
• The time-dependent Schrodinger equation is represented as

Consider two cases, when particle has positive energy (E > 0) and when it has negative energy$(E>0)$

i) $E>0$: Here $伪$is negative:

$$\begin{gathered} \cos \left(Ka\right)=\cos \left(ka\right)-\frac{m伪}{h^{2}k}sin\left(ka\right) \\ k=\sqrt{\frac{2mE}{h^2}},K=\frac{2蟺n}{Na} \end{gathered}$$

(ii) $E>0$First write Schrodinger equation for region $0<x<a-\frac{{魔}^2}{2m}\frac{{鈭�}^2蠄}{鈭�x^2}=-\left|E\right|$

$\frac{{鈭�}^2蠄}{鈭�x^2}=\frac{2m\left|E\right|}{{魔}^2}蠄=k蠄$

Since k is positive, solution is

_{${蠄}_l\left(\mathrm{x}\right)=\mathrm{A}\cosh \left(\mathrm{kx}\right)+\mathrm{B}\sin \mathrm{h}\left(\mathrm{kx}\right)$}

Using Bloch鈥檚 theorem, wave function in the 鈥渃ell鈥� to the left of the origin (in the region ₎,$-a<x<0),\mathrm{is}{\mathrm{蠄}}_{ll}\left(\mathrm{x}\right)={\mathrm{e}}^{-\mathrm{ika}}\left[\mathrm{A}\cosh \left(\mathrm{k}\left(\mathrm{x}+\mathrm{a}\right)\right)+\mathrm{B}\sin \mathrm{h}\left(\mathrm{k}\left(\mathrm{x}+\mathrm{a}\right)\right)\right]$

Now we impose boundary conditions, at x=0 wave function must be continuous but its deviation isn鈥檛 because of delta potential.$A=e^{-ika}\left(A\cos h\left(ka\right)+B\sin h\left(ka\right)\right)$

To find the derivation of wave function in point x=0 , we write Schrodinger equation:

$-\frac{h^2}{2m}\frac{{鈭�}^2蠄}{鈭�x^2}-伪未\left(x\right)蠄=-E蠄$

To find first derivation we need to integrate previous equation from$-蔚$ to $+蔚$. After taking the limit $蔚鈫�0$, RHS of equation vanishes, and we are left with:

$$\begin{gathered} \frac{{\mathrm{h}}^2}{2m}{\left[\frac{鈭�蠄}{鈭�x}\right]}_{x鈫�0^{+}}-{\left[\frac{鈭�蠄}{鈭�x}\right]}_{x鈫�0^{-}}=-伪蠄\left(0\right) \\ {\left[\frac{鈭�{蠄}_l}{鈭�x}\right]}_{x=0}-{\left[\frac{鈭�{蠄}_{ll}}{鈭�x}\right]}_{x=0}=-\frac{2m伪}{{\mathrm{h}}^2}蠄\left(0\right) \\ Bk-e^{-iKa}(A\sin h\left(ka\right)+B\cos h\left(ka\right))=-\frac{2m伪}{h^2}A \end{gathered}$$

$B=\frac{A\left(e^{iKa}-\cos h\left(魔a\right)\right)}{\sin h\left(ka\right)}$

Result in expression about it

$$\begin{gathered} Ak\left(e^{ika}-\cos h\left(ka\right)\right)-k{e}^{-ika}\left(A\sin h^2\left(ka\right)+A{e}^{ika}\cos h\left(ka\right)-A\cos h^2\left(ka\right)\right)=-\frac{2m伪A}{h^2}\sin h\left(ka\right) \\ k{e}^{ika}-k\cos h\left(ka\right)-k{e}^{-ika}\left(e^{ika}\cos h\left(ka\right)-1\right)=-\frac{2m伪A}{h^2}\sin h\left(ka\right) \\ k{e}^{ika}-2k\cos h\left(ka\right)-\frac{2m伪A}{h^2}\sin h\left(ka\right)\frac{2m伪}{{魔}^2}\sin h\left(ka\right) \\ \cos \left(Ka\right)=\cos h\left(ka\right)-\frac{m伪}{h^{2}k}\sin h\left(ka\right) \end{gathered}$$

In order to graph this function, we introduce substitute:

$$\begin{gathered} z=-ka,尾=-\frac{m伪a}{魔} \\ \cos \left(ka\right)=f\left(z\right)=\cos h\left(z\right)+尾\frac{\sin h\left(z\right)}{\left(z\right)} \end{gathered}$$

Because must be in region between and 1 that means:

$Ka=\frac{2蟺n}{Na},n=0,1,2,...,N-1$

Every band has N states.

Therefore the states in first allowed band are

role="math" localid="1658232648734" $$\begin{gathered} \mathrm{If}E>0\mathrm{then}\cos \left(\mathrm{Ka}\right)=\cos \left(\mathrm{ka}\right)-\frac{\mathrm{尘伪}}{{\mathrm{h}}^2\mathrm{k}}\sin \left(\mathrm{ka}\right) \\ \mathrm{If}\mathrm{E}<0\mathrm{then}\cos \left(\mathrm{Ka}\right)=\cosh \left(\mathrm{ka}\right)-\frac{\mathrm{尘伪}}{{\mathrm{h}}^2\mathrm{k}}\sin \left(\mathrm{ka}\right) \end{gathered}$$