量子力学第六弹——一维定态问题

时间：2015-05-06 15:12:00 阅读：129 评论：0 收藏：0 [点我收藏+]

General Discussion of Schr?dinger Equation

single particle equation

i ? ? ? t | ψ (x, t) ? = H | ψ (x, t) ?, H = ? ? 2 2 m ? 2 + V (x, t)

$i\hbar \frac{\partial}{\partial t }|\psi(\mathbf{x},t)\rangle=H|\psi(\mathbf{x},t)\rangle,H=-\frac{\hbar ^2}{2m}\nabla^2+V(\mathbf{x},t)$

multi-particle

i ? ? ? t | ψ (x 1, ?, x N, t) ? = H | ψ (x 1, ?, x N, t) ?, H = \sum i = 1 N [? ? 2 2 m ? 2 + V (x, t)] + V (x 1, ?, x N, t)

$i\hbar \frac{\partial}{\partial t }|\psi(\mathbf{x}_1,\cdots,\mathbf{x}_N,t)\rangle=H|\psi(\mathbf{x}_1,\cdots,\mathbf{x}_N,t)\rangle,H=\sum_{i=1}^N[-\frac{\hbar ^2}{2m}\nabla^2+V(\mathbf{x},t)]+V(\mathbf{x}_1,\cdots,\mathbf{x}_N,t)$

Properties

? ? t ρ + ? ? j = 0

$\frac{\partial}{\partial t}\rho+\nabla\cdot\mathbf{j}=0$

where

ρ = | ψ (x, t) 2 | = ψ (x, t) ? ψ (x, t) j = ? 2 m i [ψ ? (x, t) ? ψ (x, t) ? ψ (x, t) ? ψ ? (x, t)]

$\rho=|\psi(\mathbf{x},t)^2|=\psi(\mathbf{x},t)^*\psi(\mathbf{x},t) \\ \mathbf{j}=\frac{\hbar}{2m\mathrm{i}}[\psi^*(\mathbf{x},t)\nabla\psi(\mathbf{x},t)-\psi(\mathbf{x},t)\nabla\psi^*(\mathbf{x},t)]$

proof
We assume $V(\mathbf{x},t)$ to be real

$i ? ? ? t ψ (x, t) = ? ? 2 2 m ? 2 ψ (x, t) + V (x, t) ψ (x, t) ? ? i ? ? ? t ψ ? (x, t) = ? ? 2 2 m ? 2 ψ ? (x, t) + V (x, t) ψ ? (x, t) ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ψ ? (x, t) i ? ? ? t ψ (x, t) = ψ ? (x, t) (? ? 2 2 m ? 2 ψ (x, t) + V (x, t) ψ (x, t)) ψ (x, t) (? i ? ? ? t ψ ? (x, t)) = ψ (x, t) (? ? 2 2 m ? 2 ψ ? (x, t) + V (x, t) ψ ? (x, t))$ $i\hbar \frac{\partial}{\partial t }\psi(\mathbf{x},t)=-\frac{\hbar ^2}{2m}\nabla^2\psi(\mathbf{x},t)+V(\mathbf{x},t)\psi(\mathbf{x},t)\\ \Rightarrow -i\hbar \frac{\partial}{\partial t }\psi^*(\mathbf{x},t)=-\frac{\hbar ^2}{2m}\nabla^2\psi^*(\mathbf{x},t)+V(\mathbf{x},t)\psi^*(\mathbf{x},t)\\ \Rightarrow\begin{cases}\psi^*(\mathbf{x},t)i\hbar \dfrac{\partial}{\partial t }\psi(\mathbf{x},t)=\psi^*(\mathbf{x},t)\left(-\dfrac{\hbar ^2}{2m}\nabla^2\psi(\mathbf{x},t)+V(\mathbf{x},t)\psi(\mathbf{x},t)\right)\\ \psi(\mathbf{x},t)\left(-i\hbar \dfrac{\partial}{\partial t }\psi^*(\mathbf{x},t)\right)=\psi(\mathbf{x},t)\left(-\dfrac{\hbar ^2}{2m}\nabla^2\psi^*(\mathbf{x},t)+V(\mathbf{x},t)\psi^*(\mathbf{x},t)\right)\end{cases}$
$(1)-(2)$
$l . h . s = i ? (ψ ? ? ? t ψ + ψ ? ? t ψ ?) = i ? ? ? t | ψ (x, t) 2 | r . h . s = ? ? 2 2 m (ψ ? ? 2 ψ ? ψ ? 2 ψ ?) = ? ? 2 2 m ? ? (ψ ? ? ψ ? ψ ? ψ ?) ? i ? ? ? t | ψ (x, t) 2 | + ? 2 m i ? ? (ψ ? ? ψ ? ψ ? ψ ?) \equiv 0$ $l.h.s=i\hbar(\psi^*\frac{\partial}{\partial t}\psi+\psi\frac{\partial}{\partial t}\psi^*)=i\hbar \frac{\partial}{\partial t}|\psi(\mathbf{x},t)^2|\\r.h.s=-\dfrac{\hbar ^2}{2m}(\psi^*\nabla^2\psi-\psi\nabla^2\psi^*)\\=-\dfrac{\hbar ^2}{2m}\nabla\cdot(\psi^*\nabla\psi-\psi\nabla\psi^*)\\\Rightarrow i\hbar \frac{\partial}{\partial t}|\psi(\mathbf{x},t)^2|+\frac{\hbar}{2mi}\nabla\cdot(\psi^*\nabla\psi-\psi\nabla\psi^*)\equiv0$
let $\rho=|\psi(\mathbf{x},t)^2|,\mathbf{j}=\dfrac{\hbar}{2m\mathrm{i}}[\psi^*\nabla\psi-\psi\nabla\psi^*]$
thus
$? ? t ρ + ? ? j = 0$ $\frac{\partial}{\partial t}\rho+\nabla\cdot\mathbf{j}=0$

Some properties of 1D stationary problem

i ? ? ? t | ψ (x, t) ? = H | ψ (x, t) ?

$i\hbar \frac{\partial}{\partial t }|\psi(\mathbf{x},t)\rangle=H|\psi(\mathbf{x},t)\rangle$
Assume

|ψ(x,t)?=|ψ(x)??e?iEt/? $|\psi(\mathbf{x},t)\rangle=|\psi(\mathbf{x})\rangle \cdot e^{-iEt/\hbar}$
Then

H ψ = E ψ

$H\psi=E\psi$

$\left[-\dfrac{\hbar ^2}{2m}\nabla^2+V(\mathbf{x})\right]\psi(\mathbf{x})=E\psi(\mathbf{x})$
正问题： $V(\mathbf{x})\rightarrow E,\psi(\mathbf{x})$
逆问题： $E \rightarrow V(\mathbf{x})$
both $\psi(\mathbf{x}),\psi^*(\mathbf{x})$ are solutions.
if $\psi(\mathbf{x})$ is not degenerate, it can be chosen to be real.
if $\psi(\mathbf{x})$ is a solution and $V(\mathbf{x})=V(-\mathbf{x})$ then $\psi(-\mathbf{x})$ is another solution.
if $V(\mathbf{x})=V(-\mathbf{x})$ , a complete set of solutions.

Infinite Square Potential Well Problem

Potential

V (x) = {0, \infty, 0 < x < a e l s e

$V(x)=\begin{cases}0,&0<x<a\\ \infty ,&else\end{cases}$

Wave function

d 2 d x 2 ψ (x) + 2 m E ? 2 ψ (x) = 0

$\frac{d^2}{dx^2}\psi(x)+\frac{2mE}{\hbar^2}\psi(x)=0$

General Solution

ψ (x) = A sin (k x + δ), k = 2 m E ? ? ? ? \sqrt ?

$\psi(x)=A\sin(kx+\delta),k=\frac{\sqrt{2mE}}{\hbar}$

Properties

Bounded
Discrete（束缚态 $\rightarrow$ 离散谱，非束缚态 $\rightarrow$ 连续谱）
Exact solution

Boundary condition

ψ (0) = 0, ψ (a) = 0

$\psi(0)=0,\psi(a)=0$

Solution

ψ n (x) ? ? ? ? ? 2 a ? ? \sqrt sin (n π x a), 0, 0 < x < a e l s e

$\psi_n(x)\begin{cases}\sqrt{\dfrac 2 a}\sin\left(\dfrac{n\pi x}{a}\right),&0<x<a\\0,&else\end{cases}$

E n = ? 2 π 2 n 2 2 m a 2

$E_n=\frac{\hbar^2\pi^2n^2}{2ma^2}$

Finite Square Potential Well Problem

Potential

V (x) = {0, V 0, | x | < a 2 e l s e

$V(x)=\begin{cases}0,&|x|<\dfrac a2\\ V_0 ,&else\end{cases}$

Wave function

? ? ? ? ? ? ? ? ? d 2 d x 2 ψ (x) + 2 m E ? 2 ψ (x) = 0, d 2 d x 2 ψ (x) + (2 m E ? 2 ? V 0) ψ (x) = 0 ， | x | < a 2 e l s e

$\begin{cases}\dfrac{d^2}{dx^2}\psi(x)+\dfrac{2mE}{\hbar^2}\psi(x)=0, &|x|<\dfrac a2\\\dfrac{d^2}{dx^2}\psi(x)+\left(\dfrac{2mE}{\hbar^2}-V_0\right)\psi(x)=0，&else\end{cases}$

General Solution

? ? ? ? ? ? ? ? ? ψ (x) ～ e \pm β x, β = 2 m ( V 0 ? E ) ? ? ? ? ? ? ? ? ? ? \sqrt ? ， ψ (x) ～ sin k x, cos k x, e \pm i k x, k = 2 m E ? ? ? ? \sqrt ? ， | x | < a 2 e l s e

$\begin{cases}\psi(x)\sim e^{\pm\beta x},\beta=\dfrac{\sqrt{2m(V_0-E)}}{\hbar} ，&|x|<\dfrac a2\\\psi(x)\sim \sin kx,\cos kx,e^{\pm ikx},k=\dfrac{\sqrt{2mE}}{\hbar} ，&else\end{cases}$
Because

e±ikx $e^{\pm ikx}$ does not have a definite parity, it cannot be a solution.

Properties

Bounded
Discrete
Not exact solution

Boundary condition

? ? ? ? ? ? ? ψ (x) \to 0, [ln cos k x]' x = a 2 = [ln e ? β x]' x = a 2, [ln sin k x]' x = a 2 = [ln e ? β x]' x = a 2, | x | \to \infty (even parity) (odd parity)

$\begin{cases}\psi(x)\to 0,&|x|\to\infty\\ [\ln \cos kx]‘_{x=\frac a2}=[\ln e^{-\beta x}]‘_{x=\frac a2},&\text{(even parity)}\\ [\ln \sin kx]‘_{x=\frac a2}=[\ln e^{-\beta x}]‘_{x=\frac a2},&\text{(odd parity)}\end {cases}$

Solution

ψ n (x) = ? ? ? ? ? 2 a ? ? \sqrt sin (n π x a), 0, 0 < x < a e l s e

$\psi_n(x)=\begin{cases}\sqrt{\dfrac 2 a}\sin\left(\dfrac{n\pi x}{a}\right),&0<x<a\\0,&else\end{cases}$

E n = ? 2 π 2 n 2 2 m a 2

$E_n=\frac{\hbar^2\pi^2n^2}{2ma^2}$
The minimum energy is not zero.

Reduced Equations

? ? ? ? ? ? ? ? ? k tan (k a 2) = β ? k cot (k a 2) = β ? {ξ tan ξ = η, ? ξ cot ξ = η, even parity odd parity

$\begin{cases}k\tan\left(\dfrac {ka}2\right)=\beta\\ -k\cot\left(\dfrac {ka}2\right)=\beta\end{cases}\\\Rightarrow\begin{cases}\xi\tan\xi=\eta,&\text{even parity}\\ -\xi\cot\xi=\eta,&\text{odd parity}\end{cases}$
where

ξ=ka2,η=βa2 $\xi=\dfrac{ka}2,\eta=\dfrac{\beta a}2$

The above equation can only be solved numerically.

Finite Square Potential Barrier Problem

Potential

V (x) = {V 0, 0, 0 < x < a e l s e

$V(x)=\begin{cases}V_0,&0<x<a\\ 0 ,&else\end{cases}$

Wave function

? ? ? ? ? ? ? ? ? d 2 d x 2 ψ (x) + 2 m ( E ? V 0 ) ? 2 ψ (x) = 0, d 2 d x 2 ψ (x) + 2 m E ? 2 ψ (x) = 0, 0 < x < a e l s e

$\begin{cases}\dfrac{d^2}{dx^2}\psi(x)+\dfrac{2m(E-V_0)}{\hbar^2}\psi(x)=0,&0<x<a\\ \dfrac{d^2}{dx^2}\psi(x)+\dfrac{2mE}{\hbar^2}\psi(x)=0,&else\end{cases}$

General Solution

ψ (x) = ? ? ? A e κ x + B e ? κ x, e i k x + Q e ? i k x, S e i k x, 0 < x < a x < 0 x > a

$\psi(x)=\begin{cases}Ae^{\kappa x}+Be^{-\kappa x},&0<x<a\\e^{ikx}+Qe^{-ikx},&x<0\\Se^{ikx},&x>a \end{cases}$
where

κ=2m(V0?E)??????????√?,k=2mE????√? $\kappa=\dfrac{\sqrt{2m(V_0-E)}}{\hbar},k=\dfrac{\sqrt{2mE}}{\hbar}$ ,

Q $Q$ is the reflection coefficient amplitude and

S $S$ is the transmission coefficient amplitude.

T=|S|2,R=|Q|2=1?T $T=|S|^2,R=|Q|^2=1-T$

Properties

Bounded
Discrete（束缚态 $\rightarrow$ 离散谱，非束缚态 $\rightarrow$ 连续谱）
Exact solution

Boundary condition

? ? ? ? ? ? ? ? ? ? ? ? ? 1 + Q = A + B i k κ (1 ? Q) = A ? B A e κ a + B e ? κ a = S e i k a A e κ a ? B e ? κ a = i k κ S e i k a

$\begin{cases}1+Q=A+B\\\dfrac{ik}{\kappa}(1-Q)=A-B\\Ae^{\kappa a}+Be^{-\kappa a}=Se^{ika}\\Ae^{\kappa a}-Be^{-\kappa a}=\frac{ik}{\kappa}Se^{ika}\end{cases}$

Solution

? ? ? ? ? ? ? ? ? ? ? T = | S | 2 = 4 k 2 κ 2 ( k 2 + κ 2 ) 2 s h 2 κ a + 4 k 2 κ 2, R = | Q | 2 = ( k 2 + κ 2 ) 2 s h 2 κ a ( k 2 + κ 2 ) 2 s h 2 κ a + 4 k 2 κ 2 = 1 ? T

$\begin{cases}T=|S|^2=\dfrac{4k^2\kappa^2}{(k^2+\kappa^2)^2\mathrm{sh}^2\kappa a+4k^2\kappa^2},\\R=|Q|^2=\dfrac{(k^2+\kappa^2)^2\mathrm{sh}^2\kappa a}{(k^2+\kappa^2)^2\mathrm{sh}^2\kappa a+4k^2\kappa^2}=1-T\end{cases}$

Discussion

Quantum tunneling effect: even $E<V_0$ , $T \neq 0$ , even $E>V_0$ , $T \neq 1$

1D Harmonic Oscillator Problem

Potential

V (x) = 1 2 m ω 2 x 2

$V(x)=\frac 12 m\omega^2 x^2$

Thanks to Prof. Guo Hong

量子力学第六弹——一维定态问题

标签：量子力学波函数一维定态问题薛定谔方程

原文地址：http://blog.csdn.net/u013795675/article/details/45534983

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