User:物理极客铭/沙盒

一維解

The barriers outside a one-dimensional box have infinitely large potential, while the interior of the box has a constant, zero potential.

這是最簡單無限深方陷阱。這裡, 粒子只可以在直線上來回運動并最終撞上堅不可摧的陷阱壁. 陷阱的陷阱壁有無窮大的勢位. 反過來講，陷阱內的勢位總是為0 這意味著沒有任何力施加于陷阱內的粒子而且該粒子可以自由移動. 然而, 無窮大的力施加于觸碰到陷阱壁上的粒子以防止粒子逃離該陷阱. 勢位能被描述為

$V(x)={\begin{cases}0,&x_{c}-{\tfrac {L}{2}}<x<x_{c}+{\tfrac {L}{2}},\\\infty ,&{\text{otherwise,}}\end{cases}},$

其中L 是陷阱的長度，x_c 陷阱的中心坐標，x 陷阱內栗子的坐標. 簡單情況下包括陷阱關於原點對稱 (x_c = 0 ) 或者陷阱的端點在原點 (x_c = L/2 )。

位置的波函數方程

在量子力學中, 波動方程告訴我們粒子行为最基本的解釋; 粒子可測量的性質(例如粒子的位子, 動量和能量) 都被包括在波函數中 波函數 $\psi (x,t)$ 可以通過解薛定諤方程來得到其表達式。

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

其中 $\hbar$ 是 reduced Planck constant, $m$ 是粒子的質量, $i$ 虛數單位, $t$ 是時間。

在陷阱內, 沒有任何力施加于該粒子, 這意味著陷阱內的波函數在時間與空間中的振荡與自由粒子的形式是一樣的:

$\psi (x,t)=[A\sin(kx)+B\cos(kx)]\mathrm {e} ^{-i\omega t},$

(1)

$A$ ，$B$ 是任意複數. 空間和時間的震蕩的頻率能通過wavenumber $k$ 和 角頻率 $\omega$ 得到. 它們都和粒子總能量有關 by 這個表達式

$E=\hbar \omega ={\frac {\hbar ^{2}k^{2}}{2m}},$

which is known as the自由粒子的 dispersion relation. Here one must notice that now, since the particle is not entirely free but under the influence of a potential (the potential V described above), the energy of the particle given above is not the same thing as ${\frac {p^{2}}{2m}}$ where p is the momentum of the particle, and thus the wavenumber k above actually describes the energy states of the particle, not the momentum states (id est, it turns out that the momentum of the particle is not given by $p=\hbar k$). In this sense, it is quite dangerous to call the number k a wavenumber, since it is not related to momentum like "wavenumber" usually is. The rationale for calling k the wavenumber is that it enumerates the number of crests that the wavefuntion has inside the box, and in this sense it is a wavenumber. This discrepancy can be seen more clearly below, when we find out that the energy spectrum of the particle is discrete (only discrete values of energy are allowed) but the momentum spectrum is continuous (momentum can vary continuously) and in particular, the relation $E={\frac {p^{2}}{2m}}$ for the energy and momentum of the particle does not hold. As said above, the reason this relation between energy and momentum does not hold is that the particle is not free, but there is a potential V in the system, and the energy of the particle is $E=T+V$, where T is the kinetic and V the potential energy.

Initial wavefunctions for the first four states in a one-dimensional particle in a box

The size (or amplitude) of the wavefunction at a given position is related to the probability of finding a particle there by $P(x,t)=|\psi (x,t)|^{2}$. The wavefunction must therefore vanish everywhere beyond the edges of the box. Also, the amplitude of the wavefunction may not "jump" abruptly from one point to the next. These two conditions are only satisfied by wavefunctions with the form

$\psi _{n}(x,t)={\begin{cases}A\sin(k_{n}(x-x_{c}+{\tfrac {L}{2}}))\mathrm {e} ^{-i\omega _{n}t},&x_{c}-{\tfrac {L}{2}}<x<x_{c}+{\tfrac {L}{2}},\\0,&{\text{otherwise,}}\end{cases}}$

where

$k_{n}={\frac {n\pi }{L}}$,

and

$\omega _{n}={\frac {\pi hn^{2}}{4L^{2}m}}$,

其中n是一個正整數 (1,2,3,4...). For a shifted box (x_c = L/2), the solution is particularly simple. The simplest solutions, $k_{n}=0$ or $A=0$ both yield the trivial wavefunction $\psi (x)=0$, which describes a particle that does not exist anywhere in the system. Negative values of $n$ are neglected, since they give wavefunctions identical to the positive $n$ solutions except for a physically unimportant sign change. Here one sees that only a discrete set of energy values and wavenumbers k are allowed for the particle. Usually in quantum mechanics it is also demanded that the derivative of the wavefunction in addition to the wavefunction itself be continuous; here this demand would lead to the only solution being the constant zero function, which is not what we desire, so we give up this demand (as this system with infinite potential can be regarded as a nonphysical abstract limiting case, we can treat it as such and "bend the rules"). Note that giving up this demand means that the wavefunction is not a differentiable function at the boundary of the box, and thus it can be said that the wavefunction does not solve the Schrödinger equation at the boundary points $x=0$ and $x=L$ (but does solve everywhere else).

Finally, the unknown constant $A$ may be found by normalizing the wavefunction so that the total probability density of finding the particle in the system is 1. It follows that

$\left|A\right|={\sqrt {\frac {2}{L}}}.$

Thus, A may be any complex number with absolute value √(2/L); these different values of A yield the same physical state, so A = √(2/L) can be selected to simplify.

It is expected that the eigenvalues, i.e., the energy $E_{n}$ of the box should be the same regardless of its position in space, but $\psi _{n}(x,t)$ changes. Notice that $x_{c}-{\tfrac {L}{2}}{L}$ represents a phase shift in the wave function, This phase shift has no effect when solving the Schrödinger equation, and therefore does not affect the eigenvalue.

Momentum wave function

The momentum wavefunction is proportional to the Fourier transform of the position wavefunction. With $k=p/\hbar$ (note that this k, describing the momentum states, is not the same thing as the k_n above, which described the energy states) the momentum wave function is given by:

$\phi _{n}(p,t)={\frac {1}{\sqrt {2\pi \hbar }}}\int _{-\infty }^{\infty }\psi _{n}(x,t)e^{-ikx}\,dx={\sqrt {\frac {L}{\pi \hbar }}}\left({\frac {n\pi }{n\pi +kL}}\right)\,{\textrm {sinc}}\left({\tfrac {1}{2}}(n\pi -kL)\right)e^{-ikx_{c}}$

where sinc(.) is the sinc function (sinc(x)=sin(x)/x). for the centered box (x_c = 0), the solution is particularly simple since the phase factor on the right is unity. Note that in the problematic case of $n\pi +kL=0$ (i.e. $k=k_{n}$), the sinc function is also zero, and the limiting case is taken.

It can be seen that the momentum spectrum is continuous, and one can conclude that for energy state described by the $k_{n}$ wavenumber, the momentum can, when measured, attain also other values than $p=\pm \hbar k_{n}$, although these are the most likely momentum values (the momentum wavefunction attains its greatest values at these momentum values). Hence one also sees that since the energy is $E_{n}={\frac {\hbar ^{2}k_{n}^{2}}{2m}}$ (since we are talking about the n:th energy eigenstate) the relation $E={\frac {p^{2}}{2m}}$ does not necessarily hold; here p is the measured momentum (since the energy eigenstate $\psi _{n}$ is not a momentum eigenstate, and actually not even a superposition of two momentum eigenstates, as one could imagine from equation (1) above, it has no well-defined momentum before measurement).

Position and momentum probability distributions

In classical physics, the particle can be detected anywhere in the box with equal probability. In quantum mechanics, however, the probability density for finding a particle at a given position is derived from the wavefunction as $P(x)=|\psi (x)|^{2}.$ For the particle in a box, the probability density for finding the particle at a given position depends upon its state, and is given by

$P_{n}(x,t)={\begin{cases}{\frac {L}{2}}\sin ^{2}(k_{n}(x-x_{c}+{\tfrac {L}{2}})),&x_{c}-L/2<x<x_{c}+L/2,\\0,&{\text{otherwise,}}\end{cases}}$

Thus, for any value of n greater than one, there are regions within the box for which $P(x)=0$, indicating that spatial nodes exist at which the particle cannot be found.

In quantum mechanics, the average, or expectation value of the position of a particle is given by

$\langle x\rangle =\int _{-\infty }^{\infty }xP_{n}(x)\,\mathrm {d} x.$

For the steady state particle in a box, it can be shown that the average position is always $\langle x\rangle =x_{c}$, regardless of the state of the particle. For a superposition of states, the expectation value of the position will change based on the cross term which is proportional to $\cos(\omega t)$.

The variance in the position is a measure of the uncertainty in position of the particle:

$\mathrm {Var} (x)=\int _{-\infty }^{\infty }(x-\langle x\rangle )^{2}P_{n}(x)\,dx={\frac {L^{2}}{12}}\left(1-{\frac {6}{n^{2}\pi ^{2}}}\right)$

The probability density for finding a particle with a given momentum is derived from the wavefunction as $P(x)=|\phi (x)|^{2}.$. As with position, the probability density for finding the particle at a given momentum depends upon its state, and is given by

$P_{n}(p)={\frac {L}{\pi \hbar }}\left({\frac {n\pi }{n\pi +kL}}\right)^{2}\,{\textrm {sinc}}^{2}\left({\tfrac {1}{2}}(n\pi -kL)\right)$

where, again, $k=p/\hbar$. The expectation value for the momentum is then calculated to be zero, and the variance in the momentum is calculated to be:

$\mathrm {Var} (p)=\left({\frac {\hbar n\pi }{L}}\right)^{2}$

The uncertainties in position and momentum ($\Delta x$ and $\Delta p$) are defined as being equal to the square root of their respective variances, so that:

$\Delta x\Delta p={\frac {\hbar }{2}}{\sqrt {{\frac {n^{2}\pi ^{2}}{3}}-2}}$

This product increases with increasing n, having a minimum value for n=1. The value of this product for n=1 is about equal to 0.568 $\hbar$ which obeys the Heisenberg uncertainty principle, which states that the product will be greater than or equal to $\hbar /2$

Another measure of uncertainty in position is the information entropy of the probability distribution H_x:

$H_{x}=\int _{-\infty }^{\infty }P_{n}(x)\log(P_{n}(x)x_{0})\,dx=\log \left({\frac {2L}{e\,x_{0}}}\right)$

where x₀ is an arbitrary reference length.

Another measure of uncertainty in momentum is the information entropy of the probability distribution H_p:

$H_{p}(n)=\int _{-\infty }^{\infty }P_{n}(p)\log(P_{n}(p)p_{0})\,dp$

where p₀ is an arbitrary reference momentum. The integral is difficult to express analytically for general n, but in the limit as n approaches infinity:

$\lim _{n->\infty }H_{p}(n)=\log \left({\frac {4\pi \hbar \,e^{2(1-\gamma )}}{L\,p_{0}}}\right)$

where γ is Euler's constant. The quantum mechanical entropic uncertainty principle states that for $x_{0}\,p_{0}=\hbar$

$H_{x}+H_{p}(n)\geq \log(e\,\pi )\approx 2.14473...$ (nats)

For $x_{0}\,p_{0}=\hbar$, the sum of the position and momentum entropies yields:

$H_{x}+H_{p}(\infty )=\log(8\pi \,e^{1-2\gamma })\approx 3.06974...$ (nats)

which satisfies the quantum entropic uncertainty principle.

Energy levels

The energy of a particle in a box (black circles) and a free particle (grey line) both depend upon wavenumber in the same way. However, the particle in a box may only have certain, discrete energy levels.

The energies which correspond with each of the permitted wavenumbers may be written as

$E_{n}={\frac {n^{2}\hbar ^{2}\pi ^{2}}{2mL^{2}}}={\frac {n^{2}h^{2}}{8mL^{2}}}$.

The energy levels increase with $n^{2}$, meaning that high energy levels are separated from each other by a greater amount than low energy levels are. The lowest possible energy for the particle (its zero-point energy) is found in state 1, which is given by

$E_{1}={\frac {\hbar ^{2}\pi ^{2}}{2mL^{2}}}.$

The particle, therefore, always has a positive energy. This contrasts with classical systems, where the particle can have zero energy by resting motionlessly. This can be explained in terms of the uncertainty principle, which states that the product of the uncertainties in the position and momentum of a particle is limited by

$\Delta x\Delta p\geq {\frac {\hbar }{2}}$

It can be shown that the uncertainty in the position of the particle is proportional to the width of the box. Thus, the uncertainty in momentum is roughly inversely proportional to the width of the box. The kinetic energy of a particle is given by $E=p^{2}/(2m)$, and hence the minimum kinetic energy of the particle in a box is inversely proportional to the mass and the square of the well width, in qualitative agreement with the calculation above.