# 7.4 The quantum particle in a box  (Page 4/12)

 Page 4 / 12 The probability density distribution | ψ n ( x ) | 2 for a quantum particle in a box for: (a) the ground state, n = 1 ; (b) the first excited state, n = 2 ; and, (c) the nineteenth excited state, n = 20 .

The probability density of finding a classical particle between x and $x+\text{Δ}x$ depends on how much time $\text{Δ}t$ the particle spends in this region. Assuming that its speed u is constant, this time is $\text{Δ}t=\text{Δ}x\text{/}u,$ which is also constant for any location between the walls. Therefore, the probability density of finding the classical particle at x is uniform throughout the box, and there is no preferable location for finding a classical particle. This classical picture is matched in the limit of large quantum numbers. For example, when a quantum particle is in a highly excited state, shown in [link] , the probability density is characterized by rapid fluctuations and then the probability of finding the quantum particle in the interval $\text{Δ}x$ does not depend on where this interval is located between the walls.

## A classical particle in a box

A small 0.40-kg cart is moving back and forth along an air track between two bumpers located 2.0 m apart. We assume no friction; collisions with the bumpers are perfectly elastic so that between the bumpers, the car maintains a constant speed of 0.50 m/s. Treating the cart as a quantum particle, estimate the value of the principal quantum number that corresponds to its classical energy.

## Strategy

We find the kinetic energy K of the cart and its ground state energy ${E}_{1}$ as though it were a quantum particle. The energy of the cart is completely kinetic, so $K={n}^{2}{E}_{1}$ ( [link] ). Solving for n gives $n={\left(K\text{/}{E}_{1}\right)}^{1\text{/}2}$ .

## Solution

The kinetic energy of the cart is

$K=\frac{1}{2}m{u}^{2}=\frac{1}{2}\left(0.40\phantom{\rule{0.2em}{0ex}}\text{kg}\right){\left(0.50\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)}^{2}=0.050\phantom{\rule{0.2em}{0ex}}\text{J}.$

The ground state of the cart, treated as a quantum particle, is

${E}_{1}=\frac{{\pi }^{2}{\hslash }^{2}}{2m{L}^{2}}=\frac{{\pi }^{2}{\left(1.05\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-34}\text{J}·\text{s}\right)}^{2}}{2\left(0.40\phantom{\rule{0.2em}{0ex}}\text{kg}\right){\left(2.0\phantom{\rule{0.2em}{0ex}}\text{m}\right)}^{2}}=1.700\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-68}\text{J}.$

Therefore, $n={\left(K\text{/}{E}_{1}\right)}^{1\text{/}2}={\left(0.050\text{/}1.700\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-68}\right)}^{1\text{/}2}=1.2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{33}$ .

## Significance

We see from this example that the energy of a classical system is characterized by a very large quantum number. Bohr’s correspondence principle    concerns this kind of situation. We can apply the formalism of quantum mechanics to any kind of system, quantum or classical, and the results are correct in each case. In the limit of high quantum numbers, there is no advantage in using quantum formalism because we can obtain the same results with the less complicated formalism of classical mechanics. However, we cannot apply classical formalism to a quantum system in a low-number energy state.

Check Your Understanding (a) Consider an infinite square well with wall boundaries $x=0$ and $x=L$ . What is the probability of finding a quantum particle in its ground state somewhere between $x=0$ and $x=L\text{/}4$ ? (b) Repeat question (a) for a classical particle.

a. 9.1%; b. 25%

Having found the stationary states ${\psi }_{n}\left(x\right)$ and the energies ${E}_{n}$ by solving the time-independent Schrӧdinger equation [link] , we use [link] to write wave functions ${\text{Ψ}}_{n}\left(x,t\right)$ that are solutions of the time-dependent Schrӧdinger’s equation given by [link] . For a particle in a box this gives

${\text{Ψ}}_{n}\left(x,t\right)={e}^{\text{−}i{\omega }_{n}t}{\psi }_{n}\left(x\right)=\sqrt{\frac{2}{L}}{e}^{\text{−}i{E}_{n}t\text{/}\hslash }\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}\frac{n\pi x}{L},n=1,2,3,...$

#### Questions & Answers

in the wave equation y=Asin(kx-wt+¢) what does k and w stand for.
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No
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there is mistake in my first msg correction is 40cm=40.0×10^-2m =400.0×10^-3m =400mm. sorry for the mistake friends.
Prema
40cm=40.0×10^-2m =400.0×10^-3m =400mm.
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this msg is out of mistake. sorry friends​.
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explain l-s coupling   By Eric Crawford  By     By Subramanian Divya