# 7.4 The quantum particle in a box

 Page 1 / 12
By the end of this section, you will be able to:
• Describe how to set up a boundary-value problem for the stationary Schrӧdinger equation
• Explain why the energy of a quantum particle in a box is quantized
• Describe the physical meaning of stationary solutions to Schrӧdinger’s equation and the connection of these solutions with time-dependent quantum states
• Explain the physical meaning of Bohr’s correspondence principle

In this section, we apply Schrӧdinger’s equation to a particle bound to a one-dimensional box. This special case provides lessons for understanding quantum mechanics in more complex systems. The energy of the particle is quantized as a consequence of a standing wave condition inside the box.

Consider a particle of mass $m$ that is allowed to move only along the x -direction and its motion is confined to the region between hard and rigid walls located at $x=0$ and at $x=L$ ( [link] ). Between the walls, the particle moves freely. This physical situation is called the infinite square well    , described by the potential energy function

$U\left(x\right)=\left\{\begin{array}{cc}0,\hfill & 0\le x\le L,\hfill \\ \infty ,\hfill & \text{otherwise}\text{.}\hfill \end{array}$

Combining this equation with Schrӧdinger’s time-independent wave equation gives

$\frac{\text{−}{\hslash }^{2}}{2m}\phantom{\rule{0.2em}{0ex}}\frac{{d}^{2}\psi \left(x\right)}{d{x}^{2}}=E\psi \left(x\right),\text{for}\phantom{\rule{0.2em}{0ex}}0\le x\le L$

where E is the total energy of the particle. What types of solutions do we expect? The energy of the particle is a positive number, so if the value of the wave function is positive (right side of the equation), the curvature of the wave function is negative, or concave down (left side of the equation). Similarly, if the value of the wave function is negative (right side of the equation), the curvature of the wave function is positive or concave up (left side of equation). This condition is met by an oscillating wave function, such as a sine or cosine wave. Since these waves are confined to the box, we envision standing waves with fixed endpoints at $x=0$ and $x=L$ .

Solutions $\psi \left(x\right)$ to this equation have a probabilistic interpretation. In particular, the square $|\psi \left(x\right){|}^{2}$ represents the probability density of finding the particle at a particular location x . This function must be integrated to determine the probability of finding the particle in some interval of space. We are therefore looking for a normalizable solution that satisfies the following normalization condition:

$\underset{0}{\overset{L}{\int }}dx|\psi \left(x\right){|}^{2}=1.$

The walls are rigid and impenetrable, which means that the particle is never found beyond the wall. Mathematically, this means that the solution must vanish at the walls:

$\psi \left(0\right)=\psi \left(L\right)=0.$

We expect oscillating solutions, so the most general solution to this equation is

${\psi }_{k}\left(x\right)={A}_{k}\phantom{\rule{0.2em}{0ex}}\text{cos}\phantom{\rule{0.2em}{0ex}}kx+{B}_{k}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}kx$

where k is the wave number, and ${A}_{k}$ and ${B}_{k}$ are constants. Applying the boundary condition expressed by [link] gives

${\psi }_{k}\left(0\right)={A}_{k}\text{cos}\left(k·0\right)+{B}_{k}\text{sin}\left(k·0\right)={A}_{k}=0.$

Because we have ${A}_{k}=0$ , the solution must be

${\psi }_{k}\left(x\right)={B}_{k}\phantom{\rule{0.2em}{0ex}}\text{sin}\phantom{\rule{0.2em}{0ex}}kx.$

If ${B}_{k}$ is zero, ${\psi }_{k}\left(x\right)=0$ for all values of x and the normalization condition, [link] , cannot be satisfied. Assuming ${B}_{k}\ne 0$ , [link] for $x=L$ then gives

$0={B}_{k}\text{sin}\left(kL\right)⇒\text{sin}\left(kL\right)=0⇒kL=n\pi ,n=1,2,3\text{,...}$

We discard the $n=0$ solution because $\psi \left(x\right)$ for this quantum number would be zero everywhere—an un-normalizable and therefore unphysical solution. Substituting [link] into [link] gives

For the question about the scuba instructor's head above the pool, how did you arrive at this answer? What is the process?
as a free falling object increases speed what is happening to the acceleration
of course g is constant
Alwielland
acceleration also inc
Usman
which paper will be subjective and which one objective
jay
photo electrons doesn't emmit when electrons are free to move on surface of metal why?
What would be the minimum work function of a metal have to be for visible light(400-700)nm to ejected photoelectrons?
give any fix value to wave length
Rafi
40 cm into change mm
40cm=40.0×10^-2m =400.0×10^-3m =400mm. that cap(^) I have used above is to the power.
Prema
i.e. 10to the power -2 in the first line and 10 to the power -3 in the the second line.
Prema
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.
Prema
this msg is out of mistake. sorry friends​.
Prema
what is physics?
why we have physics
because is the study of mater and natural world
John
because physics is nature. it explains the laws of nature. some laws already discovered. some laws yet to be discovered.
Yoblaze
is this a physics forum
explain l-s coupling
how can we say dirac equation is also called a relativistic equation in one word
what is the electronic configration of Al
what's the signeficance of dirac equetion.?
what is the effect of heat on refractive index
As refractive index depend on other factors also but if we supply heat on any system or media its refractive index decrease. i.e. it is inversely proportional to the heat.
ganesh
you are correct
Priyojit
law of multiple
Wahid
if we heated the ice then the refractive index be change from natural water
Nepal
can someone explain normalization condition
Swati
yes
Chemist
1 millimeter is How many metres
1millimeter =0.001metre
Gitanjali