# 2.5: Overview of Particle in a box

- Page ID
- 340424

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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)## A One-Dimension Box

The system consists of a particle (one electron, one molecule, etc) that can only reside **inside** a one dimensional box (a line of length L). Outside the box there is an infinite potential, thus the particle cannot escape. Inside the box, no force or potential acts on the particle, thus inside the box, it is free to move.

Since the \( |\psi(x)|^2\) gives the probability of finding the particle in space, that probability must be zero outside the box. If the probability density is null, that means \(\psi(x)=0\) outside the box and

\[ \psi(x) =0 \text{ for any position } x < 0 \text{ and for any position }\ x >0. \]

Since the Probability density outside the box is \(=0\) and the probability inside the box is \(=0\), this means that at the edges,

\[ \psi(0) = 0 \text{ and } \psi(L)=0\]. This is called the **Boundary Condition**.

The Schrödinger equation for a 1D box allows us to find all the possible analytical forms of \(\psi(x)\) inside the box and the energy of each of those possible states. In this case the equation is

\[\textcolor{purple}{- \dfrac{\hbar^2}{2m}\dfrac{d^2}{dx^2}}\psi(x) = \textcolor{darkgreen}{E}\cdot \psi(x) \label{1}\]

where \(\hbar\) is Planck's constant divided by \(2\pi\) and \(m\) is the mass of the particle.

Although we don't need to know *how *to solve this equation, it is very useful to take a look and understand the solutions which are of the form

\[\psi_n(x)= \sqrt{\dfrac{2}{L}} sin(\dfrac{n\ \pi}{L}\cdot x) \label{2}\]

where \(L\) is the length of the box, \( x \) is the variable in the wavefunction, and \(n\) is the quantum number that identifies each possible state. The are infinite solutions for Equation \ref{1}, each one for a different value of \(n\). If we choose one state (i.e. \(n=1\) ) and we replace \(\psi_1(x)\) into Equation \ref{1} we obtain the \(\textcolor{darkgreen}{energy (E_1)}\) for that specific state.

\[\textcolor{darkgreen}{E_n=\dfrac{h^2 }{8mL^2}n^2} \label{3}\]

Each of these solutions will have a different probability density and a different energy.

Example \(\PageIndex{1}\)

What are the first 4 possible states for the particle in a box of length 10Å

**Solution**

\(n\) | \(\psi_n(x)\) | \(\textcolor{darkgreen}{E_n}\) | Probability Density |

1 | \(\sqrt{\dfrac{2}{L}} sin(\dfrac{\pi}{L}\cdot x)\) | \(\textcolor{darkgreen}{E_n=\dfrac{h^2 }{8mL^2}1} = 0.376eV \) | |

2 | \(\sqrt{\dfrac{2}{L}} sin(\dfrac{2\pi}{L}\cdot x)\) | \(\textcolor{darkgreen}{E_n=\dfrac{h^2 }{8mL^2}4}=1.504 eV\) | |

3 | \(\sqrt{\dfrac{2}{L}} sin(\dfrac{3\pi}{L}\cdot x)\) | \(\textcolor{darkgreen}{E_n=\dfrac{h^2 }{8mL^2}9}=3.385eV\) | |

4 | \(\sqrt{\dfrac{2}{L}} sin(\dfrac{4\pi}{L}\cdot x)\) | \(\textcolor{darkgreen}{E_n=\dfrac{h^2 }{8mL^2}16}=6.017eV\) |

The graph shows the probability density for each state. It is plotted with each curve vertically displaced to emphasized that \(E_4>E_3 >E_2>E_1\).

## The Position of the Particle

For each of those states, we can always ask the question: Where will we find the particle? or What is the probability to find the particle between two values of \(x\)? That information can obtained from the **Probability Density**.

We start by looking at the state with \(n=1\), which has the lowest energy and it is call the** ground state**.

To find the particle between \(x=3 Å\) and \(x=5 Å\) we would need to solve the integral \(\int^5_3|\psi(x)|^2dx\), which is represented in Figure \(\PageIndex{3}\) by the shaded area in red.

##### What is the probability to find the particle between \(x=5 Å\) and \(x=7 Å\) and how does it compare with the probability to find it between \(x=3 Å\) and \(x=5 Å\)?

To answer this question we can look at the shaded areas in the probability density plot.

Sometimes it is useful to compare two different regions. Figure \(\PageIndex{3}\) shows the two regions shaded with different colors, and it is clear that the probability to find the particle on one side is the same as finding it on the other side. Is the particle on the left? or is it on the right?

The particle behaves as a wave, and as such we cannot say that it is on the left or on the right. If we do a a single measurement we might find it on the left (or on the right), but if we continue doing multiple measurements in systems that are similarly prepared (always a particle in a 1D box, always in the state with \(n=1\) and energy \(0.376 eV\)) half of the times we will find it on the left side and half of the times we will find it on the right side. This is the probabilistic interpretation of the wavefunction. The probability density provides statistical information of the possible outcome of the measurements.

If we prepare the particle on a different state, let's say \(n=2\), with energy= \(1.504 eV\), the Probability Density will have a different shape, and the outcome of the experiments will yield a different result.

Figure \(\PageIndex{4}\) again shows two regions with equally shaded areas (\(x=3 \text{ to } x=5 Å\) and \(x=5 \text{ to } x=7 Å\)) indicating that the probability to find the particle is the same when we measure in regions of equal lengths on both sides of the box.

Can we ever do a measurement and find the particle at \( x=5 Å \)?

The Probability Density of \(n=2\) shows an interesting characteristic. Figure \(\PageIndex{4}\) shows that there is a **node** at that position, that is **\(|\psi(5 Å)|^2=0 \)**, indicating that the probability of finding the particle at that specific point in space is **exactly 0.** It doesn't matter how many measurements we performed, we will never find the particle at \( x=5 Å\)! But...how does the particle move from left to right if it can never go through the center? In reality, this question is wrongly posed. When the particle is in the state \( n=2 \) its standing wave has no amplitude at \( x=5 Å\), so it can never be found at that point.

Figure \(\PageIndex{3}\) shows that as the value of \(n\) increases, the number of nodes also increases. We can even read the value of \(n\) by counting how many nodes are in the Probability Density plot

## The Energies of the Particle

Table \(\PageIndex{1}\) shows the energies for the first 4 states of the particle. The generalized solution is given in Equation \ref{3}, where \(n=1,2,3,...\infty \)

This equation shows that the values of the energy do not change continuously but instead they vary in units of \(n\): the \(\textcolor{darkgreen}{\text{Energy}}\) is **quantized**.

##### How much Energy is required to transfer the particle from the ground state to the first excited state (\(n=1 \rightarrow n=2)\)?

\begin{align}

\Delta E &= E_{final}- E_{initial} \\

&= \dfrac{h^2}{8mL^2}n^2_{f} -\dfrac{h^2}{8mL^2}n^2_{i} \\

&= \dfrac{h^2}{2mL^2}(n^2_f - n^2_i) \\

&= \dfrac{h^2}{mL^2} (4-1) \\

&= \dfrac{h^2}{mL^2} 3

\end{align}

We can find a general equation that describes the difference in energy between two consecutive state considering that \(n_f=n_i +1\)

\begin{align}

\Delta E &= E_{final}- E_{initial} \\

&= \dfrac{h^2}{2mL^2}[(n+1)^2 - n^2] \\

&= \dfrac{h^2}{mL^2} (2n+1) \label{4}\\

\end{align}

Equation \ref{4} shows that transitions between two states corresponds to quantized values of energies. As the states increase (higher \(n\)), the value of the energy increases with a linear dependence in \(n\), or in other words, the energy spacing between states is not the same. The transition from \(n=7 \rightarrow n=8\) requires more energy than the transition from \(n=1\rightarrow n=2\).

The particle in a box only has Kinetic Energy (there is not potential inside the box) , therefore the calculated energy corresponds to Kinetic Energy. Of course that will not be the case if we are studying a different system like an electron in the H atom, or a molecule vibrating.

The probability density plot in Figure \(\PageIndex{3}\) shows the presence of multiple nodes for higher values of \(n\). When the value of the quantum number \(n\)increases, the Kinetic Energy increases and the number of nodes increases.

A higher number of nodes in a standing wave corresponds to a higher frequency (shorter wavelength). Recalling deBroglie, where \(\lambda =\dfrac{h}{p}\), we know that shorter \(\lambda\) is equivalent to larger momentum and larger momentum implies higher Kinetic Energy, since \(K.E. = \dfrac{1}{2}mv^2 = \dfrac{p^2}{2m}\).

Note

We can even read the value of \(n\) by counting how many nodes are in the Probability Density plot and knowing that for a Particle in a 1-D box, the number of nodes will be \(n-1\)

##### Can the particle have \(\textcolor{darkgreen}{Energy =0}\)?

The minimum energy a particle can have is for the ground state, with \(n=1\),

\[\textcolor{darkgreen}{E =\dfrac{h^2}{8mL^2}\cdot 1}=K.E.\]

Since all this energy corresponds to kinetic energy, the particle is NEVER at rest, it will always have a non-zero value of momentum.

## Correspondence Principle

In 1920 Bohr described the Correspondence Principle. The idea behind it is that there must be a set of conditions for which the "new" Theory (Quantum Mechanics) and the "old" Theory (Classical Mechanics) must provide similar results, a point of contact between both models to desribe matter.

We can understand that region of *correspondence* between the two theories by comparing the results of the particle in a 1-Dimensional box, under conditions that we would expect to use classical mechanics (heavy masses or large boxes).

Figure \(\PageIndex{5}\), left panel shows the energy levels for an electron in a box of length 10 Å and a box of length 100 Å. As the box becomes much wider, the spacing between the energy levels becomes very small. If the length of the box continues to increase, at some point that spacing will almost disappear, and the energy of the particle could be described as a continuous.

The right panel in Figure \(\PageIndex{5}\) shows the consequence of increasing the mass of the particle. Again we see how the spacing between states corresponds into a continuum ef energies as the mass becoems heavier.

When we look at the probability to find the particle in space, the correspondence principle can also be understood analyzing the results for a particle in a box.

Figure \(\PageIndex{6}\) compares the probability density for a particle in a state with \(n=3\) and \(n=75\). The probability to find the electron between \(x=3Å\) and \(x=4Å\) is very different than the probability to find the electron between \(x=4Å\) and \(x=5Å\),when the system is in a low \(n\) state, but those probabilities becomes the same for very large values of the quantum number. In the plot on the right, the probability to find the particle is the same in any region of the box, a result corresponding to the classical treatment of a particle.