Particle in a box

The particle in a box or infinite square well problem is one of the simplest non-trivial solutions to Schrödinger's wave equation. As such it is often encountered in introductory quantum mechanics material as a demonstration of the quantization of energy. In its simplest form the problem is one-dimensional, and involves a single particle living in an infinite potential well. What this means in the 1D case, which we examine here, is that there is a particle that is free to move back and forth along a line in between two points (e.g. 0 and $L$, or $-a/2$ and $a/2$). The particle is however constrained to this region, and cannot under any circumstances be found outside of it.

In this article we will rigorously define this problem and find the solutions to the corresponding Schrödinger wave equation, and then examine some of the properties and implications of these solutions.

The 1D square well

Setting up the problem

Figure 1: The classical view of a particle in a box. It sees no potential between $x=0$ and $x=L$ so can move freely there, but can never pass through the infinite potential barriers on either side.

As depicted in Figure 1, the potential can be given by

$V(x)=0\,\!$ for $0\leq x\leq L$ and $V(x)=\infty$ otherwise.

That is, in a region of length $L$ (the box or well) the potential is zero; everywhere else it is infinite. An immediate consequence of this is that the particle must be located somewhere between $x=0$ and $x=L$, since if it were anywhere else it would have to have infinite energy.

This means that we only have to solve Schrödinger's equation inside the box and since the potential does not depend on time we can use the time-independent version of the Schrödinger equation. With $V(x)=0$ in the box we have

$-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}\psi (x)=E\psi (x)\ ,$

where $\hbar$ is Planck's constant divided by 2π, and $m$ and $E$ are the mass and energy of the particle, respectively. This is nothing but the eigenvalue problem, and our task is to determine the energy eigenvalue(s) $E$ and eigenstate(s) $\psi$ that solve it. The equation is a second-order linear ODE with constant coefficients, and the general solution can be written as

$\psi (x)=a\cos \left({\sqrt {\frac {2mE}{\hbar ^{2}}}}x\right)+b\sin \left({\sqrt {\frac {2mE}{\hbar ^{2}}}}x\right)\ ,$

where $a$ and $b$ are complex constants to be determined from conditions on the system. The first condition we can use is that the wavefunction must be continuous. We know that the particle cannot exist at positions $x<0$, which tells us that the wavefunction must be identically zero there (in order for the probability of finding the particle there to also be zero). Continuity then implies that at $x=0$ the wavefunction is also zero, so we have

$0=\psi (0)=a\cos(0)+b\sin(0)=a\ ,$

meaning that the wavefunction can now be written

$\psi (x)=b\sin \left({\sqrt {\frac {2mE}{\hbar ^{2}}}}x\right)\ .$

Energy quantization

Using the same continuity argument at $x=L$ tells us that

$0=\psi (L)=b\sin \left({\sqrt {\frac {2mE}{\hbar ^{2}}}}L\right)\ .$

One solution to this is of course $b=0$. However, this would mean that the wavefunction vanishes everywhere -- implying that there is no particle! The other way to satisfy this equality is to have the sine term vanish, which will happen if

${\sqrt {\frac {2mE}{\hbar ^{2}}}}L=2n\pi \ ,$

where mathematically $n$ can be any integer. This problem is not purely mathematical though, and we know for physical reasons that $m$, $E$, $\hbar$, and $L$ are all greater than zero. This means that $2n\pi$ must also be greater than zero, so $n$ is restricted to being a natural number, $1,2,3,...$. We can now solve for the particle's energy,

$E_{n}={\frac {n^{2}\pi ^{2}\hbar ^{2}}{2mL^{2}}}\ ,$

where we have labelled the energy by the integer $n$. We have just derived energy quantization! Without the potential well, i.e. if the particle was free, its energy would be allowed to take on any real number. Once inside the box though, only a specific discrete set of energy eigenvalues is permitted.

Substituting $E_{n}$ back into $\psi$, we now have an infinite number of possible wavefunctions for the particle,

$\psi _{n}(x)=b\sin \left({\frac {n\pi x}{L}}\right)\ .$

The final step is to determine the value of $b$. This requires another condition that we can impose upon the problem.

Normalization

The probability of finding the particle somewhere must be $1$. Since the wavefunction vanishes everywhere outside $[0,L]$, the normalization condition is

$1=\int _{0}^{L}\psi ^{*}(x)\psi (x)\,\mathrm {d} x=|b|^{2}\int _{0}^{L}\sin ^{2}\left({\frac {n\pi x}{L}}\right)\,\mathrm {d} x=|b|^{2}{\frac {L}{2}}\ .$

Solving for $b$ tells us $b=e^{i\varphi }{\sqrt {2/L}}$, where $\varphi$ is some overall quantum phase (a real number). Because overall phases do not affect measurable results we are free to choose any value we want for $\varphi$ without affecting the physics. For simplicity then, we set $\varphi =0$.

The solutions

Figure 2: A plot of $\psi _{n}(x)$ for the first 7 wavefunctions, $n=1$ (darkest) to $n=7$ (lightest).

We have now solved the particle-in-a-box problem. According to the Schrödinger wave equation the particle confined to the (infinite) box can only take on certain energy eigenvalues, namely

$E_{n}={\frac {n^{2}\pi ^{2}\hbar ^{2}}{2mL^{2}}}\ ,$

and the wavefunction for the particle when it is in the $n$th energy eigenstate is

$\psi _{n}(x)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi x}{L}}\right)$

inside the box, and $\psi (x)=0$ outside it.

Figure 2 shows a plot of the first seven solutions for the wavefunction of a particle in a box. Note how as $n$ increases the number of waves that fit in the box increases as well, meaning that the wavelength has decreased and therefore frequency has increased. As with electromagnetic waves a higher frequency corresponds to a higher energy, in agreement with the expression for $E_{n}$.

Alternate description

Another common (equivalent) way to describe this problem is to shift the location of the box, so that it runs from $-L/2$ to $L/2$ instead of $0$ to $L$. The derivation of the allowed wavefunctions is very similar to what we did here, or the results can be obtained via coordinate transformation (specifically, a translation). With this different description of the square well the energy eigenvalues remain unchanged (as is to be expected) while the resulting wavefunctions are given by

$\psi _{n}(x)={\sqrt {\frac {2}{L}}}\cos \left({\frac {n\pi x}{L}}\right),$

if $n$ is odd, and

$\psi _{n}(x)={\sqrt {\frac {2}{L}}}\sin \left({\frac {n\pi x}{L}}\right),$

if $n$ is even.

Properties of the particle in a box

Completeness

The $\psi _{n}$ are complete, so form a basis for the possible wavefunctions, or states, of the particle. This means that an arbitrary state, $\psi (x)$, can be expressed as

$\psi (x)=\sum _{n=1}^{\infty }c_{n}\psi _{n}(x)$

where $\left|c_{n}\right|^{2}$ is the probability of measuring the particle's energy to be $E_{n}$, and therefore finding the particle in state $\psi _{n}$.

Non-vanishing ground state energy

The particle is not allowed to have zero energy, because if $E=0$ then $n=0$, which makes the wavefunction

$\psi _{0}(x)={\sqrt {\frac {2}{L}}}\sin \left(0\cdot {\frac {\pi x}{L}}\right)=0\ .$

But if $\psi =0$ for all $x$ then there is zero probability of finding the particle no matter where you look -- which means that there is no particle! What this implies then is that the lowest allowed energy or ground state energy for the particle in the box is

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

The free space limit

We can think of 'free space' as an infinitely wide box. Therefore if we take $L$, the width of the box, to infinity we would expect to find that in the limit we recover the result of a free particle. In fact, this is exactly what happens.

Mathematically we would find the same result for either of the solutions above, but physically free space doesn't have an infinite potential wall at $x=0$. It is therefore more intuitive to look at the second solution, in which the potential well extends from $x=-L/2$ to $L/2$.

Notice that the energy spacing between two consecutive energy levels is

$\Delta E=E_{n+1}-E_{n}={\frac {\pi ^{2}\hbar ^{2}}{2mL^{2}}}(2n+1)\ .$

It is easy to see that for any finite value of $n$ the value of $\Delta E$ goes to zero as $L$ goes to infinity. This tells us that a free particle can have any (real, positive) energy.

Perhaps surprisingly the wavefunction also appears to go to zero everywhere in the limit, which at first may appear to be a problem. As discussed above if the wavefunction vanishes everywhere then there shouldn't be a particle. The difference in this case is subtle, and has to do with the fact that we are dealing with infinities. The important point to note is that if the particle is in state $\psi _{n}$ then the probability to find it if we look in the entire box is given by

$\int _{-L/2}^{L/2}\psi _{n}(x)^{*}\psi _{n}(x)\,dx={\frac {2}{L}}\int _{-L/2}^{L/2}\sin ^{2}\left({\frac {n\pi x}{L}}\right)\,dx=1\ ,$

which is independent of $L$. This means that we would still find the particle if we looked in all of space after taking the limit $L\to \infty .$

Generalization to 3D (in progress)

The standard three dimensional particle in a box is would be more accurately described as a particle in a ball. The potential for the problem can is most easily expressed in spherical coordinates as

$V(r,\theta ,\phi )={\Big \{}$

$0\,\!$ $;\ r\leq L$ $\infty$ $;\ r>L\ .$

This potential describes the region in which the particle can exist as a sphere of radius $L$ centred at the origin; outside the ball the potential is infinite. Note that this means the potential is independent of $\theta$ and $\phi$.

The 3D time-independent Schrödinger equation for this system is

$-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}\psi (r,\theta ,\phi )=E\psi (r,\theta ,\phi )\ .$

Due to the spherical symmetry of the Hamiltonian the wavefunction can be written as

$\psi (r,\theta ,\phi )=R(r)Y(\theta ,\phi )\ ,$

and then the technique of separation of variables results in two differential equations to be solved, one for $R$ and one for $Y$. It turns out that the solutions for $R$ are given by the spherical Bessel functions, $J_{n}(r)$, and the solutions for $Y$ are the spherical harmonics, $Y_{lm}(\theta ,\phi )$. This gives us three quantum numbers, $n,l,m$, as opposed to the single quantum number $n$ obtained in the one-dimensional case above. The solution to the spherical well is

$\psi _{nlm}(r,\theta ,\phi )=J_{n}(r)Y_{lm}(\theta ,\phi )\ .$

The derivation

The Laplacian in spherical coordinates is given by

$\nabla ^{2}\psi ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial \psi }{\partial r}}\right)+{\frac {1}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial \psi }{\partial \theta }}\right)+{\frac {1}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}\psi }{\partial \phi ^{2}}}\ .$

We can solve this problem using separation of variables. Since the potential depends on $r$ but not $\theta$ or $\phi$ we postulate that the wavefunction $\psi$ can be separated into a product of two functions, one depending on $r$ and the other depending on $\theta$ and $\phi$,

$\psi (r,\theta ,\phi )=R(r)Y(\theta ,\phi )\ ,$

which turns the Laplacian of the wavefunction into

${\begin{aligned}\nabla ^{2}\psi &={\frac {Y}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial R}{\partial r}}\right)+{\frac {R}{r^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial Y}{\partial \theta }}\right)+{\frac {R}{r^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}Y}{\partial \phi ^{2}}}\\&=\left[{\frac {1}{Rr^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial R}{\partial r}}\right)+{\frac {1}{Yr^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial Y}{\partial \theta }}\right)+{\frac {1}{Yr^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}Y}{\partial \phi ^{2}}}\right]\psi \ .\end{aligned}}$

We can now cancel $\psi$ from both sides of Schrödinger equation above, giving us

${\frac {1}{Rr^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial R}{\partial r}}\right)+{\frac {1}{Yr^{2}\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial Y}{\partial \theta }}\right)+{\frac {1}{Yr^{2}\sin ^{2}\theta }}{\frac {\partial ^{2}Y}{\partial \phi ^{2}}}=-{\frac {2mE}{\hbar ^{2}}}\ .$

The variables are finally separated once we multiply through by $r^{2}$ and rearrange the equation to give

${\frac {1}{R}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial R}{\partial r}}\right)+{\frac {2mEr^{2}}{\hbar ^{2}}}=-{\frac {1}{Y}}\left[{\frac {1}{\sin \theta }}{\frac {\partial }{\partial \theta }}\left(\sin \theta {\frac {\partial Y}{\partial \theta }}\right)+{\frac {1}{\sin ^{2}\theta }}{\frac {\partial ^{2}Y}{\partial \phi ^{2}}}\right]=\alpha$

in which the left-hand side only depends on $r$ while the right-hand side depends on $\theta$ and $\phi$ instead. This means that when $r$ is varied on the left-hand side the right-hand side will remain constant, and similarly the left-hand side is unaffected by variations of the angular variables on the right-hand side. Therefore both sides must be constant in general, and we have called this constant $\alpha$. Evaluating the derivative in the first term we find

${\frac {\partial ^{2}R}{\partial r^{2}}}+{\frac {2}{r}}{\frac {\partial R}{\partial r}}+\left({\frac {2mE}{\hbar ^{2}}}-{\frac {\alpha }{r^{2}}}\right)R=0$

...

Solutions are Bessel functions...

to be continued...

Now we look to the second term. We again want to use separation of variables so let $Y(\theta ,\phi )=\Theta (\theta )\Phi (\phi )$, which gives us

$-\alpha ={\frac {1}{\Theta \sin \theta }}\left(\cos \theta {\frac {\partial \Theta }{\partial \theta }}+\sin \theta {\frac {\partial ^{2}\Theta }{\partial \theta ^{2}}}\right)+{\frac {1}{\Phi \sin ^{2}\theta }}{\frac {\partial ^{2}\Phi }{\partial \phi ^{2}}}\ .$

The $\theta$ and $\phi$ terms can indeed be separated:

$-{\frac {\sin ^{2}\theta }{\Theta }}\left(\alpha \Theta +\cot \theta {\frac {\partial \Theta }{\partial \theta }}+{\frac {\partial ^{2}\Theta }{\partial \theta ^{2}}}\right)={\frac {1}{\Phi }}{\frac {\partial ^{2}\Phi }{\partial \phi ^{2}}}=\beta \ .$

Let's deal with the DE in $\phi$ first,

${\frac {\partial ^{2}\Phi }{\partial \phi ^{2}}}-\beta \Phi =0\quad \implies \quad \Phi =\Phi _{0}e^{\phi {\sqrt {\beta }}}+\Phi _{1}e^{-\phi {\sqrt {\beta }}}\ .$

Now, we know that $\phi$ represents an angular coordinate so we should have that $\Phi (\phi )=\Phi (\phi +2\pi )$, which yields

$\Phi _{0}e^{\phi {\sqrt {\beta }}}+\Phi _{1}e^{-\phi {\sqrt {\beta }}}=\Phi _{0}e^{(\phi +2\pi ){\sqrt {\beta }}}+\Phi _{1}e^{-(\phi +2\pi ){\sqrt {\beta }}}$

or

$\Phi _{0}e^{\phi {\sqrt {\beta }}}\left(1-e^{2\pi {\sqrt {\beta }}}\right)+\Phi _{1}e^{-\phi {\sqrt {\beta }}}\left(1-e^{-2\pi {\sqrt {\beta }}}\right)=0$

Since this must hold true for all values of $\phi$, the bracketed terms must both be zero. (To see this more explicitly, consider the two specific equations that arise from $\phi =0$ and $\scriptstyle \phi ={\frac {\pi }{2{\sqrt {\beta }}}}$.) We have shown then that $\scriptstyle \exp(\pm 2\pi {\sqrt {\beta }})=1$ must be true, which tells us that $\scriptstyle {\sqrt {\beta }}=im$, or $\beta =-m^{2}$, where $m$ is any integer. This means that our equation in $\theta$ has become

${\frac {\partial ^{2}\Theta }{\partial \theta ^{2}}}+\cot \theta {\frac {\partial \Theta }{\partial \theta }}+\left(\alpha -{\frac {m^{2}}{\sin ^{2}\theta }}\right)\Theta =0$

...

Solutions are Legendre polynomials...

to be continued...