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.

The 1D square well

Setting up the problem

The classical view of a particle in a box. It sees no potential between Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=L} so can move freely there, but can never pass through the infinite potential barriers on either side.

The potential can be given by

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)=0\,\!} for Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0\le x\le L} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)=\infty} otherwise.

That is, in a region of length Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=L} , since if it were anywhere else it would have to have infinite energy.

Since Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle V(x)} does not depend on time we can use the time-independent version of the Schrödinger equation,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle -\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}\psi(x)=E\psi(x)\ , }

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar} is Planck's constant divided by 2π, and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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) Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} and eigenstate(s) Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} that solve it. The equation is a second-order linear ODE with constant coefficients, and the general solution can be written as

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)=a\cos\left(\sqrt{\frac{2mE}{\hbar^2}}x\right)+b\sin\left(\sqrt{\frac{2mE}{\hbar^2}}x\right)\ ,}

where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle a} and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=0} the wavefunction is also zero, so we have

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0=\psi(0)=a\cos(0)+b\sin(0)=a\ ,}

meaning that the wavefunction can now be written

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi(x)=b\sin\left(\sqrt{\frac{2mE}{\hbar^2}}x\right)\ .}

Energy quantization

Using the same continuity argument at Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=L} tells us that

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 0=\psi(L)=b\sin\left(\sqrt{\frac{2mE}{\hbar^2}}L\right)\ .}

One solution to this is of course Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{\frac{2mE}{\hbar^2}}L = 2n\pi\ ,}

where mathematically Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} can be any integer. This problem is not purely mathematical though, and we know for physical reasons that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle m} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E} , Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \hbar} , and Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle L} are all greater than zero. This means that Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 2n\pi} must also be greater than zero, so Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle n} is restricted to being a natural number, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1,2,3,...} . We can now solve for the particle's energy,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n=\frac{n^2\pi^2\hbar^2}{2mL^2}\ ,}

where we have labelled the energy by the integer Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle E_n} back into Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi} , we now have an infinite number of possible wavefunctions for the particle,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_n(x)=b\sin\left(\frac{n\pi x}{L}\right)\ .}

The final step is to determine the value of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b} . This requires another condition that we can impose upon the problem.

Normalization

The probability of finding the particle somewhere must be Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 1} . Since the wavefunction vanishes everywhere outside Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle [0,L]} , the normalization condition is

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle b=e^{i\varphi}\sqrt{2/L}} , where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi} without affecting the physics. For simplicity then, we set Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \varphi=0} .

The solutions

A plot of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \psi_n(x)} for the first 7 wavefunctions, Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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.

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 $L\to \infty .$