Particle in a box

In physics, the particle in a box (or the square well) is a simple idealized system that can be completely solved within quantum mechanics. It is the situation of a particle confined within a finite region of space (the box) by an infinite potential that exists at the walls of the box. The particle experiences no forces while inside the box, but is constrained by the walls to remain in the box. This is similar to the situation of a gas confined in a container. For simplicity we start with the 1-dimensional case, where all motion is constrained to a single dimension. Later we will extend the discussion to the 2 and 3 dimensional cases. See also the Particle in a spherically symmetric potential where the case is treated of a particle in a spherical box, or the particle in a ring which shows the case for a particle in a 1D ring. The statistical mechanics of many particles in a box is developed in the gas in a box article.

As we shall see, the solution of the Schrödinger equation for the particle in a box problem reveals some decidedly quantum behavior of the particle that agrees with observation but contrasts sharply with the predictions of classical mechanics. This is a particularly useful illustration because this behaviour is not "forced" on the system, it arises naturally from the initial conditions. It neatly demonstrates that quantum behaviour is a natural outcome of any wave-like system, contrary to the common concept of a "quantum leap" where the behavior is almost magical.

The quantum behavior in the box includes:

1. Energy quantization - It is not possible for the particle to have any arbitrary definite energy. Instead only discrete definite energy levels are allowed (if the state is not a steady state, however, any energy past zero-point energy is allowed on average).
2. Zero-point energy - The lowest possible energy level of the particle, called the zero-point energy, is nonzero.
3. Nodes - In contrast to classical mechanics the Schrödinger equation predicts that for some energy levels there are nodes, implying positions at which the particle can never be found.

One can solve analytically the Schrödinger equation for such a simple potential. However trivial, this case is both of great technical value for the insights it allows, and of paramount physical importance. Depending on the boundary conditions, one can use the solutions to describe two important systems. If one considers real valued solutions (of which detailed derivation is given below), one describes actual potentials of heterostructures called quantum wells which trap spatially particles, typically electrons and holes. If one considers complex valued solutions, one describes conveniently a particle propagating freely in a constrained volume (like a solid).

Contents

1 The particle in a 1-dimensional box 2 The particle in a 2-dimensional or 3-dimensional box 3 Free propagation 4 External links

The particle in a 1-dimensional box

For the 1-dimensional case in the <math>x<math> direction, the time-independent Schrödinger equation can be written as:

<math>-\frac{\hbar^2}{2 m} \frac{d^2 \psi}{d x^2} + V(x) \psi = E \psi \quad (1)<math>

where

<math>\hbar = \frac{h}{2 \pi}<math>

where h is Planck's constant, m is the mass of the particle, ψ is the (complex valued) wavefunction that we want to find, V(x) is a function describing the potential at each point x and E is the energy, a real number. For the case of the particle in a 1-dimensional box of length L, the potential is zero inside the box, but rises abruptly to infinity at x = 0 and x = L. Thus for the region inside the box V(x) = 0 and Equation 1 reduces to:

<math>-\frac{\hbar^2}{2 m} \frac{d^2 \psi}{d x^2} = E \psi \quad (2)<math>

This is a well studied eigenvalue problem with a general solution of:

<math>\psi = A \sin(kx) + B \cos(kx)\quad<math>

<math>E = \frac{k^2 \hbar^2}{2m} \quad (3)<math>

Here, A and B can be any complex numbers, and k can be any real number (k must be real because E is real).

Now in order find the specific solution for the problem at hand, we must specify the appropriate boundary conditions and find the values for A and B that satisfy those conditions. One usually resorts to one of the following two choices, describing two kind of systems. The first case, with which we shall pursue our derivation, demands that ψ equal zero at x = 0 and x = L. A handwaving argument to motivate these boundary conditions is that the particle is unlikely to be found at a location with a high potential (the potential repulses the particle), thus the probability of finding the particle, |ψ|², must be small in these regions and decreases with increasing potential. For the case of an infinite potential, |ψ|² must infinitesimally small or 0, thus ψ must also be zero in this region. In summary,

<math>\psi(0)=\psi(L)=0 \quad (4)<math>

The second case, to which solutions are given in section free propagation at the end of this article, does not compel the wavefunction to vanish at the boundary. This means that when the particle reaches one border of the well, it instantaneously disappears from this side to reappear on the opposite side, as if the well was some kind of torus. The value of the solutions are discussed in the appropriate section. We now resume derivation with vanishing boundary conditions.

Substituting the general solution from Equation 3 into Equation 2 and evaluating at <math>x= 0<math> (<math>\psi = 0<math>), we find that <math>B = 0<math> (since <math>\sin (0)=0<math> and <math>\cos(0) = 1<math>). It follows that the wavefunction must be of the form:

<math>\psi = A \sin(kx) \quad (5) <math>

and at <math>x = L<math> we find:

<math>\psi = A \sin(kL) = 0 \quad (6)<math>

One solution for Equation 6 is A = 0, however, this "trivial solution" would imply that <math>\psi = 0<math> everywhere (I.e. the particle isn't in the box.) and can be thrown out. If <math>A \neq 0<math> then <math>\sin(kL) = 0<math>, which is only true when:

<math>kL = n \pi \quad \mathrm{where} \quad n = 1,2,\ldots<math>

<math>\mathrm{or} \quad k = \frac{n \pi}{L} \quad (7) <math>

(note that <math>n = 0<math> is ruled out because then <math>\psi=0<math> everywhere, corresponding to the case where the particle is not in the box. Negative values of <math>n<math> are also neglected, since they merely change the sign of <math>\sin(nx))<math>. Now in order to find <math>A<math> we recognize that the particle must exist somewhere in space, so the integral of <math>|\psi|^2<math> over all <math>x<math>, which corresponds to the probability of finding the particle somewhere, is equal to 1:

<math>1 = \int_{-\infty}^{\infty} \left| \psi \right|^2 \, dx = \left| A \right|^2 \int_0^L \sin^2 kx \, dx = \left| A \right|^2 \frac{L}{2}<math>

or

<math>\left| A \right| = \sqrt{\frac{2}{L}} \quad (8)<math>

Thus, A may be any complex number with absolute value √(2/L); these different values of A yield the same physical state, so we choose A = √(2/L) to simplify.

Finally, substituting the results from Equations 7 and 8 into Equation 3 and defining p as the momentum of the particle, the complete set of energy eigenfunctions for the 1-dimensional particle in a box problem is:

<math>\psi_n = \sqrt{\frac{2}{L}} \sin\left(\frac{n \pi x}{L}\right) \quad (9)<math>

<math>E_n = \frac{n^2\hbar^2 \pi ^2}{2mL^2} = \frac{n^2 h^2}{8mL^2} \quad (10)<math>

with

<math>n=1,2,3,\ldots<math>

The absolute value of the momentum can also be determined from an energy eigenfunction:

<math>|p_n| = \hbar k = \frac{nh}{2L}<math>

Note, that as mentioned previously, only "quantized" energy levels are possible. Also, since n cannot be zero, the lowest energy from Equation 10 is also non-zero. This zero-point energy, as it is called, can be explained in terms of the uncertainty principle. Because the particle is constrained within a finite region, the variance in its position is upper-bounded. Thus due to the uncertainty principle the variance in the particle's momentum cannot be zero, so the particle must contain some amount of energy that increases as the length of the box, L, decreases.

Also, since ψ consists of sine waves, for any value of n greater than one, there are regions within the box for which ψ and thus ψ² both equal zero, indicating that for these energy levels, nodes exist in the box where the probability of finding the particle is zero.

The particle in a 2-dimensional or 3-dimensional box

For the 2-dimensional case the particle is confined to a rectangular surface of length L_x in the x-direction and L_y in the y-direction. Again the potential is zero inside the "box" and infinite at the walls. For the region inside the box, where the potential is zero, the two dimensional analogue of Equation 2 applies:

<math>-\frac{\hbar^2}{2m} \left( \frac{\partial^2\psi}{\partial x^2}+\frac{\partial^2 \psi}{\partial y^2} \right) =E\psi \quad(11)<math>

In this case ψ is a function of both x and y, so ψ=ψ(x,y). In order to solve Equation 11, we use the method of separation of variables. First, we assume that ψ can be expressed as the product of two independent functions, the first depending only on x and the second depending only on y; i.e.:

<math> \psi(x,y) = X(x) Y(y) \quad (12) <math>

Substituting Equation 12 into Equation 11 and evaluating the partial derivatives gives:

<math>-\frac{\hbar^2}{2m} \left( Y\frac{\partial^2X}{\partial x^2}+X\frac{\partial^2 Y}{\partial y^2} \right) =E X Y \quad(13)<math>

which upon dividing by XY and rewriting d²X/dx² as X" and d²Y/dy² as Y" becomes:

<math>-\frac{\hbar^2}{2m} \left( \frac{X''}{X}+\frac{Y''}{Y} \right) =E \quad(14)<math>

Now we note that since X"/X is independent of y, varying y can only change the Y"/Y term. However, from Equation 14 we see that changing Y"/Y without varying X"/X, would also change E, but E is a constant, so Y"/Y must also be a constant, independent of y. The same argument can be applied to show that X"/X is independent of x. Since X"/X and Y"/Y are constants, we can write:

<math>-\frac{\hbar^2}{2m}\frac{X''}{X} = E_x \quad and \quad -\frac{\hbar^2}{2m}\frac{Y''}{Y} = E_y \quad (15)<math>

where E_x + E_y = E. Expanding X" and Y" in terms of the derivatives and rearranging gives:

<math>-\frac{\hbar^2}{2m}\frac{\partial^2X}{\partial x^2} = E_x X\quad (16)<math>

<math>-\frac{\hbar^2}{2m}\frac{\partial^2Y}{\partial y^2} = E_y Y\quad (17)<math>

each of which are of the same form as the 1-dimensional Schrödinger equation (Equation 2) we solved in the previous section. Thus, adapting the results from the previous section gives:

<math>X_{n_x}=\sqrt{\frac{2}{L_x}} \sin \left( \frac{n_x \pi x}{L_x} \right) \quad (18)<math>

<math>Y_{n_y}=\sqrt{\frac{2}{L_y}} \sin \left( \frac{n_y \pi y}{L_y} \right) \quad (19)<math>

Finally, since ψ=XY and E = E_x + E_y, we obtain the solutions:

<math>\psi_{n_x,n_y} = \sqrt{\frac{4}{L_x L_y}} \sin \left( \frac{n_x \pi x}{L_x} \right) \sin \left( \frac{n_y \pi y}{L_y} \right) \quad (20) <math>

<math>E_{n_x,n_y} = \frac{h^2}{8m} \left[ \left( \frac{n_x}{L_x} \right)^2 + \left( \frac{n_y}{L_y} \right)^2 \right] \quad (21) <math>

The same separation of variables technique can be applied to the three dimensional case to give the energy eigenfunctions:

<math>\psi_{n_x,n_y,n_z} = \sqrt{\frac{8}{L_x L_y L_z}} \sin \left( \frac{n_x \pi x}{L_x} \right) \sin \left( \frac{n_y \pi y}{L_y} \right) \sin \left( \frac{n_z \pi z}{L_z} \right) \quad (22)<math>

<math>E_{n_x,n_y,n_z} = \frac{h^2}{8m} \left[ \left( \frac{n_x}{L_x} \right)^2 + \left( \frac{n_y}{L_y} \right)^2 + \left( \frac{n_z}{L_z} \right)^2 \right] \quad (23) <math>

with

<math>n_i=1,2,3,\ldots<math>

The absolute value of the momentum in direction i=x,y or z is

<math>|p_i| = \frac{hn_i}{2L_i}<math>

An interesting feature of the above solutions is that when two or more of the lengths are the same (e.g. L_x = L_y), there are multiple wavefunctions corresponding to the same total energy. For example the wavefunction with n_x = 2, n_y = 1 has the same energy as the wavefunction with n_x = 1, n_y = 2. This situation is called degeneracy and for the case where exactly two degenerate wavefunctions have the same energy that energy level is said to be doubly degenerate. Degeneracy results from symmetry in the system. For the above case two of the lengths are equal so the system is symmetric with respect to a 90° rotation.

Free propagation

If the potential is zero (or constant) everywhere, one describes a free particle. This leads to some difficulties of normalization of the momentum or energy eigenfunctions. One way around is to constrain the particle in a finite volume V of arbitrary (large) extension, in which it is free to propagate. It is expected that in the limit of V→ ∞ we recover the free particle while allowing in the intermediate calculations the use of properly normalized states. Also, when describing for instance a particle propagating in a solid, one does not expect spatially localized states but instead completely delocalized states (within the solid), meaning that the particles propagates inside it (since it can be everywhere with the same probability, conversely to the sine solutions we encountered where the particle has favored locations). This understanding follows from the solutions of the Schrödinger equation for zero potential following from the so-called Von-Karman boundary conditions; i.e., the wavefunction assumes same values on opposite sides of the box but it is not required to be zero here. One can then check that the following solutions obey eq. 1:

<math>\textrm{in~1D}:\ \psi_k(x)={1\over\sqrt L}e^{ikx},\quad k={2n\pi\over L}, n\in\mathbb{Z}<math>

<math>\textrm{in~3D}:\ \psi_{\mathbf{k}}(x)={1\over\sqrt{L^3}}e^{i\mathbf{k}\cdot\mathbf{r}},\quad k_x={2n_x\pi\over L}, k_y={2n_y\pi\over L}, k_z={2n_z\pi\over L}, n_x, n_y, n_z\in\mathbb{Z}<math>

The energy remains <math>\hbar^2 k^2/2m<math> (cf. eq. 3) but interestingly, now the k are twice as before (cf. eq. 7). This is because in the previous case, n was strictly positive whereas now it can be negative or zero (the ground state). The solutions where the sine does not superpose to itself after a translation of L can not be recovered with exponentials, since in this propagating particle interpretation, the derivative is discontinuous at the border, meaning that the particle acquires infinite velocity here. This shows how the two interpretations bear intrinsically differing behaviours.

External links

• Scienceworld (http://scienceworld.wolfram.com/physics/InfiniteSquarePotentialWell.html) (Infinite Potential Well)

• Scienceworld (http://scienceworld.wolfram.com/physics/FiniteSquarePotentialWell.html) (Finite Potential Well)

• 1-D quantum mechanics java applet (http://www.falstad.com/qm1d/) simulates particle in a box, as well as other 1-dimensional cases.
• 2-D particle in a box applet (http://www.falstad.com/qm2dbox/)fr:Particule dans une boîte