Particle in a box

In quantum mechanics, the wavefunction gives the most fundamental description of the behavior of a particle; the measurable properties of the particle (such as its position, momentum and energy) may all be derived from the wavefunction.[3] The wavefunction ${\displaystyle \psi (x,t)}$ can be found by solving the Schrödinger equation for the system

${\displaystyle i\hbar {\frac {\partial }{\partial t}}\psi (x,t)=-{\frac {\hbar ^{2}}{2m}}{\frac {\partial ^{2}}{\partial x^{2}}}\psi (x,t)+V(x)\psi (x,t),}$

where ${\displaystyle \hbar }$ is the reduced Planck constant, ${\displaystyle m}$ is the mass of the particle, ${\displaystyle i}$ is the imaginary unit and ${\displaystyle t}$ is time.

Inside the box, no forces act upon the particle, which means that the part of the wavefunction inside the box oscillates through space and time with the same form as a free particle:[1][4]

${\displaystyle \psi (x,t)=[A\sin(kx)+B\cos(kx)]\mathrm {e} ^{-i\omega t},}$

(1)

where ${\displaystyle A}$ and ${\displaystyle B}$ are arbitrary complex numbers. The frequency of the oscillations through space and time is given by the wavenumber ${\displaystyle k}$ and the angular frequency ${\displaystyle \omega }$ respectively. These are both related to the total energy of the particle by the expression

${\displaystyle E=\hbar \omega ={\frac {\hbar ^{2}k^{2}}{2m}},}$

which is known as the dispersion relation for a free particle.[1] Here one must notice that now, since the particle is not entirely free but under the influence of a potential (the potential V described above), the energy of the particle given above is not the same thing as ${\displaystyle {\frac {p^{2}}{2m}}}$ where p is the momentum of the particle, and thus the wavenumber k above actually describes the energy states of the particle, not the momentum states (i.e. it turns out that the momentum of the particle is not given by ${\displaystyle p=\hbar k}$). In this sense, it is quite dangerous to call the number k a wavenumber, since it is not related to momentum like "wavenumber" usually is. The rationale for calling k the wavenumber is that it enumerates the number of crests that the wavefunction has inside the box, and in this sense it is a wavenumber. This discrepancy can be seen more clearly below, when we find out that the energy spectrum of the particle is discrete (only discrete values of energy are allowed) but the momentum spectrum is continuous (momentum can vary continuously) and in particular, the relation ${\displaystyle E={\frac {p^{2}}{2m}}}$ for the energy and momentum of the particle does not hold. As said above, the reason this relation between energy and momentum does not hold is that the particle is not free, but there is a potential V in the system, and the energy of the particle is ${\displaystyle E=T+V}$, where T is the kinetic and V the potential energy.

Initial wavefunctions for the first four states in a one-dimensional particle in a box

The size (or amplitude) of the wavefunction at a given position is related to the probability of finding a particle there by ${\displaystyle P(x,t)=|\psi (x,t)|^{2}}$. The wavefunction must therefore vanish everywhere beyond the edges of the box.[1][4] Also, the amplitude of the wavefunction may not "jump" abruptly from one point to the next.[1] These two conditions are only satisfied by wavefunctions with the form

${\displaystyle \psi _{n}(x,t)={\begin{cases}A\sin \left(k_{n}\left(x-x_{c}+{\tfrac {L}{2}}\right)\right)\mathrm {e} ^{-i\omega _{n}t}\quad &x_{c}-{\tfrac {L}{2}}<x<x_{c}+{\tfrac {L}{2}}\\0&{\text{otherwise}}\end{cases}},}$

where [5]

${\displaystyle k_{n}={\frac {n\pi }{L}}}$,

and

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

where n is a positive integer (1,2,3,4...). For a shifted box (x_c = L/2), the solution is particularly simple. The simplest solutions, ${\displaystyle k_{n}=0}$ or ${\displaystyle A=0}$ both yield the trivial wavefunction ${\displaystyle \psi (x)=0}$, which describes a particle that does not exist anywhere in the system.[6] Negative values of ${\displaystyle n}$ are neglected, since they give wavefunctions identical to the positive ${\displaystyle n}$ solutions except for a physically unimportant sign change.[6] Here one sees that only a discrete set of energy values and wavenumbers k are allowed for the particle. Usually in quantum mechanics it is also demanded that the derivative of the wavefunction in addition to the wavefunction itself be continuous; here this demand would lead to the only solution being the constant zero function, which is not what we desire, so we give up this demand (as this system with infinite potential can be regarded as a nonphysical abstract limiting case, we can treat it as such and "bend the rules"). Note that giving up this demand means that the wavefunction is not a differentiable function at the boundary of the box, and thus it can be said that the wavefunction does not solve the Schrödinger equation at the boundary points ${\displaystyle x=0}$ and ${\displaystyle x=L}$ (but does solve it everywhere else).

Finally, the unknown constant ${\displaystyle A}$ may be found by normalizing the wavefunction so that the total probability density of finding the particle in the system is 1. It follows that

${\displaystyle \left|A\right|={\sqrt {\frac {2}{L}}}.}$

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

It is expected that the eigenvalues, i.e., the energy ${\displaystyle E_{n}}$ of the box should be the same regardless of its position in space, but ${\displaystyle \psi _{n}(x,t)}$ changes. Notice that ${\displaystyle x_{c}-{\tfrac {L}{2}}}$ represents a phase shift in the wave function. This phase shift has no effect when solving the Schrödinger equation, and therefore does not affect the eigenvalue.

If we set the origin of coordinates to the left edge of the box, we can rewrite the spacial part of the wave function succinctly as:

${\displaystyle \psi _{n}(x)={\begin{cases}{\sqrt {\frac {2}{L}}}\sin(k_{n}x)\quad {}{\text{for }}n{\text{ even}}\\{\sqrt {\frac {2}{L}}}\cos(k_{n}x)\quad {}{\text{for }}n{\text{ odd}}\end{cases}}}$.

The momentum wavefunction is proportional to the Fourier transform of the position wavefunction. With ${\displaystyle k=p/\hbar }$ (note that the parameter k describing the momentum wavefunction below is not exactly the special k_n above, linked to the energy eigenvalues), the momentum wavefunction is given by

${\displaystyle \phi _{n}(p,t)={\frac {1}{\sqrt {2\pi \hbar }}}\int _{-\infty }^{\infty }\psi _{n}(x,t)e^{-ikx}\,dx={\sqrt {\frac {L}{\pi \hbar }}}\left({\frac {n\pi }{n\pi +kL}}\right)\,{\textrm {sinc}}\left({\tfrac {1}{2}}(n\pi -kL)\right)e^{-ikx_{c}}e^{i(n-1){\tfrac {\pi }{2}}}e^{-i\omega _{n}t},}$

where sinc is the cardinal sine sinc function, sinc(x)=sin(x)/x. For the centered box (x_c= 0), the solution is real and particularly simple, since the phase factor on the right reduces to unity. (With care, it can be written as an even function of p.)

It can be seen that the momentum spectrum in this wave packet is continuous, and one may conclude that for the energy state described by the wavenumber k_n, the momentum can, when measured, also attain other values beyond ${\displaystyle p=\pm \hbar k_{n}}$.

Hence, it also appears that, since the energy is ${\displaystyle E_{n}={\frac {\hbar ^{2}k_{n}^{2}}{2m}}}$ for the nth eigenstate, the relation ${\displaystyle E={\frac {p^{2}}{2m}}}$ does not strictly hold for the measured momentum p; the energy eigenstate ${\displaystyle \psi _{n}}$ is not a momentum eigenstate, and, in fact, not even a superposition of two momentum eigenstates, as one might be tempted to imagine from equation (1) above: peculiarly, it has no well-defined momentum before measurement!

In classic physics, the particle can be detected anywhere in the box with equal probability. In quantum mechanics, however, the probability density for finding a particle at a given position is derived from the wavefunction as ${\displaystyle P(x)=|\psi (x)|^{2}.}$ For the particle in a box, the probability density for finding the particle at a given position depends upon its state, and is given by

${\displaystyle P_{n}(x,t)={\begin{cases}{\frac {2}{L}}\sin ^{2}(k_{n}(x-x_{c}+{\tfrac {L}{2}})),&x_{c}-L/2<x<x_{c}+L/2,\\0,&{\text{otherwise,}}\end{cases}}}$

Thus, for any value of n greater than one, there are regions within the box for which ${\displaystyle P(x)=0}$, indicating that spatial nodes exist at which the particle cannot be found.

In quantum mechanics, the average, or expectation value of the position of a particle is given by

${\displaystyle \langle x\rangle =\int _{-\infty }^{\infty }xP_{n}(x)\,\mathrm {d} x.}$

For the steady state particle in a box, it can be shown that the average position is always ${\displaystyle \langle x\rangle =x_{c}}$, regardless of the state of the particle. For a superposition of states, the expectation value of the position will change based on the cross term which is proportional to ${\displaystyle \cos(\omega t)}$.

The variance in the position is a measure of the uncertainty in position of the particle:

${\displaystyle \mathrm {Var} (x)=\int _{-\infty }^{\infty }(x-\langle x\rangle )^{2}P_{n}(x)\,dx={\frac {L^{2}}{12}}\left(1-{\frac {6}{n^{2}\pi ^{2}}}\right)}$

The probability density for finding a particle with a given momentum is derived from the wavefunction as ${\displaystyle P(x)=|\phi (x)|^{2}}$. As with position, the probability density for finding the particle at a given momentum depends upon its state, and is given by

${\displaystyle P_{n}(p)={\frac {L}{\pi \hbar }}\left({\frac {n\pi }{n\pi +kL}}\right)^{2}\,{\textrm {sinc}}^{2}\left({\tfrac {1}{2}}(n\pi -kL)\right)}$

where, again, ${\displaystyle k=p/\hbar }$. The expectation value for the momentum is then calculated to be zero, and the variance in the momentum is calculated to be:

${\displaystyle \mathrm {Var} (p)=\left({\frac {\hbar n\pi }{L}}\right)^{2}}$

The uncertainties in position and momentum (${\displaystyle \Delta x}$ and ${\displaystyle \Delta p}$) are defined as being equal to the square root of their respective variances, so that:

${\displaystyle \Delta x\Delta p={\frac {\hbar }{2}}{\sqrt {{\frac {n^{2}\pi ^{2}}{3}}-2}}}$

This product increases with increasing n, having a minimum value for n=1. The value of this product for n=1 is about equal to 0.568 ${\displaystyle \hbar }$ which obeys the Heisenberg uncertainty principle, which states that the product will be greater than or equal to ${\displaystyle \hbar /2}$

Another measure of uncertainty in position is the information entropy of the probability distribution H_x:[7]

${\displaystyle H_{x}=\int _{-\infty }^{\infty }P_{n}(x)\log(P_{n}(x)x_{0})\,dx=\log \left({\frac {2L}{e\,x_{0}}}\right)}$

where x₀ is an arbitrary reference length.

Another measure of uncertainty in momentum is the information entropy of the probability distribution H_p:

${\displaystyle H_{p}(n)=\int _{-\infty }^{\infty }P_{n}(p)\log(P_{n}(p)p_{0})\,dp}$

${\displaystyle \lim _{n\to \infty }H_{p}(n)=\log \left({\frac {4\pi \hbar \,e^{2(1-\gamma )}}{L\,p_{0}}}\right)}$

where γ is Euler's constant. The quantum mechanical entropic uncertainty principle states that for ${\displaystyle x_{0}\,p_{0}=\hbar }$

${\displaystyle H_{x}+H_{p}(n)\geq \log(e\,\pi )\approx 2.14473...}$ (nats)

For ${\displaystyle x_{0}\,p_{0}=\hbar }$, the sum of the position and momentum entropies yields:

${\displaystyle H_{x}+H_{p}(\infty )=\log(8\pi \,e^{1-2\gamma })\approx 3.06974...}$ (nats)

which satisfies the quantum entropic uncertainty principle.

The energy of a particle in a box (black circles) and a free particle (grey line) both depend upon wavenumber in the same way. However, the particle in a box may only have certain, discrete energy levels.

The energies which correspond with each of the permitted wavenumbers may be written as[5]

${\displaystyle E_{n}={\frac {n^{2}\hbar ^{2}\pi ^{2}}{2mL^{2}}}={\frac {n^{2}h^{2}}{8mL^{2}}}}$.

The energy levels increase with ${\displaystyle n^{2}}$, meaning that high energy levels are separated from each other by a greater amount than low energy levels are. The lowest possible energy for the particle (its zero-point energy) is found in state 1, which is given by[8]

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

The particle, therefore, always has a positive energy. This contrasts with classical systems, where the particle can have zero energy by resting motionlessly. This can be explained in terms of the uncertainty principle, which states that the product of the uncertainties in the position and momentum of a particle is limited by

${\displaystyle \Delta x\Delta p\geq {\frac {\hbar }{2}}}$

It can be shown that the uncertainty in the position of the particle is proportional to the width of the box.[9] Thus, the uncertainty in momentum is roughly inversely proportional to the width of the box.[8] The kinetic energy of a particle is given by ${\displaystyle E=p^{2}/(2m)}$, and hence the minimum kinetic energy of the particle in a box is inversely proportional to the mass and the square of the well width, in qualitative agreement with the calculation above.[8]

The wavefunction of a 2D well with n_x=4 and n_y=4

If a particle is trapped in a two-dimensional box, it may freely move in the ${\displaystyle x}$ and ${\displaystyle y}$-directions, between barriers separated by lengths ${\displaystyle L_{x}}$ and ${\displaystyle L_{y}}$ respectively. For a centered box, the position wave function may be written including the length of the box as ${\displaystyle \psi _{n}(x,t,L)}$. Using a similar approach to that of the one-dimensional box, it can be shown that the wavefunctions and energies for a centered box are given respectively by

${\displaystyle \psi _{n_{x},n_{y}}=\psi _{n_{x}}(x,t,L_{x})\psi _{n_{y}}(y,t,L_{y})}$,

${\displaystyle E_{n_{x},n_{y}}={\frac {\hbar ^{2}k_{n_{x},n_{y}}^{2}}{2m}}}$,

where the two-dimensional wavevector is given by

${\displaystyle \mathbf {k_{n_{x},n_{y}}} =k_{n_{x}}\mathbf {\hat {x}} +k_{n_{y}}\mathbf {\hat {y}} ={\frac {n_{x}\pi }{L_{x}}}\mathbf {\hat {x}} +{\frac {n_{y}\pi }{L_{y}}}\mathbf {\hat {y}} }$.

For a three dimensional box, the solutions are

${\displaystyle \psi _{n_{x},n_{y},n_{z}}=\psi _{n_{x}}(x,t,L_{x})\psi _{n_{y}}(y,t,L_{y})\psi _{n_{z}}(z,t,L_{z})}$,

${\displaystyle E_{n_{x},n_{y},n_{z}}={\frac {\hbar ^{2}k_{n_{x},n_{y},n_{z}}^{2}}{2m}}}$,

where the three-dimensional wavevector is given by:

${\displaystyle \mathbf {k_{n_{x},n_{y},n_{z}}} =k_{n_{x}}\mathbf {\hat {x}} +k_{n_{y}}\mathbf {\hat {y}} +k_{n_{z}}\mathbf {\hat {z}} ={\frac {n_{x}\pi }{L_{x}}}\mathbf {\hat {x}} +{\frac {n_{y}\pi }{L_{y}}}\mathbf {\hat {y}} +{\frac {n_{z}\pi }{L_{z}}}\mathbf {\hat {z}} }$.

In general for an n-dimensional box, the solutions are

${\displaystyle \psi =\prod _{i}\psi _{n_{i}}(x_{i},t,L_{i})}$

The n-dimensional momentum wave functions may likewise be represented by ${\displaystyle \phi _{n}(x,t,L_{x})}$ and the momentum wave function for an n-dimensional centered box is then:

${\displaystyle \phi =\prod _{i}\phi _{n_{i}}(k_{i},t,L_{i})}$

An interesting feature of the above solutions is that when two or more of the lengths are the same (e.g. ${\displaystyle L_{x}=L_{y}}$), there are multiple wavefunctions corresponding to the same total energy. For example, the wavefunction with ${\displaystyle n_{x}=2,n_{y}=1}$ has the same energy as the wavefunction with ${\displaystyle n_{x}=1,n_{y}=2}$. 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.

The wavefunction for a quantum-mechanical particle in a box whose walls have arbitrary shape is given by the Helmholtz equation subject to the boundary condition that the wavefunction vanishes at the walls. These systems are studied in the field of quantum chaos for wall shapes whose corresponding dynamical billiard tables are non-integrable.

β-carotene is a conjugated polyene

Conjugated polyene systems can be modeled using particle in a box. The conjugated system of electrons can be modeled as a one dimensional box with length equal to the total bond distance from one terminus of the polyene to the other. In this case each pair of electrons in each π bond corresponds to their energy level. The energy difference between two energy levels, n_f and n_i is:

${\displaystyle \Delta E={\frac {(n_{f}^{2}-n_{i}^{2})h^{2}}{8mL^{2}}}}$

The difference between the ground state energy, n, and the first excited state, n+1, corresponds to the energy required to excite the system. This energy has a specific wavelength, and therefore color of light, related by:

${\displaystyle \lambda ={\frac {hc}{\Delta E}}}$

A common example of this phenomenon is in β-carotene. β-carotene (C₄₀H₅₆)[10] is a conjugated polyene with an orange color and a molecular length of approximately 3.8 nm (though its chain length is only approximately 2.4 nm).[11] Due to β-carotene's high level of conjugation, electrons are dispersed throughout the length of the molecule, allowing one to model it as a one-dimensional particle in a box. β-carotene has 11 carbon-carbon double bonds in conjugation;[10] each of those double bonds contains two π-electrons, therefore β-carotene has 22 π-electrons. With two electrons per energy level, β-carotene can be treated as a particle in a box at energy level n=11.[11] Therefore, the minimum energy needed to excite an electron to the next energy level can be calculated, n=12, as follows[11] (recalling that the mass of an electron is 9.109 × 10⁻³¹ kg[12]):

${\displaystyle \Delta E={\frac {(n_{f}^{2}-n_{i}^{2})h^{2}}{8mL^{2}}}}$

${\displaystyle ={\frac {(12^{2}-11^{2})h^{2}}{8mL^{2}}}}$

${\displaystyle =2.3658\times 10^{-19}{\text{ J}}}$

Using the previous relation of wavelength to energy, recalling both Planck's constant h and the speed of light c:

${\displaystyle \lambda ={\tfrac {hc}{\Delta E}}}$

${\displaystyle =0.00000084{\text{ m}}=840{\text{ nm}}}$

This indicates that β-carotene primarily absorbs light in the infrared spectrum, therefore it would appear white to a human eye. However the observed wavelength is 450 nm,[13] indicating that the particle in a box is not a perfect model for this system.

The particle in a box model can be applied to quantum well lasers, which are laser diodes consisting of one semiconductor “well” material sandwiched between two other semiconductor layers of different material . Because the layers of this sandwich are very thin (the middle layer is typically about 100 Å thick), quantum confinement effects can be observed.[14] The idea that quantum effects could be harnessed to create better laser diodes originated in the 1970s. The quantum well laser was patented in 1976 by R. Dingle and C. H. Henry.[15]

Specifically, the quantum wells behavior can be represented by the particle in a finite well model. Two boundary conditions must be selected. The first is that the wave function must be continuous. Often, the second boundary condition is chosen to be the derivative of the wave function must be continuous across the boundary, but in the case of the quantum well the masses are different on either side of the boundary. Instead, the second boundary condition is chosen to conserve particle flux as ${\displaystyle (1/m)d\phi /dz}$, which is consistent with experiment. The solution to the finite well particle in a box must be solved numerically, resulting in wave functions that are sine functions inside the quantum well and exponentially decaying functions in the barriers.[16] This quantization of the energy levels of the electrons allows a quantum well laser to emit light more efficiently than conventional semiconductor lasers.

Due to their small size, quantum dots do not showcase the bulk properties of the specified semi-conductor but rather show quantised energy states.[17] This effect is known as the quantum confinement and has led to numerous applications of quantum dots such as the quantum well laser.[17]

Researchers at Princeton University have recently built a quantum well laser which is no bigger than a grain of rice.[18] The laser is powered by a single electron which passes through two quantum dots; a double quantum dot. The electron moves from a state of higher energy, to a state of lower energy whilst emitting photons in the microwave region. These photons bounce off mirrors to create a beam of light; the laser.[18]

The quantum well laser is heavily based on the interaction between light and electrons. This relationship is a key component in quantum mechanical theories which include the De Broglie Wavelength and Particle in a box. The double quantum dot allows scientists to gain full control over the movement of an electron which consequently results in the production of a laser beam.[18]

Quantum dots extremely small semiconductors (on the scale of nanometers).[19] They display quantum confinement in that the electrons cannot escape the “dot”, thus allowing particle-in-a-box approximations to be .[20] Their behavior can be described by three-dimensional particle-in-a-box energy quantization equations.[20]

The energy gap of a quantum dot is the energy gap between its valence and conduction bands. This energy gap ${\displaystyle \Delta E(r)}$ is equal to the gap of the bulk material ${\displaystyle E_{\text{gap}}}$ plus the energy equation derived particle-in-a-box, which gives the energy for electrons and holes.[20] This can be seen in the following equation, where ${\displaystyle m_{e}^{*}}$ and ${\displaystyle m_{h}^{*}}$ are the effective masses of the electron and hole, ${\displaystyle r}$ is radius of the dot, and ${\displaystyle h}$ is Planck's constant:[20]

${\displaystyle \Delta E(r)=E_{\text{gap}}+\left({\frac {h^{2}}{8r^{2}}}\right)\left({\frac {1}{m_{e}^{*}}}+{\frac {1}{m_{h}^{*}}}\right)}$

Hence, the energy gap of the quantum dot is inversely proportional to the square of the “length of the box,” i.e. the radius of the quantum dot.[20]

Manipulation of the band gap allows for the absorption and emission of specific wavelengths of light, as energy is inversely proportional to wavelength.[19] The smaller the quantum dot, the larger the band gap and thus the shorter the wavelength absorbed.[19][21]

Different semiconducting materials are used to synthesize quantum dots of different sizes and therefore emit different wavelengths of light.[21] Materials that normally emit light in the visible region are often used and their sizes are fine-tuned so that certain colors are emitted.[19] Typical substances used to synthesize quantum dots are cadmium (Cd) and selenium (Se).[19][21] For example, when the electrons of two nanometer CdSe quantum dots relax after excitation, blue light is emitted. Similarly, red light is emitted in four nanometer CdSe quantum dots.[22][19]

Quantum dots have a variety of functions including but not limited to fluorescent dyes, transistors, LEDs, solar cells, and medical imaging via optical probes.[19][20]

One function of quantum dots is their use in lymph node mapping, which is feasible due to their unique ability to emit light in the near infrared (NIR) region. Lymph node mapping allows surgeons to track if and where cancerous cells exist.[23]

Quantum dots are useful for these functions due to their emission of brighter light, excitation by a wide variety of wavelengths, and higher resistance to light than other substances.[23][19]