Delta potential well

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The Delta potential well is a common theoretical problem of quantum mechanics. It consists of a time-independent Schrödinger equation for a particle in a potential well defined by a delta function in one dimension.

Contents

[edit] Definition

The time-independent Schrödinger equation for the wave function \psi(x)\,\! is

H\psi(x)=\left[-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)\right]\psi(x)=E\psi(x),\,\!

where H\,\! is the Hamiltonian, \hbar\,\! is the (reduced) Planck constant, m\,\! is the mass, E\,\! the energy of the particle, and

V(x)=\lambda\delta(x)\,\!

is the delta function well with strength \lambda < 0\,\!. The potential is located at the origin. Without changing the results, any other shifted position was possible.

[edit] Derivation

The potential well splits the space in two parts (x<0, x>0\,\!). In any of these parts the potential energy is constant, and the solution of the Schrödinger equation can be written as a superposition of exponentials:

\psi_L(x)= A_r e^{i k x} + A_l e^{-ikx}\quad x<0 \,\!, and
\psi_R(x)= B_r e^{i k x} + B_l e^{-ikx}\quad x>0\,\!

where the wave vector is related to the energy via

k=\sqrt{2m E}/\hbar .\,\!

The index r/l on the coefficients A and B denotes the direction of the velocity vector (for E>0\,\!). Even though the association with propagating waves only holds for positive energies (real wave vectors), the same notation is used for E<0\,\!. The coefficients A,B have to be determined from the boundary conditions of the wave function at x=0\,\!:

\psi_L=\psi_R\,\!,
\frac{d}{dx}\psi_L = \frac{d}{dx}\psi_R - \frac{2m\lambda}{\hbar^2} \psi_R\,\!.

Note that for the second boundary condition, the derivative of wave function with respect to x is not continuous at x=0\,\! . There is a difference of \frac{2\lambda}{\hbar^2} \psi_R\,\! between the two terms. To derive this result, the trick is to integrate the Schrödinger equation around x=0, but do so over a really small range:

- \frac{\hbar^2}{2 m} \int_{ - \epsilon}^{\epsilon} \frac{d^2 \psi}{d x^2}\, dx + \int_{ - \epsilon}^{\epsilon}V(x) \psi \, dx = E \int_{ - \epsilon}^{\epsilon} \psi \, dx\,\! ;(1)

where \epsilon\,\! is a really small number.

The right-hand side of equation (1) is

E \int_{ - \epsilon}^{\epsilon} \psi \, dx \approx E \cdot 2 \epsilon \cdot \psi(0)\,\! .(2)

and this is a more accurate approximation the smaller ε is. In the limit of \epsilon \to 0 this just goes to 0.

The left-hand side of equation (1) is

- \frac{\hbar^2}{2 m} \left( \frac{d\psi_R}{dx}\bigg|_{\epsilon} - \frac{d\psi_L}{dx}\bigg|_{ - \epsilon} \right) + \lambda\int_{-\epsilon}^{\epsilon}\delta(x) \psi \, dx = 0\,\! .(3)

From the definition of Dirac Delta function,

\int_{-\epsilon}^{\epsilon}\delta(x) \psi \, dx =\psi_R(0)\,\! .(4)

In the limit of \epsilon \to 0 \,\! .

\lim_{\epsilon \to 0}\frac{d\psi_L}{dx}\bigg|_{ - \epsilon}=\frac{d\psi_L}{dx}\bigg|_0\,\! ,(5)
\lim_{\epsilon \to 0}\frac{d\psi_R}{dx}\bigg|_{\epsilon}=\frac{d\psi_R}{dx}\bigg|_0\,\! .(6)

Substitute these results (4), (5), (6) into equation (3), which, after rearranging is

\frac{d\psi_L}{dx}=\frac{d\psi_R}{dx} - \frac{2m\lambda}{\hbar^2}\psi_R\,\! .

The boundary conditions thus give the following restrictions on the coefficients

A_r+A_l=B_r+B_l\,\!
ik(A_r-A_l-B_r+B_l)= - \frac{2m\lambda}{\hbar^2}(B_r+B_l).\,\!

[edit] Transmission and Reflection

[edit] E>0\,\!

Transmission (T) and reflection (R) probability of a delta potential well. The energy is in units of . Dashed: classical result. Solid line: quantum mechanics.
Transmission (T) and reflection (R) probability of a delta potential well. The energy E>0\,\! is in units of \frac{\lambda^2}{2m\hbar^2}\,\!. Dashed: classical result. Solid line: quantum mechanics.

For positive energies, the particle is free to move in either half-space: x<0, x>0\,\!. It may be scattered at the delta-function well. The calculation is identical to the one above the only difference being that \lambda\,\! is now negative.

The quantum case can be studied in the following situation: a particle incident on the barrier from the left side (A_r\,\!). It may be reflected (A_l\,\!) or transmitted (B_r\,\!). To find the amplitudes for reflection and transmission for incidence from the left, put in the above equations A_r=1\,\! (incoming particle), A_l=r\,\! (reflection), B_l\,\!=0 (no incoming particle from the right) and B_r=t\,\! (transmission) and solve for r, t\,\!. The result is:

t=\cfrac{1}{\cfrac{im\lambda}{\hbar^2k}+1}\,\!
r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}\,\!

Due to the mirror symmetry of the model, the amplitudes for incidence from the right are the same as those from the left. The result is that there is a non-zero probability

R=|r|^2=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!

for the particle to be reflected from the barrier. This is a purely quantum effect which does not appear in the classical case.

Taking this to conclusion, the probability for transmission is:

T=|t|^2=1-R=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m \lambda^2}{2\hbar^2 E}}\,\!.

[edit] Bound state

[edit] E<0\,\!

The graph of the bound state wavefunction solution to the delta function potential is continuous everywhere, but its derivative is not at x=0.
The graph of the bound state wavefunction solution to the delta function potential is continuous everywhere, but its derivative is not at x=0.

In any one-dimensional attractive potential there will be a bound state. To find its energy, note that for E<0, k=i\sqrt{2m |E|}/\hbar \,\! is complex and the wave functions which were oscillating for positive energies in the calculation above, are now exponentially increasing or decreasing functions of x (see above). Requiring that the wave functions do not diverge at x\to\pm \infty\,\! eliminates half of the terms: A_r=B_l=0\,\!. The wave function is then

\psi_L(x)= A_l e^{|k|x}\quad x<0 \,\!, and
\psi_R(x)= B_r e^{- |k| x} \quad x>0\,\!.

From the boundary conditions and normalization conditions, it follows

A_l=B_r=\sqrt{\mid k\mid}\,\! ,
k=i\frac{m\lambda}{\hbar^2}\,\! .

The energy of the bound state is then

E=\frac{\hbar^2k^2}{2m}=-\frac{m\lambda^2}{2\hbar^2}\,\!.

[edit] Remarks

The delta function potential well is a special case of the finite potential well and follows as a limit of infinite depth and zero width of the well, keeping the product of width and depth constant equal to \lambda\,\!.

[edit] Delta potential barrier

The Delta potential barrier is a textbook problem of quantum mechanics. The problem consists of solving the time-independent Schrödinger equation for a particle in a delta function potential in one dimension.

[edit] Definition

Scattering at a delta function potential of strength . The amplitudes and direction of left and right moving waves are indicated. In red, those waves used for the derivation of the reflection and transmission amplitude.
Scattering at a delta function potential of strength \lambda\,\!. The amplitudes and direction of left and right moving waves are indicated. In red, those waves used for the derivation of the reflection and transmission amplitude.

The time-independent Schrödinger equation for the wave function \psi(x)\,\! reads

H\psi(x)=\left[-\frac{\hbar^2}{2m}\frac{d^2}{dx^2}+V(x)\right]\psi(x)=E\psi(x),\,\!

where H\,\! is the Hamiltonian, \hbar\,\! is the (reduced) Planck constant, m\,\! is the mass, E\,\! the energy of the particle and

V(x)=\lambda\delta(x)\,\!

is the delta function potential (barrier) with strength \lambda > 0\,\!. Here we have chosen the potential to be at the origin. Without changing the results, any other shifted position was possible. The first term in the Hamiltonian, -\frac{\hbar^2}{2m}\frac{d^2}{dx^2}\psi\,\! is the kinetic energy.

[edit] Derivation

The barrier divides the space in two parts (x<0, x>0\,\!). In any of these parts the particle is free, and the solution of the Schrödinger equation can be written as a superposition of left and right moving waves (see free particle)

\psi_L(x)= A_r e^{i k x} + A_l e^{-ikx}\quad x<0 \,\!, and
\psi_R(x)= B_r e^{i k x} + B_l e^{-ikx}\quad x>0\,\!

where the wave vector is related to the energy via

k=\sqrt{2m E}/\hbar .\,\!

The index r/l on the coefficients A and B denotes the direction of the velocity vector. Those coefficients have to be found from the boundary conditions of the wave function at x=0\,\!:

\psi_L=\psi_R\,\!,
\frac{d}{dx}\psi_L = \frac{d}{dx}\psi_R - \frac{2m\lambda}{\hbar^2} \psi_R\,\!.

The second of these equations follows from integrating the Schrödinger equation with respect to x\,\!. The boundary conditions thus give the following restrictions on the coefficients

A_r+A_l=B_r+B_l\,\!
ik(A_r-A_l-B_r+B_l)=-\frac{2m\lambda}{\hbar^2}(B_r+B_l).\,\!

[edit] Transmission and reflection

At this point, it is instructive to compare the situation to the classical case. In both cases, the particle behaves as a free particle outside of the barrier region. However, a classical particle having a finite energy cannot pass the infinitely high potential barrier and will be reflected from it. To study the quantum case, let us consider the following situation: a particle incident on the barrier from the left side (A_r\,\!). It may be reflected (A_l\,\!) or transmitted (B_r\,\!).

To find the amplitudes for reflection and transmission for incidence from the left, we put in the above equations A_r=1\,\! (incoming particle), A_l=r\,\! (reflection), B_l\,\!=0 (no incoming particle from the right) and B_r=t\,\! (transmission) and solve for r, t\,\!. The result is:

t=\cfrac{1}{\cfrac{im\lambda}{\hbar^2k}+1}\,\!
r=\cfrac{1}{\cfrac{i\hbar^2 k}{m\lambda} - 1}.\,\!

Due to the mirror symmetry of the model, the amplitudes for incidence from the right are the same as those from the left. The surprising result, from the classical point of view, is that there is a non-zero probability (given by the transmission coefficient)

T=|t|^2=\cfrac{1}{1+\cfrac{m^2\lambda^2}{\hbar^4k^2}}= \cfrac{1}{1+\cfrac{m\lambda^2}{2\hbar^2 E}}\,\!

for the particle to be transmitted through the barrier. This effect which differs from the classical case is called quantum tunneling.

For completeness, the probability for reflection (given by the reflection coefficient) is:

R=|r|^2=1-T=\cfrac{1}{1+\cfrac{\hbar^4k^2}{m^2\lambda^2}}= \cfrac{1}{1+\cfrac{2\hbar^2 E}{m\lambda^2}}.\,\!

[edit] Remarks, Application

The calculation presented above may at first seem unrealistic and hardly useful. However it has proved to be a suitable model for a variety of real-life systems. One such example regards the interfaces between two conducting materials. In the bulk of the materials, the motion of the electrons is quasi free and can be described by the kinetic term in the above Hamiltonian with an effective mass m\,\!. Often the surfaces of such materials are covered with oxide layers or are not ideal for other reasons. This thin, non-conducting layer may then be modeled by a local delta-function potential as above. Electrons may then tunnel from one material to the other giving rise to a current.

The operation of a scanning tunneling microscope (STM) relies on this tunneling effect. In that case, the barrier is due to the air between the tip of the STM and the underlying object. The strength of the barrier is related to the separation being stronger the further apart the two are. For a more general model of this situation, see Finite potential barrier (QM). The delta function potential barrier is the limiting case of the model considered there for very high and narrow barriers.

The above model is one-dimensional while the space around us is three-dimensional. So in fact one should solve the Schrödinger equation in three dimensions. On the other hand, many systems only change along one coordinate direction and are translationally invariant along the others. The Schrödinger equation may then be reduced to the case considered here by an Ansatz for the wave function of the type: \Psi(x,y,z)=\psi(x)\phi(y,z)\,\!.

[edit] See also

[edit] Reference

  • Griffiths, David J. (2005). Introduction to Quantum Mechanics, 2nd ed., Prentice Hall, pp. 68-78. ISBN 0-13-111892-7.