PHYS212 : Lectures

Subsections

• 7.4.1 Potential Steps
• 7.4.2 Square Barriers and Tunneling
• 7.4.3 Potential Ramps and the WKB Approximation
  • 7.4.3.1 The Method Revisited
  • 7.4.3.2 An Electron gun
  
  
• 7.4.4 Gaussian Wave Packets

7.4 Free Particles

7.4.1 Potential Steps

• An interface between two media in which particles have different but constant potential energies $V_1$ and $V_2$ is the quantum mechanical version of the planar interface between media we considered for acoustic (Section 4.3.2) and electromagnetic waves (Section 5.5).

• In practice, quantum potential steps are rather artificial constructions. We could present a ``potential ramp'' to a charged particle using a parallel plate capacitor with holes drilled in the plates allowing particles to pass through. Bringing the plates very close together would approximate a step.

• You will consider Snell's law for two dimensional reflection and refraction from such a step for homework. Here, we will continue to confine ourselves to one dimension.

• For a plane wave incident from the left with $E < V_1,V_2$, we have
  

  $$\Psi(x,t) = \left\{ \begin{array}{lc} \Psi_i + \Psi_r & (x<0)\\ \Psi_t & (x>0) \end{array}\right.$$

  $$= \left\{ \begin{array}{lc} C_i e^{i(k_1 x - \omega t)} + C_r ... ...\omega t)} & (x<0)\\ C_t e^{i(k_2 x - \omega t)} & (x>0) \end{array}\right.$$ (7.43)

  
  
  with
  

  $$k_1 = \frac{\sqrt{2m(E-V_1)}}{\hbar}$$

  $$k_2 = \frac{\sqrt{2m(E-V_2)}}{\hbar}$$

  $$\omega = \frac{E}{\hbar}$$ (7.44)

  
  

• You will show for homework that the probability current is given by
  

  $$J = \left\{ \begin{array}{rllc} J_i + J_r & = & \frac{\hbar... ...frac{\hbar k_2}{m} \vert C_t\vert^2 & (x>0) \end{array}\right.$$ (7.45)

  
  

• We define the reflection and transmission coefficients for a plane wave incident normally on a potential step in terms of probability currents
  

  $$R = \left\vert \frac{J_r}{J_i} \right\vert = \left\vert \frac{C_r}{C_i} \right\vert^2$$

  $$T = \left\vert \frac{J_t}{J_i} \right\vert = \frac{k_2}{k_1} \left\vert \frac{C_t}{C_i} \right\vert^2$$ (7.46)

  
  

• We can find relationships between the amplitudes $C_i$, $C_r$, and $C_t$ by matching the wavefunction and its slope at the step,
  

  $$C_i + C_r = C_t$$

  $$k_1 C_i - k_1 C_r = k_2 C_t$$ (7.47)

  
  
  which combine to eliminate either $C_t$ or $C_r$ to give
  

  $$R = \left( \frac{1-k_2/k_1}{1 + k_2/k_1} \right)^2$$

  $$T = \frac{4 k_2/k_1} {\left(1 + k_2/k_1 \right)^2}$$ (7.48)

  
  

• The transmission coefficient can be written in a symmetrical form
  

  $$T = \frac{4 k_1 k_2} {\left(k_1 + k_2 \right)^2}$$ (7.49)

  
  
  which leads to the conclusion that upward and downward steps between identical $V_1$ and $V_2$ have identical transmission and reflection properties.

7.4.2 Square Barriers and Tunneling

• Consider plane waves incident on a potential barrier (or well) given by
  

  $$V(x) = \left\{ \begin{array}{lc} V_o & (\vert x\vert < L/2)\\ 0 & (\vert x\vert > L/2) \end{array}\right.$$ (7.50)

  
  

• The wave function is divided into three regions,
  

  $$\Psi(x,t) = \left\{ \begin{array}{lc} C_i e^{i(k_1 x - \o... ... C_t e^{i(k_1 x - \omega t)} & (x > L/2) \end{array}\right.$$ (7.51)

  
  
  with
  

  $$k_1 = \frac{\sqrt{2mE}}{\hbar}$$

  $$k_2 = \frac{\sqrt{2m(E-V_o)}}{\hbar}$$

  $$\omega = \frac{E}{\hbar}$$ (7.52)

  
  

• The transmission and reflection coefficients of the barrier (or well) are given by Eq.'s 7.46 with the $C_i$, $C_r$, and $C_t$ of Eq. 7.51, and as with the potential step, relationships among these coefficients are found via boundary matching.

• Matching $\Psi$ and $\Psi'$ at $x=-L/2$ we find
  

  $$C_i e^{-ik_1L/2} + C_r e^{ik_1L/2} = A e^{-ik_2L/2} + B e^{ik_2L/2}$$

  $$k_1 (C_i e^{-ik_1L/2} - C_r e^{ik_1L/2}) = k_2 (A e^{-ik_2L/2} - B e^{ik_2L/2})$$ (7.53)

  
  
  and at $x=L/2$
  

  $$A e^{ik_2L/2} + B e^{-ik_2L/2} = C_t e^{ik_1L/2}$$

  $$k_2(A e^{ik_2L/2} - B e^{-ik_2L/2}) = k_1 C_t e^{ik_1L/2}$$ (7.54)

  
  

• These four equations cannot be solved for five unknowns (the amplitudes), but they can be solved for four amplitude ratios. We are interested in $\frac{C_r}{C_i}$ and $\frac{C_t}{C_i}$ in particular.

• With some algebraic exertion, Eq.'s 7.53 and 7.54 can be combined to eliminate the intermediate amplitude ratios $\frac{A}{C_i}$ and $\frac{B}{C_i}$
  

  $$\frac{C_r}{C_i} = \left[ \frac{i}{2} \frac{k_2^2 - k_1^2}{k_1 k_2} \sin(k_2 L) \right] \frac{C_t}{C_i}$$ (7.55)

  
  

• Taking the complex square of Eq. 7.55 we have
  

  $$\left\vert \frac{C_r}{C_i} \right\vert^2 = \frac{1}{4}\left... ...right)^2 \sin^2(k_2 L) \left\vert\frac{C_t}{C_i}\right\vert^2$$ (7.56)

  
  

• The wavenumber is $k_1$ on either side of the barrier (or well), so the reflection and transmission coefficients (Eq. 7.46) are simply
  

  $$R = \left\vert \frac{C_r}{C_i} \right\vert^2$$

  $$T = \left\vert \frac{C_t}{C_i} \right\vert^2$$ (7.57)

  
  
  which together with $R+T=1$ and Eq. 7.56 gives
  

  $$T = \left[1 + \frac{1}{4}\left( \frac{k_2^2 - k_1^2}{k_1 k_2} \right)^2 \sin^2(k_2 L) \right]^{-1}$$ (7.58)

  
  

• In terms of the energies,
  

  $$T = \left[1 + \frac{1}{4} \frac{V_o^2}{E(E-V_o)} \sin^2 \left( \frac{L}{\hbar} \sqrt{2m(E-V_o)} \right) \right]^{-1}$$ (7.59)

  
  

• Fig. 7.1 shows plots of $T$ vs. $E$ for electrons scattering from a barrier of height 1 eV and a well of depth -1 eV. Note the effect of narrowing the well.

• Note that in Fig. 7.1(a), the transmission probability is not zero for $E < V_o$. A quantum mechanical particle can ``tunnel'' through barriers greater than its total energy. The sine function in Eq. 7.59 has an imaginary argument for $E < V_o$, giving exponential instead of oscillatory behavior.

• Wells and barriers have $T=1$ at particular energies corresponding to the roots of the $\sin^2$ factor in Eq. 7.59,
  
  $$\frac{L}{\hbar} \sqrt{2m(E-V_o)} = n\pi$$
  
  
  which correspond to energies^7.4
  

  $$E_n = \frac{\hbar^2 \pi^2}{2m L^2} n^2 + V_o$$ (7.60)

  
  
  Hence, we can learn about the structure of the well ($L$ and $V_o$) from scattering measurements (transmission vs. energy).

  Figure 7.1: The transmission coefficient as a function of plane wave total energy $E$ for electrons scattering from a potential barrier with $V_o = 1$ eV and (b) a potential well with $V_o = -1$ eV. The solid curves correspond to width $L=1$ nm, and the dashed curves to $L=0.5$ nm.

  (a)

  (b)

• We also see tunneling through the barrier. That is, in Fig. 7.1(a) we have $T > 0$ for $E < V_o$.

7.4.3 Potential Ramps and the WKB Approximation

7.4.3.1 The Method Revisited

• The WKB method presented in Section 3.2.1 can be adapted to quantum mechanical systems. It is a suitable approach to systems, such as potential ramps, with potential energy functions which vary gradually relative to local wavelengths.

• The WKB approximation involves only the position-dependent part of the wave function. The time-inependent parts of the Scrödinger equation and the classical wave equation have the same form, and so do their solutions. Hence, the quantum mechanical version of the WKB approximation is the same as the classical version developed in Section 3.2.1.

• Plane wave WKB wavefunctions take the same form in the quantum and classical cases,
  

  $$\psi(x) = A(x) e^{i\Phi(x)}$$ (7.61)

  
  

• With the WKB wavefunction, Eq. 7.16 becomes
  

  $$\left[A'' - A \Phi'^2 + \frac{\sqrt{2m(E-V(x))}}{\hbar^2} A \right] + i \left[ 2A'\Phi' + A \Phi'' \right] = 0$$ (7.62)

  
  

• In the third term in the real part of Eq. 7.62 we recognize the local wavenumber,
  

  $$k(x) = \frac{\sqrt{2m(E-V(x))}}{\hbar}$$ (7.63)

  
  

• In classical applications of the WKB method, the problem is essentially finding the function $k(x)$. In quantum mechanical applications, we have a recipe (Eq. 7.63).

• Once $k(x)$ is defined, the real and imaginary parts of Eq. 7.62 are identical to Eq.'s 3.17 and 3.18, and we can draw the same conclusions we drew in the classical case.

• Using the explicit form of $k(x)$ given by Eq. 7.63, and noting that $p(x) = \hbar k(x)$ we can state the WKB phase and amplitude functions in terms of either energy or momentum as
  

  $$\Phi(x) = \frac{\sqrt{2m}}{\hbar} \int_{x_o}^x \sqrt{E - V(x)} dx = \frac{1}{\hbar} \int_{x_o}^x p(x) dx$$ (7.64)

  
  
  and
  

  $$A(x) = A_o \sqrt{\frac{k_o}{k(x)}} = A_o \sqrt{\frac{p_o}{p(... ... = A_o \left( \frac{E - V_o}{(E - V(x))} \right)^{\frac{1}{4}}$$ (7.65)

  
  

• Finally, the complete solutions take the form
  

  $$\Psi(x,t) = A_o \sqrt{\frac{k_o}{k(x)}} \exp\left[i \left( \int_{x_o}^x k(x) dx - \omega t \right)\right]$$ (7.66)

  
  
  or alternatively
  

  $$\Psi(x,t) = A_o \sqrt{\frac{p_o}{p(x)}} \exp\left[\frac{i}{\hbar} \left( \int_{x_o}^x p(x) dx - E t \right)\right]$$ (7.67)

  
  

• The probability density is then
  

  $$\varrho(x) = \vert A_o\vert^2 \frac{k_o}{k(x)} = \vert A_o\vert^2 \frac{p_o}{p(x)}$$ (7.68)

  
  

• The dependence of the wavefunction and probability density on the momentum indicate
  • Greater momentum $\Leftrightarrow$ smaller amplitude $\Psi$ and smaller peak $\varrho$
  • Greater momentum $\Leftrightarrow$ larger local $k$, shorter local $\lambda$

• We can state the limit of the validity of the approximation, given by Eq. 3.23 in terms of $V(x)$ as
  
  $$\frac{\vert V'(x)\vert}{(E-V(x))^{\frac{3}{2}}} « \frac{\sqrt{2m}}{\pi\hbar}$$
  
  

7.4.3.2 An Electron gun

• Linear potential ramps arise from constant, uniform force fields. The gravitational field near the surface of the earth and the electric field between two parallel charged plates are good approximations to uniform fields.

• As an example, consider the limits on the WKB approximation for electrons in a uniform electric field - an electron gun. The potential energy of an electron in an electric field oriented along $-\hat{\imath}$ is given by^7.5
  
  $$V(x) = - e \mathcal{E} x$$
  
  
  where $\mathcal{E}$ is the magnitude of the field.

• The limit of validity of the WKB approximation is
  
  $$\frac{e\mathcal{E}}{(E+e\mathcal{E}x)^{\frac{3}{2}}} « \frac{\sqrt{2m}}{\pi\hbar}$$
  
  
  which, recognizing $x = 0$ as the worst case (shortest local wavelength), can be simplified to give
  

  $$\mathcal{E} « \frac{\sqrt{2m}}{\pi \hbar e} E^{\frac{3}{2}}$$ (7.69)

  
  

• For electrons, the limit is
  
  $$\mathcal{E} « \left( 1.63 \frac{1}{\mathrm{eV^{\frac{1}{2}} \cdot e \cdot nm}} \right) E^{\frac{3}{2}}$$
  
  
  where $E$ is expressed in eV, which corresponds to 52 V/nm or 52 MV/mm for 10 eV electrons. This is not a difficult constraint to satisfy.

7.4.4 Gaussian Wave Packets

• We considered a classical Gaussian wave pulse in Section 2.2.5.

• In the quantum mechanical context, a traveling pulse represents a traveling ``particle'' with well defined position.

• Our quantum mechanical packet takes the same initial form as the classical one (Eq. 2.55 shown in Fig. 2.4),
  
  $$\Psi(x,0) = Ae^{-\frac{x^2}{2\sigma^2}}$$
  
  
  which we can normalize
  
  $$\vert A\vert^2 \int_{-\infty^\infty} e^{-\frac{x^2}{\sigma^2... ...fty}^\infty e^{-u^2} du = \vert A\vert^2 \sigma \sqrt{\pi} = 1$$
  
  
  to find
  

  $$\Psi(x,0) = \frac{1}{\sqrt{\sigma \sqrt{\pi}}} e^{-\frac{x^2}{2\sigma^2}}$$ (7.70)

  
  

• The Fourier spectrum of a quantum mechanical wavefunction is called its k-space wavefunction or momentum-space wavefunction $\phi(k)$. Following the classical calculation of Section 2.2.5, we have
  

  $$\phi(k) = \sqrt{\frac{\sigma}{\sqrt{\pi}}} e^{-\frac{\sigma^2 k^2}{2}}$$ (7.71)

  
  

• Eq. 7.71 describes a Gaussian k distribution centered on $k=0$. Examples are shown in Fig. 2.5.

• Classical, nondispersive waves all move at the same speed, independent of wavenumber, so our classical Gaussian packet could only travel at a single, well-defined speed. However, an average $k$ of zero means zero average momentum in the quantum mechanical context. We would like a traveling packet, so we will shift our wavenumber distribution so that it is centered on a nonzero average wavenumber $\bar{k}$,
  

  $$\phi(k) = \sqrt{\frac{\sigma}{\sqrt{\pi}}} e^{-\frac{\sigma^2 (k-\bar{k})^2}{2}}$$ (7.72)

  
  

• We are now prepared to find the full time dependent wavefunction via the inverse Fourier transformation^7.6
  

  $$\Psi(x,t) = \sqrt{\frac{\sigma}{\sqrt{\pi}}} \frac{1}{\sqrt{... ...)^2}{2} + i\left(kx - \frac{\hbar k^2}{2m} t \right)\right] dk$$ (7.73)

  
  

• It simplifies the mathematics (but only a little) to center the integral on $\bar{k}$ with the substitution $u = k - \bar{k}$, yielding
  

  $$\Psi(x,t) = \sqrt{\frac{\sigma}{\sqrt{\pi}}} \frac{1}{\sqrt{... ...ar{k}) x - \frac{\hbar(u+\bar{k})^2}{2m} t \right)\right] du$$ (7.74)

  
  

• Expanding the binomials and removing constant terms from the integral, we have
  

  $$\Psi(x,t) = \sqrt{\frac{\sigma}{\sqrt{\pi}}} \frac{1}{\sqrt{2 \pi}} e^{i(\bar{k}x - \bar{\omega} t)} I$$ (7.75)

  
  
  with
  

  $$I = \int_{-\infty}^\infty \exp \left[ -\left(\frac{\sigma^2}... ...u^2 +i u \left(x - \frac{\hbar \bar{k}}{m} t \right)\right] du$$ (7.76)

  
  
  where we have defined the average frequency
  

  $$\bar{\omega} = \frac{\hbar \bar{k}^2}{2m}$$ (7.77)

  
  

• To simplify completing the square in the exponent in $I$, we will further define
  

  $$W(t) = \sqrt{ \sigma^2 + \frac{i\hbar}{m} t }$$ (7.78)

  
  
  and
  

  $$X(t) = x - \frac{\hbar \bar{k}}{m} t$$ (7.79)

  
  
  giving
  

  $$I = \int_{-\infty}^\infty e^{-\left( \frac{W^2 u^2}{2} - i u X \right)} du$$ (7.80)

  
  

• Completing the square in the exponent, we find
  

  $$-\left( \frac{W^2 u^2}{2} - i u X \right)^2 = -\left( \frac... ...sqrt{2}} - \frac{i X}{\sqrt{2}W} \right)^2 - \frac{X^2}{2W^2}$$ (7.81)

  
  

• With the substitution
  
  $$v = \frac{W u}{\sqrt{2}} - \frac{i X}{\sqrt{2}W}$$
  
  
  we can write the integral
  
  $$I = \frac{\sqrt{2}}{W} e^{-\frac{X^2}{2W^2}} \int_{-\infty}^... ...y e^{-v^2} dv = \frac{\sqrt{2\pi}}{W} e^{-\frac{X^2}{2W^2}}$$
  
  

• The position space wavefunction is therefore
  

  $$\Psi(x,t) = \sqrt{\frac{\sigma}{\sqrt{\pi}}} \frac{1}{W} ... ...ft( -\frac{X^2}{2W^2} + i(\bar{k}x - \bar{\omega} t) \right)}$$ (7.82)

  
  

• Eq. 7.82 is a product of a plane wave with wavenumber $\bar{k}$ and a Gaussian envelope with time-dependent, complex width and amplitude. The peak of the Gaussian envelope follows the trajectory of the root of $X(t)$,
  

  $$x = \frac{\hbar \bar{k}}{m} t$$ (7.83)

  
  
  which moves to the right with the classical velocity of a particle of mass $m$ and momentum $\hbar \bar{k}$.

• The probability density
  

  $$\varrho(x,t) = \frac{\sigma}{\sqrt{\pi}} \frac{1}{\vert W\v... ...\frac{X^2}{2} \left( \frac{1}{W^2} + \frac{1}{W^{2*}} \right)}$$ (7.84)

  
  
  requires some manipulation. Noting that
  $$\begin{eqnarray*} \frac{1}{W^2} + \frac{1}{W^{2*}} &=& \frac{W^{2*} + W^2}{\vert... ...^2} t^2} = \frac{2}{\sigma^2 + \frac{\hbar^2}{\sigma^2 m^2} t^2} \end{eqnarray*}$$
  
  
  we find
  

  $$\varrho(x,t) = \frac{\sigma}{\sqrt{\pi}} \frac{1}{\sqrt{\s... ...ight)^2} {\sigma^2 + \frac{\hbar^2}{\sigma^2 m^2} t^2} \right]$$ (7.85)

  
  

  Figure 7.2: The probability density of a Gaussian wave packet describing a 10 eV electron with $\sigma = 1.0$ nm at $t = 0,5,10,15,20$ fs ( $v_g = \frac{\hbar \bar{k}}{m} = 1.9$ nm/fs).

  

• Finally, it is convenient to write the probability in terms of its time dependent width
  

  $$\Sigma(t) = \sigma \sqrt{1 + \frac{\hbar^2}{\sigma^4 m^2} t^2}$$ (7.86)

  
  
  as
  

  $$\varrho(x,t) = \frac{1}{\sqrt{\pi} \Sigma} \exp\left[ -\fr... ...ft( x - \frac{\hbar \bar{k}}{m} t\right)^2} {\Sigma^2} \right]$$ (7.87)

  
  
  Fig. 7.2 shows the time evolution of a 10 eV electron with $\sigma = 1.0$ nm.

Copyright © 2003-2008, Lewis A. Riley

Updated Wed Jan 18 09:51:28 2006