PHYS212 : Lectures

Subsections

• 7.2.1 Separation of Variables
• 7.2.2 Plane Waves
• 7.2.3 Aside : Convenient Units
• 7.2.4 Standing Waves : The Infinite Square Well

7.2 The Schrödinger Equation

7.2.1 Separation of Variables

• The Schrödinger equation in its most general form is
  

  $$\left[ -\frac{\hbar^2}{2m} \nabla^2 + V(\vec{r},t) \right] ... ...ec{r},t) = i\hbar \frac{\partial}{\partial t} \Psi(\vec{r},t)$$ (7.8)

  
  
  It is a second order, linear, partial differential equation. $V(\vec{r},t)$ is the potential energy function of the system.

• For one dimensional systems - systems for which $V$ is a function of only one spatial variable, Eq. 7.8 becomes
  

  $$\left[ -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} ... ...ght] \Psi(x,t) = i\hbar \frac{\partial}{\partial t} \Psi(x,t)$$ (7.9)

  
  

• With time independent potentials $V(x)$, we can separate the position and time dependences of the Schrödinger equation. With the product wavefunction
  

  $$\Psi(x)(t) = \psi(x) f(t)$$ (7.10)

  
  
  the differential equation becomes
  
  $$-\frac{\hbar^2}{2m} f(t) \frac{\partial^2}{\partial x^2} \ps... ...\psi(x) f(t) = i\hbar \psi(x) \frac{\partial}{\partial t} f(t)$$
  
  

• Dividing by the product wave function of Eq. 7.10, we obtain
  

  $$-\frac{\hbar^2}{2m} \frac{1}{\psi(x)} \frac{\partial^2}{\par... ... V(x) = i\hbar \frac{1}{f(t)} \frac{\partial}{\partial t} f(t)$$ (7.11)

  
  

• We have separated the independent variables, so we choose a separation constant $C$, and write two equations
  

  $$-\frac{\hbar^2}{2m} \frac{1}{\psi(x)} \frac{d^2}{d x^2} \psi(x) + V(x) = C$$ (7.12)

  
  
  
  

  $$i\hbar \frac{1}{f(t)} \frac{d}{d t} f(t)=C$$ (7.13)

  
  

• Eq. 7.13 can be integrated to give
  
  $$f(t) = e^{-i\frac{C}{\hbar}t}$$
  
  
  which we recognize as the time dependence of a harmonic oscillator with angular frequency
  
  $$\omega = \frac{C}{\hbar}$$
  
  

• With the help of deBroglie, we recognize via
  

  $$E = \hbar \omega$$ (7.14)

  
  
  that the separation constant $C$ must be the energy.

• The wavefunction is now
  

  $$\Psi(x,t) = \psi(x) e^{-i \omega t}$$ (7.15)

  
  
  where $\psi(x)$ depends on the potential energy function $V(x)$ of the system and is found by solving the time independent Schrödinger equation (Eq. 7.12 rearranged)
  

  $$\left[ - \frac{\hbar^2}{2m} \frac{d^2}{d x^2} + V(x) \right] \psi(x) = E \psi(x)$$ (7.16)

  
  

7.2.2 Plane Waves

• A free ``particle'' (remember, all we have are waves here, no particles to be found) subject to no forces. Hence, the potential energy is constant, $V(x,t) = V_o$.

• The time independent Schrödinger equation becomes
  
  $$-\frac{\hbar^2}{2m} \frac{d^2}{d x^2} \psi(x) = (E - V_o) \psi(x)$$
  
  
  or
  

  $$\frac{d^2}{d x^2} \psi(x) = - \frac{2m(E-V_o)}{\hbar^2} \psi(x)$$ (7.17)

  
  

• One might recognize $p^2 = 2m(E-V_o)$ in the constant on the right. Further, the wavenumber $k = \frac{p}{\hbar}$, yielding a simple form of Eq. 7.17
  

  $$\psi''(x) = - k^2 \psi(x)$$ (7.18)

  
  
  which we recognize as giving the spatial wavefunction of a plane wave.

• The general free particle solutions of the Schrödinger equation in one dimension are superpositions of plane waves
  

  $$\Psi_k(x,t) = C e^{i(kx - \omega t)}$$ (7.19)

  
  
  where particles with positive $k$ travel in the $\hat{\imath}$ direction, and particles with negative $k$ travel in the $-\hat{\imath}$ direction.

• The arbitrary constant $C$ of a classical plane wave is determined by initial conditions. In the quantum mechanical context, wave functions must be normalized. However, the plane wave presents the challenge of a uniform probability density over all space
  

  $$\varrho = \Psi_k^* \Psi_k = \vert C\vert^2$$ (7.20)

  
  
  which defies attempts at normalization via Eq. 7.1,
  

  $$\int_{-\infty}^\infty \Psi_k^* \Psi_k dx = \vert C\vert^2 \int_{-\infty}^\infty dx = \infty$$ (7.21)

  
  

• Evidently, plane waves do not represent physically reasonable states of individual particles. However, superpositions of plane waves produce wave packets which are localized, and hence can be normalized. We will also use plane waves to represent continuous beams of particles.

• If we are confined to wave packets in describing actual systems, we have bandwidth limits
  
  $$\Delta k \Delta x = \frac{\Delta p}{\hbar} \approx 2\pi$$
  
  
  and
  
  $$\Delta \omega \Delta t = \frac{\Delta E}{\hbar} \Delta t \approx 2\pi$$
  
  
  or
  

  $$\Delta p \Delta x \approx h$$

  $$\Delta E \Delta t \approx h$$ (7.22)

  
  
  which are called uncertainty principles in the context of quantum mechanics.

7.2.3 Aside : Convenient Units

• We did not focus on specific cases as we studied classical waves. We have considerably less intuition about quantum systems than we do about classical ones, so it may be helpful to be more explicit here.

• Any system has characteristic length and time scales on which significant variations occur and corresponding characteristic energy, momentum, and angular momentum scales.

• For example, consider electrons in atomic/molecular systems with sizes on the order of $10^{-10}-10^{-9}$ m.

• Taking the characteristic length $L=1$ nm as the characteristic wavelength, we have characteristic energy
  $$\begin{eqnarray*} E &=& \frac{p^2}{2m} = \frac{(\hbar k)^2}{2m} = \frac{(\hbar ... ...511\mathrm{e}{6} \mathrm{eV}} \frac{4 \pi^2}{2(1 \mathrm{nm})^2} \end{eqnarray*}$$
  
  
  yielding
  

  $$E_o = 1.5 eV$$ (7.23)

  
  
  so electron-volts are convenient units for atomic and molecular systems.

• The characteristic time follows from the energy
  $$\begin{eqnarray*} \tau_o &=& \frac{2\pi}{\omega_o} = \frac{2\hbar\pi}{E_o}\\ &=... ...6.5821\mathrm{e}{-16} \mathrm{eV \cdot s}) \pi}{1.5 \mathrm{eV}} \end{eqnarray*}$$
  
  
  which gives
  

  $$\tau_o = 2.8 \mathrm{fs}$$ (7.24)

  
  
  so femtoseconds are convenient units of time for these systems.

• You will find convenient units for atomic nuclei for homework.

7.2.4 Standing Waves : The Infinite Square Well

• We can construct quantum mechanical standing waves from plane waves traveling in opposite directions as we did in Section 1.5.3 (See Eq. 1.73),
  

  $$\Psi(x,t) = \Psi_k(x,t) - \Psi_{-k}(x,t)$$

  $$= C \left( e^{ikx} - e^{-ikx} \right) e^{-i\omega t}$$

  $$= 2iC \sin(kx) e^{-i\omega t}$$ (7.25)

  
  

• The position dependence of a standing wave can be written as a real function
  

  $$\psi(x) = C \sin(x)$$ (7.26)

  
  
  where $C$ remains arbitrary for the moment. Remember this when we consider probability current in Section 7.3.

• Taking the superposition of just two plane waves has not saved us from the problem of normalization. However, Eq. 7.25 can describe (approximately) a physically reasonable system - the infinite square well.

• Consider the quantum mechanical analog of a string of length $L$ fixed at both ends. What does it mean that the ends of the wavefunction are ``fixed'' or that $\psi(0) = \psi(L) = 0$?

• We will find that it is possible to find a finite probability density in classically forbidden regions where $E < V(x)$. The only way to ensure that $\psi$ is exactly zero is to make $V(x)$ infinite.

• Hence, the potential energy function
  

  $$V(x) = \left\{ \begin{array}{cc} 0 & (0 < x < L)\\ \infty & (x \leq 0, x \geq L) \end{array}\right.$$ (7.27)

  
  
  called an infinite square well gives the same boundary conditions as a string fixed at both ends.

• With reasoning identical to that for standing waves on a string, or for acoustic waves in a pipe closed at both ends (or open at both ends), we have
  

  $$k_n = \frac{n\pi}{L} \hspace{1 cm} (n=1, 2, 3,...)$$ (7.28)

  
  

• The frequencies of the quantum system differ from those of its classical counterparts due to nonlinear dispersion,
  
  $$\omega_n = \frac{\hbar k_n^2}{2m} = \frac{\hbar \pi^2}{2mL^2} n^2$$
  
  
  which gives
  

  $$\omega_n = \omega_o n^2$$ (7.29)

  
  
  and
  

  $$E_n = \hbar \omega_o n^2$$ (7.30)

  
  
  if we define
  

  $$\omega_o = \frac{\hbar \pi^2}{2mL^2}$$ (7.31)

  
  

• We can now state the solutions of Eq. 7.25 more explicitly,
  
  $$\Psi(x,t) = C \sin \left( \frac{n \pi x}{L} \right) e^{-i \frac{E_n}{\hbar} t}$$
  
  

• Noting that $\Psi(x,t) = 0$ outside of the well, we can normalize the wavefunction
  $$\begin{eqnarray*} 1 &=& \vert C\vert^2 \int_{0}^{L} \sin^2 \left( \frac{n \pi... ...c{n \pi}{2}\\ &\Rightarrow& \vert C\vert = \sqrt{\frac{2}{L}} \end{eqnarray*}$$
  
  
  giving
  

  $$\Psi(x,t) = \sqrt{\frac{2}{L}} \sin \left( \frac{n \pi x}{L} \right) e^{-i \frac{E_n}{\hbar} t}$$ (7.32)

  
  

• The probability density is therefore
  

  $$\varrho_n(x) = \frac{2}{L} \sin^2 \left( \frac{n \pi x}{L} \right)$$ (7.33)

  
  

• Although the wavefunctions $\Psi_n(x,t)$ oscillate in the same way as standing waves on a string, the corresponding probability densities $\varrho_n(x)$ are independent of time. For this reason, the $\Psi_n(x,t)$ are called stationary states. You will find for homework that superpositions of the stationary states are not stationary.

• For an electron confined to an infinite square well of width equal to the diameter of a hydrogen atom, $L=0.11$ nm we obtain
  $$\begin{eqnarray*} E_o &=& \frac{(\hbar c)^2 \pi^2}{2 (mc^2) L^2}\\ &=& \frac{(1... ...m{eV}} \frac{\pi^2}{2(0.11 \mathrm{nm})^2}\\ &=& 31 \mathrm{eV} \end{eqnarray*}$$
  
  
  for the ground state energy.

• Unlike hydrogen, an infinitely deep well has an infinite set of energy states. The first few energies are
  
  $$E = 31, 124, 279, 496 \mathrm{eV}, ...$$
  
  

• It is a mistake to think of these solutions as particles bouncing back and forth between the ``walls'' of the well.
  • Plane waves don't describe particles with well defined position and momentum.
  • These waves are used to calculate probability densities, and the densities they produce are quite different from that of a classical particle confined in one dimension.

Copyright © 2003-2009, Lewis A. Riley

Updated Wed Jan 18 09:51:28 2006