Wednesday, April 16, 2008

Solving the Schrödinger Equation

To understand simple quantum systems and solving the appropriate Schrödinger equation, we move through increasingly complicated systems. Whenever such an equation is solved, we typically acquire the wavefunction ψ and expressions for the energy E.

system I
A free particle in a potential of V=0. The general solution to the SE can be expressed two ways, both of which are commonly used: ψ = A'sin(kx) + B'cos(kx) or ψ = Aeikx+Be-ikx (the latter used when we care about which direction the particle is moving, the former when we want mathematical nicety). The wavevector k is equal to √2mE/h_bar.

system II
A particle in a potential of V=Vo. The general solutions are the same as in system I, except here k=√2mT/h_bar, where T is the kinetic energy. Since k = 2π/λ, we can see that the wavelength decreases as T increases.

system III
A particle confined to an infinite one-dimensional square well (V=0 inside). Here the wavefunctions are ψn=√2/L sin(nπx/L) and the energies are En=n2h2/8mL2 where n = 1, 2, ...

Three nonclassical results arise: (a) quantization of energy, which arises from putting constraints on the wavefunction (requiring it to go to zero at x=0 and x=L), (b) appearance of nodes, which limit positions at which a particle can exist in a box, and (c) nonzero ground-state (also known as the zero point energy). This "particle in a box" problem has been used to model, among other things, electrons in conjugated molecules and electrons in wires. Moreover it is one of the easier quantum systems to solve that simply demonstrates the important concepts of quantization of energy, nodes, normalization. and the correspondence principle.

system IV
A particle confined to a two-dimensional infinite box (V=0, ∞ outside). Using the method of separation of variables, we assume that ψ=X(x)Y(y), put it back into the SE and, while crossing our fingers, hope that it will crack into two equations. Fortunately it does just that, giving multiplicative wavefunctions ψ(x,y) = 2/√LxLy sin(nxπx/Lx)sin(nyπy/Ly) and additive energies E=h2/8m(nx2/Lx2 + ny2/Ly2). The solution clearly gives two quantum numbers (arising from two dimensions/coordinates) and the possibility of degeneracy arises, where two or more distinct wavefunctions have the same energy. The extension to three-dimensional boxes (or higher) should be straightforward to write down without solving the Schrödinger Equation from scratch.

system V
The barrier problem, in which a particle experiences a discontinuity in potential; for example, going from V=0 to V=Vo at x=0. Fortunately we know these solutions from systems I and II and can write them down immediately: ψI = Aeikx+Be-ikx and ψII = Ceikx+De-ikx. We note, however, since there is nothing to reflect the particle back once it passes the barrier, that D=0.

To make these wavefunctions plausible we must "glue" them together; in other words, we must make them connect [ψI(x=0)=ψII(x=0)] and connect smoothly [dψI/dx(x=0)=dψII/dx(x=0)].

When we calculate the transmission probability T=C*C/A*A, we find that it can never equal unity (meaning that there will always be some amount of reflection from a barrier. Also, when the particle energy E is less than the potential Vo we find that the particle can exist in the barrier to a finite, though exponentially decaying degree. Such penetration into the classical forbidden area is called tunneling, a very important phenomenon in quantum chemistry and the subject of the next lecture.


Aliya said...

yay! thank you for continuing the blog :)

Anonymous said...

can i just say that p chem III blows my mind