4.1: Infinite Potential Well
( \newcommand{\kernel}{\mathrm{null}\,}\)
Consider a particle of mass m and energy E moving in the following simple potential: V(x)={0for 0≤x≤a∞otherwise. It follows from Equation ([e5.2]) that if d2ψ/dx2 (and, hence, ψ) is to remain finite then ψ must go to zero in regions where the potential is infinite. Hence, ψ=0 in the regions x≤0 and x≥a. Evidently, the problem is equivalent to that of a particle trapped in a one-dimensional box of length a. The boundary conditions on ψ in the region 0<x<a are ψ(0)=ψ(a)=0. Furthermore, it follows from Equation ([e5.2]) that ψ satisfies d2ψdx2=−k2ψ in this region, where k2=2mEℏ2. Here, we are assuming that E>0. It is easily demonstrated that there are no solutions with E<0 which are capable of satisfying the boundary conditions ([e5.4]).
The solution to Equation ([e5.5]), subject to the boundary conditions ([e5.4]), is ψn(x)=Ansin(knx), where the An are arbitrary (real) constants, and
kn=nπa, for n=1,2,3,⋯. Now, it can be seen from Equations ([e5.6]) and ([e5.8]) that the energy E is only allowed to take certain discrete values: that is, En=n2π2ℏ22ma2.
In other words, the eigenvalues of the energy operator are discrete. This is a general feature of bounded solutions: that is, solutions for which |ψ|→0 as |x|→∞. According to the discussion in Section [sstat], we expect the stationary eigenfunctions ψn(x) to satisfy the orthonormality constraint
∫a0ψn(x)ψm(x)dx=δnm. It is easily demonstrated that this is the case, provided An=√2/a. Hence,
ψn(x)=√2asin(nπxa) for n=1,2,3,⋯.
Finally, again from Section [sstat], the general time-dependent solution can be written as a linear superposition of stationary solutions:
ψ(x,t)=∑n=0,∞cnψn(x)e−iEnt/ℏ, where
Contributors and Attributions
Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)