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Physics LibreTexts

4.1: Infinite Potential Well

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Consider a particle of mass m and energy E moving in the following simple potential: V(x)={0for 0xaotherwise. 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 x0 and xa. 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=2mE2. 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=n2π222ma2.

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)eiEnt/, where

cn=a0ψn(x)ψ(x,0)dx.

Contributors and Attributions

  • Richard Fitzpatrick (Professor of Physics, The University of Texas at Austin)


This page titled 4.1: Infinite Potential Well is shared under a not declared license and was authored, remixed, and/or curated by Richard Fitzpatrick.

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