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Length of an infinite square well? 
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#1
Oct908, 07:50 PM

P: 86

Actually, this is more of a general question relating to a homework problem I already did. I was given the initial wavefunction of a particle in an infinite square well:
[tex] \Psi(x,0) = Ax[/tex] if [tex](0 \leq x \leq \frac{a}{2})[/tex], and [tex] =A(ax)[/tex] if [tex](\frac{a}{2} \leq x \leq a) [/tex] And of course [tex]\Psi(0,0) = \Psi(a,0) = 0[/tex] I was asked to find [tex]\Psi(x,t)[/tex] , which I did, and I was also asked to find the probability that "a measurement of the energy would yield the value [tex] E_1[/tex]", the ground state energy, which I also did. However, this probability is dependent on the length of the well, given by 'a'. I was curious about this, and I found that the probability that "a measurement of the energy would yield the value [tex] E_1[/tex]" increases as the value of 'a' increases. Taking the limit as 'a' approaches infinity gives a probability of finding the energy in the ground state to be approx. 0.986, which means that there is a nonzero probability of finding the particle in another energy level. OK, as far as I can tell for a given wavefunction, increasing the length of the infinite square well increases the probability of finding the particle in the lowest energy state. Mathematically I understand this, but I am still lacking physical intuition about what is actually happening when the length of the well increases. So, for a given wavefunction, WHY does increasing the length of the infinite square well increases the probability of finding the particle in the lowest energy state? Thanks, I hope this question makes sense! *melinda* 


#2
Oct908, 09:42 PM

P: 369

Are you sure? Did you normalize the wavefunction properly?
The probability is a dimensionless number: but a is a length, and it is the only length in the problem here: there is no way to make a dimensionless number that depends on a. So I'd think, without actually calculating anything, that the probability to find E_1 should be independent of a. 


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