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Let's say we have a particle in an infinite well, and let's also say it is in the ground state. Now we make the well bigger by very quickly moving one of the boundaries of the well.

How do we compute the probability of the particle being in the ground state of the new well? Also, what is the prob of the particle being in the first excited state of the new well?

Assume for concreteness that the inital (1D)well is from x=0 to x=L and the new well is from x=0 to x=2L. We simply pull the right wall of the well to x=2L.

Alex1994
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1 Answers1

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See. The expression for eigen-energies of an infinitely deep potential well or a box potential is $E_n=\frac{(n^2)(\pi^2)(ℏ^2)}{2mL^2}$

Ground state energy eigen value = $E_1=\frac{(\pi^2)(ℏ^2)}{2mL^2}$

Clearly the energy eigen values are inversely proportional to the width of well. Now if you are increasing the width size from L to $2L$ that makes the eigenenergy equation

$E_n=\frac{(n^2)(\pi^2)(ℏ^2)}{2m(2L)^2}$

For the first excited state of new wall $\implies$ $E_1=\frac{(\pi^2)(ℏ^2)}{8mL^2}$

How do we compute the probability of the particle being in the ground state of the new well? Also, what is the prob of the particle being in the first excited state of the new well?

To answer that we need to look at the wave function of the particle at the first excited state of well width L that is $\psi_1(x)=\sqrt(2/a) \sin(\pi x/L)$

Now the wave function of the particle at the first excited state of well width 2L becomes

$\psi_1(x)=\sqrt(2/a) \sin(\pi x/2L)$

The probability of finding the particle is directly proportional to the squared modulus of the wave function which is Born's rule of probabilistic interpretation. Hope it helped.

ostrichguy
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