Calculate the probability that the electron is in the range

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SUMMARY

The discussion focuses on calculating the probability of finding an electron in the range (0, L/2) using the wave function defined as ψ(x,t) = (1/sqrt(2))[u2(x)e^(-iE2t/ħ) + u3(x)e^(-iE3t/ħ)]. The probability density function (PDF) is derived as |ψ(x,t)|^2 = |u2(x)|^2 + |u3(x)|^2 + 2Re[u2(x)u3*(x)e^(-i(E3-E2)t/ħ)]. The integration of this PDF over the specified range yields a time-dependent probability that oscillates based on the energy difference between states n=2 and n=3. The period of oscillation can be determined from the cosine term in the probability expression.

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harpua
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Was just wondering if anyone could help me w/ this question.

Consider the wave function
(psi (x,t)) = (1/sqrt(2))[u2(x)exp(-iE2t/h(bar) + u3(x)exp(-iE3t/h(bar)}. calculate the probability that the electron is in the range (0,L/2) as as function of time. What is the period of oscillation of the probability? NOTE: the wavefuctions u2(x) and u3(x) refer to the n=2 and n=3 states of the infinite well located at 0<x<L.

any help would be appreciated.
 
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I'm guessing that [tex]u_3(x)[/tex] is supposed to have a coefficient of [tex]\frac{1}{\sqrt{2}}[/tex] (so that the wavefunction is normalised).

What you have to do is do calculate the probability integral for 0 < L < L/2. That is, do the integral from 0 < x < L/2 of [tex]\Psi^{*}\Psi[/tex]. You'll notice that the probability density [tex]\Psi^{*}\Psi[/tex] is no longer time-independent! It will wobble between being the u2 probability density and being the u3 probability density - figuring out the period of oscillation shouldn't be too hard once you have the density function sorted out.

Cheerio!

Kane
 
Last edited:


Sure, I'd be happy to help with this question. To calculate the probability that the electron is in the range (0,L/2), we can use the probability density function (PDF) given by:

|psi(x,t)|^2 = |u2(x)|^2 + |u3(x)|^2 + 2Re[u2(x)u3*(x)e^(-i(E3-E2)t/h(bar)]

where u2(x) and u3(x) are the wave functions for the n=2 and n=3 states of the infinite well, respectively. The term in the exponential represents the oscillation of the probability with time.

To find the probability in the range (0,L/2), we need to integrate the PDF over this range, which gives us:

P = integral from 0 to L/2 of |psi(x,t)|^2 dx

= integral from 0 to L/2 of [|u2(x)|^2 + |u3(x)|^2 + 2Re[u2(x)u3*(x)e^(-i(E3-E2)t/h(bar)]] dx

= integral from 0 to L/2 of |u2(x)|^2 dx + integral from 0 to L/2 of |u3(x)|^2 dx + integral from 0 to L/2 of 2Re[u2(x)u3*(x)e^(-i(E3-E2)t/h(bar)] dx

Since the wave functions for the n=2 and n=3 states are normalized, their integrals over the range (0,L/2) are equal to 1. This means that the first two terms in the above equation are equal to 1. The third term can be simplified using the identity:

2Re[u2(x)u3*(x)e^(-i(E3-E2)t/h(bar)] = |u2(x)||u3(x)|cos[(E3-E2)t/h(bar)]

Substituting this into the integral and using the fact that the wave functions are real (meaning their complex conjugates are equal to themselves), we get:

P = 1 + 1 + 2 cos[(E3-E2)t/h(bar)] integral from 0 to L/2 of |u2(x)||u3(x)| dx

The integral in the last term is just a constant, so it can be fact
 

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