Quantum Mechanics (potential well)

[roam]

Homework Statement

I need help mostly with part (c) of the following problem:

http://img402.imageshack.us/img402/6376/question5.jpg

Homework Equations

Schrödinger equations

Correspondence principle

The Attempt at a Solution

(a)

$\int^{0.2}_0 |\psi|^2 \ dx = \int^{0.2}_0 \frac{2}{L}\sin^2 \left( \frac{\pi x}{L} \right) \ dx$

$= \int^{0.2}_0 \frac{2}{2L} \left( 1-\frac{cos 2 \pi x}{L} \right) \ dx$

$= \frac{1}{L} |x- \frac{L}{2 \pi} \sin \frac{2 \pi x}{L}|^{0.2}_0$

$=0.15084$

(b) For the 99^th excited energy state

$\psi_{100} = \sqrt{\frac{2}{L}} \sin \left( \frac{100 \pi x}{L} \right)$

$= \int^{0.2}_0 \frac{2}{L} \sin^2 \left( \frac{100 \pi x}{L} \right) dx$

$= \frac{1}{L} |x-\frac{L}{100 \pi} \times \sin \left( \frac{200\pi x}{L} \right)|^{0.2}_0$

$=1.98346 \times 10^{-10}$

Hopefully my working so far is correct. And we can repeat the process in (a) and (b) for when the electron is within 0.100 nm of the centre of the well (x= 0.4).

(c) So, comparing my results from there is a greater probability of finding the electron within 0.2 nm of the left wall when it is in its ground state, than when it is in the 99th excited state. And the probability of finding a particle at a position is always less than one. So how does this illustrate the "correspondence principle"? I looked it up on wikipedia but I still don't know how to explain this. Also For a "classical electron" how exactly would the probabilities would be different?

Any help would be greatly appreciated.

 

[TSny]

$\int^{0.2}_0 |\psi|^2 \ dx = \int^{0.2}_0 \frac{2}{L}\sin^2 \left( \frac{\pi x}{L} \right) \ dx$

$= \int^{0.2}_0 \frac{2}{2L} \left( 1-\frac{cos 2 \pi x}{L} \right) \ dx$

$= \frac{1}{L} |x- \frac{L}{2 \pi} \sin \frac{2 \pi x}{L}|^{0.2}_0$

$=0.15084$

Looks good except for the final numerical value. I get a somewhat larger value.

(b) For the 99^th excited energy state

$\psi_{100} = \sqrt{\frac{2}{L}} \sin \left( \frac{100 \pi x}{L} \right)$

$= \int^{0.2}_0 \frac{2}{L} \sin^2 \left( \frac{100 \pi x}{L} \right) dx$

$= \frac{1}{L} |x-\frac{L}{100 \pi} \times \sin \left( \frac{200\pi x}{L} \right)|^{0.2}_0$

$=1.98346 \times 10^{-10}$

Check the factor of 100 in the denominator of the second term in the last expression. Also, the numerical value is way off. You can get a good estimate for this case by noting that the large denominator in the second term makes that term very small compared to the first term. So, if you dropped the second term, the probability would just be x/L|$^{.2}_{0}$

When thinking about the classical case, note that a classical particle would move at constant speed. So, if you divided the box into two equal halves, the particle would spend the same amount of time in each half. So, the probability of finding the particle in the left half would be 1/2. Likewise, if you divided the box into 5 equal parts, the particle would spend the same time in each part. So, the probability of finding the particle in anyone of those parts would be 1/5.

 

[roam]

Looks good except for the final numerical value. I get a somewhat larger value.

Thank you. Oops, here's what I've got now:

$\frac{1}{0.6 \times 10^{-9}} ((0.2 \times 10^{-9}) - \frac{0.6 \times 10^{-9}}{2 \pi} \ sin \left( \frac{2 \pi (0.2 \times 10^{-9}))}{0.6 \times 10^{-9}} \right)) = 0.1955025$

Check the factor of 100 in the denominator of the second term in the last expression. Also, the numerical value is way off. You can get a good estimate for this case by noting that the large denominator in the second term makes that term very small compared to the first term. So, if you dropped the second term, the probability would just be x/L|$^{.2}_{0}$

Sorry. It should have been

$= \frac{1}{L} \left( x- \frac{L}{200 \pi} \ sin \left( \frac{200 \pi x}{L} \right) \right)$

$= \frac{1}{0.6 \times 10^{-9}} \left((0.2 \times 10^{-9})- \frac{0.6 \times 10^{-9}}{200 \pi} \ sin \left( \frac{200 \pi (0.2 \times 10^{-9})}{0.6 \times 10^{-9}} \right) \right) = 0.331955$

Now my results are different, when in the 99th excited state there is a higher probability of finding the electron there.

When thinking about the classical case, note that a classical particle would move at constant speed. So, if you divided the box into two equal halves, the particle would spend the same amount of time in each half. So, the probability of finding the particle in the left half would be 1/2. Likewise, if you divided the box into 5 equal parts, the particle would spend the same time in each part. So, the probability of finding the particle in anyone of those parts would be 1/5.

I see. Thank you for clarifying the way the classical electron would behave. But how does this, and the comparison of the results in part (a) and (b) illustrate the correspondence principle? I think the correspondence principle is supposed to relate how quantum mechanics is in agreement with classical mechanics on the macroscale but I am not sure how to relate it to the situation in this problem.

 

[roam]

Also, in part (a) it asks to find the probability 'within 0.100 nm of the centre of the well (i.e., in the middle third of the well)'. However when I change the limit of integration I get a negative probability:

$\psi_1 = \frac{1}{L} |x- \frac{L}{2 \pi} \sin \frac{2 \pi x}{L}|^{0.4}_0$

$\frac{1}{0.6 \times 10^{-9}} ((0.4 \times 10^{-9}) - \frac{0.6 \times 10^{-9}}{2 \pi} \ sin \left( \frac{2 \pi (0.4 \times 10^{-9}))}{0.6 \times 10^{-9}} \right)) = -0.4395$

So, what does a negative probability mean?

 

[voko]

Also, in part (a) it asks to find the probability 'within 0.100 nm of the centre of the well (i.e., in the middle third of the well)'. However when I change the limit of integration I get a negative probability:

$\psi_1 = \frac{1}{L} |x- \frac{L}{2 \pi} \sin \frac{2 \pi x}{L}|^{0.4}_0$

What should the lower integration limit really be here?

$\frac{1}{0.6 \times 10^{-9}} ((0.4 \times 10^{-9}) - \frac{0.6 \times 10^{-9}}{2 \pi} \ sin \left( \frac{2 \pi (0.4 \times 10^{-9}))}{0.6 \times 10^{-9}} \right)) = -0.4395$

So, what does a negative probability mean?

$\sin \left( \frac{2 \pi (0.4 \times 10^{-9})}{0.6 \times 10^{-9}} \right) = \sin \frac{4 \pi}{3} < 0$

So the entire expression cannot be negative.

As for the correspondence principle, observe that the probability expression has two terms: one linear and another oscillatory. Observe what happens with the coefficient at the oscillatory term as you go to the higher states. What does the entire expression tend to?

 

[TSny]

I see. Thank you for clarifying the way the classical electron would behave. But how does this, and the comparison of the results in part (a) and (b) illustrate the correspondence principle? I think the correspondence principle is supposed to relate how quantum mechanics is in agreement with classical mechanics on the macroscale but I am not sure how to relate it to the situation in this problem.

For a problem like this, the correspondence principle implies that the quantum system should behave like a classical system as the quantum number n gets large. For n = 1 you found that quantum mechanics predicts that the probability of finding the particle between x = 0 and x = 0.2 is 0.196 whereas for n = 100 the probability is 0.332. What would be the probability for a classical system?

 

[roam]

For a problem like this, the correspondence principle implies that the quantum system should behave like a classical system as the quantum number n gets large. For n = 1 you found that quantum mechanics predicts that the probability of finding the particle between x = 0 and x = 0.2 is 0.196 whereas for n = 100 the probability is 0.332. What would be the probability for a classical system?

So for the classical system if we divide the well into 6 equal parts then the probability of finding the particle in anyone of those parts 1/6=0.166. But as the quantum number gets larger, why does the probabillity not converge to this number?

0.332 is a lot larger than 0.166. I divided the well into 6 parts because we are considering the probability in 0.1 from the left-hand wall and the well is 0.6 nm in width.

 

[TSny]

So for the classical system if we divide the well into 6 equal parts,

Rethink that 6 The distance from x = 0 to x = .2 is what fraction of the total width of the well? Remember, you are calculating the probability of the particle being anywhere in that interval.

 

[roam]

Rethink that 6 The distance from x = 0 to x = .2 is what fraction of the total width of the well? Remember, you are calculating the probability of the particle being anywhere in that interval.

Thank you very much for your help. I understand it now. Thanks! :)