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Particle in an infinite potential box- expected values of energy

  1. Jun 2, 2013 #1
    1. The problem statement, all variables and given/known data
    I think this is a very easy problem, I will try to show you guys what I tried to come up with:
    A particle is in an infinite potential box and is described in a certain moment of the normalized wavefunction ##\psi(x)=\sqrt{\frac{8}{3a}}sin^2(\frac{\pi x}{a})## for (0<x<a) and 0 otherwise.
    a) Calculate the expected value of the energy. Compare with the ordinary (stationary) energy levels.
    b) What is the probability of finding the particle in the ground state and the first excited state?
    c) Determine <p> and Δp by leveraging development coefficients (no integrals to be solved).


    2. Relevant equations
    ##ΔxΔp≥\frac{\hbar}{2}##
    ##Δx=\frac{a}{2}##
    ##<E_kin>=\frac{(Δp_x)^2}{2m}##

    3. The attempt at a solution
    Ok, so if I understood everything correctly, estimating the value goes like this:
    ##p_x=\frac{\hbar}{2} \frac{2}{a}=\frac{\hbar}{a}##
    ##<E_kin>=\frac{(Δp_x)^2}{2m}=\frac{\hbar^2}{2ma^2}##
    and comparison with first state: ##\frac{\pi^2 \hbar^2}{2ma^2}>\frac{\hbar^2}{2ma^2}##
    as for the probability of those levels- I thought I am supposed to calculate the integral
    ##\int_{-\frac{a}{2}}^{+\frac{a}{2}}\psi^*_0 (x)\psi(x)\,dx##, but now I don't know. What do you guys think?
     
  2. jcsd
  3. Jun 2, 2013 #2

    TSny

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    Hello.

    What does ##p_x## represent and how did you get a value for it? You want to find the expectation value of the energy: ##<E> = \frac{<p^2>}{2m}##, so you need ##<p^2>##.

    Check the limits on your integral in the last line of your post.

    [Edit: Also, I don't think ##\Delta x = a/2##]
     
  4. Jun 2, 2013 #3
    ##p_x## represents the momentum of the particle, and I got it from solving the uncertainty equation ##ΔxΔp=\frac{\hbar}{2}##, where ##Δx## comes from the equation i found in a book for problems with infinite potential well- it was supposed to be width of the well divided by 2. Also, I checked the formula for the expectation value of the energy, but in the book it is ##<E>=\frac{(Δp)^2}{2m}##. I am not sure about the correctness of the ground level probability integral, or the limits of it.
     
  5. Jun 2, 2013 #4
    Remember you are given limits 0<x<a. A quick read through of chapter 2.2 of Griffiths Introduction to Quantum Mechanics will help enormously with this problem. To solve part (b) you first calculate [itex] c_{n}=∫ψ_{n}(x)^{*}ψ(x)dx [/itex]. The probability that a measurement of the energy would yield the value [itex] E_{n} [/itex] is [itex] |c_{n}|^2 [/itex]. Also, listen to TSny to solve part (a), the equation you've given is, although true in this case because <p>=0, in general not true.
     
  6. Jun 2, 2013 #5

    TSny

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    The momentum of the particle does not have a definite value, so you cannot write down a value for it.

    Here, Δp represents the uncertainty in momentum, not the momentum itself. Δp is usually defined as

    ##\Delta p = \sqrt{<\hat{p}^2> - <\hat{p}>^2}## where ##\hat{p}## is the momentum operator.

    Like Δp, Δx is the uncertainty in x and is defined by ##\Delta x = \sqrt{<x^2> - <x>^2}## and generally does not equal a/2. When you looked up Δx in a book, for what quantum state was the book considering? Maybe what you found was a value for <x> for one of the energy eigenstates. Anyway, you will not need Δx or the uncertainty principle for this problem.

    The generally correct expression for the infinite well is ##<E>=\frac{<\hat{p}^2>}{2m}##. If ##<\hat{p}^2>## happens to equal ##(Δp)^2##, then what you wrote is ok. But you would need to show ##<\hat{p}^2> = (Δp)^2##

    The expression for the integral is correct except for the limits. Note that the potential well extends from x = 0 to x = a. The integral gives the probability amplitude, not the probability.
     
    Last edited: Jun 2, 2013
  7. Jun 3, 2013 #6
  8. Jun 3, 2013 #7
  9. Jun 3, 2013 #8
    Remember ΔxΔp ≥ h-bar/2, it doesn't have to be equal - and usually is not!
     
  10. Jun 3, 2013 #9

    TSny

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    Your integration looks correct. From that result you can get <E>.

    Why do you want to use Heisenberg's uncertainty relation which deals with uncertainties (Δx and Δp) rather than expectation values?

    What should be equal to ##\frac{a}{\hbar}##? This quantity has dimensions of the inverse of momentum.
     
  11. Jun 3, 2013 #10
    The result I obtained for energy is weird, I am not sure if it is correct: ##\frac{2\pi^2\hbar^2}{3a^2 m}##. Not sure if it supposed to be like this...
     
  12. Jun 3, 2013 #11
    Looks perfectly reasonable. Why do you think its 'weird'?

    Edit: The source of confusion may be that the wavefunction you are considering is not an energy eigenstate of the ISW.
     
    Last edited: Jun 3, 2013
  13. Jun 3, 2013 #12
    I was just expecting something with ##2ma^2## in the denominator and natural number in the nominator, like in the formula ##E_n=\frac{n^2 \pi^2 \hbar^2}{2ma^2}##
     
  14. Jun 3, 2013 #13
    but why can't I obtain ##a## free value for probability of the energies?
     
  15. Jun 3, 2013 #14
    Read the edit to my last post. This form is true for energy eigenstates of the infinite square well [itex] ψ(x)=√\frac{2}{L}sin(k_{n}x) [/itex]. The wavefunction given to you in the problem is not of this form.
     
  16. Jun 3, 2013 #15

    TSny

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    I believe that's the correct answer. As tannerbk noted, you should not expect the answer to correspond to one of the energy eigenstates because your wavefunction is not an energy eigenstate.

    But you can see that your answer is just 4/3 times the ground state energy.
     
  17. Jun 3, 2013 #16
    ok, but I still have the problem with the probabilities- what do I do wrong?
     
  18. Jun 3, 2013 #17

    TSny

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  19. Jun 3, 2013 #18
    What have you done? Read my first post (post 4). Calculate [itex] c_{n} [/itex] for the ground state and first excited state. My notation was bad in that post, remember you must use both the wavefunction given to you in the problem and the energy eigenstates.
     
  20. Jun 3, 2013 #19
    Those stupid typos... thanks for the notice:) Is it okay that for the ground state the probability is equal to 1 and for the first excited state it is equal to 4/9?
     
  21. Jun 3, 2013 #20
    No, the sum over all the individual probablities should equal 1. Show your work and we can help.
     
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