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Infinite square well

  1. Mar 1, 2015 #1

    Maylis

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    1. The problem statement, all variables and given/known data
    upload_2015-3-1_15-43-29.png
    upload_2015-3-1_15-43-11.png

    2. Relevant equations


    3. The attempt at a solution
    (a)
    $$ \int_{0}^{a} \mid \Psi (x,0) \mid^{2} \hspace {0.02 in} dx = 1 $$
    $$ \int_{0}^{a} \mid A[ \psi_{1}(x) + \psi_{2}(x) ] \mid^{2} \hspace {0.02 in} dx = 1 $$
    Since the ##\psi_{1}## and ##\psi_{2}## are orthonormal (I don't know how to call ##\psi(x)##)
    $$ \int \psi_{m}^{*} \psi_{n} \hspace {0.02 in} dx = \delta_{mn} $$
    Where ##\delta_{mn}## is the kronecker delta. I suppose that the probability density will be
    $$ \mid A^{2} \mid (\psi_{1} + \psi_{2})(\psi_{1}^{*} + \psi_{2}^{*}) $$
    $$ A^{2} \int_{0}^{a} (\psi_{1} + \psi_{2})(\psi_{1}^{*} + \psi_{2}^{*}) \hspace {0.02 in} dx = 1 $$
    $$ A^{2} \int_{0}^{a} \psi_{1} \psi_{1}^{*} + \psi_{1} \psi_{2}^{*} + \psi_{2} \psi_{1}^{*} + \psi_{2} \psi_{2}^{*} \hspace {0.02 in} dx = 1 $$
    $$ \int_{0}^{a} (1 + 0 + 0 + 1) \hspace {0.02 in} dx = \frac {1}{A^{2}} $$
    Thus I evaluate ##A = \frac {1}{\sqrt {2a}}##

    (b)
    I know for the infinite square well, the constants ##c_{n}## are found from this equation
    $$ c_{n} = \sqrt {\frac {2}{a}} \int_{0}^{a} sin(\frac {n \pi}{a} x) \Psi(x,0) \hspace {0.02 in} dx $$
    $$ c_{n} = \sqrt {\frac {2}{a}} \int_{0}^{a} sin(\frac {n \pi}{a} x)\frac {1}{\sqrt {2a}}(\psi_{1}(x) + \psi_{2}(x)) \hspace {0.02 in} dx $$
    $$c_{n} = \frac {1}{a} \int_{0}^{a} sin(\frac {n \pi}{a} x)(\psi_{1}(x) + \psi_{2}(x)) \hspace {0.02 in} dx $$
    But since I don't know what ##\psi_{1}## and ##\psi_{2}## are as functions of ##x##, I can't integrate.
     

    Attached Files:

  2. jcsd
  3. Mar 1, 2015 #2
    Well, the first line of the question states that ##\psi_{1}## and ##\psi_{2}## are the first two stationary states of the infinite square well - so, the ground state and first excited state, respectively.
     
  4. Mar 1, 2015 #3

    CAF123

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    Hi Maylis,
    You made a slight error here in evaluating the normalisation constant. Two ways to see it is incorrect is to note that you have it expressed in terms of a dimensionful quantity and that conservation of probability ##\sum_{i=1}^2 |c_i|^2 = 1## is violated where ##\Psi(x,0) = c_1 \psi_1(x) + c_2\psi_2(x)##.
     
  5. Mar 1, 2015 #4

    Maylis

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    Then that means I am probably wrong about my definition of ##\mid \Psi (x,0) \mid^{2}##. What should it be?
     
  6. Mar 1, 2015 #5

    DrClaude

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    Your error is that the second line here doesn't follow from the first.
     
  7. Mar 1, 2015 #6
    The real error you made in the normalization resides in the use you made of the orthonormality relations of the eigenfunctions: it is not the simple product of the two eigenfunctions that must result 1 or 0.
     
  8. Mar 1, 2015 #7

    BvU

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    In a): If ##\psi_1## is normalized, that means that ## \int_{0}^{a} \psi_{1} \psi_{1}^{*} \; dx = 1##

    In d): Funny book, Griffiths. Quoting Peter Lorre means being from way long ago. But it says 1995.

    In b) you in fact embark on re-calculating the two ##{1\over 2}\sqrt 2## that you know already from a) !

    The ##\sqrt{2\over a} \sin \left ( {n\pi \over a }\;x\right ) ## you multiply with is ##\psi_n## ! (p. 26, (2.24) ).​

    That's not the idea! The idea for b) is that you know how the ##\psi_n## develop in time, and therefore you also know how a linear combination develops !
     
  9. Mar 2, 2015 #8

    Maylis

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    But wait, Isn't that the point in saying that two eigenfunctions are orthonormal? Their product is equal to the kronecker delta, which is either 0 or 1 depending on the indices of the eigenfunctions.
     
  10. Mar 2, 2015 #9

    DrClaude

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    No. The fact that
    $$
    \int_0^a \psi_m^* \psi_n dx = \delta_{mn}
    $$
    does not imply that
    $$
    \psi_m^* \psi_n = \delta_{mn}
    $$
    As an example, take
    $$
    \int_0^{2\pi} \cos (x) \cos (2x) dx = 0
    $$
    Obviously, ##\cos (x) \cos (2x) \neq 0##.
     
  11. Mar 2, 2015 #10

    BvU

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    I'm waiting ...

    Would you have known what to answer in (b) if they had given you ##\Psi(x, 0) = \psi_1(x)## ?
     
  12. Mar 3, 2015 #11

    Maylis

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    @DrClaude
    Okay, I see my error, thanks

    @BvU
    I was just using this example problem as a basis for learning how to solve ##\Psi (x,t)##. I just try to copy their technique.

    And to answer you question, I would do the exact same thing as the example problem did no matter what they told me ##\Psi (x,0)## is equal to.

    First they find A, then they start doing stuff with ##c_{n}##, so I tried to copy that
    upload_2015-3-3_20-56-54.png
    upload_2015-3-3_20-57-35.png upload_2015-3-3_20-58-3.png
    upload_2015-3-3_20-58-18.png
     

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    Last edited: Mar 3, 2015
  13. Mar 3, 2015 #12

    DrClaude

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    The thing is that finding the ##c_{n}##'s is trivial in the case of problem 2.5.
     
  14. Mar 3, 2015 #13

    Maylis

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    I need some sort of methodical way of solving these problems. If every problem has a different way of finding ##\Psi (x,t)##, I will have no clue what to do. So after solving for A, what is the next step? I try and do what is reasonable and use example problems as springboards for solving problems in the textbook.
     
  15. Mar 3, 2015 #14

    DrClaude

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    You need to understand what the different steps are, and why they are used. The time-dependent wave function is given by eq. [2.17]. To solve it for a particular problem, you need to know the coefficients ##c_n##. In example 2.2, where you have the functional form of ##\Psi(x,0)##, you find the coefficients ##c_n## using eq. [2.34]. In problem 2.5, you do not need to carry out the integral of eq. [2.34]: you should be able to make a link between the given ##\Psi(x,0)## and eq. [2.16].
     
  16. Mar 3, 2015 #15

    Maylis

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    Okay, finally now I am beginning to be able to separate the wheat from the chaff.

    So then ##c_{1}## and ##c_{2}## are both ##\frac {1}{\sqrt{2}}## by virtue of the equation
    $$\Psi(x,0) = \sum_{n=1}^{\infty} c_{n} \psi_{n}(x) $$
    And by inspection of the given wave function at ##t=0##

    Then I use the equation
    $$ \Psi(x,t) = \sum_{n=1}^{\infty} c_{n} \sqrt {\frac{2}{a}} sin \Big ( \frac {n \pi}{a} x \Big )e^{\frac {-i(n^{2} \pi^{2} \hbar) t}{2ma^{2}}} $$
    Which gives me
    $$ \Psi(x,t) = \frac {1}{\sqrt{a}} \Big [sin \Big ( \frac {\pi}{a} x \Big )e^{-i \omega t} + sin \Big (\frac {2 \pi}{a} x \Big )e^{ -4i \omega t} \Big ] $$

    Now for ##\mid \Psi(x,t) \mid^{2}##, I will use ##\Psi^{*} \Psi##. I think ##\Psi^{*}## should be
    $$ \Psi^{*}(x,t) = \frac {1}{\sqrt{a}} \Big [sin \Big ( \frac {\pi}{a} x \Big )e^{i \omega t} + sin \Big (\frac {2 \pi}{a} x \Big )e^{ 4i \omega t} \Big ] $$

    $$ \mid \Psi(x,t) \mid^{2} = \frac {1}{\sqrt{a}} \Big [sin \Big ( \frac {\pi}{a} x \Big )e^{-i \omega t} + sin \Big (\frac {2 \pi}{a} x \Big )e^{ -4i \omega t} \Big ]\frac {1}{\sqrt{a}} \Big [sin \Big ( \frac {\pi}{a} x \Big )e^{i \omega t} + sin \Big (\frac {2 \pi}{a} x \Big )e^{ 4i \omega t} \Big ]$$

    $$ \mid \Psi(x,t) \mid^{2} = \frac {1}{a} \Big [sin^{2} \Big ( \frac {\pi}{a} x \Big ) + sin \Big ( \frac {\pi}{a} x \Big )sin \Big ( \frac {2 \pi}{a} x \Big )e^{3i \omega t} + sin \Big ( \frac {\pi}{a} x \Big )sin \Big ( \frac {2 \pi}{a} x \Big )e^{-3i \omega t} + sin^{2} \Big ( \frac {2 \pi}{a} x \Big )\Big ] $$

    $$ \mid \Psi(x,t) \mid^{2} = \frac {1}{a} \Big [sin^{2} \Big ( \frac {\pi}{a} x \Big ) + sin \Big ( \frac {\pi}{a} x \Big )sin \Big ( \frac {2 \pi}{a} x \Big ) \Big [e^{3i \omega t} + e^{-3i \omega t} \Big ] + sin^{2} \Big ( \frac {2 \pi}{a} x \Big )\Big ] $$

    $$ \mid \Psi(x,t) \mid^{2} = \frac {1}{a} \Big [sin^{2} \Big ( \frac {\pi}{a} x \Big ) + 2sin \Big ( \frac {\pi}{a} x \Big )sin \Big ( \frac {2 \pi}{a} x \Big ) cos(3 \omega t) + sin^{2} \Big ( \frac {2 \pi}{a} x \Big )\Big ] $$
     
    Last edited: Mar 3, 2015
  17. Mar 3, 2015 #16

    BvU

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    I second Claude's reply. Is there a fellow student or an instructor to talk it over with ? (Otherwise you're stuck with us :smile: )

    "All" problems have the same way of finding ##\Psi(x,t)##

    • In the case of a time-independent V the time dependent Schroedinger equation (SE) allows separation of variables: ##\Psi(x,t) = \psi(x)\; e^{-iEt/ \hbar}##
    And all effort is directed at solving ##H\psi=E\psi## (the TISE) with solutions that are eigenfunctions ##\psi_n## for which ##H\psi_n=E_n\psi_n## .​

    The corresponding ##\Psi_n## are steady states in the sense that (because ##\Psi_n(x,t) = \psi_n(x) \; e^{-iE_nt/ \hbar}## it follows that ##\Psi_n^*(x,t) \Psi_n(x,t)=\psi_n^*\psi_n## is time-independent.​

    But the ##\Psi_n## are not time independent ! ​

    The SE is linear, so if $$\Psi(x,0) = \Sigma c_n\;\psi_n \qquad (1)\quad $$ then $$\Psi_n(x,t) =\Sigma c_n\;\psi_n(x) \; e^{-iE_nt/ \hbar} \qquad (2)\quad $$ and that's really all there's to it.

    Doesn't mean it's simple: everything worth knowing has to do with forms like $$\Psi^*(x,t) \;\Psi(x,t) = \Sigma_{n,m} c_n^* c_m \psi_n^*(x) \;\psi_m(x) \; e^{-i(E_n-E_m)t/ \hbar}\qquad (3)\quad$$ and there you do see a time dependence emerge. (The subject of your exercise).

    You also see that normalization is maintained over time: the ##\psi_n## are orthonormal, so integrating over x gives ##\Sigma_n c_n^* c_n e^0 ## !​


    My question
    "Would you have known what to answer in (b) if they had given you ## \Psi(x, 0) = \psi_1(x) ## ?" led you to reply
    from which we can deduce you have a newer edition than I have available herer now (no example 2.1 mentioned and the exercise is problem 2.6) and we can also deduce that some more understanding is to be transferred:

    The game plan consists of two steps: 1. Find the ##c_n## 2. Work out (3).

    The first step was already taken in part (a),
    at least I hope that you now would write something like $$
    A^*A \int_{0}^{a} \psi_{1} \psi_{1}^{*} + \psi_{1} \psi_{2}^{*} + \psi_{2} \psi_{1}^{*} + \psi_{2} \psi_{2}^{*} \hspace {0.02 in} dx = 1 \Leftrightarrow \\
    A^*A (1 + 0 + 0 + 1) = 1 \Leftrightarrow |A| = {1\over 2}\sqrt 2$$ and we can take the ##c_n## to be real, so ##c_1=c_2={1\over 2}\sqrt 2##.​



    [edit] well, you yourself beat me to it ! I'm impressed and really happy.
    (and now you can also guess how slowly I type !)​
     
  18. Mar 3, 2015 #17

    Maylis

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    BvU, I've been done with that for over an hour, now I just keep reediting my post as I make screw ups and catch it o_O. I am in a foreign country as an exchange student with people whose english skills are not up to par for me to have intelligent conversation on the subject, and my professor has no office hours, and the TAs are useless. So yes I'm stuck with you guys, and that is way better than being on my own...
     
  19. Mar 3, 2015 #18

    BvU

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    Good things come slowly. Understanding, for example. Don't hurry -- no need ! :smile:

    (he said wisely, with a long history of frequently stumbling over his own feet)
     
  20. Mar 3, 2015 #19

    Maylis

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    Yeah, well I am not at the same pace as the class (we are already in ch.3) and I've already forgone studying for all my other classes just for this stupid quantum mechanics...I need to hurry :H
     
  21. Mar 3, 2015 #20

    BvU

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    Good luck ! (QM took me an extra year ... never regretted it)
     
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