1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Wave function and infinite square well potential

  1. Oct 19, 2013 #1
    1. The problem statement, all variables and given/known data
    An electron in a one-dimensional infinite square well potential of length L is in a
    quantum superposition given by ψ = aψ1+bψ2, where ψ1 corresponds to the n = 1 state, ψ2 corresponds to the n = 2 state, and a and b are constants. (a) If a = 1/3, use the
    normalization requirement for ψ to determine the value of b. (b) If we perform a
    measurement of the energy of the electron, what is the probability we will measure E1?
    What is the probability we will measure E2?

    2. Relevant equations
    Don't know.

    3. The attempt at a solution
    So basically I have to find the ψ of an electron in the ground state and then in the n = 2 state? Do I solve for that? And how do I normalize a wave function that doesn't have "i" in the exponent? I've only learned that normalizing is making the "i" a "-i" as the conjugate, and then multiplying it with the original function. How do I normalize something without "i" in its exponent?

    And then
  2. jcsd
  3. Oct 19, 2013 #2

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    No - that is not what it says.
    The n states are stationary states - the problem is telling you that the particle is not in a stationary state. It is in a single non-stationary state that happens to have the same shape as a sum of the first two stationary states.

    The stationary states are not the only solutions to the schodinger equation for the infinite square well ... any linear superposition of them are also solutions.

    Exactly the same way:

    Consider an arbitrary complex number z in terms of a,b which are real and positive:
    ##z=a+ib \Rightarrow z^\star = a-ib## which you are used to... but what if z=a alone?
    ##z=a \Rightarrow z=a+i0## now do it.
    ##z^\star = a-i0## but zero times i is zero so ##z^\star = a##

    ...thus: the complex conjugate of a real number is itself.

    Tread carefully: it says to use the "normalization condition" - what does that mean for a and b?
    Last edited: Oct 19, 2013
  4. Oct 20, 2013 #3
    I don't understand what to use as ψ. Do I use ei(kx-wt)? That's a wave function I see in this book. It's confusing because whenever they need to do something to a wave function, they pull one out of thin air and operate on it.
    I do need a wave function for the electron in order to do this problem, right? So how do I create one?

    The normalization requirement is the "requirement" that the probability of the particle existing within the well is 100% right? So I set ∫ψ*ψ=1, plug in the wave function that I find for the electron, and then normalize it?

  5. Oct 20, 2013 #4


    User Avatar
    Gold Member

    I don't think you have to sub anything for ##\psi_1## and ##\psi_2##. The orthogonality of the ##\psi_i## will allow you to cancel terms.
  6. Oct 20, 2013 #5
    So they cancel and I end up with 0?
  7. Oct 20, 2013 #6


    User Avatar
    Gold Member

    No, what does the orthogonality condition state? i.e what is ##\langle \psi_i | \psi_j \rangle = \int \psi_i^* \psi_j \,dx## equal to?
  8. Oct 20, 2013 #7
    Oh, so the conjugate of a wave function times the wave function itself are orthogonal? Ok.
    I believe that it's equal to 1, because 1 is the probability of the particle existing in the well. So orthogonality condition is the same as the probability?
  9. Oct 20, 2013 #8


    User Avatar
    Gold Member

    That is correct if ##i = j##, but what happens if ##i \neq j##? Have you come across orthogonality in your course yet?
    The physics is that the ##\psi_i## are orthogonal meaning they represent different physical states of the wave function. This means you cannot express each of the eigenstates as a linear combination of each other, but the wave function can be expressed as a linear combination of the eigenstates, which is precisely what you have in your problem.
    Last edited: Oct 20, 2013
  10. Oct 20, 2013 #9


    User Avatar
    Staff Emeritus
    Science Advisor
    Homework Helper
    Education Advisor

    As CAF123 noted, you don't actually need expressions for the eigenstates to solve this problem. Nevertheless, you should know what they are. The infinite square well problem is a standard problem in intro quantum mechanics. It's probably worked out in your textbook.
  11. Oct 20, 2013 #10

    Simon Bridge

    User Avatar
    Science Advisor
    Homework Helper

    Lets not get ahead of ourselves - look at one thing at a time or it will get confusing.
    ... what is the "normalization requirement"?

    if ##\psi = \sum_i c_1\psi_i##
    if each ##psi_i## are normalized, then there is a requirement on the ##c_i##'s that make ##\psi## normalized as well.

    That is the normalization requirement.

    Why it works (that "orthogonality stuff) is another question - which should be covered in your text and your course notes along with the requirement itself. I want you to get used to using them. Meantime - when you see it you will also see what you are expected to do.
    Last edited: Oct 20, 2013
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted

Similar Discussions: Wave function and infinite square well potential