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Infinite square well problem

  1. Sep 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Regarding the wave function in an infinite square well extending from -L to L:
    If the position is measured at time t, what results can be found and with what probabilities will this results be found?

    2. Relevant equations

    the wave function is a superposition of the ground and first excited state.

    |psi(o)> = N[|1>+|2>]

    N = normalization constan

    3. The attempt at a solution

    I'm having trouble conceptually understanding the question. my current thought process is that when the position is measured at time t the possible outcomes are the two eigenfunctions with probabilities equivalent to the corresponding eigenfunctions expansion coefficient squared.

    However for some reason this doesn't make any sense to me. another line of thinking is that because the basis of our operator is all x between -L and L, the possible outcomes of measuring position at time t are all x between -L and L. Is it correct to say then that the probability of measuring a certain position x` is given by the wave functions probability distribution evaluated at x`:confused:

    I am in an agonizing state of confusion so any help is much appreciated.
  2. jcsd
  3. Sep 15, 2009 #2


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    Were you told to assume the particle was in this state as part of the original question? And by |psi(o)>, I assume you mean [itex]\vert\psi(t=0)\rangle[/itex]?

    Not quite. When you measure an observable, the measurement will always yield one of the operator's eigenvalues (not eigenfunctions).

    In addition, if [itex]\vert1\rangle[/itex] and [itex]\vert2\rangle[/itex] are the ground and first excited states, doesn't that mean that they are energy eigenfunctions? In order to figure out where the particle can be located at [itex]t=0[/itex], you need to express [itex]\vert\psi(t=0)\rangle[/itex] in terms of the position eigenfunctions. And in order to figure out where the particle can be located at some later time [itex]t[/itex], you need to calculate [itex]\vert\psi(t)\rangle[/itex] and express it in terms of the position eigenfunctions
  4. Sep 15, 2009 #3
    my apologies, i found the position eigenfunctions. that is what i was referring to. So then, would the possibilities when measuring position be any -L < x < L , and the probabilities be the value of the wave function, in position space, at any x between -L and L?
  5. Sep 15, 2009 #4


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    I'm not 100% sure what you are trying to tell me here. Are you saying that the [itex]\vert1\rangle[/itex] and [itex]\vert2\rangle[/itex] in your expression for [itex]|\psi(t=0)\rangle[/itex] are the position eigenstates? Or are are you telling me that they are energy eigenstates, and you've used that to calculate [itex]|\psi(t=0)\rangle[/itex] and [itex]|\psi(t)\rangle[/itex] in terms of the position eigenstates?

    No, the possibilities will be the position eigenvalues, and their corresponding probabilities will be given by the coefficient of the corresponding eigenstate times it's complex conjugate.

    For example, If I gave you the state [itex]|\psi(t)\rangle=\left(\frac{1}{\sqrt{3}}\vert x=\frac{a}{2}\rangle+\sqrt{\frac{2}{3}}\vert x=-\frac{a}{3}\rangle\right)e^{iEt/\hbar}[/itex], the possible outcomes would be [itex]x=a/2[/itex] and [itex]x=-a/3[/itex], and their corresponding probabilities would be 1/3 and 2/3.
    Last edited: Sep 15, 2009
  6. Sep 16, 2009 #5
    I'm saying that they are energy eigenstates, and I've used that to calculate |[tex]\right\psi(t=0)\rangle[/tex] and |[tex]\right\psi(t)\rangle[/tex] in terms of the position eigenstates.

    my wave equation in position space is:


    in your example you actually gave values for x. what's confusing me is that x isn't specified in my problem. my eigenstates are functions of x (sines and cosines) and the problem just asks the possibilities and probabilities from measuring position, but i am not told what the position eigenvalues are. So should i assume the position eigenvalues can be any x between the well boundaries?
  7. Sep 16, 2009 #6


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    Surely, you must mean

    [tex]\Psi(x,t)=\langle x\left|\right\Psi(t)\rangle=\left\{\begin{array}{lcr}\frac{1}{\sqrt{2L}}\left(\cos\left(\frac{x\pi}{2L}\right)e^{\frac{-i\hbar\pi^{2}t}{8mL^{2}}}+\sin\left(\frac{x\pi}{L}\right)e^{\frac{-i\hbar\pi^{2}t}{2mL^{2}}}\right) & , & |x|< L \\ 0 & , & |x| \geq L\end{array}\right.[/tex]

    ...right?:wink: (Writing [itex]\Psi(x,t)[/itex] inside a Ket makes no sense notationally)

    Yes, unlike the energy and momentum states, the position eigenstates are continuous over the Reals for a particle in a box. You can measure any value for the position, and the probability of a particular measurement [itex]x[/itex], is of course [itex]P(x,t)=|\langle x\vert \Psi(t)\rangle|^2=|\Psi(x,t)|^2[/itex].

    Clearly, all [itex]|x|\geq L[/itex] will have zero probability of being measured, so those are impossible positions.
    Last edited: Sep 16, 2009
  8. Sep 16, 2009 #7
    Right, thanks a lot for clearing that up for me, gabbagabbahey!:smile:
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