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Normalizing a State Function for an Infinite Square Well

  1. Mar 13, 2014 #1
    1. The problem statement, all variables and given/known data

    Normalize: [tex]\Psi_1 (x,t) = N_1 \cos(\frac{\pi x}{L}) e^{-\frac{iE_1t}{\hbar}}[/tex]
    Where [itex]N_1[/itex] and [itex]E_1 [/itex] are the normalization constant and energy for the ground state of a particle in an infinite square well.

    2. Relevant equations

    Normalization Condition:
    [tex]\int_\infty^\infty P(x,t) dx = \int_\infty^\infty \Psi(x,t)^* \Psi(x,t) dx = 1[/tex]
    (sorry... that’s –inf to +inf for the integrals, but I can’t quite get it right in latex :grumpy:)

    3. The attempt at a solution
    So, I apply the normalization condition. My exponential term and its complex conjugate cancel each other out, which leaves me with:
    [tex]N^2_1 \int_\infty^\infty {\cos(\frac{\pi x}{L})}^2 dx = 1[/tex]

    I use a handy trig power reducing identity, and when I integrate I get:
    [tex]N^2_1 (\frac{x}{2} +\frac{L}{4\pi}\sin(\frac{2\pi x}{L}))[/tex]

    But... then things start to go off the rails.

    I don't know if I got my calculus wrong (pretty sure I didn't) or just completely misunderstood the physics (probably), but I don't know how to evaluate this integral without it blowing up to infinity! And it's absolutely not supposed to, because it's described as a 'physically admissible' state function.

    I know my normalization constant should equal [itex]\sqrt{\frac{2}{L}}[/itex], and I assumed that my terms in psi*psi could commute, and that my imaginary exponential terms should cancel each other out prior to integration. But just looking at a graph of cos^2 shows my integrand clearly won't converge to a finite value.

    Does anybody know what I'm doing wrong?

    Much thanks in advance!!!
     
    Last edited: Mar 13, 2014
  2. jcsd
  3. Mar 13, 2014 #2
    Uh... sin(2*pi) = 0 and sin(0) = 0. Nothing is going to infinity. Reduces to L/2.
     
  4. Mar 13, 2014 #3
    Oh I just realized what you must not be understanding: the integral is between 0 and L because the wave function is zero outside that interval.
     
  5. Mar 13, 2014 #4
    Hi, thanks for the response!

    I get that the state function (and the prob. density) for a particle in an infinite well are bounded. I think that relates to the stationary states? But do you know how I can extend that property to the normalization condition?

    As in... how do I know the function = 0 outside the interval of 0 to L? Because my integrand cos^2 definitely doesn't = 0 outside those bounds.

    As far as I know, in order to normalize I need to integrate over the entire space (which is +/- inf), and I need to do that to determine a normalization constant, expectation value, and other values.

    In short: Can you describe a way that I can reduce my (+/- inf) conditions to the (0 to 2pi) conditions you describe?

    Thanks!
    CQ~
     
  6. Mar 13, 2014 #5
    The particle can't exist outside the well. The potential is infinite there.
     
  7. Mar 13, 2014 #6
    So based on what you're saying... does that imply (+/- inf) --> (0 to L) for infinite square wells?

    I understand the particle can't exist outside the well. But when I'm asked to integrate from (+/- inf) does that mean I should reduce my integration to values of (0-L)? And what about multiples of that value? They act pretty much like multiples of pi. How do I ignore higher multiples of L?

    Thanks
    CQ~
     
  8. Mar 13, 2014 #7
    What is the integral of zero?
     
  9. Mar 14, 2014 #8
    Geez... I guess a constant? But this is sorta sidetracking my original question... As in, I got my state function, I applied my normalization condition, and then I integrated. Nothing in that procedure told me to reduce my bounds from (+/- inf) to (0 to L), which is apparently what you're asking me to do. So what are you asking me to do, and why are you asking me to do this? Because it definitely doesn't sound like you are helping me solve my problem at this point.
     
  10. Mar 14, 2014 #9
    It's zero. It makes not difference if you include everything outside the well. Its contribution is zero.
     
  11. Mar 14, 2014 #10

    jtbell

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    Staff: Mentor

    Your wave function is actually

    $${\Psi _1}\left( {x,t} \right) = \left\{ {\begin{array}{*{20}{l}}
    {0,}&{x \le 0}\\
    {{N_1}\cos \left( {\frac{{\pi x}}{L}} \right){e^{ - i{E_1}t/\hbar }},}&{0 < x < L}\\
    {0,}&{x \ge L}
    \end{array}} \right.$$

    because, as tman12321 noted, the potential is infinity outside the box, so the wave function must be zero outside the box. Now, think of your normalization integral as

    $$\int_{ - \infty }^{ + \infty } {\Psi _1^*\Psi _1^{}dx} = \int_{ - \infty }^0 {\Psi _1^*\Psi _1^{}dx} + \int_0^L {\Psi _1^*\Psi _1^{}dx} + \int_L^{ + \infty } {\Psi _1^*\Psi _1^{}dx} $$

    That is, split up the integral into "pieces" corresponding to the "pieces" used in the definition of the wave function.
     
  12. Mar 16, 2014 #11
    if the "Well" is in [0,L], the wave function in the well should be a sine function,not cos, because we need the wave function equals to zero in the boundary . the integral is like jtbell do
     
  13. Mar 16, 2014 #12
    Oops. The interval should have been [-L/2, L/2].
     
  14. Mar 16, 2014 #13

    Simon Bridge

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    ... which tells you that the potential well in post #1 probably goes from -L/2 to +L/2.

    The notes in post #10 need to be modified accordingly.
    [edit] oh you saw - cool!

    I'd like to go back to:
    ... the wavefunction you have is for a stationary state.
    ... the normalization is part of the stationary state - therefore the boundary must apply there too.

    I suspect you don't understand what the normalization condition is for.

    The point of an infinite square well is that the particle cannot be found outside the well.
    Therefore the probability of finding the particle inside the well is 1.

    In general: the probability of finding the particle inside a<x<b is $$P(a<x<b)=\int_a^b\Psi^\star\Psi\;\text{d}x$$ ... if a=-L/2 and b=+L/2, then that is the whole well - which probability must be 1 - so we write:
    $$\int_{-L/2}^{L/2}\Psi^\star\Psi\;\text{d}x=1$$

    In general, ##\Psi=Nf(x,t)## where N is the normalization coefficient.
    What you have been given is f(x,t) and you are tasked to find N.

    In general, it is possible to find the particle beyond the classical limits of the well, so you should divide the region into different parts as per post #10. You'll get to this later in the course.

    What's tripping you up right now, I suspect, is that you are focussed too much on the equations and not enough on what they are telling you. Relate it to probabilities.
     
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