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Energy of an Infinite Square Well

  1. Nov 4, 2013 #1


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    Hey guys, this is my first post so go easy on me.

    I was looking over the simple case of a 1D particle restrained inside an infinite square well potential ("particle in a box") and was having some difficulty understanding the relationship between the energy states and the expectation value for the energy.

    Using the time independent Schrödinger equation and normalizing the wave function I get:

    ψ(x) = sqrt(2/L) * sin (n*pi*x / L)

    Which implies:

    k = n*pi / L = p / hbar = sqrt(2mE) / hbar

    E = (n*pi*hbar)^2 / 2m*L^2

    Then I try calculating the expectation value for the energy. (Here is where I have trouble.)

    <E> = ∫ψ* i hbar ∂ψ/∂t dx = i hbar ∫ψ* 0 dx = 0

    [Where the bounds of the integral are from -∞ to ∞]

    How can both of these statements about the energy of the particle be true? I feel like I am missing something fundamental. Does the uncertainty principle play a role here? Or is the Energy operator simply not valid in the time independent case of Schrödinger's equation?
  2. jcsd
  3. Nov 4, 2013 #2
    Welcome to Physics Forums.

    The expectation value of energy is
    \langle E\rangle = \int_{0}^L dx \psi^* \hat{H} \psi,
    where your Hamiltonian is
    \hat{H} = \hat{p}^2/(2m) + V(\hat{x}).

    Yup. If you want to use that equation then use the full wavefunction (which includes time).
    \Psi_n(x,t) = \sqrt{2/L}\sin(n\pi x/L)\exp(-\iota E_nt/\hbar).
    So the expectation value can be evaluated as
    \langle E(t)\rangle = \int_{0}^L dx \Psi_n^*(x,t)\iota \hbar\frac{\partial}{\partial t}\Psi_n(x,t).
  4. Nov 4, 2013 #3
    Trust me, it is not a simple case at all.
  5. Nov 4, 2013 #4
    I have a qüestion:

    How can be possible \langle E(t)\rangle to be only the expected energy and not the real energy of the particle? I mean, if the particle it's in a superposition of states, its energy should be [itex]\sum\E_{n}*P_{n}[/itex] where E_{n} is the energy of the n-ieth eigenvalue and P_{n} its probability (its amplitude squared).

    Sorry if this comment can un-focus the main qüestion...
  6. Nov 4, 2013 #5


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    Staff: Mentor

    No, if the particle is in a superposition of energy eigenstates, then when we measure the energy, we get one of the energy eigenvalues. We cannot predict which of the energy eigenvalues we will get, only the probability for each eigenvalue. Your sum is the expectation value, which usually does not equal any of the individual eigenvalues.
  7. Nov 4, 2013 #6
    Here's your problem. The energy is given by ##E_n = \frac{n^2 \pi ^2 \hbar ^2 }{2mL^2 }## which was derived from the boundary conditions of the wave function. If you want the statistical mean of the energy of a system in an infinite square well.

    There are two ways to find the expectation value of the energey, the first which has been shown is:

    $$\int ^L _0 \Psi (x,t)^* [ \frac{\hbar ^2 }{2m} \nabla ^2 \Psi (x,t)]dx$$

    Where you'll notice the limits of integration are ##l## and ##0##
    Last edited: Nov 5, 2013
  8. Nov 5, 2013 #7


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    Homework Helper

    Yeah, that's true. TBW said that his wavefunction is: ψ(x) = sqrt(2/L) * sin (n*pi*x / L) But that's not quite right, because his wavefunction is ψ(x) = sqrt(2/L) * sin (n*pi*x / L) from x=0 to x=L and ψ(x) = 0 outside this range.

    edit: or instead I guess you can define
    [tex]\int_0^L \ dx |x\rangle \langle x|[/tex]
    to be the identity operator for this 'space', but I feel this is a bit more confusing. (And you'd still need the boundary conditions anyway).
    Last edited: Nov 5, 2013
  9. Nov 5, 2013 #8
    This is quiet wrong. [itex] \psi \neq 0 [/itex] outside the well because it doesn't exist outside the well.
  10. Nov 5, 2013 #9


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    Homework Helper

    the wavefunction is non-zero outside the well because it doesn't exist outside the well? ... I think you have this the wrong way around. If the wavefunction is non-zero outside of the well, then there is some non-zero probability to be found outside the well. But we don't want this because we have an infinite potential outside the well.
  11. Nov 5, 2013 #10
    What I mean is that wavefunction is not defined outside the well (and hence certainly it is not equal to zero). The Hilbert Space for this particular problem is [itex] L^2([0,l],dx)[/itex] in which functions like you are mentioning do not exist at all.
  12. Nov 5, 2013 #11


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    Staff: Mentor

    I'm uncomfortable with this way of approaching the problem, although my objection may be just aesthetic. If there's a region of space in which the probability of finding the particle is zero, I'd like my mathematical formalism to deliver a probability amplitude of zero for that region, not "undefined". It's easier to explain the physical interpretation that way.
  13. Nov 5, 2013 #12


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    Homework Helper

    I see what you mean now. Yeah, that is what I was talking about when I said you could have an identity operator like
    [tex]\int_0^L dx \ |x\rangle \langle x| [/tex]
    I was probably not very clear about that. Anyway, I think that the square-well problem is usually introduced with x going from ##-\infty## to ##\infty## because that is more intuitive for people (especially who are new to quantum mechanics). To add to what Nugatory said, if we use [itex] L^2([0,l],dx)[/itex] as the Hilbert space as you said, then we also need to include boundary conditions. But if we have x from ##-\infty## to ##\infty## and also specify that there is a very large potential for ##0>x## and ##x>L##, this tells us that the wavefunction needs to go to zero at the boundary of the well. I think this idea of the potential constricting the wavefunction to the well is more intuitive.
  14. Nov 6, 2013 #13
    I agree with Nugatory. Any box with particle(s) inside must be inside the universe and therefore we have to define the amplitude of the particle outside the box (as 0). In fact, the amplitude is 0 in the edges in order to avoid a discontinuity.
  15. Nov 6, 2013 #14


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    Homework Helper

    I think using ##L^2([0,l],dx)## (as Ravi said) is useful mathematically, because then we are guaranteed an orthonormal basis of energy eigenfunctions. But the 'extremely large potential' for a particle outside the box is a more intuitive interpretation than using ##L^2([0,l],dx)## with boundary conditions.
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