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Infinite potential well

  1. Jun 30, 2009 #1
    infinite potential well and the uncertainty principle

    the solution for Schroedinger equation in infinite potential well satisfy the following

    energy levels:

    5464deee159d922f51c081d408951169.png

    where l is the width of the well.

    E can't be zero since then [tex]\psi[/tex]=0 so there isn't any particle in the well . i read in

    a book that "there is a tight connection between this fact (E[tex]\neq[/tex]0) and the

    uncertainty principle", what exactly is the connection?
     
    Last edited: Jun 30, 2009
  2. jcsd
  3. Jun 30, 2009 #2
    I don't know what the author meant, but my guess would be:
    - since the well has finite width, the uncertainty in position is always finite, i.e. [tex]\Delta X < \infty[/tex]
    - now, if you take time-independent Schroedinger equation [tex]H\Psi=E\Psi \Leftrightarrow \frac{\partial^2 \Psi}{\partial x^2}=\frac{-2mE}{\hbar^2}\Psi[/tex] and put this into momentum uncertainty: [tex]\langle P^2 \rangle - \langle P \rangle^2[/tex] and calculate the integrals, you obtain that [tex](\Delta P)^2[/tex] is something like [tex]\frac{-2mE}{\hbar^2}[/tex]. So [tex]E=0[/tex] would violate uncertainty principle, since for [tex]E=0[/tex] [tex]\Delta P = 0[/tex], [tex]\Delta X < \infty[/tex]
     
    Last edited: Jun 30, 2009
  4. Jun 30, 2009 #3
    Thanks for your time, i really appreciate it.

    but i didn't understand two things:

    1. how can you assume that [tex]H\Psi=E\Psi [/tex]?

    2. why does [tex]\Delta X < \infty[/tex] imply for contradiction in the uncertainty principle?
     
  5. Jun 30, 2009 #4
    Ad. 1. When trying to determine possible energy values, we look first for a separated solution to the Schrodinger equation, i.e. solution of the form [tex]\psi (x, t) = \Psi (x) \phi (t)[/tex]. If the Hamiltonian [tex]H[/tex] is time-independent, separation of variables proves that [tex]H\Psi = E\Psi[/tex] for some constant [tex]E[/tex] (look this part up in any textbook). Now we have to determine possible values of [tex]E[/tex], and the argument above show that we cannot have [tex]E=0[/tex]. This is actually a bit of an overkill, since the equation [tex]\frac{\partial^2 \Psi}{\partial x^2}=0[/tex], along with boundary conditions [tex]\Psi (0)=\Psi(L)=0[/tex], gives [tex]\Psi = 0[/tex] instantly, but I guess this is what the author had in mind.

    Ad. 2. The uncertainty principle says that [tex]\Delta X \Delta P \geq \frac{\hbar}{2}[/tex]. For [tex]E=0[/tex] we get [tex]\Delta P=0[/tex] as well, so the uncertainty principle could be satisfied only if [tex]\Delta X = \infty[/tex]. But the well has finite width, so position uncertainty is also finite.

    Obviously, this is a very roundabout way of proving that [tex]E \neq 0[/tex].
     
  6. Jun 30, 2009 #5
    Thanks,

    now it's all clear
     
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