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I Energy quantization of the particle in a box

  1. Mar 15, 2016 #1
    I'm self-learning quantum mechanics and I'm reading the famous Shankar book (Principles of Quantum Mechanics - second edittion). At page 161 of the book, I don't understand the following part of this page:

    "Let us restate the origin of energy quantization in another way. Consider the search for acceptable energy eigen-functions, taking the finite well as an example. If we start with some arbitrary values ##\psi(x_0)## and ##\psi'(x_0)##, at some point ##x_0## to the right of the well, we can integrate Schrodinger's equation numerically. (Recall the analogy with the problem of finding the trajectory of a particle given its initial position and velocity and the force on it.) As we integrate out to ##x \to \infty##, ##\psi## will surely blow up since ##\psi_{III}## contains a growing exponential. Since ##\psi(x_0)## merely fixes the overall scale, we vary ##\psi'(x_0)## until the growing exponential is killed. [Since we can solve problem analytically in region III, we can even say what the desired value of ##\psi'(x_0)## is: it is given by ##\psi'(x_0) = -\kappa \psi(x_0)##. Verify, starting with Eq. (5.2.4), that this implies ##B=0##.] We are now out of the fix as ##x \to \infty##, but we are committed to whatever comes out as we integrate to the left of ##x_0##. We will find that ##\psi## grows exponential till we reach the well, whereupon it will oscillate. When we cross the well, ##\psi## will again start to grow exponentially, for ##\psi_I## also contains a growing exponentially in general. Thus there will be no acceptable solution at some randomly chosen energy. It can, however, happen that for certain values of energy, ##\psi## will be exponentially damped in both regions I and III. [At any point ##x_0'## in region I, there is a ratio ##\psi'(x_0')/\psi(x_0')## for which only the damped exponential survives. The ##\psi## we get integrating from region III will not generally have this feature. At special energies, however, this can happen.] These are the allowed energies and the corresponding functions are the allowed eigen-functions. Having found them, we can choose ##\psi(x_0)## such that they are normalized to unity. For a nice numerical analysis of this problem see the book by Eisberg and Resnick.$".

    I need the formula details to understand it. Another question: "overall scale", what does it mean?
     
  2. jcsd
  3. Mar 15, 2016 #2

    bhobba

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    I can't follow it either - maybe with work but not easily.

    However I can point you to a series of lectures that does carefully explain what's going on and much more besides:


    Thanks
    Bill
     
  4. Mar 15, 2016 #3

    Nugatory

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    The "formula" is just the time-independent Schrodinger's equation for the infinite square well potential. It will have been covered somewhere in previous 160-odd pages. There is also an assumption here that you know what numerical integration is; you don't have to actually do it, but you have to know what it is.

    The bit about "overall scale" is just referring to the fact that if ##\psi## is a solution to Schrodinger's equation, then any constant multiple of ##\psi## is also a solution. Thus we can choose ##\psi## to have any value we please at any single point without losing any generality. We then work from there to find the ##\psi## and ##\psi'## that when numerically integrated give us the desired declining exponential on both sides of the well. That's an unnormalized eigenfunction, but as a final step we can multiply it by whatever constant is necessary to normalize it.
     
  5. Mar 15, 2016 #4

    stevendaryl

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    To answer the last question first, Schrodinger's equation for an energy eigenstate is:

    [itex](\frac{-\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V(x)) \psi(x) = E \psi(x)[/itex]

    Obviously, if [itex]\psi(x)[/itex] solves this equation, then so does [itex]\tilde{\psi}(x) \equiv C \psi(x)[/itex] for any constant [itex]C[/itex]. So we can choose [itex]\psi(x_0)[/itex] to be anything at all, and solve the equation, then at the end, we can multiply by a constant [itex]C[/itex] to make [itex]\int |\psi(x)|^2 dx = 1[/itex]. So he's just saying it doesn't matter what choice we make for [itex]\psi(x_0)[/itex].

    As far as the details for his claim, I'm not sure what extra details you want. Mathematically, Schrodinger's equation implicitly gives a function [itex]\psi(x) = F(x,E,x_0, \psi_0,\psi'_0)[/itex] that depends on 5 quantities: (1) [itex]x[/itex], of course, (2) the energy [itex]E[/itex], (3) [itex]x_0[/itex], your starting point, (4) [itex]\psi_0 \equiv \psi(x_0)[/itex], the value at [itex]x=x_0[/itex], and (5) [itex]\psi'_0[/itex], the value of [itex]\psi'[/itex] at [itex]x=x_0[/itex]. Let's hold [itex]\psi_0[/itex] and [itex]x_0[/itex] fixed, so we get a function of three arguments: [itex]\psi(x) = F(x,E,\psi'_0)[/itex]. There are various ways to compute [itex]F[/itex], either numerically or in a power series.

    Now, the first claim being made is that for almost all values of [itex]\psi'_0[/itex], [itex]lim_{x \rightarrow \infty} F(x,E,\psi'_0) = \infty[/itex]. Only for one very specific value of [itex]\psi'_0[/itex] will it be the case that [itex]lim_{x \rightarrow \infty} F(x,E,\psi'_0) = 0[/itex]. The author claims that that value is [itex]\psi'_0 = -\kappa \psi_0[/itex]. So let's pick that value for [itex]\psi'_0[/itex]

    The second claim being made is that for almost all values of [itex]E[/itex], if [itex]\psi'_0 = -\kappa \psi_0[/itex], then [itex]lim_{x \rightarrow -\infty} F(x,E,\psi'_0) = \infty[/itex]. An energy eigenvalue is some value of [itex]E[/itex] such that [itex]lim_{x \rightarrow -\infty} F(x,E,\psi'_0) = 0[/itex] when [itex]\psi'_0 = -\kappa \psi_0[/itex].

    So if you are solving the Schrodinger equation numerically, then you could do it this way:
    1. Fix [itex]x_0[/itex] and [itex]\psi_0[/itex]
    2. Make a guess for a starting value for [itex]E[/itex]
    3. Make a guess for a starting value for [itex]\psi'_0[/itex]
    4. Check if [itex]F(x,E,\psi'_0)[/itex] blows up for [itex]x[/itex] large and positive.
    5. If so, adjust [itex]\psi'_0[/itex] and go back to 4.
    6. If not, then check if [itex]F(x,E,\psi'_0)[/itex] blows up for [itex]x[/itex] large and negative.
    7. If so, adjust [itex]E[/itex] and go back to 3.
    8. If not, then you've found an approximate eigenvalue.
     
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