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Why must E exceed Vmin(x) for normalizable solutions?

  1. Nov 2, 2015 #1
    Hi! I've been studying the time-independent Schrödinger equation and the infinite square well and was faced with this problem from Griffith's "Introduction to Quantum Mechanics". Rewriting the equation this way $$\frac{d^2\psi}{dx^2}=\frac{2m}{\hbar^2}[V(x)-E]\psi$$, I have to show that E must exceed the minimum value of V(x), otherwise $$\psi$$ and its second derivative will always have the same sign, which would compromise its normalization. Why is that? I've been thinking for awhile and can't come up with a good proof that such function cannot be normalized.
     
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  3. Nov 2, 2015 #2

    jfizzix

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    As part of an answer, we could imagine the following.

    If the function [itex]\psi(x)[/itex] and its second derivative [itex]\psi''(x)[/itex] had the same sign, then what would happen at a critical point [itex]x_0[/itex] where the first derivative [itex]\psi'(x_0)=0[/itex]?

    If [itex]\psi(x)[/itex] were positive at this critical point, then [itex]\psi''(x)[/itex] would have to be as well, so the wavefunction would be concave up.
    Since [itex]\psi''(x_0)>0[/itex], the values of [itex]\psi(x)[/itex] near [itex]x_0[/itex] would be larger than [itex]\psi(x_0)[/itex], and [itex]\psi''(x)[/itex] would never change sign.

    As a result, the wavefunction would just get larger and larger farther away from the critical point. Since the wavefunction couldn't reach zero for extreme values of [itex]x[/itex], the wavefunction would not be normalizeable.

    For a more complete proof, you could also consider whether a wavefunction could exist that had no critical points, and also what happens if the critical point(s) has a negative or zero sign.
     
  4. Nov 6, 2015 #3
    I see! Makes perfect sense! Thank you very much!
     
  5. Nov 7, 2015 #4

    vanhees71

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    It's much easier to understand in the operator formalism, i.e., not expressing it in terms of the position representation (wave functions).

    The Hamiltonian reads
    $$\hat{H}=\hat{\vec{p}^2}{2m} + V(\hat{\vec{x}}).$$
    If now ##V## is bounded from below, then so is the Hamiltonian, because the kinetic energy is positive semidefinite, and there's a ground state. Without loss of generality we can assume that the minimum value of ##V## is ##0##. Assume there's an energy eigenstate with ##E=0##, it must then be an eigenvector of ##\hat{\vec{p}}## with eigenvalues ##\vec{p}=0##, but a momentum eigenstate cannot be a normalizable state, because the momentum operator has a purely continuous spectrum. Thus if there is a bound ground state, its energy eigenvalue cannot be 0 but must be >0.
     
  6. Nov 7, 2015 #5

    PeroK

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    Just to add to @jfizzix's answer. The following properties of a real-valued ##\Psi## are sufficient:

    1) ##\Psi## is continuous on ##\mathbb{R}##.

    2) ##\lim_{x \to \pm \infty} \Psi(x) = 0##

    One way to start a formal proof is to note that, unless ##\Psi = 0##, ##\exists x_0## such that ##\Psi(x_0) \ne 0##. And, without loss of generality, you can assume ##\Psi(x_0) > 0##. From there, you can use the intermediate and mean-value theorems to show the existence of a local maximum with a positive value - and that contradicts the "same-sign" property.

    The case for a complex-valued ##\Psi## follows from the real-valued case.
     
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