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Show E must exceed minimum value of V(x) to be normalizable

  1. May 18, 2015 #1
    1. The problem statement, all variables and given/known data

    Show that E must exceed the minimum value for V(x) for every normalizable solution to the TISE, Hψ = Eψ

    Hint: rewrite the TISE as d2ψ/dx2 = (2m/ħ)(V(x) - E)

    2. Relevant equations

    I understand how to get the TISE Hψ = Eψ into the form of the hint by using the hamiltonian and the differential form of the momentum operator. That's all fine

    3. The attempt at a solution

    The solution to the differential equation given in the hint is:

    ψ = Ae√(2m/ħ)(V(x) - E) + Be-√(2m/ħ)(V(x) - E)

    For the solution to be normalizable, it has to go to zero at infinity. This means it can't be oscillatory, therefore the powers of the exponents have to be real.

    But doesn't this mean that V has to be greater than E and not the other way round?

    If E is greater than V anywhere, then in that region there will be an imaginary exponent and the solution will be oscillatory.
     
  2. jcsd
  3. May 18, 2015 #2

    PeroK

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    Note that you have ##V(x)## here as a function of ##x##. It's not a constant. So, you can't get exponential solutions to the equation.

    Instead, think about the relationship between ##\psi## and its second derivative in elementary calculus terms.
     
  4. May 18, 2015 #3
    Ah okay. So if E > V, then the second derivative is negative, which means the gradient is decreasing? I understand that if E > V everywhere, then the function is always "turning downwards" so will vanish at infinity and negative infinity. But if it only exceeds the minimum value at one point then the function could easily be non-normalizable, no?
     
  5. May 18, 2015 #4

    PeroK

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    I don't know whether you noticed that you dropped ##\psi## from the RHS of your equation. If ##E < V_{min}## then ##\forall x \ E < V(x)## and ##\forall x \ V(x) - E > 0## hence:

    ##\psi''(x) = K(x)\psi(x)## where ##K(x)## is always positive. Now, think about maxima and minima for ##\psi##
     
  6. May 18, 2015 #5
    I'm not really sure what you've written there. If E<Vmin then then any turning point is a minimum and the function can't be normalised. But if E>Vmin at any point is the function necessarily normalisable? It doesn't strike me as such, it seems like there must be more conditions.
     
  7. May 18, 2015 #6

    PeroK

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    You're forgetting about the sign of the function itself ##\psi(x)## can be +ve or -ve.
     
  8. May 18, 2015 #7
    Right, yes I was. So what really needs to happen is that the wavefunction must tend to zero for large values. I think this entails the gradient going to 0 and the value going to 0 at large x. Looking at d2ψ/dx2 = (2m/ħ)(V(x) - E)ψ I can't really integrate the RHS because V(x) is generalised. Any thoughts? Assuming V<E always then for positive ψ then the gradient is decreasing with x which is good and for negative ψ it is increasing. But I really need to be able to say something about how the gradient varies with x, which I can't.

    I also don't know what this symbol means ∀ (you used it earlier).
     
  9. May 18, 2015 #8
     
  10. May 18, 2015 #9

    PeroK

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    I guess you don't do many proofs by contradiction on a physics course! Here's a flavour:

    Let ##x## be a local maximum for ##\psi##. If ##\psi(x) > 0## then ##\psi''(x) > 0## but that implies x is a local minimum. Which is a contradiction. Continue in this vein.

    ##\forall## means "for all"
     
  11. May 18, 2015 #10
    I can't really continue that. You've just said that x is a local maximum and then that it's a minimum, so something must be inconsistent, but the TISE is definitely correct since it's derived from the TDSE which is axiomatic. I think I might have to give up.
     
  12. May 18, 2015 #11
    Okay hang on, if ψ′′(x)=K(x)ψ(x) where K(x) is always positive for E<V, then if E<V the turning points are always minima so you can't have any maxima? Is that what you're saying? I can't really see how proof by contradiction can be used here...
     
  13. May 18, 2015 #12
    Does the contradiction thing show that you only have minima when ψ is negative and maxima when ψ is positive? If it does, then it doesn't really imply much because I can think of several functions that can and can't be normalised with that condition such as sin(x) which isn't normalisable and xe-x2 which is.
     
  14. May 18, 2015 #13

    PeroK

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    Not quite.

    The point is that any function whose sign is always the same as the second derivative cannot be normalisable. To show this, you need a bit of real analysis involving derivatives and some mathematical logic in the shape of a "proof".

    This same thing cropped up in another thread recently - nothing to do with QM - but it was the same thing: the second derivative always had the same sign as the function. The function can't start and end at 0. In your case, for large +ve and -ve x. But, also over a finite interval.

    The next step in the proof would be:

    Let ##x## be a local minimum with ##\psi(x) < 0## again you can get a contradiction.

    So, the two valid possibilities are:

    a) A single local minimum with ##\psi(x) \ge 0##

    or

    b) A single local maximum with ##\psi(x) \le 0##


    In either case, the function can't get back to 0. If you were doing a real analysis course, there might be a bit of work yet! But for QM, that's a sufficiently rigorous argument!

    or

    c) ##\psi(x) = 0 \ \ \forall x##
     
    Last edited: May 18, 2015
  15. May 18, 2015 #14
    Great, I understand, thank you. I'm just going to recap here:

    ψ'' = (2m/ħ)(V(x) - E)ψ.

    Show that E > Vmin to get a normalisable solution.

    To do this, consider the case where E<Vmin and show that it can't lead to normalisation.

    Well if ψ is positive, then ψ'' is positive too, and if ψ is negative, then ψ'' is negative too. So the function can't get back to zero if the function goes either side of 0 if E<Vmin.

    So every normalisable case must have the other condition, where E>Vmin.

    Thanks again!
     
  16. May 18, 2015 #15

    PeroK

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    Try this exercise. Try to draw a function defined on [0, 1] which is 0 at both 0 and 1. But, doesn't have a maximum > 0 or a minimum < 0.

    That's effectively what the 2nd derivative condition is preventing.

    And, of course, a function which must be 0 at ##\pm \infty## would have the same problem. It must have a maximum > 0 or a minimum < 0 or both.
     
  17. May 18, 2015 #16
    Great, yeah, I've just done that. So that's what the deal is if V>E always, and it just can't go back to zero. So to be normalisable, something's got to break, and it's the condition V>E that changes. With that, the prohibitive second derivative condition is broken and the function can be normalised.

    In case anybody from the future is wondering, this is question 2 from chapter 2 of Griffith's introduction to quantum mechanics.
     
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