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What must Psi be continuous?

  1. Mar 28, 2010 #1
    At least physically, why must Psi be continuous? Sorry if this question has been asked before. Most of the things I read however just state that it is, & leave it at that.
     
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  3. Mar 28, 2010 #2

    Vanadium 50

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    Because the derivative is momentum: a discontinuity means infinite momentum.
     
  4. Mar 29, 2010 #3

    clem

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    If Psi were discontinuous, there would be two different probabilities for a the particle to be at the same point in space.
     
  5. Mar 29, 2010 #4

    Fredrik

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    What Vanadium50 and Clem said are the physical reasons. I'll add that mathematically, the fact that the function satisfies the Schrödinger equation means that it's differentiable, and differentiable functions are continuous.

    There's a related question that I still don't know the answer to: Is there a mathematical reason why [itex]|\psi(x)|\rightarrow 0[/itex] as [itex]x\rightarrow \pm\infty[/itex]?

    (There exist square integrable smooth functions that don't satisfy this requirement, so the standard answer to this question is wrong).
     
    Last edited: Mar 29, 2010
  6. Mar 29, 2010 #5

    SpectraCat

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    I guess you meant as x--> +/- infinity above? Otherwise I don't understand what you wrote at all.

    If I am right and it was supposed to be x--> +/- infinity, then I would be interested to see an example of a square integrable smooth function that doesn't tend to zero in that limit.
     
  7. Mar 29, 2010 #6

    Fredrik

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    Yes, that's what I meant. Sorry about that. I have edited my post to correct it.

     
  8. Mar 29, 2010 #7

    George Jones

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    Maybe:

    1) [itex]\psi[/itex] is continuous and in the domain of the position operator, or;

    2) (much stronger) [itex]\psi[/itex] is continuous and in the domain of all positive integral powers of the position operator?
     
  9. Mar 29, 2010 #8

    Fredrik

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    I like that general idea, but shouldn't we take it even further? E.g. say that the physical states are the ones such that [itex]f(Q,P)\psi[/itex] is square integrable for any polynomial f.

    (Q and P are of course the position and momentum operators).

    I think something very much like this (or exactly this) is standard in the rigged Hilbert space formalism.
     
  10. Mar 29, 2010 #9

    strangerep

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    Indeed. We want an exponentiated momentum operator to get a representation of
    finite translations, but we define an exponentiated operator in terms of a Taylor
    series. Hence all powers of P must be accommodated sensibly. This notion (and a
    similar one for Q) basically motivates how one defines the "small" nuclear space
    in a Gelfand triple (aka rigged Hilbert space). It must be such that all powers
    of the relevant operators are accommodated in the space. For the case being
    discussed here, the elements of the space must not merely go to 0 at spatial
    infinity, but must do so faster than any power of x increases.
     
  11. Mar 29, 2010 #10
    We know from earlier experience with the delta potential, that the derivative of the wave function can be discontinuous, and ill defined at some points. So your demands for physical states don't look good to me.

    ---

    Another interesting fact is that the domain of the time evolution operator [itex]\exp(-\frac{it}{\hbar}H)[/itex] is usually larger than the domain of the Hamilton's operator [itex]H[/itex]. This is because always

    [tex]
    \|\exp\Big(-\frac{it}{\hbar}H\Big)\psi\|_2 = \|\psi\|_2
    [/tex]

    but sometimes

    [tex]
    \|H\psi\|_2 = \infty
    [/tex]

    even though [itex]\|\psi\|_2=1[/itex]. This is why I would not give the domain of [itex]H[/itex] too fundamental role. The actual time evolution is more fundamental, than the Hamilton's operator, and sometimes the time evolution can be well defined even when the Schrödinger's equation is not satisfied in vector space sense.
     
  12. Mar 29, 2010 #11
    If Psi is described as a function, then the probability for a particle to be in a point will always be zero. Discontinuity doesn't lead to a probability contradiction.
     
  13. Mar 30, 2010 #12
    One enlightening point of view is to simply look at some example.

    Suppose that the wave function is a following square with discontinuities

    [tex]
    \psi(x) =\frac{1}{\sqrt{2L}}\chi_{[-L,L]}(x)
    [/tex]

    Here chi is the characteristic function which equals 1 on the interval [-L,L] and equals 0 outside it.

    The Fourier transform is

    [tex]
    \hat{\psi}(p) = \sqrt{\frac{2}{L}} \frac{\sin(pL)}{p}
    [/tex]

    This might look like that the momentum is approximately zero, but the expectation value of the momentum is actually

    [tex]
    \frac{1}{2\pi} \int\limits_{-\infty}^{\infty} p|\hat{\psi}(p)|^2 dp = \frac{1}{L\pi} \int\limits_{-\infty}^{\infty}
    \frac{\sin^2(pL)}{p} dp
    [/tex]

    Oh! Its [itex]+\infty - \infty[/itex] :cool:

    But how serious is this? I'm not sure. If it is decided that this is the wave function at some instant [itex]t=0[/itex], then the time evolution would be well defined. IMO it could be an acceptable situation where the expectation value of a momentum doesn't exist. The probability density would still be well defined, so isn't everything ok then?
     
  14. Mar 30, 2010 #13
    Continuum wavefunctions do not go to 0 at infinity :smile:
    Who said our beloved plane waves are not physical? :biggrin:
     
  15. Mar 30, 2010 #14
    I was left slightly disturbed by my own comment, because I'm not sure how to give a rigor definition for the delta potential from the operator point of view. It could be that the delta potential problem must be defined in some other way than as an eigenvalue problem for some operator [itex]D(H)\to L_2(\mathbb{R})[/itex], where [itex]D(H)\subset L_2(\mathbb{R})[/itex].

    But notice that even if a potential [itex]V[/itex] is a well defined measurable function so that the multiplication operator [itex]M_V[/itex] exists, then if the potential contains discontinuities, the second derivatives of the eigenfunctions [itex]\partial_x^2\psi[/itex] will contain discontinuities too! And points of non-existence.

    So this idea for "physical states" is too strict.
     
  16. Mar 30, 2010 #15

    strangerep

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    But how do you define the exponential of an operator, if not via a Taylor series:
    [tex]
    \exp(A) ~:=~ 1 + A + \frac{1}{2}A^2 + ..... ~~~~~(?)
    [/tex]
    If so, then all terms in the series must make sense when acting on the state.
    Hence, the domain of the exponentiated operator is restricted to the much
    smaller intersection of the domains of all powers of A.
     
  17. Mar 30, 2010 #16
    If one knows how to diagonalize an operator [itex]A[/itex], then it is possible to avoid the exponential series, and define [itex]\exp(A)[/itex] by defining its action on the eigenvectors.

    For a free non-relativistic particle this means that the definition comes most nicely in the Fourier basis, like this:

    [tex]
    \hat{\psi} \mapsto \exp\Big(\frac{-it}{\hbar}H\Big)\hat{\psi},\quad\quad
    \Big(\exp\Big(-\frac{it}{\hbar}H\Big) \hat{\psi}\Big)(p) = \exp\Big(-\frac{it}{\hbar} \frac{p^2}{2m}\Big) \hat{\psi}(p)
    [/tex]

    I'm not sure what's the most general definition for [itex]\exp(A)[/itex]. It seems that there exists different definitions with different domains of validity, such that these definitions still agree under some simple conditions.
     
  18. Mar 31, 2010 #17

    Fredrik

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    Strangerep's argument about exponentials sounds convincing to me. (Thanks Strangerep, it's always good to get input from you). But now Wigner's symmetry representation theorem confuses me instead. We define a symmetry as a probability-preserving bijection on the set of unit rays of some Hilbert space H, and we impose the requirement that there's a symmetry for each member of the Galilei group (when we want to end up with non-relativistic QM) or Poincaré group (when we want to end up with special relativistic QM). The theorem says (roughly) that there exists a unitary representation of the group of symmetries into GL(H') where H' is a Hilbert space. I always thought that H'=H, but now it seems to me that H' must be a subset of H, at least if the H we started with is the whole space L^2(R) (let's continue to talk about only one spatial dimension to keep the notation simple).

    Does anyone know if the nuclear space H' (i.e., the space of all [itex]\psi\in L^2(\mathbb R)[/itex] such that [itex]f(Q,P)\psi\in L^2(\mathbb R)[/itex] for all polynomials f) is a Hilbert space too? Is H' a "big" or a "small" supspace of H? I mean, is it dense in H or something like that?

    Joostpuur, I don't see how the things you did change anything. Since p^2/2m is an eigenvalue of H, you can replace p^2/2m with H on the right, and now we have recovered the problem you tried to avoid.
     
  19. Mar 31, 2010 #18
    The use of a mapping

    [tex]
    L_2(\mathbb{R})\to L_2(\mathbb{R}),\quad \hat{\psi}\mapsto e^{- itH}\hat{\psi},\quad \Big(e^{-itH}\hat{\psi}\Big)(p) = e^{-it\frac{p^2}{2m}}\hat{\psi}(p)
    [/tex]

    avoids all problems, related to situations where

    [tex]
    \int\limits_{-\infty}^{\infty} |p|^n |\hat{\psi}(p)|^2 dp = \infty
    [/tex]

    with some [itex]n[/itex].

    ----

    Why do you keep this interest to the condition where [itex]f(Q,P)\psi\in L_2[/itex] with all polynomials [itex]f[/itex]? If a potential [itex]V[/itex] contains discontinuities, then the eigenstates, which will be physical states, will satisfy following conditions:

    [tex]
    \psi \in C,\quad \psi \in C^1,\quad \psi\notin C^2
    [/tex]

    This demand [itex]f(Q,P)\psi\in L_2[/itex] may produce some interesting function space, but it doesn't make sense to pose this demand for "physical states".
     
    Last edited: Mar 31, 2010
  20. Apr 1, 2010 #19

    Fredrik

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    Why would you consider solutions to Schrödinger equations with discontinuous potentials to be physical states? I'd say those are just as unrealistic as plane waves. Where do you find a discontinous potential in the real world? How would you design an experiment that prepares a system in such a state?

    I don't see the point of defining the exponential exp(-iHt) by

    [tex]\Big(e^{-itH}\hat{\psi}\Big)(p) = e^{-it\frac{p^2}{2m}}\hat{\psi}(p)[/tex]

    because if [tex]\hat\psi[/tex] is the momentum space wavefunction,

    [tex]e^{-it\frac{p^2}{2m}}\hat{\psi}=\sum_{n=0}^\infty \frac{1}{n!}\Big(-it\frac{p^2}{2m}}\Big)^n\hat{\psi}=\sum_{n=0}^\infty \frac{1}{n!}\Big(-itH\Big)^n\hat{\psi}[/tex]

    So we end up with exp(-iHt) being equal to a power series in H whether we want it or not.

    I got that polynomial condition from an article about the rigged Hilbert space formalism that I read some time ago. I think it was the article that Strangerep posted in the thread I started (a year ago?) about unbounded operators.
     
    Last edited: Apr 1, 2010
  21. Apr 1, 2010 #20
    The series

    [tex]
    e^{it\frac{p^2}{2m}} = \sum_{n=0}^{\infty} \frac{(it)^n}{n!}\Big(\frac{p^2}{2m}\Big)^n
    [/tex]

    will converge towards a complex number such that [itex]|z|=1[/itex]. So

    [tex]
    \int\limits_{-\infty}^{\infty} \Big| e^{it\frac{p^2}{2m}} \hat{\psi}(p)\Big|^2 dp = \int\limits_{-\infty}^{\infty} |\hat{\psi}(p)|^2 dp
    [/tex]

    will be guaranteed. Even if

    [tex]
    \int\limits_{-\infty}^{\infty} |p|^n |\hat{\psi}(p)|^2 dp = \infty
    [/tex]

    with some n, the equation

    [tex]
    \int\limits_{-\infty}^{\infty} \Big|\Big(\sum_{n=0}^{\infty} \frac{(it)^n}{n!} \Big(\frac{p^2}{2m}\Big)^n\Big)
    \hat{\psi}(p)\Big|^2 dp = \int\limits_{-\infty}^{\infty} |\hat{\psi}(p)|^2 dp
    [/tex]

    is still going to remain true. :rolleyes:

    I was thinking about the finitely deep square potential

    [tex]
    V(x) = -|V_0| \chi_{[-L,L]}(x)
    [/tex]

    but if you raise doubts about its physicalness, then I'm not sure how to proceed. I think you might as well ask where do you find smooth potentials in the real world? All mathematical potentials are approximations anyway.

    I remember this thread. I think we clashed already there...

    I didn't like the physicists' way of insisting that the unboundedness leads to continuous spectra.
     
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