Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Wavefunctions that don't satisfy Schrödinger equation

  1. Jan 23, 2015 #1
    If there exists some normalized wavefunction ##\psi## that is not a solution to the Schrödinger equation (1D), what does this mean? You can still presumably use the square of the wave function to ascertain the probability it exists at some interval in space, but does it provide any other useful information about the state of the particle? Why or why not?

    Note: I am learning about QM in a physical chemistry course, so I am not familiar with the mathematical nuances of the physics.
     
  2. jcsd
  3. Jan 23, 2015 #2
    It can still describe the particle at t=0, but can not tell you how the wave-function will evolve. Consider what happens when a particle encounters a change in potential and the wavefunction changes. The old wavefunction applies at the boundary, but only the new wavefunction can describe what happens on the other side of the boundary.
     
  4. Jan 23, 2015 #3
    So when you find the probability of finding the particle by integrating the square of the wave function in such a case, you are only getting the probability at t=0? Whereas if the wave function satisfied the Schrodinger equation, it would allow you to find the probability (as well as other attributes) at all times?
     
  5. Jan 23, 2015 #4

    strangerep

    User Avatar
    Science Advisor

    What matters is the dynamical algebra of observables that characterize a particular (class of) physical system(s). The only purpose of the Hilbert space and the wave functions therein is to be a framework for constructing a unitary representation of that algebra of observables.

    Consider a similar question in classical mechanics: one has a phase space (positions and momenta), and observables are functions on that phase space. The equation(s) of motion for a particular classical system define a solution space which is subspace of the larger phase space. Dynamical symmetries of that system map solutions into solutions, hence they leave the solution space invariant. The point here is that the larger phase space is just a mathematical framework. To describe a particular class of systems, we must specify equations of motion, and the observables of that system then corresponding to generators of the symmetries which preserve that equation of motion.

    The QM case follows similarly, except that instead of representing observables as functions on phase space, we represent them as operators on a particular Hilbert space of wavefunctions which satisfy a particular Schrodinger equation. I.e., you can think of the Hilbert space analogously to the classical solution space: we're only interested in wave functions which are relevant to a particular physical situation. In general, the Hilbert spaces of wave functions relevant to distinct physical situations are different. Compare, e.g., free motion with the bound motion in a (subthreshold) Kepler/Hydrogen problem.

    So... if you have a normalized wave function, but no Schrodinger equation, you're not describing any actual physics.
     
  6. Jan 23, 2015 #5
    Thanks strangerep,

    Some of that went over my head, but this is what I gathered from your explanation

    Having a wave function that is not a solution to Schrödinger's equation is like having a classical object that doesn't obey Newton's second law: in both cases, you cannot predict the future state (or, in other words, how the particle will evolve over time).

    Is this correct?
     
  7. Jan 23, 2015 #6

    strangerep

    User Avatar
    Science Advisor

    Yeah, I became a bit worried by that possibility halfway into my post.

    I was really just trying to counter the common misconception that it's the wavefunction that's primary. In fact, it's the observables characterizing a (class of) systems which are primary.

    ("Observables" are things like: energy, momentum, angular momentum/spin, etc. For more complicated systems there may be more, e.g., the Runge-Lenz vector for Kepler/Hydrogen.)


    I'd phrase it more strongly: such "objects" are merely in your imagination, with no correspondence to reality.
     
  8. Jan 23, 2015 #7

    TeethWhitener

    User Avatar
    Science Advisor
    Gold Member

    A somewhat less abstract answer:
    The word "normalized" is very important here. The normalization condition (in 1D, for example) is [tex]\int_{-\infty}^{\infty}{|\psi|^2 dx}=1[/tex]Functions that fulfill this condition are called square integrable, or [itex]L^2[/itex], and the set of all such functions forms a vector space (this vector space has particular properties, for which it gets a special designation as a "Hilbert space," but that's not terribly important for us right now). A vector space is a lot like ordinary 3D space, except it can have an arbitrary number of dimensions (even infinite), and whereas each point in 3D space represents a position, each point in the [itex]L^2[/itex] vector space represents a function (how about we call it [itex]\psi[/itex]?).

    One important property of vector spaces, which might be familiar to you from linear algebra, is that every vector space is spanned by a basis. Every point in 3D can be represented as a vector of the form [itex]x \mathbf{\hat{i}} +y \mathbf{\hat{j}}+z \mathbf{\hat{k}}[/itex], so we can say that [itex]\mathbf{\hat{i}}[/itex], [itex]\mathbf{\hat{j}}[/itex], and [itex]\mathbf{\hat{k}}[/itex] form a basis which spans 3D space. In exactly the same way, we can choose a basis set [itex]\phi_i, i \in \mathbb{N}[/itex] that spans [itex]L^2[/itex] space, such that an arbitrary square integrable function [itex]\psi[/itex] can be represented as a vector of the form [tex]\psi =\sum_i{a_i \phi_i}[/tex]The above framework is important because the Schrodinger equation is linear in each of its variables. This means that any linear combination of solutions is also itself a solution to the Schrodinger equation. So if you have a set of eigenfunctions [itex]\phi_i[/itex] that solve the Schrodinger equation for an arbitrary potential [itex]V(x)[/itex], then any function [tex]\psi =\sum_i{a_i \phi_i}[/tex]also solves the Schrodinger equation for that potential. It also just so happens that any set of eigenfunctions which solve the Schrodinger equation also span the relevant vector space.

    So to sum up, if you have a normalized wavefunction, it must be a solution to the Schrodinger equation, with arbitrary potential. It likely won't be an eigensolution, but it will definitely be a linear combination of them.

    [Technical note: I'm not a mathematician, so I don't know if these statements are true everywhere, or merely almost everywhere (everywhere except a set of measure zero). I expect the latter is true, but I don't know enough about measure theory to speak to it.]
     
  9. Jan 23, 2015 #8

    Matterwave

    User Avatar
    Science Advisor
    Gold Member

    I feel like there's a few points that have been missed or glossed over so far.

    1) Which "Schroedinger equation" are you talking about? The time-independent SE or the time-dependent SE?

    2) The time-independent SE is really just an eigenvalue equation. The solutions of this equation, using the normal terminology, usually means the eigenfunctions. We solve for the eigenfunctions in order to make solving the time-dependent SE easier. That's the entire point of the time-independent SE. A wave function which is not "a solution" (i.e. not an eigenfunction) of the TISE, as TeethWhitener points out, can be represented by linear superpositions of the eigenfunctions. This is because, as mentioned, the eigenfunctions span the space (a proof of this; however, is non-trivial).

    3) The time-dependent SE is where the physics actually lies. Solving the TDSE is the goal of quantum mechanics. A solution of the TDSE is a wave function which evolves in time according to the TDSE. In other words, given some initial wave function, the wave function at any subsequent time must be found by solving the TDSE forward in time. Therefore, a wave function which does not evolve in time as the TDSE dictates is not a solution of the TDSE. Such a wave function would be unphysical. The Schroedinger equation is a postulate of quantum mechanics, so if you work with wave functions which are not solutions of the (TD) Schroedinger equation, then you are not doing quantum mechanics.
     
  10. Jan 23, 2015 #9
    "Not a mathematician"? Close enough.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Wavefunctions that don't satisfy Schrödinger equation
  1. Schrödinger equation (Replies: 11)

Loading...