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I Form of wavefunctions

  1. Apr 29, 2016 #1
    Hi all!
    I'm currently watching MIT 8.04 (Quantum Physics I) on MIT open courseware. I have just finished lecture 5. In the past 2 lectures, they introduced opperators, specifically momentum, energy and position. To prove/derive (I'm not sure what the correct term is) the momentum and energy opperators, they operated on ##e^{ikx}## and got back to the De Broglie equations. So, my question is, do wavefunctions have to be in the form ##Ae^{ikx}##? I would think no, but if I am correct, would the momentum and energy opperators work for other types of wavefunctions? Why is it enough to prove it just with ##e^{ikx}##? This especially confuses me because ##e^{ikx}## isn't even normalizeable, and thus can't even be a wavefunction.
  2. jcsd
  3. Apr 29, 2016 #2


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    A wavefunction is any functions satisfying the Schroedinger equation. The specific form you have there is an eigenfunction of a free particle (##V(x)=C<E##).
    By "work", do you mean operate or act on? Yes they can, except for a special case of infinite well where the momentum operator is not defined.
    The free particle's eigenfunction does not correspond to a realizable wavefunction, but it can serve as the eigenfunction of a realizable (normalizable) wavefunction.
    Last edited: Apr 29, 2016
  4. Apr 29, 2016 #3
    Ok, so my big question is "why can you say that if it works on ##e^{ikx}## it works on any wavefunction (with a few exceptions but I don't want to focus on those)?"
  5. Apr 29, 2016 #4


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    I'm not exactly sure what you mean by "work", but every wave function [itex]\psi(x)[/itex] can be written as a superposition of exponentials:

    [itex]\psi(x) = \sqrt{\frac{1}{2\pi}} \int dk \tilde{\psi}(k) e^{ikx}[/itex]

    Since the operators of quantum mechanics are linear (typically), you can (usually) take them inside the integral:

    [itex]\hat{p} \psi(x) = \sqrt{\frac{1}{2\pi}} \int dk \tilde{\psi}(k) \hat{p} e^{ikx} = \sqrt{\frac{1}{2\pi}} \int dk \tilde{\psi}(k) (\hbar k) e^{ikx} [/itex]
  6. Apr 29, 2016 #5
    Oh, so if I'm getting you correctly, if something applies to ##e^{ikx}## then it applies to all wavefunctions because a wavefunction can be expressed in terms of ##e^{ikx}##.
  7. Apr 29, 2016 #6


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    Yes, in principle, if you know how an operator acts on [itex]e^{ikx}[/itex], you can figure out how it works on any wavefunction.
  8. Apr 29, 2016 #7


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    Wave equations were important in physics long before the invention of quantum mechanics. A wave equation is a function which satisfies the second order differential equation ∂²F/∂t² = c²∇²F. Probably, you haven't seen the ∇² ( ) operator before. An operator acts on something (a number or a vector or a function or...) You're familiar with the addition "operator" +, and the cosine operator cos( ) and probably the derivative operator d( ). The Laplacian operator ∇² (del squared) is also symbolized as ∇⋅∇ or sometimes Δ but since it must act on something, it isn't something that should be thought of as existing by itself (unlike a number, vector or function). This is simplified, of course. Anyway ∇²(F) for 1 dimension is ∂²F/∂x² (assuming x is the dimension) while for 3 dimensions it is ∂²F/∂x² + ∂²F/∂y² + ∂²F/∂z² (assuming cartesian coordinates (dimensions) x,y,& z). Functions of the form e^x have derivatives of every order, so they are a "natural" for use with differential equations. The function e^ix is, by definition, cos(x) + i*sin(x) (and you probably know that sin & cos also have derivatives of all orders...). Also the imaginary constant i here can pretty much be treated as any other number (with the obvious point that i*i = -1), so shouldn't throw your understanding. Since either ∑An*sin(x) where An is a set of constants with n ranging from 1 to ∞, or ∑Bn*cos(x) can (in the limit) exactly equal any function, F(x). The utility of expressing any function in terms of its (fourier series) basis should be apparent. Note that the notation can be confusing since it is abbreviated short-hand. For instance in the above sums, "x" is meant to be short-hand to imply "for the variable x in its range of validity, for any x..." So, a lot is implied and lecturers try to start introducing that to the student gently. I've never seen an advanced text (or paper) write out in full what exactly the authors mean, they all use shorthand, so get used to it. And get used to expanding their equations into a form meaningful to you. I suppose I should also note that classical wave equations usually were differentials with respect to spatial dimensions (coordinates) and time, but in relativistic QM time and space are mixed so ∇² may range from 1 to 4 (or from 0 to 3)... As already answered, using the form e^ix for wave functions makes working with them much much easier and you can do it without losing the universality of application to any wave equation...they're general purpose. Plus, for the simple cases (infinite potential well, harmonic oscillator, etc.) they make things neat and tidy....and easy!
  9. Apr 29, 2016 #8
    I actually have experience with the Laplacian opperator but I generally think of it as div(grad(F)) and not del squared. That is interesting though. I have worked with it in classical mechanics but not in quantum mechanics.
  10. Apr 29, 2016 #9


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    EDIT: Sorry a mistyping, in the last sentence in post #2 I should have said basis function, rather than eigenfunction.
  11. Apr 30, 2016 #10


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    ... and, very important, a wave function, representing a pure quantum state, must be square integrable, i.e., the integral
    $$\|\psi \|^2=\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} |\psi(\vec{x})|^2$$
    must exist. In order to represent a probability amplitude one must be able to normalized the wavefunction such that ##\|\psi \|=1##, and then ##|\psi(\vec{x})|^2## is the probability distribution of the particle's position.

    The plane-wave functions are generalized functions (or distributions) in the sense of functional analysis. They cannot represent states, but you can represent wave functions in terms of these generalized functions. In the case of the momentum eigenstates (I set ##\hbar=1## for convenience)
    $$u_{\vec{p}}(\vec{x})=\frac{1}{(2 \pi)^{3/2}} \exp(\mathrm{i} \vec{p} \cdot \vec{x}).$$
    You can represent any position-space wave function in terms of the corresponding momentum-space wave function. Since the momentum eigenstates are represented by the exponential function that's just Fourier transforming between the two representations:
    $$\psi(\vec{x})=\int_{\mathbb{R}} \mathrm{d}^3 \vec{p} \tilde{\psi}(\vec{p}) u_{\vec{p}}(\vec{x})$$
    and the other way around
    $$\tilde{\psi}(\vec{x})=\int_{\mathbb{R}} \mathrm{d}^3 \vec{x} \psi(\vec{x}) u_{\vec{p}}^*(\vec{x}).$$
    The momentum eigenstates are of course not square integrable, but they are "renormalized to a Dirac ##\delta## distribution" in the sense that
    $$\int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} u_{\vec{p}_1}^*(\vec{x}) u_{\vec{p}_2}(\vec{x})=\delta^{(3)}(\vec{p}_1-\vec{p}_2).$$
  12. Apr 30, 2016 #11


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    To the OP the full resolution to this requires what are called Rigged Hilbert spaces that was developed by some of the greatest mathematicians of the middle 20th century eg Schwatz (distribution theory), Grothendieck (Nuclear Spaces) and Gelfland (Gelfland Triple). It will take a lot of study to come to grips with that.

    But as a start I highly recommend the following:

    Last edited by a moderator: May 7, 2017
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