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How you come up with the wave func. in the first place?

  1. Jan 3, 2012 #1
    How do you come up with the wave function ?
    I mean in most of the places where I read or watch video, they come up with all these standard wave functions for position or momentum... but I can't find an explanation how you come up with a specific wave function ?
    My understanding is that you build it on some experimental procedure or results ?

    Please give example with something that is not position or momentum if you can.

  2. jcsd
  3. Jan 4, 2012 #2
    It makes correct experimental predictions. What more do you want?
  4. Jan 4, 2012 #3


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    Staff: Mentor

    You solve Schrödinger's equation for the situation at hand.
  5. Jan 4, 2012 #4
    :) with the same success I may just come with a simple formula that explain my test.. and don't bother with any-QM at all, complicating my life.
    Not that this proves anything but just f.e. if I want to calculate the double slit experiment, you can probably concoct a formula with oscillator (sin) and a switch(slit-count) and it will probably work.
  6. Jan 4, 2012 #5
    Hello mraptor!

    I don't think the answer "the experiments say so" is a good one in this case: the experiment justifies the formula in the end, but it doesn't explain what the reasoning behind it was.

    I don't know what you're asking for specifically. Can you give the formula you are wondering about?

    But assuming that it's the formula [itex]\lambda = \frac{\hbar}{mv}[/itex] (this becomes the more general [itex]\lambda = \frac{\hbar}{mv}\sqrt{1-\frac{v^2}{c^2}}[/itex], mentioned above, if you replace m with the relativistic mass), it was indeed de Broglie who came up with this formula and as I understood it, it was because he had heard of light (usually thought of as waves) behaving like particles, with momentum given by [itex] p = \frac{\hbar}{\lambda}[/itex] with lambda the wavelength of the light. de Broglie then took the wild guess that particles may behave like waves, and he postulated the same formula for particles, but the idea is reversed: whereas for light it gave you the impulse (characteristic of a particle) if you know the frequency (characteristic of a wave), for a particle it gives you the frequency if you know the impulse.

    You might now well ask: but where does [itex] p = \frac{\hbar}{\lambda}[/itex] come from for light? It follows from [itex]E = \hbar \omega[/itex] and the relativistic formula [itex]E = pc[/itex].

    Now you ask: and where does [itex]E = \hbar \omega[/itex] come from? From Einstein (originally)! Based on the Planck's solution to the famous black body problem, Einstein deduced that light is made of energy packets with energy [itex]E = \hbar \omega[/itex]. Now you ask: but how did Einstein deduce that? Well, actually, finding that formula wasn't hard: the bold/creative step of Einstein, was the conceptual step of light being particles instead of waves; once you've made this conceptual step and you look at Planck's mathematical work, it follows quite naturally that if light is a particle, it must have energy [itex]E = \hbar \omega[/itex].

    So the formula [itex]\lambda = \frac{\hbar}{mv}[/itex] is a byproduct of Planck's solution to the black body problem, Einstein's bold move of saying light is a particle, and de Broglie's bold move of saying a particle is a wave.
  7. Jan 4, 2012 #6
    hi mr.vodka,

    thanx for you thoughtful explanation.
    OK.. here is one example look f.e. this explanation :

    http://www.ipod.org.uk/reality/reality_quantum_casino.asp [Broken]

    Why do they use "traveling wave equation" ? Why not for example use simply sin() which is also a wave ? Or some other combination to make a wave ?

    What if I want to measure something which does not have formula yet worked out in QM, something different than position or momentum ?
    F.e: I've seen they use different formulas for electrons in the hydrogen atom !? bunch of wave functions which is OK, but someone have invented them ? How ?
    I don't need the formulas I just want to know what is the process, what they take into account, etc...and so on..
    In a way that newbie can understand it... :)
    Last edited by a moderator: May 5, 2017
  8. Jan 4, 2012 #7
    You're right, it's not the most general expression for a wave. They simply took that one as it's not too complex. One could even take a standing wave! For example

    [itex]\psi(x,t) \propto cos(kx-\omega t) + cos(kx+\omega t) [/itex]

    But they don't do this because it wouldn't have a well-defined momentum (check that by using the momentum operator), and they first want to treat a simple example where the wave has a specific momentum associated to it.

    Is this answer in the right direction? Or am I missing your point?

    EDIT: I see you added
    to your post.

    First let me note that any [itex]\psi[/itex] is allowed! There are no restrictions! (except such things as continuity) This is perhaps not stressed enough. Once you've chosen a certain wave function at a certain time, it then evolves according to the Schrödinger equation [itex] i \hbar \frac{\partial }{\partial t} \psi(x,t) = -\frac{\hbar^2}{2m} \frac{\partial^2 }{\partial t^2} \psi(x,t) + V(x,t) \psi(x,t) [/itex].

    For convenience, one is often interested in wave functions satisfying [itex] E \psi(x,t) = -\frac{\hbar^2}{2m} \frac{\partial^2 }{\partial t^2} \psi(x,t) + V(x,t) \psi(x,t) [/itex] where E is a certain constant which hasn't been fixed yet. Those kind of psi functions are called eigenstates of the hamiltonian. Note that they're not more likely than any other psi function; they're simply more useful in certain contexts.

    The [itex]\psi_{n,l,m}(\textbf r,t)[/itex] wave functions you have heard of in the context of the hydrogen atom is such a psi: an eigenstate of the hamiltonian (with the potential V that corresponding to the coulomb field of the proton). How to find it's exact form: solving the equation with [itex] E \psi[/itex] I wrote above. How to do that? Lots and lots of math :p Not by guessing, that's for sure! There are certain methods...
    Last edited: Jan 4, 2012
  9. Jan 4, 2012 #8
    hmm.. thanx alot... I seem to see the light at the end of the tunnel...
    What I get is that I try to solve the above equation...to find a function...
    It is in a sense trying to solve algebraic equation finding the unknown, but in this case the unknown is a function which could be almost any wave-like function, but some are better than other..
    One very quick question : Doesn't different wave functions give different results ? Or because we use Hermetian operator they will give similar answers !!

    Thanx again I think I get it now, even if I don't know the math to do it.. I don't expect it to know it anytime soon..I just want to understand QM in are more deeper level...
    Most of the movies, books or articles on the Internet are either for math whiz's or for total noobs afraid of formulas.
    There is no information for "middle ground" people which are not afraid of little bit of math.
    Wikipedia in most of the cases seems like a good reference which are already familiar with the matter, but not good for learning.
    I'm currently watching on youtube the lectures of prof. Susskind's .... finally know what braket-complex-vectors and Hermetian operators mean... ;)

    thanx again
  10. Jan 4, 2012 #9
    Check out the third volume of the feynman lectures :)

    As for
    Can you rephrase? I think you're misunderstanding something, and as a result I don't understand your question, so try to explicitly state what you think is true and why you think it might be a problem. The question "don't different wave functions give different results" seems to imply that you think all wave functions must give the same predictions? This is not true! Think of classical mechanics: a particle can be fixed, or have a certain velocity, those are two different states, which will lead to different predictions.
    Choosing a wave function in QM is analogous to choosing the x,y,z coordinates and the momentum of a particle.

    EDIT: actually, based on the level you're looking for (i.e. formulas if needed to better understand a concept but not more), I warmly advise https://www.amazon.com/Sneaking-Look-Gods-Cards-Revised/dp/069113037X/ref=tmm_pap_title_0
    Maybe your local library has a copy if you're not certain. Maybe it's not enough math for you though. But at least it's a very decent book which doesn't go for the easy sell but really wants you to understand it :)
    Last edited: Jan 4, 2012
  11. Jan 4, 2012 #10
  12. Jan 4, 2012 #11
    Are you asking WHY nature is quantum mechanical to begin with? Are you wondering what fundamental principles give rise to quantum mechanics and the wave function? Are you expressing dissatisfaction with what appears to be just curve fitting efforts of physicists to find whatever equations match the experimental data without a complete logical explanation for the math? Those are good questions.
  13. Jan 4, 2012 #12
    Ah, but this isn't a problem: different wave functions will indeed give different eigenvalues. This is not so strange: say you have two functions with different E's, then you have two states of the system with different energies, much like in classical mechanics: your system can have different energies depending on its variables, right?

    Why did you think that all the eigenstates of the hamiltonian must give the same predictions?

    @ friend: I don't think he's asking that, while they are good questions. He's asking other good questions: how exactly does QM work, mathematically speaking (without learning a terse QM book).
  14. Jan 4, 2012 #13
    I seem to think that what you do in an experiment is you setup a scenario. Say for example two slit experiment (suppose the physicist don't know anything yet about it).
    They do the experiment and see this wave interference and say.... hmm this is some sort of wave.. we have this QM theory which worked for all those other cases.
    Let us pick a wave function.. that solves the equiation (I think you call it Hamiltonian, havent gotten to that yet), this will guarantee us a correct results.

    Now let's shoot a photon and solve the " Op*Psi(x)=e-val*Psi(x) " for it, this will give us the probability of detecting this photon at position 'x'

    m'I even close ;)...

    If I'm you can see why I can seem to feel confused of getting different values using different wave-functions. This will mean depending on what wave-function I pick I will get different probabilities for finding the particle in that particular position 'x'.

    On the other hand when you talk of getting different values when u have different energies seem logical.
  15. Jan 4, 2012 #14
    I'm expressing dissatisfaction with the math abstractions they come up ... I'm joking :).. I'm not dissatisfied with it It will be more correct to say I'm puzzled of the whoops the QM scientists have to go trough to calculate even a simple interaction.
    Doesn't bother me to much, if nature is this way, it is this way...

    Just hoped in my dreams that after 100 years mathematicians would be able to come up with simple abstractions to calculate those things for the mere mortals ;))
  16. Jan 4, 2012 #15
    Hm, you seem to have some misconceptions. Let me try to get rid of some:

    First of all, you seem to think the scientist would calculate the psi function directly at the detection plate. This is however wrong. To be clear: it is what he eventually wants, but he cannot get it so cheaply. It's good to think about the classical analog: if a scientist wants to calculate where a particle will be, it will not have an equation that it has to solve which immediately gives the prediction: he has to put in an initial condition!

    Let me state it more clearly: in the case of the two-slit experiment, you don't know (a priori) where the system will eventually be, but you do know how it starts out: let's suppose that you shoot the particles horizontally with a fixed momentum p. Hence we know the initial wave function (classical analogy: we know the initial position and momentum). The wave function that is directed to the right and has a fixed momentum is proportional to [itex]e^{ipx / \hbar}[/itex]. This gives us [itex]\psi(x,y,z,0)[/itex]: the initial condition.

    Now this initial wave will evolve according to the Schrödinger equation (classical analogy: a particle with a certain position and momentum at time zero will evolve according to Newton's second law). Much like in classical mechanics we have to solve Newton's law to find the time evolution, we now have to solve [itex]i \hbar \frac{\partial}{\partial t} \psi(x,y,z,t) = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} \psi(x,y,z,t) + V(x,y,z) \psi(x,y,z,t)[/itex]. This equation is all you need. This equation is fundamental. You also know another equation, where the [itex]i \hbar \frac{\partial}{\partial t} \psi(x,y,z,t) [/itex] is replaced by [itex]E \psi(x,y,z,t) [/itex]. That equation is not fundamental but just happens to be convenient for some purposes at some times, but for now: forget it!

    What is important is that we have our initial condition, and our Schrödinger equation, which determines the time evolution of that wave function. With some math you thus get the correct wave function at all times. You then look at [itex]x = L[/itex] (assuming the screen is at x = L) at a later time, when the wave has reached the screen, and then and only then (after taking the square modulus) have you obtained the probability distribution across the plate.

    This finishes the double slit experiment.

    Now what about the [itex]E \psi(x,y,z,t) [/itex] equation?

    Well, first remember the [itex]e^{ipx / \hbar}[/itex] wave function. I said it represents a wave function with momentum p. Why did I say that? Well, we say that a wave has a certain value x of the observable X if the wave function is an eigenvector of the corresponding operator [itex]\hat X[/itex] with eigenvalue x (I know that's a tricky sentence, but read it carefully: x, X, and [itex]\hat X[/itex] are all different things, but intimately connected!). An example will help: lets denote the momentum observable by P (by "momentum observable", I mean the physical concepts of momentum). The specific value I mentioned I called small p. Now the observable P has an associated operator that works on wave functions. For momentum it's [itex]\hat P = -i \hbar \frac{\partial}{\partial x}[/itex] (keep it one-dimensional at the moment). This formula might look mysterious, but let's accept this formula at the moment, okay? It's a well-defined operator: if you give it a function, it returns another. Now again take my earlier sentence
    or applied to momentum:
    I.e. given a function psi we say it has a certain momentum p if [itex]\hat P \psi = p \psi[/itex] (because this is by definition what it means to be an eigenvector of [itex]\hat P[/itex] with eigenvalue p).

    You see that [itex]e^{ipx/ \hbar}[/itex] indeed satisfies this equation! Hence we say that this wave has momentum p.

    Now back to the [itex]E \psi(x,y,z,t) [/itex] equation! It's completely analogous! Above was for momentum, but we can do the same for energy:
    and traditionally [itex]\hat E[/itex] is denoted by [itex]\hat H[/itex]. The form of this operator (again: accept it for now!) [itex]\hat H = -\frac{\hbar^2}{2m} \frac{\partial^2}{\partial x^2} + V[/itex]. We say that a wave function has energy E if and only if [itex]E \psi = \hat H \psi[/itex]. Analogous to the above with momentum!

    Note that the Schrödinger equation can be rewritten as [itex]i \hbar \frac{\partial}{\partial t} \psi = \hat H \psi[/itex]. So the fact that the Schrödinger equation looks so much like the [itex]E \psi[/itex] equation is simply because it so happens that the operator corresponding to energy appears in the Schrödinger equation. But you see they're two very different equations: the Schrödinger equation is fundamental and determines the evolution of your system. The [itex]E \psi[/itex] equation is simply used to find wave functions with certain energies, much like [itex]\hat P \psi = p \psi[/itex] is used to find psi's with certain momenta (like [itex]e^{ipx/ \hbar}[/itex]).

    (It would perhaps be more logical to denote a value of energy by a small e instead of a big E, to be consistent, but again, tradition dictates we use a big letter for a value of energy, and luckily so, because it would be confusing with euler's constant!)
  17. Jan 4, 2012 #16
    I think it is getting clearer... You use [itex]e^{ipx / \hbar}[/itex] because it is part of your initial conditions. This initial wave function eventually changes according to the Shredinger equation, which you have to solve at the time you want to know how this function evolved.
    Once you have it (for this particular time) you then use it to calculate the probability distribution.

    The thing I didn't know and I first asked is how you decided to use [itex]e^{ipx / \hbar}[/itex], but now I see it is more a general physics question, rather than a QM question (if I may say so).
    In this case I understand why you used it, cause it is the formula to use for momentum wave.
    My error was that I imagined that when a physicist setups an experiment he doesnt know the initial wave function and has to do some other calculation and/or thinking/experimenting before he decides on one from the infinite of possible wave-functions. But from your explanations it seem that this should be part of the initial preconditions. When you start an experiment you should have at least vague idea of what the initial wave function will be. How it evolves later is a matter of calculations.

    thank you very much.
  18. Jan 4, 2012 #17
    I agree completely with what you say, I think you're getting the hang of it :)

    Indeed, to figure out the wave function, you first have to think physically: okay what am I doing, I'm shooting something to the right with a certain momentum. And once you know that you get into the quantum realm and try to figure out what wave function fits such a description (by using the operator and the eigenvector and eigenvalue equation).

    So this concerns the link between the physical reality and the quantum formalism. The exact link is actually still controversial, even after 100 years! When people say "we don't really understand QM" (quoting Feynman), what he is talking about specifically is the link between physical reality and our formulas.

    There are lots of different interpretations of the formulae, giving different physical pictures. The usual picture is "Copenhagen Interpretation" but it's hard to characterize it, as it has different varieties. But mainly, it's characterized by a certain pragmatism: "don't think too much about what it means" and in most cases this indeed suffices (for example I could find the above momentum state without putting too much thought into it), but many people are dissatisfied with this view as it doesn't give one big picture; it's not fool proof, in a way, in the sense that there aren't a clear set of correspondence rules which are always true, but it is sufficient to solve practical problems. It is dubbed the "shut up and calculate method" by Mermin.

    Other interpretations include multiverse, pilot wave theory, consistent histories, etc. There is a lot of interesting literature on this too. But I advise you first get acquainted with the math, in the sense that you're doing at the moment.

    Good luck and enjoy!
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