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

Basic QM questions

  1. Sep 10, 2004 #1

    I have just started an introductory course on quantum mechanics, and there are some things that I can't figure out. Similar questions have probably been answered several times before, so links to other threads are very welcome (I can't seem to find them using the search-function, I'm sorry).

    Here goes:

    My textbook starts out by stating that all matter posess both particle- and wave-properties. It introduces something called the "de Broglie wavelength" for particles, which is [tex] \lambda = h/p = h/mv [/tex].
    Later on, it introduces "wave functions" for particles (denoted [tex] \psi(x,y,z) [/tex] or [tex] \Psi(x,y,z,t) [/tex]) which must satisfy Schrödingers equation. As I understand, the (complex) wave function, in itself, has no physical meaning, but [tex] |\psi(x,y,z)|^2 [/tex] is somehow (?) related to the probability of finding the particle near (x,y,z).

    My questions here are:

    1) What is the relation between the de Broglie wavelength for a particle and the wave function for that particle? If the wave function has no physical counterpart, then what is the wavelength? In classical physics the wavelength is the distance between two points of a wave that are in the same "state" - is that also the case with the de Broglie wavelength? If so, what does this "state" refer to (in sound waves it is related to pressure, in em-waves it is related to the E- and B-fields etc.)? Bottomline: In what sense are particles waves?

    2) [tex] |\psi(x)|^2dx [/tex] (we are working in 1-d here) is said to be (proportional to?) the probability of finding a particle in an interval dx around x. But this probability must depend on how long you are looking at this interval, so what exactly is meant by this?

    3) Like I mentioned above, my book talks about two wave functions - a time-dependent [tex] \Psi(x,y,z,t) [/tex] and a time-independent [tex] \psi(x,y,z) [/tex]. Even though there is a time-dependence they also consider (and use) the time-independent version. What meaning does this have in a situation where there is time-dependence?

    I have several other questions, but these are the main ones I guess (besides the ones I forgot).
  2. jcsd
  3. Sep 10, 2004 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Hello !

    You ask good questions. I won't attempt to answer them, because there are different ways to answer them, people can disagree on what is important and not (for example, "is the wavefunction physical") and in the end it will be confusing.
    Allow me to give you an advice. Read up, different books and texts, with different viewpoints. Get technically fluent, do a lot of calculations, exercises and so on, and try, in the beginning, to take on the "shut up and calculate" attitude ; meaning: don't stop on the questions you ask above (and others), but accept what you read and get through the technicalities.
    After you have acquired enough technical skills, it is time to come back to all these issues.

  4. Sep 10, 2004 #3

    I would like to give an answer to question two.

    The wavefunction you give is time-independent. this means that the probability you wrote is onely dependent of position. Basically you gotta look at it like this : suppose you have an x-value and you put it into the probability. Then you wil get the "chance" that the particle is located there. That's all.

    The point you made about the duration of looking is very true. Now you are referring to the uncertainty of energy-interval and time-interval. If the period is very big, then the corresponding energy-interval is very small. Now like stated in the above post you can interprete these results in different ways. When one talks about intervals, it is important to know how they are defined, so what is the initial and final-value of the quantity at hand.

    Also, remember that stationary solutions of the Schrödinger equation can be factored out in a time dependent and a time-independent part. something like f(x,t) = a(x)*b(t)

  5. Sep 10, 2004 #4
    1) With the periodic behaviour in the spatial coordinates of the wavefunction you can associate a wavelength. The de Broglie wavelength

    2) if you measure in an interval dx at a certain time, this measures the probability that you find the particle there at that time. This probablity depends (in the time dependent case) on time, but not on 'how long you are looking'. When you look, the particle is there or it isnt!

    3)The time independent solutions are complete, in the sense that every function can be constructed as a linear combination of them, involving (in most cases) an infinite amount of terms (a Fourier series). So also the general time-dependent wavefunction can be written as a linear combination of the time-independent functions. If there is only one possible energy E for the particle you are looking at, the general time dependent solution is [itex]\Psi (x,t)=\psi (x) e^{-iEt/ \hbar} [/itex]. If there are more possible energies the general wavefunction is:

    [tex]\Psi (x,t)= \sum_{n=1} ^{ \inf} c_n \psi_n (x) e^{-iE_n t/ \hbar} [/tex]

    You only have to figure out the time-independent solutions (the TISE yields an infinite amount of them!), with the related energies, and determine the coefficients [itex]c_n[/itex]. When you do a measurement the probability of finding a certain energy [itex]E_n[/itex] is given by [itex]|c_n|^2[/itex].
  6. Sep 12, 2004 #5
    Thanks for your answers. I think I have a clue on 1) and 2) now - for all the rest I think I'll adopt the "shut-up-and-calculate"-strategy :)
  7. Sep 12, 2004 #6


    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Hi there. Good questions.

    The real answer is that, strictly speaking, the wavfunctions *always* have a time dependence. However, in the case of wavefunctions corresponding to state with a well-defined energy ("eigenstates of the hamiltonian", in physics speak), the time dependence is a pure phase (i.e. an imaginary exponential), so [itex] \Psi(x,t) = e^{-i E t/ \hbar} \psi(x) [/itex]. Now, a pure imaginary phase like this is totally unimportant in quantum mechanics because it drops out of most expectation values, such as [itex] \int \psi^* x \psi [/itex] or the same thing with the momentum operator and so on. So people usually just drop the time dependence and just write the spatial part [itex] \psi(x)[/itex]. However, when a state is linear combination of states of different energies, one cannot ignore those exponentials and they must be included explicitly.

    Hope this makes sense. Keep asking questions until everything is clear!

  8. Sep 13, 2004 #7


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    Number 3 isn't too difficult to answer. If you express the wavefunction as ψ(x,t)=u(x)T(t), the Schrödinger equation separates into two equations, one for u and one for T.

    When you solve the time-independent equation Hu=Eu (where H is the Hamiltonian and E is an eigenvalue of H), you will find all possible energy levels of the system (eigenvalues of H), and the wavefunctions that have a well-defined energy (eigenfunctions of H). A wavefunction is in general a superposition of such energy eigenstates. (See below).

    When you solve the other equation, you will find that you can express the time-dependent wave-function as ψ(x,t)=exp(-iHt)ψ(x,0). This result can be interpreted in two different ways:

    1. The wavefunction is time-dependent, and its time-dependence is described by ψ(x,t)=exp(-iHt)ψ(x,0). This interpretation is called the Schrödinger picture.

    2. The wavefunction is time-independent, so we should write ψ(x) instead of ψ(x,0). The function f defined by f(x)=exp(-iHt)ψ(x) is the wavefunction that would be used by another observer, who is at your position t seconds after you. This interpretation is called the Heisenberg picture.

    I suspect you will only be working in the Schrödinger picture for a while. In the Heisenberg picture, wavefunctions are time-independent, so they can be thought of as representing all information about the system, in the past, present and future (at least until somebody makes a measurement, but that's a whole different story).

    I will say a few words about 1 and 2 as well. Only an energy eigenstate, i.e. a wavefunction of the form ψ(x)=exp(-ipx) (where p is the momentum) has a well-defined wavelength. But any wavefunction can be expressed as superposition of energy eigenstates:


    If the function f is sharply peaked around some value of p, that value corresponds to an approximate wavelength.

    I should probably also mention that I'm using units in which


    In these units, the wavelength is just 1/p.

    Question 2 is more difficult to answer. This might be a good time to use the "shut up and calculate" strategy. :smile: A partial answer is that in the formulation of quantum mechanics, measurements are treated as if they are instantaneous processes (which they of course aren't). I'm not sure I can give you a much better answer than that.
  9. Sep 15, 2004 #8
    Thank you, excellent feedback.. I still have some questions though:

    nrqed>> What do you mean by this: "a pure imaginary phase like this is totally unimportant in quantum mechanics because"??

    Fredrik>> Does ψ(x)=ψ(x,0) mean that ψ(x) is the wave function corresponding to t = 0 - and that this may or may not change with time? And, if the wavelength is undefined in most situations, what does the de Broglie wavelength h/p represent? What value of p should one use, when p isn't really well-defined?? Finally I'm not sure I'm getting the Heisenberg picture ;)
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Basic QM questions