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Just a quicky! (Schrodinger's equation)

  1. Apr 2, 2009 #1
    1. The problem statement, all variables and given/known data

    Currently im doing a question where its asked me to show that the probability per unit length of finding a particle is independent of space and time, and is just a constant.


    2. Relevant equations



    3. The attempt at a solution


    The plane wave state ive been given to solve is:

    [tex]\psi(x) = exp(ikx)[/tex]

    I separated the variables of the SE and got a formula for [tex]\Psi(x,t) = \psi(x)exp(-iEt/hbar)[/tex]

    Assume lowercase psi here is constant, i cant be bothered to write it out :P

    Obviously the probability per unit length is the square of the modulus of this function.

    For some reason it shows in my notes that when you do this, the exponential part of the wavefunction dissapears when you do the square of the modulus, so you are just left with the square of the modulus of psi.

    Now that would leave me with the answer, as ive already worked out psi is a constant independent of x, but I cant figure out why the exponential part dissapears. Is it just some basic maths that im not getting here?
     
  2. jcsd
  3. Apr 2, 2009 #2

    Pengwuino

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    Gold Member

    Well for one, you already know [tex] \psi [/tex]isn't constant from the fact that you already have [tex]\psi (x) = e^{ikx} [/tex]. The reason the exponential with the energy goes away is when you take the conjugate of [tex]\Psi [/tex], you take the conjugate of [tex] \psi [/tex] along with the conjugate of [tex]e^{-iEt/\hbar } [/tex] which will eliminate the time evolution term.
     
  4. Apr 3, 2009 #3

    turin

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    Homework Helper

    What do you mean, "solve". In what sense is the plane wave state unsolved?

    I presume that the potential term in your SE is constant???

    Please explain why that is obvious. That is nothing more than an interpretation of nonrelativistic QM, and it follows from nothing. There are in fact QM interpretations in which this is not true.

    This is a rather round about way of saying ... something ... but I did not quite follow. I think what you are trying to say is that the modulus of a complex number is independent of its phase.

    No, it's not! What are you talking about?

    Indeed: complex analysis. Do you know what a complex number is?
     
  5. Apr 4, 2009 #4
    Oh sorry, the question asks me to show that [tex]\psi(x) = exp(ikx)[/tex] is a solution of the time-independent schrodinger equation.

    The potential is zero, yes. So I get the equation for Psi as:

    [tex]\Psi(x,t) = \exp(ikx).exp(-iEt/\hbar)[/tex]

    From the separation of variables stuff.

    Well yeah ok, its not obvious on that sort of scope, im just doing an exam paper here.
    The question asks me to work out the energy in terms of k, having shown that [tex]\psi(x) = exp(ikx)[/tex] is a solution to the TISE.
    By mistake ive assumed this energy, which is constant, was the same value as [tex]\psi(x)[/tex]. This is obviously wrong which ive realised.
    The energy I worked out was

    [tex]E = \frac{\hbar^{2}k^{2}}{2m}[/tex]

    I shouldn't have said "obvious", but the probability per unit length is [tex]|\Psi(x,t)|^{2}[/tex], is it not?

    Well I dont know, i'll show you exactly what im talking about:


    [tex]|\Psi(x,t)|^{2} = |exp(-iEt/\hbar)|^{2}.|\psi(x)|^{2}
    = |\psi(x)|^{2}[/tex]

    I didnt understand why this was so.

    Yeah this follows on from my previous mistake, assuming that smal phi was the energy constant above.
    The question says: "What is the corresponding time-dependent solution,
    [tex]\Psi(x,t)[/tex]?
    Show that the probability per unit length of finding a particle is constant (independent of both space and time)."


    This, I assume is where the above workings come into place, basically showing that
    [tex]|\Psi(x,t)|^{2}[/tex] is constant. Is this right?


    Yes of course, but im unfamiliar with whats going on with this modulus business.
    Where im confused now is that how can the probability per unit length be constant if these complex terms are dissapearing? Wouldnt [tex]|\psi(x)|^{2}[/tex] dissapear also? And even if it didnt how can it be independent of space?
    Can you explain to me or link me to somewhere which explains this:


    [tex]|\Psi(x,t)|^{2} = |exp(-iEt/\hbar)|^{2}.|\psi(x)|^{2}
    = |\psi(x)|^{2}[/tex]
     
  6. Apr 4, 2009 #5

    turin

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    Homework Helper

    Actually, no. If that were true, then what would you get for the total probability of finding the particle in a length L, where L here is an arbitrary length? (Please calculate this and report back.)

    The squared modulus is usually interpretted as proportional to probability density, and, when properly normalized, it is equal to probability density. Based on the result of the previous excercise, what normalization constant should you divide by in order to interpret the squared modulus of a free particle wave function as a probability density?

    You cannot understand ordinary QM without understanding this.

    Complex terms are not disappearing; imaginary terms are cancelling.

    Do you understand that |e| = 1 , where φ is any arbitrary real number? Look up Euler's identity.
     
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