1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

QM. eigenfunction help!

  1. Feb 22, 2009 #1
    1. The problem statement, all variables and given/known data

    find eigenfunction of the the given operator. compute the vaule of px and x for this eigenfunction

    2. Relevant equations
    O= p + x
    O is made up by adding the momentum operator(p) and the position operator(x); it is one dimensional


    3. The attempt at a solution
    i tried doing Oѱ=ѱp+ѱx=kѱ , h'=h bar
    or, -ih' *(d/dx ѱ) + x * ѱ= kѱ
    or, k = -ih' *(d/dx ln ѱ) + x =k
    or, kx+c=-ih' * ln ѱ +x2 /2 [after integrating]
    and then solved for ѱ and i am stuck.
     
    Last edited: Feb 22, 2009
  2. jcsd
  3. Feb 22, 2009 #2

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    Well this one is ok:

    -ih' *(d/dx ln ѱ) + x =k

    which give you:

    (d/dx ln ѱ) = (i/hbar)(k-x)

    the integration is also good:

    kx+c=-ih' * ln kx+c=-ih' * ln ѱ +x2 /2 +x^2 /2

    Why you got stucked at solving for ѱ? What did you try? How can we help you if you not tell us what you tried?
     
  4. Feb 22, 2009 #3
    after doing all that i get an equation for ѱ, that look slike
    ѱ= (exp) ((2kx+ 2c -x2)/-2ih)
    i do have an equation but its got 2 unknown k and c, how do i proceed from here to find the eigenfunciton and the values of px and x for the eigenfunction
     
  5. Feb 22, 2009 #4

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    k and c CAN be arbitrary constants, they are determined by normalization condition.

    The eigenvalue of position operator is x|x'> = x'|x'>

    The eigenvalue of momentum operator is p|x'> = -ihbar(d/dx')|x'>
     
  6. Feb 22, 2009 #5
    so does that mean i just take the two constants out and get an equation for the eigenfunction that looks like:
    ѱ= (exp) ((2kx+ 2c -x2)/-2ih) without the k and c,
    so,
    ѱ= (exp) ((2x -x2)/-2ih)


    and is x|x'> = x'|x'> mean x divided by x' where x' is the derivative? or is it just performing the eigenfunction(ѱ) on x?
     
  7. Feb 22, 2009 #6

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    nono sorry, i used the notation of sakurai. This is better:

    [tex]\hat{x}\psi(x) = x\psi(x)[/tex]

    [tex] \hat{p} \psi (x) = -i\hbar\dfrac{d}{dx}\psi(x) [/tex]

    hat- is operator.

    Should not have introduced that sloppy notation for a beginner of Qm.. sorryy

    There is not ONE unique eigenfunction to this operator, the constants are arbitrary.
     
  8. Feb 22, 2009 #7
    yeah. i am sort of a beginner. may i bother u once more. i am asked to normalize the eigenfunction (ѱ) after finding out what it is; if there are no particular solutions for it, how do i normalize it for a given range like say (-L to L)? is it logical? and if i cant then how do i find px and x for the eigenfuntion(ѱ)
     
  9. Feb 22, 2009 #8

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    In general, you normalize from -infinity to +infinity if nothing else is stated.

    You would have normalized to -L to L if you were given this wavefunction for instance:

    Psi (X) = psi(x) if -L < x < L
    Psi (X) = 0 elsewhere

    A hint regarding that normalization, your obtained function is a Gaussian function.
     
    Last edited: Feb 22, 2009
  10. Feb 22, 2009 #9
    so normalize the equation for the eigenfunctin ѱ of O(with a hat)=p+x(both with hats) leaving the two constants k and c?? on the equation ѱ= (exp) ((2kx+ 2c -x2)/-2ih).
    and px and x for the eigenfunction (ѱ) of O(with a hat) is the same thing as the eigenfunction of x on px and x on x?
     
  11. Feb 22, 2009 #10

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    I would keep the constants k and c and then just normalize the wave function ѱ= (exp) ((2kx+ 2c -x2)/-2ih)
     
  12. Feb 22, 2009 #11
    okay. so after all this. does it mean that the eigen functino of ѱ is also the eigenfunctin of the momentum operator as px?
    refering to post number 6, and when trying to find the value of px and x for ѱ(the eigenfunctin of O), why are we just executing the eigenvalue on the momentum and position operator, rather than plugging it on the eigenfunction(ѱ)?
     
  13. Feb 22, 2009 #12

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    Yes it is also an eigenfunction of the momentum operator. Remember the linearity requirement for operators.
     
  14. Feb 22, 2009 #13
    yeas, i get that part. its just the values of px and x for the eigenfunction(ѱ) of O thats not clear to me?
     
  15. Feb 22, 2009 #14

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    why is it hard to just evaluate them by operating on the wavefunction?

    Give it a try
     
  16. Feb 22, 2009 #15
    yea, i was about to get to that; to find the value of x in ѱ, i just operate for x; but what do i do for px; i can solve of x2 but what about px2??
    can ѱ represent any physical state, if at all?
     
  17. Feb 22, 2009 #16

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    so you don't know how to take a derivative of a general function?

    [tex]
    \hat{p} \psi (x) = -i\hbar\dfrac{d}{dx}\psi(x)
    [/tex]

    You will find out that the momentum is the generator of translations..

    now just do this, don't think so much, be axiomatic.

    Yes this is a Gaussian wavefunction and have many nice properties.

    "but what do i do for px; i can solve of x2 but what about px2"
    I can't understand what you are thinking about here.. just evaluate the derivative...
     
  18. Feb 22, 2009 #17
    i menat px^2, px raised to the power of two? when finding px for ѱ, i first operate x on the momentum operator and then operate the result on ѱ??
     
  19. Feb 22, 2009 #18

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    It is still an eigenfunction.....
     
  20. Feb 22, 2009 #19
    so if it were pxn then i would just apply the momentum function on any given function n times, and plug it in for on ѱ to find the value of pxn for ѱ??
     
  21. Feb 22, 2009 #20

    malawi_glenn

    User Avatar
    Science Advisor
    Homework Helper

    what is the "momentum function"? and "plug it in"?

    But yes, if you want to evaluate the eigenvaule of the momentum operator to the n-th power for some wavefunction in position space, you take the n-th derivative with respect to x and then multiply it with (-i\hbar) to the n-th power.

    Now that has nothing to do with the normalization.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook