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Eigenstates and eigenfunctions and Schrodinger - HELP PLEASE!

  1. Nov 14, 2006 #1
    A couple of things first - Whats an eigenfunction? Whats an eigenvalue? I've been doing a course on quantum mechanics for nearly 2 months and whilst these words have popped up in both notes and lectures no-one has actually bothered to explain what they are or what they mean!

    Next thing. Can someone point me in the right direction for solving these questions?

    Question:
    The even parity energy eigenstates for the infinite square well potential, defined by V(x) = 0 for -L/2<x<L/2 and V(x) = infinite otherwise are [tex]\psi_{n}(x) = \sqrt{\frac{2}{L}}cos(\frac{n\pi x}{L})[/tex] with n odd.

    a) If the momentum is measure, what values are obtained and with what probability? (Hint: Expand [tex]\psi_{n} (x)[/tex] as a sum of momentum eigenstates)

    b) Using the results from a) deduce average p.

    c) using symmetry, deduce average x.

    Now I know for a) that [tex]\hat{p}u(x) = pu(x)[/tex], but whats u(x)? Where did this come from? What is it? Does it relate to [tex]\psi(x)[/tex] at all? But apart from that I can't work out any other relationships between anything! Any help on these questions would be greatly appreciated. Especially if it can be put really simplisticaly, as I'm struggling somewhat with this course, and neither or my notes or text book seem to make much sense.
     
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  3. Nov 14, 2006 #2
    So you have an vector space linear endomorphism i.e. operator A and a vector v in that vector space. When you operate with the operator A to v you get:
    Av=a_i sum <a_i|v>, where |a_i> are the eigenstates of A and a_i are the corresponding eigenvalues. This means that, if you operate to the operators eigenstate the state remains unchanged, other states will change (measurements affecting the system etc. stuff). Man, you should really know this if you are on a quantum mechanics course. What do they teach there?
    Easiest way to approach this mathematically is to think of matrices and their presentation in the eigenbasis.

    a)
    You already have the solution of the Schrödinger equation. And what is this equation, squared momentum operator on the other side and energy on the other, maybe little integration would do good for the second order derivatives...
    b)
    By definition, propabilities come from the squared norm of the state.
     
  4. Nov 14, 2006 #3

    ZapperZ

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    I think someone may need to tell you that you do not have the mathematics background to be taking a class in QM. It appears that you do not have an adequate preparation on differential equation, since you have not even heard of "eigenfunctions" and "eigenvalues". If this is true, then I'm guessing that you may be in for quite an "adventure" the rest of your class because there are bound to be a ton more mathematics that you may not have seen before.

    My advice here is to talk to your instructor and see if you might be suited to a different class first, or maybe shore up on your mathematics first before taking this class. If that isn't possible, then maybe a crash course in mathematical physics is the only possible solution. Get one of those mathematical physics text (I recommend Boas's text) and teach yourself the tools that you are expected to know in that QM class.

    Zz.
     
  5. Nov 14, 2006 #4

    I actually have more of a maths education than the rest of my class! I've done differntial equations until it comes out of my ears, just those words have never turned up before.
     
  6. Nov 14, 2006 #5

    quasar987

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    Have you not had a linear algebra class before? It would really help you understand the ideas of QM. In linear algebra, we learn that if M is some matrix, then we call eigenvector of the matrix M the vectors v and eigenvalue associated with the vector v the scalars c such that

    [tex]\vec{v}M=c\vec{v}[/tex]

    i.e. the vectors v such that when multiplied with the matrix M give back v up to a factor c.

    This notion of eigenstuff is easily applicable to the world of operators and functions. Namely, if [tex]\hat{O}[/tex] is some operator, then the functions f and scalars c such that

    [tex]\hat{O}f=cf[/tex]

    are the eigenfunctions and their associated eigenvalues of the operator [tex]\hat{O}[/tex].

    the problem of solving the (time-independant) Schrödinger equation is an eigenvalue problem. It consists of finding the eigenfunctions [itex]\psi[/itex] and their associated eigenvalues E of the hamiltonian operator [tex]\hat{H}=\hat{p}/2m+V(x)[/tex].

    Then there are the observables; those are the things you can measure in a laboratory. To each observable 'O' is associated an operator [tex]\hat{O}[/tex], and QM says that ithe result of a measure on O can only be an eigenvalue of the operator [tex]\hat{O}[/tex]. In particular, if you try to measure the momentum (an observable) of a particle, then the result of your measure will be a number 'p' such that there exists a function [itex]u_p(x)[/itex] satisfying the eigenvalue equation

    [tex]\hat{p}u_p(x)=pu_p(x)[/tex]

    That's not all. Suppose you know that your wave function is [itex]\psi(x)[/itex] prior to the measurement. When you make the measurement of the momentum, the probability of measuring the eigenvalue p is given by the (squared norm of the) inner product of [itex]\psi(x)[/itex] with [itex]u_p(x)[/itex], where the inner product is defined for wave functions by

    [tex]<u_p, \psi>=\int_{-\infty}^{+\infty}u_p^*(x)\psi(x)dx[/tex]


    Btw - What book are your class using?
     
    Last edited: Nov 14, 2006
  7. Nov 14, 2006 #6
    As far as books go, I'm using Introduction to Quantum Mechanics by A.C. Phillips published by John Wiley and Sons. And the course notes that are handed out in lectures.

    I think I get eigenfunctions and values now.

    Now, the wave function is given to me, correct? And it is used in the TISE solution for energy, correct? But it is not the same as the u(x) that is in the TISE solution for momentum, correct? Can I get to u(x) from [tex]\psi(x)[/tex]? From there I feel I may be able to get what I want. Although I could be wrong.
     
  8. Nov 14, 2006 #7

    quasar987

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    I just now finished editing my post, so passages that were not so clear or downright nonsensical should be better now if you want to reread it.
     
  9. Nov 14, 2006 #8

    quasar987

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    The order in which the material is presented in my post does not make this clear but note that the TISE is just the eigenvalue problem associated with measuring the observale "energy" and the hamiltonian operator is just the energy in "operator form".
     
  10. Nov 14, 2006 #9

    quasar987

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    I don't get get what you mean by the word "used" in "And it is used in the TISE solution for energy, correct?". [itex]\psi[/itex] and [itex]u[/itex] are just letters; it does not matter which you choose. But the u(x) in

    [tex]\hat{p}u(x)=pu(x)[/tex]

    is as much a wave function as the [itex]\psi[/itex] in

    [tex]\hat{H}\psi=E\psi[/itex].

    I will show you how it works explicitely.

    Suppose you have a system composed of a particle swiming in some potential V(x). Now say you measure the energy of that particle and it gives E. By the postulates of QM, this number E is an eigenvalue of the TISE and it is the eigenvalue associated with some wave function(=eigenfunction) [itex]\psi_E(x)[/itex]. By the postulates of QM also, the wave function, whatever it was prior to the measurement, is now [itex]\psi_E[/itex] after the measurement, and it will remain [itex]\psi_E[/itex] until you measure something else.

    If you now decide to measure the momentum, you will measure any eigenvalue 'p' of the eigenvalue equation [tex]\hat{p}u_p(x)=pu_p(x)[/tex], but because you know that the wave function is [itex]\psi_E[/itex] prior to the measurement, you can also associate a probability to every possible value of p. These probabilities are given by [tex]|<u_p,\psi_E>|^2[/tex] (one probability for each possible p).

    After you've made your measurement one momentum, the wave function is now [itex]u_p(x)[/itex] and will remain so until a new measurement is made, etc.


    Important: You should note that in everything I said above, I always assumed that to every eigenvalue 'O' of an operator [tex]\hat{O}[/tex], there was only one eigenfunction. But that is not necessarily the case. Some eigenvalues have many eigenfunctions. This complicates things a little but not really once you understand how whole eigenstuff work.
     
    Last edited: Nov 14, 2006
  11. Nov 15, 2006 #10

    dextercioby

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    Well, generally speaking for an arbitrary endomorphism A of the linear topological space V

    [tex] A:D(A) \subseteq V \rightarrow \mbox{Im}(A)\subseteq V [/tex](1)

    the equation

    [tex] Av=\lambda v [/tex] (2)

    is called a spectral equation. If there is [itex] v\in D(A)\cap \mbox{Im}(A) [/itex] , solution to the eq. (2), corresponding to an [itex] \lambda\in \sigma_{p}(A) [/itex], then "v" is caled an eigenvector (or eigenfunction if the topological vector space is a function space) corresponding to the eigenvalue [itex] \lambda [/itex].

    Daniel.
     
    Last edited: Nov 15, 2006
  12. Nov 19, 2006 #11
    Right I think I'm making headway on this problem (possibly)

    If [tex]\psi_{n}(x) = \sqrt{\frac{2}{L}}cos(\frac{n\pi x}{L})[/tex] then by expanding as the sum of momentum eigenstates I think I get:

    [tex]\frac{1}{\sqrt{2}}\bar{h}k(\frac{1}{\sqrt{L}}(e^{ikx}-e^{-ikx}))[/tex]

    Does this look correct? I don't think its far away from the correct answer, but I'm not sure how right it actually is. I have a feeling that the last part (the part with the exponentials) should have a plus instead of a minus between the two exponentials (thus making it cos), as therefore I could see where [tex]p=\bar{h}k[/tex] comes from. Correct?
     
  13. Nov 21, 2006 #12

    dextercioby

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    It definitely should be a plus, if you were to find cosine. And it's [ itex ] \hbar [ /itex] not [ itex ]\bar{h}[/itex].

    Daniel.
     
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