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If ##V(x)## is an even function then the energy eigenfunc...

  1. Mar 19, 2016 #1
    1. The problem statement, all variables and given/known data
    If ##V(x)## is an even function [i.e. ##V(-x)=V(x)##], then the energy eigenfunctions ##\phi_E(x)## can always be taken to be either even or odd. i.e. show ##\psi_{odd}(x)\equiv\frac{\phi_E(x)-\phi_E(-x)}{2}## and ##\psi_{even}(x)\equiv\frac{\phi_E(x)+\phi_E(-x)}{2}##. The question is from Here

    2. Relevant equations
    ##[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)]\phi_E(x)=E\phi_E(x)##

    3. The attempt at a solution

    The even wave function can be described by
    $$
    [-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)]\phi_E(x)=E\phi_E(x)-----(1)
    $$
    And the odd wave function can be described by
    $$
    [-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial (-x)^2}+V(-x)]\phi_E(-x)=E\phi_E(-x)=[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)]\phi_E(-x)=E\phi_E(-x)---(2)$$

    Now adding (1) & (2) I obtain
    $$
    [-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)][\phi_E(x)+\phi_E(-x)]=E[\phi_E(x)+\phi_E(-x)]
    $$

    Unfortunately I can't seem to see where to go from here. Also what exactly is the difference between the energy eigenstates ##\phi_E## and wavefunction ##\psi##?
     
  2. jcsd
  3. Mar 19, 2016 #2

    Simon Bridge

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    Last question first: all energy eigenstates are described by an energy eigenfunction.
    They are solutions to ##\hat H\phi_E = E\phi_E## ... i.e. it is an eigenfunction of the hamiltonian operator.
    All energy eigenfunctions are wavefunctions, but not all wavefunctions are energy eigenfunctions.
    i.e. ##\psi = c_1\phi_1 + c_2\phi_2## is a wavefunction that is not an energy eigenfunction (..for ##c_1,c_2\neq 0##).

    In your setup you have used the same notation to refer to two different eigenfunctions ... one odd and one even.
    You should start with definitions ... the potential and wavefunctions will be peicewise. Take care to keep track of the peices.
     
  4. Mar 19, 2016 #3
    Hmm if the potential is symmetric is symmetric will it not just be a ##V(x)## everywhere?

    Edit: Perhaps that isn't the best way of stating what I'm trying to say. I don't see why the potential will be piecewise is what I'm trying to say
     
  5. Mar 19, 2016 #4

    Simon Bridge

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    Any potential is V(x) everywhere, that is what V(x) means.
    You need to work out how to handle the distinction.
    Try writing out what you know for x>0 first, assume that ##\phi## is a solution to that; then try it for x<0 ... in order for ##\phi## to also be a solution in this case, what property must it have?
     
  6. Mar 19, 2016 #5
    x>0 is ##-\frac{\hbar^2}{2m}\frac{\partial^2 \psi(x)}{\partial x^2}+V(x)\psi(x)=E\psi(x)##, if ##\phi## is a solution then we have ##[-\frac{\hbar^2}{2m}\frac{\partial^2}{\partial x^2}+V(x)]\phi(x)=E\phi(x)##, for x<0 do you want me to use ##x\to -x##?

    Edit: I understand now, the next step was to do (1)-(2) and then to observe that (1)+(2) is an even function and (1)-(2) is an odd function.
     
    Last edited: Mar 19, 2016
  7. Mar 20, 2016 #6

    Simon Bridge

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    That would work. Well done.
     
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