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How to show a wave function satisfies the shrodinger eqn

  1. Dec 9, 2013 #1
    1. The problem statement, all variables and given/known data

    If a state is a linear combination of two energy eigenstates, does its wavefunction satisfy the time independent shrodinger equation? I guess in general, how would I show if the wavefunctions of a state does or does not satisfy the time-independent schrodinger equation?

    2. Relevant equations

    [tex]H\left| E \right\rangle=E\left|E\right\rangle[/tex]
    [tex]\psi_\alpha(x)=\left\langle x |\alpha\right\rangle[/tex]

    3. The attempt at a solution

    Honestly have no idea what to do.
     
  2. jcsd
  3. Dec 9, 2013 #2

    BruceW

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    to start with, what is the time-independent Schrodinger equation?
     
  4. Dec 9, 2013 #3
    [tex]H\left| E \right\rangle=E\left|E\right\rangle[/tex]
     
  5. Dec 9, 2013 #4

    cepheid

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    Okay, first of all, by analogy with eigenvectors/values, an energy eigenstate is just a state (wavefunction) for which this is true: when you apply the Hamiltonian operator to it, you just get a constant (the eigenvalue, E) multiplied by it. So, if |E> is an energy eigenstate, then:$$H|E\rangle = E|E\rangle$$So energy eigenstates are states that satisfy the time-independent Schrodinger equation (which is the eigenvalue equation for the Hamiltonian).

    Now, suppose we have two such eigenstates ##|E_1\rangle## and ##|E_2\rangle##. Since they are energy eigenstates, they must each separately satisfy the eigenvalue equation. So: ##H|E_1\rangle = E|E_1\rangle## AND ##H|E_2\rangle = E|E_2\rangle##

    Now suppose we have a more general state ##|\psi\rangle## that is a linear combination of ##|E_1\rangle## and ##|E_2\rangle##:$$|\psi\rangle = \textrm{some combination of}~|E_1\rangle~\textrm{and}~|E_2\rangle$$What does it mean for psi to be a linear combination of the two eigenstates? I.e. what is a linear combination? I.e. can you fill in the right-hand side of the above equation?

    Based on your answer to the above, what happens when you apply the Hamiltonian operator to psi? ##H|\psi\rangle =~?##
     
  6. Dec 9, 2013 #5
    ok if I have two states [itex]\left| a\right\rangle[/itex] and [itex]\left| b\right\rangle[/itex] and they are energy eigenstates with energies [itex]E_a[/itex] and [itex]E_b[/itex], then a general linear combination of these states would be [tex]\left| \psi \right \rangle = c_1\left| a\right\rangle+c_2\left| b\right\rangle[/tex]
     
  7. Dec 9, 2013 #6
    so if I wanted to show that the wavefunction [itex]\phi(x)=\left\langle x|\psi\right\rangle[/itex] satisfied the SE, how would I do this, and how would I find its eigenvalue?
     
  8. Dec 9, 2013 #7

    BruceW

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    (in response to post 5) yep. so does psi satisfy the time-independent Schrodinger equation? i.e. what happens when you use the Hamiltonian operator on psi?
     
  9. Dec 9, 2013 #8

    cepheid

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    Yeah, that's right. That is a linear combination. So what happens when you apply H to this state? Remember that H is a linear operator.
     
  10. Dec 9, 2013 #9
    [tex]H\left| \psi \right \rangle = H\left( c_1\left| a\right\rangle+c_2\left| b\right\rangle\right)[/tex]
    [tex]H\left| \psi \right \rangle = c_1 H\left| a\right\rangle+c_2H\left| b\right\rangle[/tex]
    [tex]H\left| \psi \right \rangle = c_1 E_1\left| a\right\rangle+c_2E_2\left| b\right\rangle[/tex]
     
  11. Dec 9, 2013 #10
    the part that is really messing me up is the "wave equation" part.
     
  12. Dec 9, 2013 #11

    BruceW

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    yep nice. so is this the same as the time-independent Schrodinger equation for the state psi?
     
  13. Dec 9, 2013 #12
    I guess?
     
  14. Dec 9, 2013 #13

    BruceW

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    you said the time-independent Schrodinger is ##H|E\rangle = E|E\rangle## and for your situation, you have the state psi, so does psi fit this equation?
     
  15. Dec 9, 2013 #14
    I have no idea.
     
  16. Dec 9, 2013 #15

    cepheid

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    What BruceW is saying is that for psi to satisfy the time-independent Schrodinger equation, it would have to be true that ##H|\psi\rangle = E|\psi\rangle## where E is *some constant.* Look at your above result for ##H|\psi\rangle##. Is it equal to "some constant" multiplied by psi?
     
  17. Dec 9, 2013 #16
    no? I don't think it is?
     
  18. Dec 9, 2013 #17

    cepheid

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    The answer is indeed no. But why aren't you sure about your answer? This is math after all, there is a definite answer. So don't guess, do the algebra. Take your psi and multiply it by some constant C (or E, or whatever). Is this the same as what you got for H|psi>?
     
  19. Dec 9, 2013 #18
    are there any cases where it would be though? like what if one of the constants is imaginary, or if the two constants are equal?
     
  20. Dec 9, 2013 #19

    cepheid

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    Use a more physical argument to motivate your answer: eigenstates (of the Hamiltonian) are states of definite energy. So if the particle is in an eigenstate, a measurement of its energy is sure to yield the same answer always. But if the particle is in a linear combination of two eigenstates, can this wavefunction psi tell you with certainty what the energy will be, or can it only tell you about the probability of measuring a given value? Based on that, would you expect the linear combination of two eigenstates to also be an eigenstate, or not?
     
  21. Dec 9, 2013 #20

    cepheid

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    If the constants are equal (i.e. two distinct states have the same value -- they are said to be degenerate states), then yes, it would work. Their linear combination would also be an eigenstate with that same energy.
     
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