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The lowest energy eigenvalue

  1. Jul 17, 2007 #1
    Shankar 163

    1. The problem statement, all variables and given/known data

    Show that for any normalized |psi>, <psi|H|psi> is greater than or equal to E_0, where E_0 is the lowest energy eigenvalue. (Hint: Expand |psi> in the eigenbasis of H.)

    2. Relevant equations

    3. The attempt at a solution

    I think the question assumes the V = 0, so H = P^2/2m. The eigenvalues for the equation P^2/2m|p> = E|p> are then p = +/- (2mE)^(1/2) and the eigenkets are of the form | p = + (2mE)^(1/2)> and | p = (2mE)^(1/2)> (or in energy terms the eigenvalues are of the form |E,->, |E,+> and the eigenkets are the same as the momentum ones. I'm not really sure how you can expand anything with these though...
  2. jcsd
  3. Jul 17, 2007 #2
    No, this statement is valid for any Hamiltonian.
    To spare your pencil, assume the system has only two basis states and develop <psi|H|psi>, you will see how obvious this result is.
    (take the basis states as H-eigenvectors for simplicity, then imagine th consequence if you don't know these eigenvectors but would like to find the fundamental state, very useful statement indeed!)
  4. Jul 17, 2007 #3
    This is how we expand [tex] \Psi [/tex] in terms of the eigenbasis of H :

    Use the fact that the eigenvalues of H which is [tex]\{\Psi_0,\Psi_1,\Psi_2 ...\}}[/tex] forms the basis of a Hilbert space. Notice Hibert space is completed.
    That means any vector [tex]\Psi[/tex] could be written as a linearly combinition of [tex]\Psi_0,\Psi_1,\Psi_2 ...[/tex]
    i.e. [tex]\Psi = c_0\Psi_0+c_1\Psi_1+c_2\Psi_2+...[/tex].

    The problem should be trivial from this point ( use the orthogonality of the eigenvector, although the eigenvectors in degenerate state are not orthogonal, they do not affect the result. )
  5. Jul 18, 2007 #4
    Of course the end result is

    <H> = p1 H1 + p2 H2 + ...

    where pi is the probability of being in state i,
    and 0<pi<1
    and p1 + p2 + ... = 1
  6. Jul 18, 2007 #5
    OK. So we use the time-independent version of the Schrodinger equation H * psi = E * psi.

    So [tex]|\psi> = p0 * |H0> +p1 * |H1 > + ...= a0 * |E0> + a1 * | E1> + .... [/tex]

    p (and my choice of a) are arbitrary letters, right? You did not use p for momentum did you lalbatros?

    Now, [tex] <\psi| H | /psi> = (E0*a0 + E1*a1 + ...)</psi|/psi> = E0*a0 + E1*a1 + ... [/tex]

    So the minimum of that expression would occur when a0 = 1 and ai = 0 for all i > 1.

    Does that explanation work?
  7. Jul 18, 2007 #6


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    Your notation is confusing. I'll confuse the waters a little more by posting ugly, non-TeX. Which I prefer to garbled TeX. |psi>=sum(ci*|Ei>). ci are the amplitudes, |Ei> is the eigenstate corresponding to eigenvalue Ei. So <psi|H|psi>=sum(ci*conjugate(ci)*<Ei|H|Ei>)=sum(ci*conjugate(ci)*Ei*<Ei|Ei>).
    Since we'll take |Ei> normalized <Ei|Ei>=1.
    So <psi|H|psi>=sum(pi*Ei) where pi=ci*conjugate(ci). Since we'll also take |psi> normalized, sum(pi)=1.

    So if sum(pi)=1, pi>=0, do you believe sum(pi*Ei)>=E0 where E0 is the smallest of the Ei's? Can you prove it?
    Last edited: Jul 18, 2007
  8. Jul 19, 2007 #7
    I believe it.

    sum(pi*Ei) >= sum(pi*E0) = sum(pi)*E0 = E0
  9. Jul 19, 2007 #8
    How do you go from |psi>=sum(ci*|Ei>) to <psi|H|psi> = sum(ci*conjugate(ci)*<Ei|H|Ei>)?

    Shouldn't it be something like <psi|H|psi> = sum(conjugate(ci)*<Ei|) H sum(ci*|Ei>) just by replacing |psi> with sum(ci*|Ei>) and < psi| with sum(conjugate(ci)*<Ei|)?
    Last edited: Jul 19, 2007
  10. Jul 19, 2007 #9


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    We've taken the |Ei> to be orthonormal, so <Ei|Ej>=delta(i,j) And H(ci*|Ei>)=ci*Ei*|Ei>. And your proof is perfect.
    Last edited: Jul 19, 2007
  11. Jul 19, 2007 #10
    I see. Thanks.
  12. Jul 19, 2007 #11


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    You really do? You're not kidding, right?
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