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Time independant perturbation - Difficulty understanding derivation

  1. May 15, 2013 #1
    Hamiltonian is in the form ##H = H_0 + \lambda W##, where ##\lambda \ll 1## and ##W## is the perturbation. Assume the eigenstates ##\mid \psi(\lambda) \rangle## and engenenergies ##E(\lambda)## can be expanded in a power series of ##\lambda##.
    $$\mid \phi(\lambda) \rangle = \mid 0 \rangle + \lambda \mid 1 \rangle + \dots + \lambda^q \mid q \rangle$$
    $$ E(\lambda) = \epsilon_0 + \lambda \epsilon_1 + \dots + \lambda^q \epsilon_q $$
    Substituting in to the Schrodinger equation,
    $$(H_0 + \lambda W) \left[ \sum_{q=0}^\infty \lambda^q \mid q \rangle \right] = \left[ \sum_{r=0}^\infty \lambda^r \epsilon_r \right] \left[ \sum_{s=0}^\infty \lambda^s \mid s \rangle \right]$$
    By equating coefficients of successive powers of ##\lambda##,
    $$ H_0 \mid 0 \rangle = \epsilon_0 \mid 0 \rangle $$
    $$ (H_0 - \epsilon_0 ) \mid 1 \rangle + ( W - \epsilon_1 ) \mid 0 \rangle = 0 $$
    $$ (H_0 - \epsilon_0 ) \mid 2 \rangle + ( W - \epsilon_1 ) \mid 1 \rangle - \epsilon_2 \mid 0 \rangle = 0$$
    For the nth order,
    $$ (H_0 - \epsilon_0) \mid n \rangle + (W - \epsilon_1) \mid n-1 \rangle - \epsilon_2 \mid n-2 \rangle + \dots - \epsilon_n \mid 0 \rangle = 0$$
    The next part is what I don't understand. The following is what it says.

    Note that we are free to choose the norm and the phase of ##\mid \psi ( \lambda ) \rangle ##, so we require that ##\mid \psi ( \lambda ) \rangle ## is normalized and that its phase is such that the inner product ##\langle 0 \mid \psi (\lambda) \rangle## is a real number. This implies that,
    $$\langle 0 \mid 0 \rangle = 1$$
    $$\langle 0 \mid 1 \rangle = \langle 1 \mid 0 \rangle = 0$$
    $$\langle 0 \mid 2 \rangle = \langle 2 \mid 0 \rangle = -\frac{1}{2} \langle 1 \mid 1 \rangle$$
    For the nth order we obtain,
    $$\langle 0 \mid n \rangle = \langle n \mid 0 \rangle = -\frac{1}{2} \left( \langle n-1 \mid 1 \rangle + \langle n-2 \mid 2 \rangle + \dots + \langle 2 \mid n-2 \rangle + \langle 1 \mid n-1 \rangle \right)$$
    How is this obtained?
     
    Last edited: May 15, 2013
  2. jcsd
  3. May 15, 2013 #2

    Bill_K

    User Avatar
    Science Advisor

    Take the form you have, |φ(λ)> = |0> + λ|1> + ... Write out <φ(λ)|φ(λ)> = <0|0> + ... Set it equal to 1, and equate powers of λ. The λ0 term gives you 1 = <0|0>, the λ1 term gives you 0 = <0|1> + <1|0>, and so on.
     
  4. May 15, 2013 #3
    Got it. Thanks.
     
  5. May 15, 2013 #4
    OK. I still don't get it.

    What does "We require ... its phase is such that the inner product ##\langle 0 \mid \psi \rangle## is a real number." mean? Why does ##\langle 0 \mid \psi \rangle## need to be a real number?

    Also,
    $$\langle \psi (\lambda) \mid = \langle 0 \mid + \lambda^* \langle 1 \mid + (\lambda^*)^2 \langle 2 \mid + \dots + (\lambda^*)^q \langle q \mid$$, right?

    Then,
    $$
    \begin{eqnarray}
    \langle \psi (\lambda) \mid \psi (\lambda) \rangle = 1 & = & \langle 0 \mid 0 \rangle + \lambda \langle 0 \mid 1 \rangle + \lambda^2 \langle 0 \mid 2 \rangle + \dots + \lambda^q \langle 0 \mid q \rangle\\
    & & + \lambda^* \langle 1 \mid 0 \rangle + \lambda^* \lambda \langle 1 \mid 1 \rangle + \lambda^* \lambda^2 \langle 1 \mid 2 \rangle + \dots + \lambda^* \lambda^q \langle 1 \mid q \rangle\\
    & & \vdots\\
    & & + (\lambda^*)^q \langle q \mid 0 \rangle + (\lambda^*)^q \lambda \langle q \mid 1 \rangle + (\lambda^*)^q \lambda^2 \langle q \mid 2 \rangle + \dots + (\lambda^*)^q \lambda^q \langle q \mid q \rangle\\
    \end{eqnarray}
    $$
    Is this correct? If it is, what do I do with all these ##\lambda^*##s?
     
  6. May 15, 2013 #5
    I would assume that ##\lambda## is real; after all it is a dummy parameter that characterises the strength of the perturbation. In most calculations, ##\lambda## is set to 1 at the end - it is mainly used for keeping track of the orders only.
     
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