- #1
omoplata
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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 into 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?
$$\mid \phi(\lambda) \rangle = \mid 0 \rangle + \lambda \mid 1 \rangle + \dots + \lambda^q \mid q \rangle$$
$$ E(\lambda) = \epsilon_0 + \lambda \epsilon_1 + \dots + \lambda^q \epsilon_q $$
Substituting into 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?
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