# Time independant perturbation - Difficulty understanding derivation

1. May 15, 2013

### omoplata

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. May 15, 2013

### Bill_K

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.

3. May 15, 2013

### omoplata

Got it. Thanks.

4. May 15, 2013

### omoplata

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?

5. May 15, 2013

### Fightfish

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.