# Question regarding time independent perturbation theory

1. Feb 4, 2015

### Kidiz

Let's say we've a system which can be described by the Hamiltonian:

$$H_0 = \dfrac{p^2}{2m} + V(x)$$

Now suppose we introduce a perturbation given by:

$$H_1 = \lambda x^2$$

Our total hamiltonian:

$$H = H_0 + H_1 = \dfrac{p^2}{2m} + V(x) + \lambda x^2$$

Normally, the perturbation doesn't have the factor $\lambda$ inside of it. It's usually written has $H = H_0 + \lambda H_1$, which gives us the energy $E = E_n ^{0} + \lambda E_n ^{1} + \lambda ^2 E_n ^{2}$.

Now, my question is:

Is the energy given by:

$$E = E_n ^{0} + \lambda E_n ^{1} + \lambda ^2 E_n ^{2}$$

even if we've $\lambda$ inside the $H_1$? Or should it be something like $E = E_n ^{0} + E_n ^{1} + E_n ^{2}$, since the $E_n ^{1}$ and $E_n ^{2}$ will have $\lambda$ as a factor?

PS: My title is horrible, but I couldn't think of something better.

2. Feb 4, 2015

### Staff: Mentor

This. If the parameter that makes the perturbation small is inside the Hamiltonian (as it often is), then there is no λ appearing explicitly.

3. Feb 4, 2015

### Chopin

However, it's important to remember that the sequence $E^0_n + E^1_n + E^2_n + ...$ will still have increasing powers of $\lambda$ in each term. This fact, plus the fact that $\lambda$ is small, is what makes each term of the sequence get progressively smaller, which is the thing that makes perturbation theory work in the first place. Because of how central this concept is, it's often more informative to pull out the factor $\lambda$ from the interaction term and put it explicitly into the sequence, so that it's more easily seen which parameter you are expanding the series in terms of. That's why you hear people describe a QED calculation with terms like "to fourth order in $e$", even though the electric charge $e$ is built directly into the QED Hamiltonian.

Last edited: Feb 4, 2015