- #1

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

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.