- #1
Kidiz
- 21
- 4
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.
$$H_0 = \dfrac{p^2}{2m} + V(x)$$
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.