# Tag Info

1

Yes, the summation is taking over all possible integer value of $m=0,1,2,...$ except $m=n$. It can be easily seen by following the derivation of the first order perturbation theory. In your example $\psi_1^{(1)}$, it is sum over $m=0,2,3,4,...$. Note that sometimes the index start from 1 instead of 0 such as infinite square well, then you should skip 0.

2

You can work it out from the Taylor series $$\frac{1}{1 + x} = 1 - x + x^2 - x^3 + \cdots$$ where $x = \lambda a_n^{(1)} + \lambda^2 a_n^{(2)} + \cdots$. Each term can then be expanded in a power series in $\lambda$: \begin{align} -x &= -\lambda a_n^{(1)} - \lambda^2 a_n^{(2)} - \lambda^3 a_n^{(3)} - \cdots \\ x^2 &= \lambda^2 ... 2 The free field, non-interacting Lagrangian is given by\mathcal{L}=-\frac12\partial^\mu\phi\partial_\mu\phi-\frac12 m^2\phi^2. In the limit $g\rightarrow0$, the interacting Lagrangian you wrote down is required to reduce to the free field version. This means that the counterterm part has to vanish, which is only possible if this limit leads to ...

1

There will always be solutions that can't be analytical. For example, any model of more than two bodies without any special constraints, cannot be solved analytically. From the gravitational interactions between three planets to three particles interacting (electromagnetically or otherwise) in quantum theory. To have mathematically analytical solutions, ...

7

No. There is nothing wrong with perturbation theory, or with theories with known, restricted accuracy. The point of theory is to explain the results of observation from as simple an initial theoretical standpoint as possible. Therefore: Since experiment always has a finite uncertainty, one can only ask that theory match the experimental value within its ...

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