# Time-dependent Perturbation theory -maths problems

1. Oct 11, 2015

### Imperatore

Hey guys,

I signed up here because I needed some information on some quantum physics problems.

My question is related to quantum physics, and more precisely the derivation of time dependent perturbation theory. First of all, I am not able to understand all the maths structures and formulas, and it is in my opinion the biggest difficult I need to overcome to fully understand it. I checked some books and scripts just to find some more information about it, but unfortunately my doubts weren't stirred up. I hope that here I could find some more help and I really appreciate it.

So to begin with:

We get a hamiltonian, that can be written as
$$H(t)=H _{0}+V(t)$$, where $$V(t)$$ is a small time-dependent perturbation.

Let $$\psi ^{(0)} _{k}$$ be the wave functions of stationary states of unperturbed system with time multiplier. Any solution to the unperturbed wave equation can be expressed as the sum :
$$\psi= \sum_{k}a _{k} \psi ^{(0)} _{k}$$

The solutions to perturbed equation:
$$ih \frac{\delta}{\delta t}\psi=(H _{0}+V)\psi$$ (1) (where h is Dirac constant, Planck constant divided by 2 Pi)
are in the form of:

$$\psi= \sum_{k}a _{k}(t)\psi ^{(0)} _{k}$$ (2)

Substituting (2) to the (1) and knowing that functins $$\psi^{(0)}_{k}$$ satisfy and equation :

$$ih \frac{\delta}{\delta t }\psi _{k} ^{(0)}=H_{0}\psi _{k} ^{(0)}$$

somehow we obtain: (here is the problem for me. I don't understand how they solve below mentioned equations step by step)

$$ih \sum_{k}\psi _{k} ^{(0)} \frac{d}{dt}a _{k}(t)= \sum_{k} a_{k}V\psi _{k} ^{(0)}$$

and then multiplying this equation from left side by $$\psi _{m} ^{(0)}*$$ and integrating we get :

$$ih \frac{d}{dt}a _{m}(t) = \sum_{k}V _{mk}(t)a _{k}(t)$$

where $$V _{mk}(t)=\int \psi _{m} ^{(0)*} V\psi _{k} ^{(0)}dq=V _{mk}e ^{i\omega _{mk}t}$$

and $$\omega _{mk}= \frac{E _{m} ^{(0)}-E _{k} ^{(0)} }{h}$$

I will be very thankful if someone could explain it to me. I want to mention, that the majority of this text come from Landau book "Quantum mechanics. Non relativistic theory". At the end I want to add that I am Polish, so my English may be not perfect, but I still hope it is enough communicative ;)

Last edited: Oct 11, 2015
2. Oct 11, 2015

### blue_leaf77

It's just product rule of differentiation
$$\frac{\partial\psi}{\partial t} = \sum_{k} \frac{\partial }{\partial t} \left(a _{k}(t)\psi ^{(0)} _{k}\right) = \sum_{k} a _{k}(t) \frac{\partial \psi^{(0)} _{k}}{\partial t} + \psi ^{(0)} _{k} \frac{\partial a _{k}(t)}{\partial t}$$
Is that not what you can't understand?

3. Oct 11, 2015

### Imperatore

Ye,it is obvoius relation from mathematial analysis ;)
the derivative of a sum is the sum of the derivatives.
So it looks like that (step by step):

$$ih\frac{\delta}{\delta t}\sum_{k}a _{k}(t)\psi _{k}^{(0)}=(H _{0}+V)\sum_{k}{a _{k}(t)\psi _{k}^{(0)}$$

$$\Leftrightarrow ih\left( \sum_{k}a_{k}(t) \frac{\delta}{\delta t} \psi_{k}^{(0)}+\sum_{k} \frac{d}{dt} a_{k}(t) \psi_{k}^{(0)} \right)=H_{0} \sum_{k}a_{k}(t)\psi_{k}^{(0)}+V\sum_{k}a_{k}(t)\psi_{k}^{(0)}$$

From $$ih\frac{\delta}\delta t\psi_{k}^{(0)}=H_{0}\psi_{k}^{(0)}$$ we obtain by multiplying boths sides by$$\left(\frac{-i}{h}\right) \Rightarrow \frac{\delta}{\delta t}\psi_{k}^{(0)}=-\frac{i}{h}H_{0}\psi_{k}^{(0)}$$. Substituting this, gives us the final result
$$ih \sum_{k}\psi_{k}^{(0)}\frac{d}{d t}a_{k}(t)=\sum_{k}a_{k}V\psi_{k}^{0}$$

But the problem is to multiply this equation by $$\psi_{m}^(0*)$$ and then taking integral:

$$ih\int\sum_{k}\psi_{m}^{(0*)}\psi_{k}^{(0)}\frac{d}{dt}a_{k}(t)=\int\sum_{k}\psi_{m}^{(0)*}V\psi_{k}^{(0)}$$

4. Oct 11, 2015

### Imperatore

Ok, so I would retype this onto form $$ih\sum_{k}\int\psi_{m}^{(0)*}\psi_{k}^{(0)}\frac{d}{dt}a_{k}(t)=\sum_{k}\int\psi_{m}^{(0)*}V\psi_{k}^{(0)}$$
and then knowing that the integral $$\int \psi_{m}^{(0)*}\psi_{k}^{(0)}$$ can be expressed as the inner product $$(\psi_{m}^{(0)*},\psi_{k}^{(0)})=\delta_{mn}$$
So for all $$k\neq m$$ components of the sum are zero, and finally we obtain: $$ih\frac{d}{dt}a_{m}(t)=\sum_{k}\int\psi_{m}^{(0)*}V\psi_{k}^{(0)}$$

What do you think about it ?

5. Oct 11, 2015

### blue_leaf77

There should be $a_k(t)$ in the right hand side, other than that I don't see anything wrong with your derivation.