# Time dependant perturbation theory

1. Dec 29, 2009

### DickThrust

Hi, im basically trying to put a wavefunction into the Time Dependant Schrodinger Eqn, as shown in my lecture notes, but i dont understand one of the steps taken...

$$|\right \Psi (t)\rangle=\sum c_n (t) |\right u_n\rangle e^-(\frac{E_n t}{\hbar})$$
into
$$i\hbar \frac{\delta}{\delta t}|\right \Psi\rangle = H |\right \Psi \rangle$$

gives the LHS of the TDSE as:

$$i \hbar \sum \left[ c^. _n (t) - \frac{i E_n}{\hbar}c_n \right] |\right \psi \rangle = ...$$
however, I dont understand the steps taken to get:

$$\left[ c^. _n (t) - \frac{i E_n}{\hbar}c_n \right]$$

i.e. how the time dependant/independant coefficients are separated.

if anyone could help it would be greatly appreciated, thanks!

2. Dec 29, 2009

### cepheid

Staff Emeritus
Hi DickThrust, welcome to PF,

This is just the product rule for differentiation. The two functions of time being multiplied together (under the summation sign) are cn(t) and |un〉exp(-(iEnt)/ℏ). Thus, applying the product rule:

$$i\hbar \frac{\partial \Psi}{\partial t} = i\hbar \frac{\partial}{\partial t} \sum_n c_n(t) u_n e^{-i\frac{E_n t}{\hbar}} = i\hbar \sum_n \frac{\partial}{\partial t} (c_n(t) u_n e^{-i\frac{E_n t}{\hbar}})$$

$$= i\hbar \sum_n \left[c_n(t) \frac{\partial}{\partial t}(u_n e^{-i\frac{E_n t}{\hbar}}) + u_n e^{-i\frac{E_n t}{\hbar}} \frac{\partial}{\partial t}(c_n(t)) \right]$$

$$= i\hbar \sum_n \left[c_n(t) \left(\frac{-iE_n}{\hbar}u_n e^{-i\frac{E_n t}{\hbar}}\right) + u_n e^{-i\frac{E_n t}{\hbar}} \frac{\partial}{\partial t}(c_n(t)) \right]$$

$$= i\hbar \sum_n \left[-c_n \left(\frac{iE_n}{\hbar}\right) + \frac{\partial c_n}{\partial t} \right]u_n e^{-i\frac{E_n t}{\hbar}}$$​

If you add back in the bra-ket notation, (| 〉 which I excluded to remove clutter), and use the the "dot" notation for the time derivative of c, and my expression becomes exactly like yours.

3. Jan 1, 2010

### DickThrust

typical of me, always forgetting the product rule...