How Are Time-Dependent and Time-Independent Coefficients Separated in TDSE?

[DickThrust]

Hi, I am basically trying to put a wavefunction into the Time Dependant Schrodinger Eqn, as shown in my lecture notes, but i don't 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 don't 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!

 

[cepheid]

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 c_n(t) and |u_n〉exp(-(iE_nt)/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.

 

[DickThrust]

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

thanks for your help!