# General solution for the time-dependent Schrödinger equation

1. Jun 12, 2015

### Dyatlov

Hello! I have two uncertainties (hehe) about two concepts from a QM time-dependent Schrödinger equation video.
The video is
I cannot move on further if I don't fully grasp everything he explains in the video. My two issues are:
1) The general solution for the time-dependent Schrödinger equation starts at 55:20.He uses αi for a set of coefficients of states. What exactly are these? the probability densities? probability amplitudes? the absolute probabilities? He mentions any state vector can be written as a superposition of eigenvectors of the energy, so α should be a probability amplitude, so you can calculate it's squared modulus.
2) At 1:03:04 he comes up with the exponential solution e^-iEjt. Where exactly from does he gets α(0) from?
Appreciating any kind of answer which would help me shed some light on these two things. Cheers!

2. Jun 12, 2015

### blue_leaf77

1) Usually one calls the square modulus of $\alpha$ as the probability amplitude.
2) Didn't he say that he assumes he knows the wavefunction at t=0, and hence the corresponding expansion coefficients?

Last edited: Jun 12, 2015
3. Jun 12, 2015

### Dyatlov

1) Fair enough but my question was what does α stand for (he calls it coefficient of states).

4. Jun 12, 2015

### blue_leaf77

If your problem is simply being confused as what to call $\alpha$, you can also adopt what he said in the video. I personally would call it expansion coefficient, the way to obtain it is through the equation $\alpha_i(t) = \langle i | e^{-iHt/\hbar} | \psi(0) \rangle$. I guess you know about it already.

5. Jun 13, 2015

### samalkhaiat

When we write $$|\Psi(t) \rangle = \sum_{n} a_{n}(t) |n\rangle , \ \ \ \ a_{n}(t) = \langle n|\Psi(t)\rangle$$ we are simply representing the abstract state vector $|\Psi\rangle$ by a set of numbers $a_{n}$ which have all the information content in the state Psi. You should know that from vector algebra. Recall the similarity with the vector relations $$\vec{V}=\sum_{i} v_{i} \hat{e}_{i}, \ \ \ \ v_{i} = \hat{e}_{j} \cdot \vec{V} .$$ So, you say that $a_{n}(t)$ are the components of the “vector” $|\Psi\rangle$ in the $|n\rangle$ “direction”, projection of $|\Psi\rangle$ on $|n\rangle$, how much of the state $|n\rangle$ one can find in the state $|\Psi\rangle$, transition (or probability) amplitude from $|\Psi\rangle$ to $|n\rangle$, or (even better) matrix representation of the state vector $|\Psi\rangle$. Notice that in $a_{n}$-representation, the Schrodinger equation becomes $$i \frac{d}{d t} \langle m | \Psi \rangle = \sum_{n} \langle m |H| n \rangle \ a_{n}(t) ,$$ or $$i \frac{d a_{m}}{d t} = \sum_{n} H_{m n} \ a_{n}(t) .$$ This was the starting equation of the so-called Matrix Mechanics of Heisenberg. It is just an equivalent, matrix form of the differential equation of Schrodinger. Now, suppose that $|n\rangle$ form eigen-states of the Hamiltonian, $H|n\rangle = E_{n}|n\rangle$, then $$H_{m n} = \langle m | H | n \rangle = E_{n} \delta_{n m} ,$$ and the above matrix equation becomes (no sum on m) $$i \frac{d a_{m}(t)}{d t} = E_{m} a_{m}(t) .$$ Integrating this from $t=0$ to $t$, we find $$\int_{t=0}^{t} \frac{d a_{m}(t)}{a_{m}(t)} = - i E_{m} \int_{0}^{t} dt ,$$$$\ln |\frac{a_{m}(t)}{a_{m}(0)}| = - i E_{m} t ,$$ which we normally write as $$a_{m}(t) = a_{m}(0) e^{ - i E_{m} t}.$$

6. Jun 13, 2015

### Dyatlov

Thanks a lot, that helped.