I read some notes saying,

$$i\hbar \frac{dC_{i}(t)}{dt} = \sum_{j}^{} H_{ij}(t)C_{j}(t)\tag{1}$$

where $C_{i}(t) = \langle i|\psi(t)\rangle$ and $H_{ij}$ is hamiltonian matrix.

However, what is obscure to me is the way to deduce

$$H^*_{ij} = H_{ji}\tag{2}$$

in the notes.

They explain (2) as

(2) follows from the condition that the total probability that the system is in some state does not change. If you start with a particle then you have still got it as time goes on. The total probability of finding it somewhere is

$$\sum_{i}^{}\left|C_{i}^2 \right|$$

which must not vary with time. If this is to be true for any starting condition $\phi$, then (2) must also be true.

I cannot understand the relationship between the total amplitude and (2).

Well.. though I omit other contents/derivative of (1) in the notes to simplify my question, it will be very helpful for me if you figure out the explanation of the notes above.


1 Answer 1


Work out the derivative of your formula for the total probability (after changing the faulty $|C_i^2|$ to the correct$|C_i|^2$), using (1) to eliminate all derivatives. Then the conclusion should stare into your eyes.

  • $\begingroup$ Alright. I succeeded in deriving the relationship. Thanks. But if you don't mind, could you let me know how we can think that the total amplitude should be independent of time? Is that just an assumption that the notes suggest? $\endgroup$
    – true
    Jul 25, 2012 at 7:26
  • $\begingroup$ @true: Not the amplitude is constant in time, but the squared absolute value. The latter is the sum of individual squared absolute values, which represent (Born interpretation) the probability that upon measuring, the $i$the result is obtained. In an ideal measuremnt, some result must be obtained, which means that these probabilities must sum to 1. $\endgroup$ Jul 25, 2012 at 10:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.