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I searched quite a lot textbooks to answer this question, but I couldn't find anything satisfactory. This could be more general question, but it will be easier to just restrain to particle (say electron) in the infinite potential well.

When it comes to the general solution of Schroedinger equation, the wavefunction is a superposition of many stationary states:

$$\Psi(x,t) =\sum\limits_{n}{c_n\psi _n(x)e^{\frac{-iEt}{ \hbar } }}$$

And from this point, textbooks tend to continue to method of calculating the values of the $c_n$ coefficients by using Fourier transform and given $\Psi(x,t=0)$.

My question is, is it possible, to calculate these values from first principles, i.e. using some other related theories? I can imagine the task like: Given that electron with energy $E_e$ is in the one-dimensional infinite potential well of length $L$, give the probability that measured energy will correspond to the energy of lowest possible state in this box.

I doubt that when modeling such a electron, it is neccessary to make up all the c-coefficients, and just adjust it to have $\sum\limits_{n}{c_n^2}=1$ and $\sum\limits_{n}{c_n^2E_n}=E_e$. I suppose that there have to be some theory that is capable of providing answers. I was considering Fermi-Dirac distribution, which in principle seems to be created for this, but I couldn't find a straightforward answer (which probably do not exist), and I know too little to make use of it.

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    $\begingroup$ If you only know the average energy, you can't find out the coefficients, so the problem you propose cannot be answered without further information. This is not particular to QM; you could just as well ask "given that a ball is somewhere on Earth, what is its velocity?" and I wouldn't have an answer either. $\endgroup$
    – Javier
    Feb 11, 2016 at 18:12
  • $\begingroup$ I do not understand your question. If you put an electron with energy $E_e$ into a box, it will have energy $E_e$ forever, that's what stationary state means! If you are trying to say that $E_e$ is not one of the allowed stationary energies, then this question is ill-defined. There is no single "state with energy $E_e$" if $E_e$ does not occur as an eigenvalue of $H$. $\endgroup$
    – ACuriousMind
    Feb 11, 2016 at 18:13
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    $\begingroup$ The Fourier transform of $\Psi(x,0)$ is first principles... $\endgroup$ Feb 11, 2016 at 18:25
  • $\begingroup$ @ACuriousMind Ok, I wrote energy $E_e$, I meant something like energy expectation value. As far as I understand, it is possible for particle to have any expectation value of energy, but when making a measurement, I will get only energies of stationary states, with some probability attached to each of these. @Javier@AFT But where (istead of the text of exercise) I can get the REAL wavefunction for t=0? What other should i know, if I want to model electron in a box, and want to figure out the coefficients? $\endgroup$
    – user106877
    Feb 11, 2016 at 18:34
  • $\begingroup$ If your measuring apparatus measures the energy, you would have to prepare a large number of identical states, make many measurements of the energy, and then use your favorite statistical method to estimate $|c_n|^2$ and its uncertainty. Of course, then you have to ask how to prepare the states identically... physics.stackexchange.com/q/13329 $\endgroup$ Feb 12, 2016 at 17:52

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Given only $<E_e>$ and a 1-d infinite square well, without any other constraints there is no way to find the general wave function. The particular $c_n$ coefficients of the eigenfunctions will depend on how you prepare the total system, i.e., what are the constaints/forces involved in the preparation.

Usually what happens in real-world physics is that carefully prepared systems are measured and the distribution of measurement (energy?) values are used to extract $c_n$ values. Then an explanation of why those $c_n$ values occur is deduced. That gives theorists more information about how other slightly modified systems might behave. Measurements are repeated and compared to the predictions. Think about how qubits and quantum computing is being developed.

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  • $\begingroup$ Thank you Bill, that was exactly what I was asking about. $\endgroup$
    – user106877
    Feb 15, 2016 at 10:43

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