0
$\begingroup$

To calculate the probability of eigenvalues from a superposition of two wave-functions do you normalize the overall wave-funtion or the individual wave-functions.

I am aware that one can write $\Psi = \sum_n c_n \psi_n $ and that the probability of getting the eigenvalue (of an operator) is $=|c_n|^2$. However I am a little lost at how to work out what $c_n$ is.

For example say I have $\Psi =A\sin(2\pi x/a)e^{-iE_1t/\hbar} + B\sin(4\pi x/a)e^{-iE_2t/\hbar}$ that is already normalized - If I wanted to find the values of $E_1$ and $E_2$ by applying the Hamiltonian operator and then wanted to find the probabilities of $E_1$ and $E_2$ respectively would I need to normalize both $A\sin(2\pi x/a)$ and $B\sin(4\pi x/a)$ before I use the Hamiltonian on each individually and include the normalization constant in $C_n$?

$\endgroup$
1
$\begingroup$

$\phi_n$ should be a complete basis set so that any wavefunction $\Phi$ can be expanded on the basis, and the modulus square sum of the coefficients is normalized automatically.

It looks like your situation here is that there are two particles in the system, and what is the probablity of finding one particle at a position $r$? $E_1$ and $E_2$ are energy eigenvalue from solving Schrodinger equation of each particle in a one-dimensional well of width $a$. $A$ and $B$ can be $1/\sqrt{2}$ if these two particles are indistinguishable.

The orignial two wavefunctions are eigenfunctions. If you put two wavefunctions in superposition, the sum wavefuntion is also a eigenfunction. The square sum of the coefficients should be normalized. The probability density at each r is $|\Phi(r)|^2= \Phi^*(r) \Phi(r)$. It includes two cross terms of $\phi_1$ and $\phi_2$, e.g. $\phi_1^*\phi_2$ and $\phi_2^*\phi_1$, which can be explicitly calculated to give the proprobability of finding one particle at $r$.

The superposition situation can be the superposition of two energy levels, or a two-slit interference.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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