Normalizing a Discrete Sum: Is the Wavefunction Fully Normalized?

[Isaac0427]

Say you have two energy eigenstates $\phi_1$ and $\phi_2$, corresponding to energies $E_1$ and $E_2$. The particle has a 50% chance of having each energy. The wavefunction would thus be
$\psi=\frac{\phi_1}{\sqrt{2}}+\frac{\phi_2}{\sqrt{2}}$
Even though the coefficients are normalized (i.e. $\sum_n c_n=1$), is the wavfefunction normalized? My thought would be no. If $|\psi|^2=\frac{|\phi_1|^2}{2}+\frac{|\phi_2|^2}{2}$ then it would be, but it seems like we would have an extra term. So, the wavefunction would really be
$\psi=A\left(\frac{\phi_1}{\sqrt{2}}+\frac{\phi_2}{\sqrt{2}}\right)$
Where A is a constant satisfying normalization. But, even with the A out front, the probability of the particle having the energy $E_1$ would still be $\left|\frac{1}{\sqrt{2}}\right|^2$ and the same thing for $E_2$. Is this all correct?

If this is all true, would it be common practice to leave the wavefunction in that form so you can distinguish the coefficients representing the probability amplitude of a particle having a certain energy (in this case both were $\frac{1}{\sqrt{2}}$) from the coefficient normalizing the wavefunction (in this case denoted by A)?

Thanks!

 

[jtbell]

it seems like we would have an extra term.

Write out $\int{\psi^* \psi\, dx}$ in terms of $\phi_1$ and $\phi_2$. What do the extra terms look like, and what do they evaluate to?

(Hint: look in your textbook or other source for the word "orthogonal" in connection with energy eigenstates.)

 

[Isaac0427]

Write out $\int{\psi^* \psi\, dx}$ in terms of $\phi_1$ and $\phi_2$. What do the extra terms look like, and what do they evaluate to?

(Hint: look in your textbook or other source for the word "orthogonal" in connection with energy eigenstates.)

Thank you!

I can't believe I missed that.

Although, that does lead me to another question, but I will ask that one tomorrow.