Question:$$\psi(x)=A(3u_1(x)+4u_2(x))$$where $u_1(x)$ and $u_2(x)$ are energy eigenfunctions. How to normalize function $\psi(x)$?

My intuitive solution:
I got $$\int^{\infty}_{-\infty}|\psi(x)|^2dx=A^2\int^{\infty}_{-\infty}(9u_1^2(x)+16u_2^2(x)+12u_1(x)^{*}u_2(x)+12u_2(x)^{*}u_1(x))dx$$ Since $u_1(x)$ and $u_2(x)$ are eigenfunctions, the total area under each of them squared must be 1, thus: $$\int^{\infty}_{-\infty}|\psi(x)|^2dx=A^2(9+16+\int^{\infty}_{-\infty}12u_1(x)^{*}u_2(x)dx+\int^{\infty}_{-\infty}12u_2(x)^{*}u_1(x)dx)$$

Since in the remaining integral, $u_1(x)^{*}u_2(x)$ and $u_2(x)^{*}u_1(x)$ are the definition of the inner products of $u_1(x)$ and $u_2(x)$, which, in quantum mechanics, is the square-root of probabilities of getting $u_1(x)$ or $u_2(x)$ knowing that the initial state function is the other one, inside the integral it's just constants, which will make the integral become infinite. If so, the area under the absolute squared wavefunction will be infinite regardless of the value of A.

I am kind of stuck at the infinite integrals. Could someone please give me a hint?

  • 2
    $\begingroup$ Eigenfunctions of hermitian operators (to different eigenvalues) are orthogonal. Does that help? $\endgroup$ – CuriousOne Oct 15 '14 at 5:52
  • $\begingroup$ Um...I don't know what is a hermitian operator...but you're saying that the two eigenfunctions' inner product will get a zero? Does that apply to all energy eigenfunctions? $\endgroup$ – Ruihong Yuan Oct 15 '14 at 6:08
  • $\begingroup$ ...only to those that correspond to self-adjoint operators. $\endgroup$ – Danu Oct 15 '14 at 6:09
  • $\begingroup$ The Hamilton operator is hermitian (i.e. all its eigenvalues are real numbers). Eigenfunctions to different energy eigenvalues are orthogonal. Multiple eigenfunctions to the same (degenerate) eigenvalue can be orthogonalized (by forming an orthogonal set of linear combinations of non-orthogonal functions). $\endgroup$ – CuriousOne Oct 15 '14 at 6:15
  • $\begingroup$ Okay. But my problem doesn't say anything about the eigenfunctions themselves, how can we know if the function is complex or real? Or the energy eigenfunctions are all real? $\endgroup$ – Ruihong Yuan Oct 15 '14 at 6:18

We are told that $\left|1\right>$ and $\left|2\right>$ are energy eigenstates, meaning they are eigenstates of the Hamiltonian $\hat{H}$, a Hermitian operator: ${\hat H}^\dagger = H$. Eigenstates of a Hermitian operator with different eigenvalues are orthogonal (see this): $\left<m|n\right> = \delta_{mn}$ (using Kronecker delta), assuming the eigenstates are normalized. So, normalizing the state $\left|\psi\right>$ gives $$ \begin{eqnarray} 1 = \left<\psi | \psi \right> &=& \left[A(3 \left|1\right> + 4 \left|2\right>)\right]^\dagger A(3 \left|1\right> + 4 \left|2\right>) \\ &=& \left|A\right|^2 (3 \left<1\right| + 4 \left<2\right|) (3 \left|1\right> + 4 \left|2\right>) \\ &=& \left|A\right|^2 (3^2 \left<1|1\right> + 3*4 \left<1|2\right> + 4*3 \left<2|1\right> + 4^2 \left<2|2\right> ) \\ &=& 25 \left|A\right|^2 \end{eqnarray} $$


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