I am having some conceptual issues with understanding how a wave function is normalized using simple harmonic oscillators.

If you need to find $A$ for a simple harmonic oscillator of mass $m$ and spring constant $k$ that starts out in the state:

$$\psi(x, 0) = A(2\psi_1(x) + \psi_2(x))$$

Should this be represented as a free particle in the form

$$\psi(x, 0)=Ae^{ikx}$$

Finding $A$ means normalizing the wave function which means

$$\int_{-\infty }^{\infty } \left | \psi (x) \right |^{2}dx=1$$

But here I'm guessing I need to multiply $A$ for both $\psi_1 (x)$ and $\psi_2 (x)$ and then normalize them separately.

I am trying to understand exactly how this is suppose to be represented in 1-D simple harmonic oscillators for Schrodinger Equation. Any assistance is greatly appreciated.

  • $\begingroup$ The harmonic oscilator wavefunctions are not plane waves, Nor is any finite linear combination of of them. A particle in a harmonic potential is not a free particle by definition $\endgroup$ Oct 2, 2018 at 14:19
  • $\begingroup$ @BySymmetry Thank you for clarifying that. Looking at this a now. I guess I can just use complex conjugates and simply it without worrying about how the wave is actually represented. Does that seem right? $\endgroup$
    – user30558
    Oct 2, 2018 at 14:32

1 Answer 1


Uh, okay, first, forget about $A e^{ikx}$. The eigenfunctions of an harmonic oscillator are well known and they can be found in any table. They are proportional to hermite polynomials.

Secondly, do not doubt: stick to the notation.

$$1=\int_{-\infty}^{+\infty} |\psi(x)|^2 dx$$

It says to do the integral of the squared modulus of the whole wavefunction.

including $A$. The value will depend on $A$, and that's how you get its value. $1=$something that depends on $A$.

So yes, you have to do

$$1=\int_{-\infty}^{+\infty} \left|\ \ A\left[\ 2\ \psi_1(x)+\psi_2(x)\ \ \right]\ \right|^2 dx $$

Now, the trick is using the complex conjugate:

$$1=\int_{-\infty}^{+\infty} \ \ A^*\left[\ 2\ \psi_1^*(x)+\psi_2^*(x)\ \ \right] \cdot A\left[\ 2\ \psi_1(x)+\psi_2(x)\ \ \right] dx $$

So you'll have $A^* A$, which is $|A|^2$ inside the integral. The rest are 4 terms, result of the distributive property. That's why we split that into 4 integrals.

Since constants can be extracted outside, the integral of $\psi_1^*\psi_2$ can be joined with the one of $\psi_2^*\psi_1$ to create twice the real part.

Anyways, that can reduce 1 integral, but there are 3 more integrals to do.

You've got two options:

  1. ("Expert"): use recurrence formulae and other identities so that you end up having integrals of the form

$\int |\psi|^2 dx$, which are 1.

This is not always possible.

  1. Replace the wavefunctions. You need to know the wwavefunctions (they're Hermite polynomials), and do the integrals manyally. Then, find the value of $A$.
  • 1
    $\begingroup$ While the OP doesn't state this, it might be that $\psi_1$ and $\psi_2$ are the normalized eigenfunctions of the quantum HO, in which case they are orthogonal. Then you don't have to join the $\psi_1^*\psi_2$ integrals. They are known to be zero, unless the instructor forces them to do that work, too. Eigenfunctions (with discrete eigenvalues) of hermitian operators must be orthogonal. $\endgroup$
    – Bill N
    Oct 2, 2018 at 14:50
  • 1
    $\begingroup$ True, my mind swtched to the general case. The wavefunctions of the HO are normalized and ortogonal. What's more, they have a parity, so, if the interval is symmetric, as usual, you can get rid of the whole integral (odd), or do twice from 0 to the end (even). $\endgroup$
    – FGSUZ
    Oct 2, 2018 at 14:57
  • $\begingroup$ Thank you guys this definitely helps with my understanding. I do think ψ∗1ψ2 are orthonormal and go to zero. $\endgroup$
    – user30558
    Oct 2, 2018 at 15:20

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